SOLUTIONS MANUAL &TEST BANK for Interplanetary Astrodynamics 1st Edition by David B. Spencer and Dav

TEST BANK & SOLUTIONS MANUAL for Interplanetary Astrodynamics. 1st Edition by David B. Spencer and Davide Conte

Taking another time-derivative: 𝑥̈ = 𝜌̈ cos 𝜃 sin 𝜙 − 𝜌𝜙̇ 2 cos 𝜃 sin 𝜙 − 𝜃̇ 2 cos 𝜃 sin 𝜙 + 𝜌𝜙̈ cos 𝜙 cos 𝜃+ ̈ sin 𝜙 sin 𝜃 + 2𝜌̇ 𝜙̇ cos 𝜙 cos 𝜃 − 2𝜌̇ 𝜃̇ sin 𝜙 sin 𝜃 − 2𝜌𝜙̇ 𝜃̇ cos 𝜙 sin 𝜃 − 𝜃𝜌 𝑦̈ = 𝜌̈ sin 𝜙 sin 𝜃 − 𝜌𝜙̇ 2 sin 𝜙 sin 𝜃 − 𝜌𝜃̇ 2 sin 𝜙 sin 𝜃 + 𝜌𝜙̈ cos 𝜙 sin 𝜃+ + 𝜌𝜃̈ cos 𝜃 sin 𝜙 + 2𝜌̇ 𝜙̇ cos 𝜙 sin 𝜃 + 2𝜌̇ 𝜃̇ cos 𝜃 sin 𝜙 + 2𝜌𝜃̇ 𝜙̇ cos 𝜙 cos 𝜃 𝑧̈ = 𝜌̈ cos 𝜙 − 2𝜌̇ 𝜙̇ sin 𝜙 − 𝜌𝜙̈ sin 𝜙 − 𝜌𝜙̇ 2 cos 𝜙 Equating each 𝑥, 𝑦, and 𝑧 acceleration expressed in spherical coordinates with its respective acceleration terms gives us the equations of motion for the two-body problem in terms of spherical coordinates 𝜌, 𝜙, and 𝜃 𝜌̈ cos 𝜃 sin 𝜙 − 𝜌𝜙̇ 2 cos 𝜃 sin 𝜙 − 𝜃̇ 2 cos 𝜃 sin 𝜙 + 𝜌𝜙̈ cos 𝜙 cos 𝜃+ ̈ sin 𝜙 sin 𝜃 + 2𝜌̇ 𝜙̇ cos 𝜙 cos 𝜃 − 2𝜌̇ 𝜃̇ sin 𝜙 sin 𝜃 − 2𝜌𝜙̇ 𝜃̇ cos 𝜙 sin 𝜃+ − 𝜃𝜌 𝜇 sin 𝜙 cos 𝜃 =0 𝜌2 𝜌̈ sin 𝜙 sin 𝜃 − 𝜌𝜙̇ 2 sin 𝜙 sin 𝜃 − 𝜌𝜃̇ 2 sin 𝜙 sin 𝜃 + 𝜌𝜙̈ cos 𝜙 sin 𝜃+ + 𝜌𝜃̈ cos 𝜃 sin 𝜙 + 2𝜌̇ 𝜙̇ cos 𝜙 sin 𝜃 + 2𝜌̇ 𝜃̇ cos 𝜃 sin 𝜙 + 2𝜌𝜃̇ 𝜙̇ cos 𝜙 cos 𝜃 +

+

𝜇 sin 𝜙 sin 𝜃 =0 𝜌2

𝜇 cos 𝜙 𝜌̈ cos 𝜙 − 2𝜌̇ 𝜙̇ sin 𝜙 − 𝜌𝜙̈ sin 𝜙 − 𝜌𝜙̇ 2 cos 𝜙 + =0 𝜌2 √ where we used the fact that 𝜌 = 𝑟 = 𝑥 2 + 𝑦 2 + 𝑧 2 .

2

Problem 2 Prove that for the unperturbed two-body problem, orbital energy is constant.

Solution 2

Start with the vis-viva equation, Equation (2.50): 𝐸 = 𝑣2 − 𝜇𝑟 To prove that energy is constant, we need to take its time derivative and show that it is equal to zero: 𝑑 𝐯⋅𝐯 𝜇 𝑑𝐸 𝑑 = ( − ) 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 [ (𝐫 ⋅ 𝐫)1/2 ] =

1 𝐯̇ ⋅ 𝐯 + 𝐯 ⋅ 𝐯̇ ̇ − 𝜇 − 𝐫−3 (2𝐫 ⋅ 𝐫) ) [ 2 ) ( 2

Recall that 𝐯̇ = 𝐫̈ = −𝜇𝐫 and 𝐫̇ = 𝐯, so 𝑟3 −𝜇𝐫 𝜇𝐫 𝑑𝐸 =𝐯⋅( 3 )+ 3 ⋅𝐯 𝑑𝑡 𝑟 𝑟 𝜇𝐫 𝜇𝐫 = −𝐯 ⋅ ( 3 ) + 𝐯 ⋅ 3 = 0 𝑟 𝑟 Thus, 𝑑𝐸 = 0 which means that orbital energy is constant. 𝑑𝑡

3

Problem 3 Prove that the angular momentum vector and eccentricity vector are orthogonal to each other.

Solution In order to prove that two vectors are orthogonal, or perpendicular, to each other, one must show that their dot product is zero.

Starting with the definitions of eccentricity, Equation (2.27), 𝐞=

𝐯×𝐡 𝐫 − 𝜇 𝑟

and angular momentum, Equation (2.41), 𝐡=𝐫×𝐯 we take the dot product between angular momentum and eccentricity, 𝐡⋅𝐞=𝐡⋅ =

𝐯×𝐡 𝐫 − ( 𝜇 𝑟)

1 1 𝐡 ⋅ (𝐯 × 𝐡) − (𝐫 × 𝐯) ⋅ 𝐫 𝜇 𝑟

where we used the definition of angular momentum for the second term. We then use the scalar triple product on the above equation, which, for three generic vectors 𝐀, 𝐁, and 𝐂 is 𝐀 ⋅ (𝐁 × 𝐂) = 𝐁 ⋅ (𝐂 × 𝐀) = 𝐂 ⋅ (𝐀 × 𝐁) This helps us simplify the first term as 1 1 1 𝐡 ⋅ (𝐯 × 𝐡) = 𝐯 ⋅ (𝐡 × 𝐡) = 𝐯 ⋅ 𝟎 = 0 𝜇 𝜇 𝜇 and the second term as 1 1 1 − 𝐫 ⋅ (𝐯 × 𝐫) = − 𝐯 ⋅ (𝐫 × 𝐫) = − 𝐯 ⋅ 𝟎 = 0 𝑟 𝑟 𝑟 which proves that 𝐡 ⋅ 𝐞 = 0 and thus 𝐡 ⟂ 𝐞 = 0.

4

Problem 4 Prove that for the unperturbed two-body problem, orbital motion is confined to a plane (this is the orbital plane).

Solution Planar motion means that the angular momentum (𝐡) is conserved. Thus, showing that the time derivative of 𝐡 is zero is sufficient to show that the motion is constrained to a plane. This derivation is shown in Section 2.3, and summarized here.

Taking the cross product of 𝐫 with Equation (2.9) gives 𝐫 × (𝐫̈ +

𝜇𝐫 =0 𝑟3 )

𝐫 × 𝐫̈ + 𝐫 ×

𝜇𝐫 =0 𝑟3

The second term is zero (cross product of a vector with itself), simplifying the above equation to 𝐫 × 𝐫̈ = 0 which is equivalent to 𝑑 𝑑𝐡 𝑑 ̇ = (𝐫 × 𝐯) = =0 (𝐫 × 𝐫) 𝑑𝑡 𝑑𝑡 𝑑𝑡 which proves that angular momentum is constant, and thus orbital motion is planar for the unperturbed two-body problem.

5

Problem 5 Show that the speed of a spacecraft in a circular orbit is √ 𝜇 𝑣𝑐 = 𝑟 Solution Starting with the vis-viva equation (Equation (2.50)), 𝑣2 𝜇 −𝜇 − = 2 𝑟 2𝑎 we let 𝑎 = 𝑟 since a circular orbit has a constant radius, which is equal to its semimajor axis. This gives 𝑣2 𝜇 −𝜇 − = 2 𝑟 2𝑟 𝑣2 𝜇 1 𝜇 = − 2 𝑟 2𝑟 Solving for 𝑣 gives √ 𝑣𝑐 =

6

𝜇 𝑟

Problem 6 A spacecraft in Earth orbit passes through its apogee (apoapsis with respect to Earth) at a radius of 𝑟𝑎 = 9, 500 km, and with speed 𝑣𝑎 = 5.95 km/s. Determine how much time (in hours) elapses before the spacecraft returns to apogee.

Solution Computing this time means computing orbital period. From Equation (2.57), √ 𝑎3 𝑇 = 2𝜋 𝜇 which means that we need to first compute the semimajor axis of the orbit.

Using the vis-viva equation, Equation (2.50), 𝑣2 𝜇 −𝜇 − = 2 𝑟 2𝑎 we solve for 𝑎 to get 𝑎=

−𝜇 𝑣2 − 2𝜇𝑟

which is equivalent to Equation (2.61).

We can use the radius and speed that we are given, 𝑟𝑎 and 𝑣𝑎 , to compute semimajor axis using the above equation, resulting in 𝑎 = 8, 216.31 km, which corresponds to an orbital energy of 𝐸 = −24.2566 𝑘𝑚2 /𝑠 2 .

Using the equation for orbital period gives √ 𝑇 = 2𝜋

𝑎3 = 7, 411.8 𝑠 𝜇

which corresponds to 8.05884 hours.

7

Problem 7 Starting with any known relationships, show that the orbital eccentricity can be expressed as √ 2𝑝𝐸 𝑒 = 1+ 𝜇 where 𝑝 is semi-latus rectum, 𝐸 is orbital energy, and 𝜇 is the gravitational parameter.

Solution Starting with Equation (2.35) 𝑝 = 𝑎(1 − 𝑒 2 ) and plugging Equation (2.50) 𝑎= into it, results in 𝑝= Simplifying,

−𝜇 2𝐸

−𝜇 𝑎(1 − 𝑒 2 ) 2𝐸

2𝑝𝐸 = 𝑒2 − 1 𝜇 2𝑝𝐸 +1 𝜇 √ 2𝑝𝐸 𝑒 = 1+ 𝜇 𝑒2 =

where we took only the positive root since 𝑒 > 0.

8

Problem 8 Radar measurements give the following data for a spacecraft in Earth orbit at some time 𝑡1 : • 𝑟 = 8, 000 km • 𝑟̇ = −0.8201 km/s • 𝜃̇ = 5.635 × 10−2 deg/s Calculate: a Orbital energy, in km2 /s2 b Semimajor axis, in km c Angular momentum, in km2 /s d Semi-latus rectum, in km e Eccentricity f Radii of periapsis and apoapsis, in km g Orbital period, in hours h Flight-path angle at time 𝑡1 in rad Solution Note that the 𝜃̇ measurement must be converted into rad/s, which gives 𝜃̇ = 9.835 × 10−4 rad/s.

Computing the requested quantities gives: a Orbital energy: using Equation (2.46), 𝐸 = −18.5364 km2 /s2 b Semimajor axis: using Equation (2.50), 𝑎 = 10, 751.8 km c Angular momentum: using Equation (2.47), ℎ = 62, 943.6 km2 /s d Semi-latus rectum: using Equation (2.30), 𝑝 = 9, 939.52 km e Eccentricity: using Equation (2.35), 𝑒 = 0.27486 f Radii of periapsis and apoapsis: using Equation (2.37), 𝑟𝑝 = 7, 796.55 km and 𝑟𝑎 = 13, 707 km g Orbital period: using Equation (2.57), 𝑇 = 3.08198 hours h Flight-path angle (FPA) at time 𝑡1 : FPA is the current spacecraft heading, i.e. the angle between its velocity and the local horizon, which can be computed as 𝜙 = ± arccos ( 𝑟𝑣ℎ ) = −0.103858 rad. In order to resolve the ± in this equation, refer to Equation (2.74)

9

Problem 9 At time 𝑡 = 0, a spacecraft on an Earth orbit with 𝑎 = 50, 612 km and 𝑒 = 0.63174 is located at 𝜃 = 42.319◦ . At what time did the satellite previously pass through periapsis?

Solution Here, we are asked to solve for time of periapse passage, 𝑡𝑝 . In order to do so, we use Kepler’s Time Equation, or Equations (2.96) and (2,97). However, we must first convert true anomaly into eccentric anomaly using Equation (2.91) keeping in mind that all calculations must be done using radians, not degrees, √ 𝐸 = 2 arctan

𝜃 1−𝑒 tan = 0.363695 𝑟𝑎𝑑 ( 2 )] [ 1+𝑒

where no quadrant check is needed since 𝐸2 and 𝜃2 always reside in the same quadrant.

From Equations (2.96) and (2.97), √ 𝐸 − 𝑒 sin 𝐸 =

𝜇 (𝑡 − 𝑡𝑝 ) 𝑎3

Solving for 𝑡𝑝 with the fact that 𝑡 = 0 gives √ 𝜇 𝑡𝑝 = − 3 (𝐸 − 𝑒 sin 𝐸) = −2, 506.23 𝑠 𝑎 which is equal to approximately -41.77 minutes. In other words, the spacecraft passed through periapsis 41.77 min prior to 𝑡 = 0.

10

Problem 10 A satellite on an Earth orbit with 𝑎 = 12, 587 km and 𝑒 = 0.42 passed through periapsis at 10:04:06 (time here is given in the form in 24-hour format). Calculate the time, in 24-hour format, at which the satellite will reach 𝜃 = 297◦ .

Solution Here, we are asked to solve for elapsed time since periapse passage, (𝑡 − 𝑡𝑝 ). In order to do so, we use Kepler’s Time Equation, or Equations (2.96) and (2,97). However, we must first convert true anomaly into eccentric anomaly using Equation (2.91) keeping in mind that all calculations must be done using radians, not degrees, √ 𝐸 = 2 arctan

𝜃 1−𝑒 = −0.74628 𝑟𝑎𝑑 tan ( 2 )] [ 1+𝑒

where no quadrant check is needed since 𝐸2 and 𝜃2 always reside in the same quadrant. However, since we are asked to find when the spacecraft will arrive at the given true anomaly, we must add 2𝜋 to 𝐸 to compute the future spacecraft passage, resulting in 𝐸 = 5.537 rad.

From Equations (2.96) and (2.97), √ 𝐸 − 𝑒 sin 𝐸 = Solving for (𝑡 − 𝑡𝑝 ) gives

√

𝜇 (𝑡 − 𝑡𝑝 ) 𝑎3

𝜇 (𝐸 − 𝑒 sin 𝐸) = 13, 022 𝑠 𝑎3 So, adding 13,022 s to 10:04:06 means adding 3 hours (10,800 s), 37 minutes (2,220 s) and 2 s, resulting in a time of 13:41:08. 𝑡 − 𝑡𝑝 =

11

Problem 11 A satellite in Earth orbit passes through perigee at 𝑇0 = 978 s on an orbit with 𝑒 = 0.7 and 𝑎 = 30, 500 km. Using the Newton-Raphson iteration method and an absolute tolerance of Δ𝐴 = 0.001, calculate the true anomaly 𝜃 at 𝑡 = 35, 000 s. Do not implement a relative tolerance check. This problem must be done by hand and calculator (i.e., no computer programs or use of calculator solver routines). Show each step in the iteration – 𝐸𝑜𝑙𝑑 , 𝐸𝑛𝑒𝑤 , and tolerance check.

Solution We must use Kepler’s Time Equation (Equations (2.96) and (2.97)) to solve for eccentric anomaly 𝐸 at the given time 𝑡 √ 𝜇 (𝑡 − 𝑡𝑝 ) = 𝐸 − 𝑒 sin 𝐸 𝑀= 𝑎3 √ where 𝑀 = 𝑎𝜇3 (𝑡 − 𝑇0 ) = 4.03254 rad.

Following the Newton-Raphson method discussed in Section 2.6.1, we set up 𝑓 (𝐸) and 𝑓 ′ (𝐸) as follows 𝑓 (𝐸) = 𝐸 − 𝑒 sin(𝐸) − 𝑀 𝑓 ′ (𝐸) = 1 − 𝑒 cos(𝐸) where only 𝐸 is unknown. For Newton-Raphson, the update step for each iteration is 𝐸𝑛𝑒𝑤 = 𝐸𝑜𝑙𝑑 −

𝑓 (𝐸𝑜𝑙𝑑 ) 𝑓 ′ (𝐸𝑜𝑙𝑑 )

We then use 𝐸𝑜𝑙𝑑 = 𝑀 as the initial guess. To check the absolute tolerance for every step, we check if |𝐸𝑛𝑒𝑤 − 𝐸𝑜𝑙𝑑 | < Δ𝐴 . Iterations are carried on until the absolute tolerance is met. This results in: Iteration 𝐸𝑜𝑙𝑑 (rad) 1 4.03254 2 3.65453 3 3.67606

𝐸𝑛𝑒𝑤 (rad) 3.65453 3.67606 3.67601

|𝐸𝑛𝑒𝑤 − 𝐸𝑜𝑙𝑑 | (rad) 0.3780 0.0215 <0.001

Thus, 𝐸 = 3.67601 rad from which 𝜃 can be computed using Equation (2.91) as √ 𝜃 = 2 arctan

𝑒+1 𝐸 tan = 3.3706 𝑟𝑎𝑑 ( 2 )] [ 𝑒−1

which is also equivalent to 𝜃 = 193.12◦ .

12

Problem 12 Write a computer program that a takes as input • A position vector (in km) and a velocity vector (in km/s) in ECI (Earth-Centered Inertial) coordinates.

• An initial time, 𝑡1 • A later time, 𝑡2 • Tolerances 𝛿𝑅 (relative tolerance) and 𝛿𝐴 (absolute tolerance) for the iterative procedure used to solve Kepler’s time equation. b Outputs the input data, with labels and units, and calculates the following: Classical orbital elements 𝑎, 𝑒, 𝑖, 𝜔, and Ω (Note that these are 5 of the 6 COEs) c The DCM 𝐶 𝐸𝑃 that converts from ECI to perifocal coordinates. d Position (in km) and velocity (in km/s) vectors in perifocal coordinates at time 𝑡1 . e True and eccentric anomalies 𝜃1 and 𝐸1 at time 𝑡1 . f Time of periapsis passage, 𝑡𝑝 . g True and eccentric anomalies 𝜃2 and 𝐸2 at time 𝑡2 . h Position (in km) and velocity (in km/s) vectors in perifocal coordinates at time 𝑡2 . i Position (in km) and velocity (in km/s) vectors in ECI coordinates at time 𝑡2 . j Outputs the values in items (b) – (i) with labels and units.

Run the program with the input values: 𝐫𝟏 = 5, 214.1𝐼̂ + 6, 322.9𝐽̂ − 3, 001.7𝐾̂ 𝑘𝑚 𝐯𝟏 = −4.1549𝐼̂ + 3.1666𝐽̂ − 4.5044𝐾̂ 𝑘𝑚/𝑠 𝑡1 = 0 𝑠 𝑡2 = 12, 500 𝑠 Δ𝑅 = 10−6 Δ𝐴 = 10−6 Solution The resulting quantities for each computation are summarized in the table below

13

Part Quantity

b

Value and Unit

Equation(s) Used

𝑎

9,108.95 km

(2.61)

𝑒

0.201529

(2.62)

𝑖

0.759947 rad

(2.64)

𝜔

2.09983 rad

(2.71), (2.72)

Ω

3.62717 rad

(2.68), (2.69)

𝐫𝐏𝐐𝐖 𝟏

⎡ 0.7384 −0.3179 0.5947 ⎤ ⎥ ⎢ ⎢ 0.5928 0.7265 −0.3477⎥ ⎥ ⎢ ⎢−0.3215 0.6093 0.7249 ⎥ ⎦ ⎣ 55.10𝑝̂ + 8, 727.72𝑞̂ + 0𝑤̂ km

(2.78), (2.79); or (2.152)

𝐯𝐏𝐐𝐖 𝟏

−6.7535𝑝̂ + 1.40369𝑞̂ + 0𝑤̂ km/s

(2.152)

𝜃1

1.56448 rad

(2.73)

𝐸1

1.3617 rad

(2.91)

f

𝑡𝑝

-1,603.6 s or 7,048.3 s

(2.96), (2.97)

g

𝜃2

3.71099 rad

(2.73) (compute after 𝐸2 )

𝐸2

3.83091 rad

(2.96), (2.97) (must iterate)

𝐫𝐏𝐐𝐖 𝟐

−8, 864.9𝑝̂ − 5, 678.5𝑞̂ + 0𝑤̂ km

(2.78), (2.79); or (2.152)

𝐯𝐏𝐐𝐖 𝟐

3.64103𝑝̂ − 4.32705𝑞̂ + 0𝑤̂ km/s

(2.152)

𝐫𝐄𝐂𝐈 𝟐

−9, 909.7𝐼̂ − 1, 304.3𝐽̂ − 3, 299.1𝐾̂ km

(𝐶 𝐸𝑃 )𝑇 𝐫𝐏𝐐𝐖 𝟐

𝐯𝐄𝐂𝐈 𝟐

0.123732𝐼̂ − 4.30091𝐽̂ + 3.66979𝐾̂ km/s

(𝐶 𝐸𝑃 )𝑇 𝐯𝐏𝐐𝐖 𝟐

c

d e

h i

𝐶 𝐸𝑃

𝐶3 (𝜔)𝐶1 (𝑖)𝐶3 (Ω)*

*the notation 𝐶𝑗 (𝛼) indicates a direction cosine matrix (DCM) about the j-axis (j = 1, 2, 3) through an angle 𝛼. DCMs should not be confused with rotation matrices. A rotation matrix is the inverse of a DCM.

14

Problem 13 Consider a spacecraft in orbit around the Earth with the following Initial Conditions (ICs; position and velocity vectors) in an Earth-Centered Inertial (ECI) frame at some time 𝑡0 : 𝐫𝟎 = 666.780𝐼̂ + 7, 621.32𝐽̂ + 2, 338.97𝐾̂ 𝑘𝑚 𝐯𝟎 = −7.22541𝐼̂ + 0.266320𝐽̂ + 1.52177𝐾̂ 𝑘𝑚/𝑠 Consider the Cartesian formulation of the J2-perturbed Equations of motion for the two-body problem: { } 3 𝑅𝑒2 𝑧2 𝜇𝑥 𝑥̈ = − 3 1 − 𝐽2 5 −1 ] 𝑟 2 ( 𝑟 2 )[ 𝑟 2 { } 3 𝑧2 𝑅𝑒2 𝜇𝑦 𝑦̈ = − 3 1 − 𝐽2 5 −1 ] 𝑟 2 ( 𝑟 2 )[ 𝑟 2 { } 𝜇𝑧 3 𝑅𝑒2 𝑧2 𝑧̈ = − 3 1 − 𝐽2 5 −3 ] 𝑟 2 ( 𝑟 2 )[ 𝑟 2 where: 𝜇 = 398, 600 km3 /s2 is the gravitational parameter of the Earth, 𝑅𝑒 = 6, 378.14 km is the equatorial radius of the Earth, 𝐽2 = 1.08263 × 10−3 the coefficient of Earth’s oblateness, and √ 2 2 𝑟 = 𝑥 + 𝑦 + 𝑧 2 is the magnitude of the position of the spacecraft with respect to Earth’s center.

Set up a state 𝐒 = {𝑥 𝑦 𝑧 𝑥̇ 𝑦̇ 𝑧} ̇ and rewrite the above EOMs in state-space form using these given states. Then, using a computer program such as MATLAB, numerically integrate the ICs of this spacecraft using the above EOMs for 30 days. Provide two figures: 1. Figure 1: a 3D plot (x vs. y vs. z) of the spacecraft orbit (in red) with respect to Earth’s center in ECI coordinates. Also, for reference plot the Earth as a blue sphere having radius equal to 𝑅𝑒 .

2. Figure 2: subplots showing: (a) Semimajor Axis (SMA) in km vs. time (b) Eccentricity vs. time (c) Inclination in degrees vs. time (d) Orbital period in seconds vs. time (e) Perigee in km vs. time (f) Apogee in km vs. time If using MATLAB, compare the difference between using different integration schemes, such as ode45 vs. ode113. What do you notice? Which one is faster for a given accuracy?

Solution Following the numerical integration methods discussed in Section 2.8, we obtain the graphs shown in Figures 1 and 2. Numerical integration using both MATLAB’s ode45 and ode113 shows that ode113 is faster than ode45.

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Figure 1: Spacecraft Orbit in ECI

Figure 2: Deviation of Orbital Parameters 16

Interplanetary Astrodynamics Chapter 3 Problem Solutions

Problem 1 Compute the period, time unit (in seconds), and dimension unit (in kilometers) of the systems listed in Table 3.1. You may find it useful to consult the appendix for various constants you will need, such as semi-major axes and gravitational parameters.

Solution The value of dimension (or length) unit (LU) is simply the value of semimajor axis 𝑎, while the period is computed as given by Equation (3.10) √ 𝑎3 𝑇 = 2𝜋 𝐺 (𝑚1 + 𝑚2 ) where 𝐺 = 6.6743 × 10−20 km3 kg-1 s-2 is the universal gravitational constant. Then, the time unit (TU) can be computed as given by Equation (3.11) 1 𝑇𝑈 =

𝑇 2𝜋

Compiling the results for all systems listed in Table 3.1 gives the following numerical results

1

System

Mass Ratio, 𝜇

1 LU (km)

Period (s)

1 TU (s)

Earth-Moon

1.215060e-02

3.844000e+05 2.356987e+06 3.751261e+05

Sun-Mercury

1.660148e-07

5.790931e+07 7.599252e+06 1.209459e+06

Sun-Venus

2.447835e-06

1.082041e+08 1.940945e+07 3.089110e+06

Sun-Earth

3.003460e-06

1.495978e+08 3.155262e+07 5.021756e+06

Sun-Mars

3.227137e-07

2.279422e+08 5.934516e+07 9.445075e+06

Sun-Jupiter

9.536922e-04

7.783274e+08 3.742697e+08 5.956687e+07

Sun-Saturn

2.857260e-04

1.426983e+09 9.294258e+08 1.479227e+08

Mars-Phobos

1.660513e-08

9.376000e+03 2.755916e+04 4.386176e+03

Mars-Deimos

2.299699e-09

2.345800e+04 1.090626e+05 1.735786e+04

Jupiter-Ganymede

7.803794e-05

1.070400e+06 6.180722e+05 9.836925e+04

Saturn-Titan

2.365667e-04

1.221870e+06 1.377519e+06 2.192389e+05

Pluto-Charon

1.085112e-01

1.959140e+04 5.516461e+05 8.779721e+04

Note about the pronunciation of Pluto’s moon, Charon: the pronunciation "Sharon" is commonly accepted among many English speakers. However, in Greek mythology, the name originates from Charon or Kharon (𝑋 𝛼𝜌𝜔𝜈) ́ and was the name of the ferryman who would carry the dead across the river Styx. The Greek letter X (capitalized) or 𝜒 (lower case) in the name 𝑋 𝛼𝜌𝜔𝜈 ́ has a sound similar to the Spanish j in j𝑎mon. ́ All of Pluto’s moons (Charon, Hydra, Kerberos, Nix, and Styx) are named after Greek mythology.

2

Problem 2 Compute the x- and y-coordinates of the collinear Lagrange points for the following systems: (a) Earth-Moon (b) Sun-Mercury (c) Sun-Venus (d) Sun-Earth (e) Sun-Mars (f) Sun-Jupiter (g) Sun-Saturn (h) Mars-Phobos (i) Mars-Deimos (j) Jupiter-Ganymede (k) Saturn-Titan (l) Pluto-Charon Solution The y-coordinate of all collinear points is zero, so it is only necessary to solve for the x-coordinate.

In order to do so, we use Equation (3.39) to set up a function of the form 𝑓 (𝑥) =

(1 − 𝜇)(𝑥 − 𝜇) 𝜇(𝑥 + 1 − 𝜇) + −𝑥 |𝑥 − 𝜇|3 |𝑥 + 1 − 𝜇|3

Solving for 𝑓 (𝑥) = 0 gives three real-valued solutions which correspond to the x-coordinates of 𝐿1 , 𝐿2 , and 𝐿3 , which we here denote as 𝑥𝐿1 , 𝑥𝐿2 , and 𝑥𝐿3 , respectively. Solving for the zeroes of this algebraic function can be cumbersome due to the presence of the absolute values. MATLAB’s fzero command was used here to obtain the results shown in the following table.

3

System

𝜇

𝑥𝐿1 (ND)

𝑥𝐿2 (ND)

𝑥𝐿3 (ND)

Earth-Moon

1.215060e-02 -8.369151e-01 -1.155682e+00

1.005063e+00

Sun-Mercury

1.660148e-07 -9.961939e-01 -1.003815e+00

1.000000069e+00

Sun-Venus

2.447835e-06 -9.906823e-01 -1.009371e+00

1.0000010199e+00

Sun-Earth

3.003460e-06 -9.900266e-01 -1.010034e+00

1.000001251e+00

Sun-Mars

3.227137e-07 -9.952513e-01 -1.004763e+00

1.000000e+00

Sun-Jupiter

9.536922e-04 -9.323699e-01 -1.068826e+00

1.000397e+00

Sun-Saturn

2.857260e-04 -9.547492e-01 -1.046069e+00

1.000000134e+00

Mars-Phobos

1.660513e-08 -9.982321e-01 -1.001770e+00 1.00000000691e+00

Mars-Deimos

2.299699e-09 -9.990851e-01 -1.000915e+00 1.00000000095e+00

Jupiter-Ganymede

7.803794e-05 -9.705872e-01 -1.029842e+00

1.000033e+00

Saturn-Titan

2.365667e-04 -9.575005e-01 -1.043252e+00

1.000099e+00

Pluto-Charon

1.085112e-01 -5.930430e-01 -1.262516e+00

1.045139e+00

In this table, we used additional significant figures for some 𝑥𝐿3 values that are extremely small.

4

Problem 3 How do the values you computed in Problem 2 compare with the approximations for 𝐿1 , 𝐿2 , and 𝐿3 given by Equation (3.55)?

Solution Equation (3.55) gives an approximation for the x-coordinates of the 𝐿1 , 𝐿2 , and 𝐿3 , which we here denote as 𝑥𝐿1 , 𝑥𝐿2 , and 𝑥𝐿3 , respectively, so that 𝜇 1/3 𝑥𝐿1 ≈ ( ) − 1 3 𝜇 1/3 𝑥𝐿2 ≈ − ( ) − 1 3 5𝜇 𝑥𝐿3 ≈ 1 + ( 12 ) The following table summarizes the results of computing approximate values for 𝑥𝐿1 , 𝑥𝐿2 , and 𝑥𝐿3 .

System

𝜇

Earth-Moon

1.215060e-02

-8.405986e-01 -1.159401e+00 1.005063e+00

Sun-Mercury

1.660148e-07

-9.961893e-01 -1.003811e+00 1.000000e+00

Sun-Venus

2.447835e-06

-9.906556e-01 -1.009344e+00 1.000001e+00

Sun-Earth

3.003460e-06

-9.899962e-01 -1.010004e+00 1.000001e+00

Sun-Mars

3.227137e-07

-9.952441e-01 -1.004756e+00 1.000000e+00

Sun-Jupiter

9.536922e-04

-9.317511e-01 -1.068249e+00 1.000397e+00

Sun-Saturn

2.857260e-04

-9.543323e-01 -1.045668e+00 1.000119e+00

Mars-Phobos

1.660513e-08

-9.982311e-01 -1.001769e+00 1.000000e+00

Mars-Deimos

2.299699e-09

-9.990848e-01 -1.000915e+00 1.000000e+00

Jupiter-Ganymede 7.803794e-05

-9.703702e-01 -1.029630e+00 1.000033e+00

𝑥𝐿1 (ND)

𝑥𝐿2 (ND)

𝑥𝐿3 (ND)

Saturn-Titan

2.365667e-04

-9.571178e-01 -1.042882e+00 1.000099e+00

Pluto-Charon

1.085112e-01

-6.692871e-01 -1.330713e+00 1.045213e+00

Computing the percentage difference between the numerical solutions (Problem 2) and the approximate solutions using Equation (3.55) can be done as follows | 𝑥𝐿 , 𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑙 − 𝑥𝐿𝑖 , 𝑎𝑝𝑝𝑟𝑜𝑥. | | × 100% Percentage Difference = || 𝑖 | 𝑥𝐿𝑖 , 𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑙 | | with 𝑖 = 1, 2, 3. Doing so results in the following table.

5

System

𝜇

% Difference 𝑥𝐿1

% Difference 𝑥𝐿2

% Difference 𝑥𝐿3

Earth-Moon

1.215060e-02

0.438203905350070

0.320785222575412 0.000009769945384

Sun-Mercury

1.660148e-07

0.000469844938868

0.000465080575677 0.000000000000022

Sun-Venus

2.447835e-06

0.002699586399057

0.002632645018871 0.000000000000022

Sun-Earth

3.003460e-06

0.003076736262845

0.002995083495722

< 10−14

Sun-Mars

3.227137e-07

0.000726291929042

0.000717105059948

< 10−14

Sun-Jupiter

9.536922e-04

0.066416675305139

0.054036055092968 0.000000004715131

Sun-Saturn

2.857260e-04

0.043688710896332

0.038405000317756 0.000000000126795

Mars-Phobos

1.660513e-08

0.000102884310336

0.000102399485460

< 10−14

Mars-Deimos

2.299699e-09

0.000027723644489

0.000027656019694

< 10−14

Jupiter-Ganymede 7.803794e-05

0.022356697113660

0.020608597303759 0.000000000002576

0.039989328222425

0.035454070329668 0.000000000071958

Saturn-Titan

2.365667e-04

Pluto-Charon

1.085112e-01 11.391846098281569 5.124834377092897 0.007098841803256

From the above table, one can see how the approximation introduces errors on the order of << 1%, except for the Pluto-Charon system, which has a relatively large 𝜇 value. The second-worst approximation is for the Earth-Moon system, where errors are on the order of 10−1 %.

6

Problem 4 The nonlinear equations of motion in the restricted three-body problem are: (1 − 𝜇)(𝑥 − 𝜇) 𝜇(𝑥 + 1 − 𝜇) − 𝑟13 𝑟23 (1 − 𝜇)𝑦 𝜇𝑦 𝑦̈ + 2𝑥̇ − 𝑦 = − − 3 𝑟13 𝑟2 (1 − 𝜇)𝑧 𝜇𝑧 − 3 𝑧̈ = − 𝑟13 𝑟2 𝑥̈ − 2𝑦̇ − 𝑥 = −

where

√ 𝑟1 = 𝑟2 =

√

(𝑥 − 𝜇)2 + 𝑦 2 + 𝑧 2 (𝑥 + 1 − 𝜇)2 + 𝑦 2 + 𝑧 2

For the restricted three-body problem, the complete set of equations of motion for the linear solutions about the Lagrange points are found as: { } (𝑥𝑒 − 𝜇)2 (𝑥𝑒 + 1 − 𝜇)2 1 1 𝛿 𝑥̈ − 2𝛿 𝑦̇ − 𝛿𝑥 = −𝛿𝑥 (1 − 𝜇) 3 − 3 +𝜇 3 −3 + [ 𝑟1𝑒 ] [ 𝑟2𝑒 ] 𝑟1𝑒5 𝑟2𝑒5 (𝑥𝑒 − 𝜇)𝑦𝑒 (𝑥𝑒 + 1 − 𝜇)𝑦𝑒 + 𝛿𝑦 3(1 − 𝜇) + 3𝜇 5 [ ] 𝑟1𝑒 𝑟2𝑒5 (𝑥𝑒 − 𝜇)𝑦𝑒 (𝑥𝑒 + 1 − 𝜇)𝑦𝑒 𝛿 𝑦̈ + 2𝛿 𝑥̇ − 𝛿𝑦 = 𝛿𝑥 3(1 − 𝜇) + 3𝜇 + 5 [ ] 𝑟1𝑒 𝑟2𝑒5 } { 1 𝑦𝑒2 1 𝑦𝑒2 − 𝛿𝑦 (1 − 𝜇) 3 − 3 5 + 𝜇 3 − 3 5 [ 𝑟1𝑒 [ 𝑟2𝑒 𝑟2𝑒 ] 𝑟2𝑒 ] Using these equations, assess the stability of 𝐿1 for the systems listed in Problem 2.

Solution We can simplify the equations given in the problem statement since the y- and z-coordinates for 𝐿1 are zero, so that √ 𝑟1𝑒 = (𝑥𝐿1 − 𝜇)2 = 𝑥1𝑒 − 𝜇 √ 𝑟2𝑒 = (𝑥𝐿1 + 1 − 𝜇)2 = 𝑥1𝑒 + 1 − 𝜇 where the subscript ‘𝑒’ here denotes the fact that we are considering equilibrium points of the system. Here, we will show a detailed solution for the Earth-Moon system. For all other systems, the solution method is identical, and while the numerical values are slightly different, the end result (that 𝐿1 is unstable) is the same.

For the Earth-Moon system, 𝑥1𝑒 = −0.8369 (refer to the solution for Problem 2), so that the perturbed equations provided in the problem statement simplify to 𝛿 𝑥̈ − 2𝛿 𝑦̇ − 𝛿𝑥 = −𝛿𝑥 (−10.2935) 𝛿 𝑦̈ + 2𝛿 𝑥̇ − 𝛿𝑦 = −𝛿𝑦 (5.1468) 7

In matrix form { } { } { } { } 1 0 𝛿 𝑥̈ 0 −2 𝛿 𝑥̇ −11.2935 0 𝛿𝑥 0 + + = 0 4.1468] 𝛿𝑦 0 [0 1] 𝛿 𝑦̈ [2 0 ] 𝛿 𝑦̇ [ We assume that this linear system of ODEs has a solution of the form { } 𝛿𝑥 = 𝐴 𝑒 𝜆𝑡 𝛿𝑦 so that {

}

{

} 𝛿𝑥 =𝜆 𝛿𝑦 { } { } 𝛿 𝑥̈ 𝛿𝑥 = 𝜆2 𝛿 𝑦̈ 𝛿𝑦 𝛿 𝑥̇ 𝛿 𝑦̇

Substituting these into the matrix equation for 𝛿𝑥 and 𝛿𝑦 we get { } { } 𝜆2 − 11.2935 −2𝜆 𝛿𝑥 0 = 2 2𝜆 𝜆 + 4.1468] 𝛿𝑦 0 [ Here, the solutions are the trivial solution (which gives no information on the system) or the matrix in the above equation is not invertible, i.e. its determinant is zero, so det

𝜆2 − 11.2935 −2𝜆 =0 2 2𝜆 𝜆 + 4.1468] [

This gives us the following characteristic equation (𝜆2 − 11.2935)(𝜆2 + 4.1468) + 4𝜆2 = 0 𝜆4 − 3.1467𝜆2 − 46.832 = 0 which is a quadratic equation in 𝜆2 . Solving for 𝜆2 using the quadratic equation and then taking further square roots results in the following four values of 𝜆 𝜆1 = 2.932 𝜆2 = −2.932 𝜆3 = 2.334 i 𝜆4 = −2.334 i In order for a dynamical system to be stable, all roots of the characteristic equation must have non-positive real parts. However, since 𝜆1 is a positive real number, the system is unstable at the computed equilibrium point, which means that the Lagrange point 𝐿1 is unstable.

8

Problem 5 Compute the value of the Jacobi integral at each Lagrange point computed in Problem 2 assuming zero-velocity components in the x-, y-, and z-directions.

Solution From Equation (3.63), the Jacobi constant can be computed as follows 1−𝜇 𝜇 1 − 𝐽 = − (𝑥 2 + 𝑦 2 ) − 2 𝑟1 𝑟2 where we assumed that all velocity components are zero, as stated by the problem. Furthermore, 𝑟1 and 𝑟2 can be simplified since the y- and z-components of the collinear Lagrange points are zero, so 𝑟1 = 𝑥𝐿𝑖 − 𝜇 𝑟2 = 𝑥𝐿𝑖 + 1 − 𝜇 where 𝑖 = 1, 2, 3 for 𝐿1 , 𝐿2 , and 𝐿3 , respectively. Thus, the Jacobi constant simplifies to 1 1−𝜇 𝜇 𝐽 = − 𝑥𝐿2𝑖 − − 2 |𝑥𝐿𝑖 − 𝜇| |𝑥𝐿𝑖 + 1 − 𝜇| where we need to use absolute values for the denominators since 𝑟1 > 0 and 𝑟2 > 0. Computing the Jacobi constant using the x-coordinates of each collinear point (refer to Problem 2) results in the following table System

𝜇

𝐽 (𝐿1 )

𝐽 (𝐿2 )

𝐽 (𝐿3 )

Earth-Moon

1.215060e-02 -1.594170625224589

-1.586080287273612

-1.506073582531158

Sun-Mercury

1.660148e-07 -1.500065070997697

-1.500064960320992

-1.500000083007400

Sun-Venus

2.447835e-06 -1.500388858029825

-1.500387226125325

-1.500001223917438

Sun-Earth

3.003460e-06 -1.500445344888560

-1.500443342561633

-1.500001501729906

Sun-Mars

3.227137e-07 -1.500101245712571

-1.500101030569587

-1.500000161356849

Sun-Jupiter

9.536922e-04 -1.519378038491350

-1.518742117111655

-1.500476836539595

Sun-Saturn

2.857260e-04 -1.508910362635931

-1.508719852663865

-1.500142862147275

Mars-Phobos

1.660513e-08 -1.500014053103714

-1.500014042033623

-1.500000008302565

Mars-Deimos

2.299699e-09 -1.500003765337075

-1.500003763803943

-1.500000001149850

Jupiter-Ganymede 7.803794e-05 -1.503820911405483

-1.503768882392033

-1.500039018906516

Saturn-Titan

2.365667e-04 -1.507882004358728

-1.507724273455416

-1.500118282765730

Pluto-Charon

1.085112e-01 -1.810171662165219

-1.739669120616945

-1.553995665696784

As discussed in Chapter 3, various correct definitions of the Jacobi constant exist. Using a definition that is different from what we used here will result in different Jacobi constant values.

9

Problem 6 Using the nonlinear equations of motion in the restricted three-body problem: (a) if the state vector is: 𝐗 = (𝑥 𝑦 𝑧 𝑥̇ 𝑦̇ 𝑧) ̇ 𝑇 , and using the values of 𝜇 = 0.16 and the following initial conditions: 𝐗(0) = (−1.024114 0 0.528576 0 0.381137 0)𝑇 , numerically integrate the equations of motion for one period (you will have to experimentally determine what the period is (hint: think of the symmetry of the problem and where does the orbit cross the plane of symmetry again?)). Show plots of x vs. y, x vs. z, y vs. z and a three-dimensional plot of x vs. y vs. z. You should get a periodic, halo orbit. Identify the period of the orbit. (b) Confirm the invariance of these equations of motion under the transformation 𝑦 → −𝑦 and 𝑡 → −𝑡. (c) For another case, we don’t have the exact initial conditions for 𝑥0 and 𝑧0 . If 𝜇 = 0.2 and the initial conditions are 𝐗(0) = (−1.01 0 0.58 0 0.413250 0)𝑇 , use a differential correction process to correct the initial conditions to obtain a periodic halo orbit. Show three iterations of your differential correction results with the correct-ed initial conditions and period (or half period) after each iteration.

Solution (a) Through trial and error, it is possible to determine that the period of the orbit is 2.52 TU. Alternatively, one can set an event flag to stop the integration once the state returns to the initial state (within some tolerance). The plots we obtain from this numerical integration are shown in Figure 1. The resulting orbit is an 𝐿2 Northern halo orbit. Notice that the value of 𝜇 used here (0.16) is considered to be rather large – compare it with 𝜇 for Earth-Moon and Pluto-Charon.

Although not requested by the problem, one can compute the Lagrange point locations for 𝐿1 , 𝐿2 , and 𝐿3 (refer to Problem 2) resulting in the following x-coordinates 𝑥𝐿1 = −0.502942319513751 𝑥𝐿2 = −1.271094441718751 𝑥𝐿3 = 1.066420639638059 (b) Integrating the equations of motion backwards in time using the ICs in part (a) results in the same orbit. However, a more rigorous proof is done by replacing 𝑦 with −𝑦 and 𝑡 with −𝑡 in the CR3BP equations of motion: 𝑥̈ − 2𝑦̇ − 𝑥 = Ω𝑥 𝑦̈ + 2𝑥̇ − 𝑦 = Ω𝑦 𝑧̈ = Ω𝑧 Looking at the above equations term-by-term, and performing the required substitutions (𝑦 → −𝑦

10

Figure 1: Orbital Plot for Part (a) and 𝑡 → −𝑡), gives 𝑥̈ → 𝑥̈ −𝑑𝑦 𝑑𝑦 = = 𝑦̇ 𝑦̇ → −𝑑𝑡 𝑑𝑡 𝜔𝑥 → Ω 𝑥 𝑑 𝑦̇ 𝑦̈ → − = −𝑦̈ 𝑑𝑡 𝑥̇ → −𝑥̇ Ω𝑦 → −Ω𝑦 𝑧̈ → 𝑧̈ Ω𝑧 → Ω𝑧

11

Thus, the new equations of motion become 𝑥̈ − 2𝑦̇ − 𝑥 = Ω𝑥 −𝑦̈ − 2𝑥̇ + 𝑦 = −Ω𝑦 𝑧̈ = Ω𝑧 The x- and z-equations are identical to the original equations of motion, while the y-equation can be multiplied by −1 in order to get the original y-equation of motion. This proves that the CR3BP equations of motion are invariant (do not change) under the transformation 𝑦 → −𝑦 and 𝑡 → −𝑡. (c) Using a differential corrector (Section 3.7.1) to correct for 𝑥0 and 𝑧0 for the given fixed value of 𝑦̇ 0 means using Equations (3.112) and (3.119). After three iterations, the new 𝑥0 and 𝑧0 are within ∼ 10−5 accuracy of the desired values, and these numerical results are shown in the table below.

Iterating for more steps resulting in 𝑥0 and 𝑧0 with even higher accuracy, if desired.

Element

Iteration 1

Iteration 2

Iteration 3

𝑇 2

1.3121

1.3466

1.3465

𝑥0

-1.0100

-1.0141

-1.0145

𝑦0

0

0

0

𝑧0

0.5800

0.5879

0.5876

𝑥̇ 0

0

0

0

𝑦̇ 0

0.4133

0.4133

0.4133

𝑧̇ 0

0

0

0

𝑥(𝑇 /2)

-0.8115

-0.8143 −6

-7.1126×10

-0.8144 −6

1.5844×10−6

𝑦(𝑇 /2)

-2.6553×10

𝑧(𝑇 /2)

-0.0851

-0.0912

-0.0912

𝑥(𝑇 ̇ /2)

-0.0150

-0.0033

-7.6783×10−7

𝑦(𝑇 ̇ /2)

-2.0501

-1.9706

-1.9707

𝑧(𝑇 ̇ /2)

0.0206

-3.3031×10−4

-1.0398×10−4

𝛿𝑥0

0.0041

3.6758×10−4

-1.4878×10−5

𝛿𝑦0

-0.0079

2.3535×10−4

3.9651×10−5

12

Problem 7 Numerically integrate the equations of motion of the circular restricted three-body problem using the mass ratio of the Earth-Moon system (𝜇 = 0.01215060) for the following sets of initial conditions: (a) 𝐗(0) = (−1.180732845 0 0.013007284 0 0.156830543 0)𝑇 (b) 𝐗(0) = (−1.178004826 0 0.052029136 0 0.169784129 0)𝑇 (c) 𝐗(0) = (−1.125032004 0 0.182101977 0 0.225430661 0)𝑇 (d) 𝐗(0) = (−1.014568158 0 0 0 − 0.9385765507 0)𝑇 (e) 𝐗(0) = (−0.750000000 0 0.6204002141 0 − 0.3228789168 0)𝑇 (f) 𝐗(0) = (−1.156957 0 0 0 0.48000000 0)𝑇 You will need to determine the period of these orbits to correctly plot them (hint: think of the symmetry of the problem). What orbit types do you get? Match and verify that the orbits you obtained are the same as the orbits plotted on Figure 3.55.

Figure 3.56: Various Periodic Orbits in the CR3BP Solution Performing the required numerical integration and plotting each orbit reveals that the given initial conditions (ICs) correspond to the plotted orbits as summarized in the following table 13

ICs

Orbit

(a)

Small halo (blue)

(b)

Medium halo (red)

(c)

Large halo (orange)

(d)

Planar Lyapunov (purple)

(e)

Vertical Lyapunov (green)

(f)

Distant retrograde orbit (light blue)

14

Problem 8 In the planar restricted three-body problem, the equations for the zero-velocity curve can be written as: 𝐶=−

(1 − 𝜇) 2 2 𝜇 2 2 𝜇(1 − 𝜇) 𝑟1 + − 𝑟2 + + 2 ( 𝑟1 ) 2 ( 𝑟2 ) 2

and 1 1−𝜇 𝜇 𝐶 = − (𝑥 2 + 𝑦 2 ) − − 2 𝑟1 𝑟2 (a) Show that these two expressions are equivalent. (b) Using the first equation above, show that the zero-velocity curves disappear where 𝐶=

1 [−3 + 𝜇(1 − 𝜇)] 2

(c) For a system with 𝜇 = 0.25, choose the value of the Jacobi constant half-way between 𝐶(𝐿1 ) and 𝐶(𝐿2 ), i.e. the arithmetic average between 𝐶(𝐿1 ) and 𝐶(𝐿2 ) (be sure to state this value of the Jacobi constant) and develop a computer program to plot the zero-velocity curve for this value. Be sure to annotate the location of the collinear libration points. Test your code at one more location between 𝐶(𝐿2 ) and 𝐶(𝐿3 ) and identify the Jacobi constant value you use.

Solution Note that 𝐶 and 𝐽 are used interchangeably in the literature to both denote the Jacobi constant. (a) We start by setting the 𝐶 equations equal to each other −

1−𝜇 𝜇 𝜇 2 2 𝜇(1 − 𝜇) 1 (1 − 𝜇) 2 2 𝑟1 + − 𝑟2 + + = − (𝑥 2 + 𝑦 2 ) − − 2 ( 𝑟1 ) 2 ( 𝑟2 ) 2 2 𝑟1 𝑟2

and we expand terms and simplify 𝜇 𝜇 𝜇) −(1 − 𝜇) (1 − 𝜇) 2 −(1 − 𝜇(1 − 𝜇) 1 2 2 𝜇 2

− 𝑟1 + − 𝑟2 − + = − (𝑥 + 𝑦 ) + −

2 𝑟 2 𝑟2 2 2 𝑟 𝑟2 1 1 (1 − 𝜇) 2 𝜇 2 𝜇(1 − 𝜇) 1 2 𝑟1 − 𝑟2 + = − (𝑥 + 𝑦 2 ) − 2 2 2 2

Recall that the position vectors are defined as 𝐫𝟏 = (𝑥 − 𝜇)𝐢 + 𝑦𝐣 𝐫𝟐 = [𝑥 + (1 − 𝜇)]𝐢 + 𝑦𝐣 so 𝑟12 = 𝑥 2 − 2𝑥𝜇 + 𝜇2 + 𝑦 2 𝑟22 = 𝑥 2 + 2𝑥(1 − 𝜇) + (1 − 𝜇)2 + 𝑦 2 15

Substituting the equations for 𝑟12 and 𝑟22 into the prior equations gives −

𝜇 𝜇(1 − 𝜇) 1 1−𝜇 2 (𝑥 − 2𝑥𝜇 + 𝜇2 + 𝑦 2 ) − [𝑥 2 + 2𝑥(1 − 𝜇) + (1 − 𝜇)2 + 𝑦 2 ] + = − (𝑥 2 + 𝑦 2 ) 2 2 2 2

Multiplying all by −2 and simplifying (1 − 𝜇)(𝑥 2 − 2𝑥𝜇 + 𝜇2 + 𝑦 2 ) + 𝜇[𝑥 2 + 2𝑥(1 − 𝜇) + (1 − 𝜇)2 + 𝑦 2 ] − 𝜇(1 − 𝜇) = (𝑥 2 + 𝑦 2 Considering the left-hand side of the equation, we can expand it and simply as 2 2

− + 𝜇𝑦 2 + 𝑥 2 −2𝑥𝜇 𝜇 2 + 𝑦 2 − 𝜇𝑦 2 + 𝜇𝑥 2 + 2𝑥𝜇 2𝑥𝜇 + 𝜇 − 2𝜇 2 + 𝜇𝑥 2 + 2𝑥𝜇 − 𝜇 3 −

𝑦 2 𝜇 − 𝜇+ 𝜇 2 = 𝑥 2 + 𝑦 2 + 𝜇 3 + which is equal to the right-hand side and thus proves that the two given expressions for the Jacobi constant are equivalent. (b) We can show that zero-velocity curves disappear at 𝐿4 and 𝐿5 , where 𝑟1 = 𝑟2 = 1. We start with 𝐶=−

𝜇 2 2 𝜇(1 − 𝜇) (1 − 𝜇) 2 2 𝑟1 + − 𝑟2 + + 2 ( 𝑟1 ) 2 ( 𝑟2 ) 2

and substitute 𝑟1 = 𝑟2 = 1 into it to get (1 − 𝜇) 𝜇 𝜇(1 − 𝜇) (1 + 2) − (1 + 2) + 2 2 2 3 3 𝜇(1 − 𝜇) = − (1 − 𝜇) − 𝜇 + 2 2 2 3 3 3 𝜇(1 − 𝜇) = − + 𝜇 − 𝜇 + 2 2 2 2 1 = [−3 + 𝜇(1 − 𝜇)] 2

𝐶=−

which is equal to a constant and proves the fact that zero-velocity curves disappear at 𝐿4 and 𝐿5 . (c) For 𝜇 = 0.25 the x-coordinates of the Lagrange points can be computed as 𝑥𝐿1 = −0.360743428367017 𝑥𝐿2 = −1.265858102510350 𝑥𝐿3 = 1.103166848822924 Refer to the solution for Problem 2 for computations related to these values. Furthermore, 𝑟1 and 𝑟2 can be simplified since the y- and z-components of the collinear Lagrange points are zero, so 𝑟1 = 𝑥𝐿𝑖 − 𝜇 𝑟2 = 𝑥𝐿𝑖 + 1 − 𝜇 where 𝑖 = 1, 2, 3 for 𝐿1 , 𝐿2 , and 𝐿3 , respectively. Thus, the Jacobi constant simplifies to 1−𝜇 𝜇 1 𝐶(𝐿𝑖 ) = − 𝑥𝐿2𝑖 − − 2 |𝑥𝐿𝑖 − 𝜇| |𝑥𝐿𝑖 + 1 − 𝜇| 16

where we need to use absolute values for the denominators since 𝑟1 > 0 and 𝑟2 > 0. The corresponding Jacobi constant values to each Lagrange point are 𝐶(𝐿1 ) = −1.935329401439718 𝐶(𝐿2 ) = −1.780597028114743 𝐶(𝐿3 ) = −1.622470510138496 Note that in the literature 𝐽 and 𝐶 are used interchangeably to denote the Jacobi constant. The halfway values of the Jacobi constant between 𝐿1 and 𝐿2 and between 𝐿2 and 𝐿3 are therefore 1 𝐶𝐿1 /𝐿2 = [𝐶(𝐿1 ) + 𝐶(𝐿2 )] = −1.857963214777230 2 1 𝐶𝐿2 /𝐿3 = [𝐶(𝐿2 ) + 𝐶(𝐿3 )] = −1.701533769126619 2 Figures 2 and 3 show zero-velocity surfaces where the perimeter of the surfaces represent the Jacobi constants computed above, 𝐶𝐿1 /𝐿2 and 𝐶𝐿2 /𝐿3 , respectively. The physical sizes of 𝑚1 and 𝑚2 in the figures are arbitrary, and are used only to visualize the position of the primaries. The Lagrange points are also plotted for reference.

17

Figure 2: Zero-velocity surface plot for 𝐶 values up to 𝐶𝐿1 /𝐿2

18

Figure 3: Zero-velocity surface plot for 𝐶 values up to 𝐶𝐿2 /𝐿3

19

Interplanetary Astrodynamics Chapter 4 Problem Solutions

Problem 1 Using Equation (4.8), plot a graph of Earth’s obliquity vs. time between January 1, 2000 and January 1, 3000. How do the initial and final values of obliquity compare? What is their difference?

Create another plot using the same approximation, but this time from January 1, 2000 for 10,000 years. When does the approximation start diverging significantly from the prediction given in Figure 4.4?

Solution Plotting Earth’s using Equation (4.8) from January 1, 2000 to January 1, 3000 (10 centuries) gives Figure 1. The difference between final and initial values is -0.12952˝ , i.e. the Earth’s obliquity on January 1, 3000 is estimated to be 0.12952˝ less than that on January 1, 2000. Repeating this process for 100 centuries gives Figure 2. The final value of obliquity estimated with this approximation is 22.637279˝ which is visually very similar to that reported in Figure 4.4. The computed approximation seems to start diverging from Figure 4 around 90 centuries, where the minimum value of obliquity happens if using Equation (4.8), but it occurs later (around 100 centuries) in Figure 4.4.

1

Figure 1: Earth’s Obliquity for 10 centuries since J2000

Figure 2: Earth’s Obliquity for 100 centuries since J2000 2

Problem 2 Consider an interplanetary Hohmann transfer from Earth to Mars (refer to Chapter 5 if needed).

Using the resultant Mars arrival 𝑣8 , compute the impact parameter and eccentricity of the arrival trajectory assuming that your targeted periapsis with respect to Mars has an altitude of 300 km.

Solution As provided in Appendix A, the symbol for Mars is D, and the orbital and physical parameters of interest for this problem are also taken from Appendix A 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝜇D “ 43, 050 km3 /s2 𝑅D “ 3, 397 km 𝑟D{@ “ 227, 942, 167.9 km 𝑟C{@ “ 149, 597, 800 km Thus, the targeted orbit at Mars arrival has periapsis radius 𝑟𝑝{D “ 𝑅D ` 300km “ 3,697 km.

In order to compute the 𝑣8{D , we need to solve part of this interplanetary Hohmann transfer (as discussed in Chapter 5). We compute the semimajor axis and energy of the Hohmann transfer orbit (subscript ‘H’) as ˘ 1` 𝑎𝐻 “ 𝑟C{@ ` 𝑟D{@ “ 188, 769, 983.9 km 2 ´𝜇@ 𝐸𝐻 “ “ ´351.486 km2 {s2 2𝑎𝐻 Then, we compute the heliocentric speeds that the spacecraft would have at apoapsis (i.e. right before encountering Mars) and that Mars has as d 2𝜇@ 𝑣𝑎{@ “ 2𝐸𝐻 ` “ 21.47926 km/s 𝑟D{@ c 𝜇@ 𝑣D{@ “ “ 24.12810 km/s 𝑟D{@ Thus, 𝑣8{D is just the difference of these two speeds 𝑣8{D “ 𝑣D{@ ´ 𝑣𝑎{@ “ 2.64884 km/s To compute the impact parameter, we use Equation (4.15) d 2𝜇D Δ “ 𝑟𝑝{D 1 ` “ 7, 683.42 km 2 𝑟𝑝{D 𝑣8{D To compute the eccentricity of the Mars arrival hyperbolic orbit, we use Equations (2.50) and (2.37), so that the hyperbolic semimajor axis and eccentricity become ´𝜇D 𝑎 “ 2 “ ´6, 135.66 km 𝑣8{D 𝑟𝑝{D 𝑒 “1´ “ 1.60254 𝑎 3

Problem 3 Compute the Julian Date (JD) and Modified Julian Date (MJD) for a given time (e.g., when your homework is due). Make sure to use enough decimal places so that your accuracy is within 1 ms (10´3 s). Then, compute the Greenwich location angle corresponding to this time.

Solution For this problem, we will use the date: January 1, 2025 00:00:00 UTC. To compute JD, using Equation (4.45) gives 2,460,676.5000000.

To compute MJD, we use the computed JD and Equation (4.46), giving 60,676.0000000.

4

Problem 4 Create a ground track similar to that of Figure 4.7 for an Earth orbiting spacecraft that has classical orbital elements with respect to ECI as follows: (a) semimajor axis: 6671 km (b) eccentricity: 0.3 (c) inclination: 45˝ (d) RAAN: 20˝ (e) argument of periapsis: 95˝ Assume that the spacecraft starts at periapsis (true anomaly = 0˝ ) on January 1, 2030, and propagate its orbit using the unperturbed two-body model for 5 full orbits.

Solution Following the steps given in Sections 4.3 and 4.3.1 gives the ground track shown in Figure 3, where tick marks are each separated by 60 seconds. For reference, 𝐽 𝐷 “ 2462502.50000 and 𝜃𝑔 “ 1.75069204 rad.

Figure 3: Ground Track

5

Problem 5 Compute the coordinates of your school/institution using the topocentric equatorial coordinates (IJK) discussed in Section 4.3.2 for a given time (e.g., when your homework is due). Compute these coordinates assuming that the Earth is a perfect sphere (flattening parameter = 0) and using a flattening parameter of 0.003353. How do the results differ?

Solution For this example, we will use Embry-Riddle Aeronautical University – Prescott Campus as the institution of choice. This location has the following altitude 𝐻 , latitude 𝜙, and longitude 𝜃 𝐻 “ 1.572 km 𝜃 “ ´112.450644˝ 𝜙 “ 34.617177˝ Additionally, from Appendix A we know that Earth’s equatorial radius is 𝑅C “ 6, 378.140 km and we are given Earth’s flattening as 𝑓 “ 0.003353.

Using Equation (4.23) with 𝑓 “ 0 gives ˜ ¸ 𝑅C 𝑋“ a ` 𝐻 cos 𝜙 cos 𝜃 “ ´2, 005.0185 km 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 ˜ ¸ 𝑅C ` 𝐻 cos 𝜙 sin 𝜃 “ ´4, 852.3614 km 𝑌 “ a 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 ¸ ˜ 𝑅C p1 ´ 𝑓 q2 ` 𝐻 sin 𝜙 “ 3, 624.2537 km 𝑍“ a 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 while using 𝑓 “ 0.003353 gives ˜

¸

𝑅C a `𝐻 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙

𝑋“ ˜

¸ 𝑅C a `𝐻 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙

𝑌 “ ˜ 𝑍“

cos 𝜙 cos 𝜃 “ ´2, 007.1875 km cos 𝜙 sin 𝜃 “ ´4, 857.6105 km ¸

2

𝑅C p1 ´ 𝑓 q a `𝐻 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙

sin 𝜙 “ 3, 603.8906 km

The percent differences in the x-, y-, and z- components are 0.1081%, 0.1081%, and 0.5650%, respectively.

6

Problem 6 Compute the coordinates of the NASA Perseverance landing location using the topocentric equatorial coordinates (IJK) discussed in Section 4.3.2 for a given time (e.g., when your homework is due). Compute these coordinates assuming that Mars is a perfect sphere (flattening parameter = 0) and using a flattening parameter of 0.00648. How do the results differ?

Solution For this problem, we will use the landing date of Perseverance, i.e. February 18, 2021, where the rover was located at altitude 𝐻 , latitude 𝜙, and longitude 𝜃 𝐻 “ 2.478 km 𝜃 “ 77.451˝ 𝜙 “ 18.445˝ This data can be obtained directly from JPL Horizons. Additionally, from Appendix A we know that Mars’s equatorial radius is 𝑅D “ 3, 389.5 km and we are given Earth’s flattening as 𝑓 “ 0.00648.

Using Equation (4.23) with 𝑓 “ 0 gives ˜ ¸ 𝑅D 𝑋“ a ` 𝐻 cos 𝜙 cos 𝜃 “ 699.1294 km 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 ˜ ¸ 𝑅D ` 𝐻 cos 𝜙 sin 𝜃 “ 3, 140.8550 km 𝑌 “ a 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 ¸ ˜ 𝑅D p1 ´ 𝑓 q2 ` 𝐻 sin 𝜙 “ 1, 073.2021 km 𝑍“ a 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 while using 𝑓 “ 0.00648 gives ˜

¸

𝑅D a `𝐻 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙

𝑋“ ˜ 𝑌 “

¸ a

˜ 𝑍“

cos 𝜙 cos 𝜃 “ 699.5816 km

𝑅D `𝐻 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙 ¸

2

a

cos 𝜙 sin 𝜃 “ 3, 142.8863 km

𝑅D p1 ´ 𝑓 q `𝐻 1 ´ p2𝑓 ´ 𝑓 2 q sin2 𝜙

sin 𝜙 “ 1, 060.0337 km

The percent differences in the x-, y-, and z- components are 0.0646%, 0.0646%, and 1.2423%, respectively.

7

Problem 7 Compute the JD and MJD for your next birthday. If you know it, include the hours and minutes correspond to your time of birth, otherwise, use midnight UTC. Solution For this problem, we will use the date: January 22, 2025 15:05:00 UTC. To compute JD, using Equation (4.45) gives 2,460,698.12847.

To compute MJD, we use the computed JD and Equation (4.46), giving 60,697.62847.

8

Problem 8 Using Equation (4.42), compute the date on which TDB and TCB will start differing by more than a minute.

Solution From Equation (4.42), we find the difference between TCB and TDB as 𝑇 𝐶𝐵 ´ 𝑇 𝐷𝐵 “ 1.550519768 ˆ 10´8 Δ𝑇 ` 6.55 ˆ 10´5 where Δ𝑇 is 𝐽 𝐷𝑇 𝐶𝐵 ´𝑇0 , i.e. the time elapse from Jan 1, 1977 00:00:00 UTC in seconds (we multiplied the 86,400 to turn this quantity into seconds). Setting 𝑇 𝐶𝐵 ´ 𝑇 𝐷𝐵 “ 60 s and solving for Δ𝑇 gives Δ𝑇 “

60 ´ 6.55 ˆ 10´5 « 44, 788 days or 122.7 years 1.550519768 ˆ 10´8

Adding these days to Jan 1, 1977 00:00:00 UTC leads to Aug 17, 2099, which is the day upon which TCB and TDB will be 60 seconds apart from each other.

9

Problem 9 Compute the ephemeris of the Moon using Simpson’s approximation (discussed in Section 4.5) for a given time, e.g. from when your homework is due for the following 10 years. Plot the orbital elements of the Moon with respect to ECI for this time frame.

Solution For this problem, we will use the initial and final dates Jan 1, 2025 and Jan, 1, 2035, respectively.

Computing the lunar ephemeris for these dates using Simpson’s approximation results in the classical orbital elements shown in Figures 4 and 5.

Figure 4: Classical Orbital Elements of the Moon with respect to ECI (1 of 2)

10

Figure 5: Classical Orbital Elements of the Moon with respect to ECI (2 of 2)

11

Problem 10 Using JPL Horizons, compute the position of Earth and Mars for a given time (e.g., when your homework is due). Compute the OWLT for this time. Then, obtain the positions of Earth and Mars from JPL Horizons for one synodic period and plot the OWLT for this time.

Solution For this problem, we will use the initial date of Jan 1, 2025. As provided in Appendix A, the synodic period of Earth and Mars is 2.1351 years (or 780 days). This results in a final date of Feb 20, 2027.

The positions of Earth and Mars in the J2000 ecliptic reference frame at the initial date are listed in the following tables. 𝑥C (AU)

𝑦C (AU)

𝑧C (AU)

-0.184413 0.962073

0.000129

𝑥D (AU)

𝑦D (AU)

𝑧D (AU)

-0.527416 1.520324

0.044935

The OWLT between Earth and Mars for this synodic period is plotted on Figure 6. The maximum and minimum OWLT computed are 20.158436 minutes and 5.341873 minutes, respectively.

12

Figure 6: OWLT between Earth and Mars for one Synodic Period (High-Fidelity Ephemeris)

13

Problem 11 Repeat Problem 10, but use the low-fidelity ephemeris model discussed in this chapter (Standish and Williams’ method).

Solution For this problem, we will use the initial date of Jan 1, 2025. As provided in Appendix A, the synodic period of Earth and Mars is 2.1351 years (or 780 days). This results in a final date of Feb 20, 2027.

The positions of Earth and Mars in the J2000 ecliptic reference frame at the initial date are listed in the following tables. 𝑥C (AU)

𝑦C (AU)

𝑧C (AU)

-0.178693 0.966957 -0.000055 𝑥D (AU)

𝑦D (AU)

𝑧D (AU)

-0.521810 1.525229

0.044761

The OWLT between Earth and Mars for this synodic period is plotted on Figure 7. The maximum and minimum OWLT computed are 20.160159 minutes and 5.341875 minutes, respectively. Note that OWLT is almost identical to what was computed in the previous problem using a high-fidelity ephemeris model.

14

Figure 7: OWLT between Earth and Mars for one Synodic Period (Low-Fidelity Ephemeris)

15

Problem 12 Using the method explained in Section 4.6, compute the position of the Sun with respect to the Earth for a given time (e.g., when your homework is due).

Solution For this problem, we will use Jan 1, 2025 as the given date. The following tables shows the parameters, values, and referenced equations used to compute solar ephemeris with respect to ECI for the given time, and the resulting Sun’s coordinates at the given time, respectively.

Parameter

Value

Reference

𝜆𝑀@

281.39198554˝

Equation (4.52)

𝑀@

358.03080941˝

Equation (4.53)

𝜆𝑒𝑐𝑙𝑖𝑝𝑡𝑖𝑐

281.32482029˝

Equation (4.54)

𝜙𝑒𝑐𝑙𝑖𝑝𝑡𝑖𝑐

0˝

Equation (4.54)

𝑟@

0.98330568 AU Equation (4.55)

𝜀C

23.43602619˝

𝑥@ (AU) 0.19311875

𝑦@ (AU)

Equation (4.8) 𝑧@ (AU)

-0.88474131 -0.38352241

Although not requested by the problem statement, computing the Sun’s position with respect to the Earth for 10 years from the given date results in the components and magnitudes shown in Figure 8.

16

Figure 8: Sun’s Position with respect to the Earth

17

Problem 13 Using JPL Horizons and the low-fidelity model discussed in Section 4.7.1, generate the ephemeris data for Venus from a given time (e.g., when your homework is due) for 10 years. How do the two methods compare? Plot the difference in distance of the planet with respect to the Sun and the planet’s orbital speed.

Solution In order to use Standish and William’s approximation, we need to refer to Appendix C, and use the appropriate values corresponding to the celestial body of interest in combination with Equations (4.57) - (4.59). After obtaining the orbital elements of the body of interest, we can use the methods and equations in Chapter 2 to obtain position and velocity of any given body with respect to the gravitational source (here, the Sun).

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Venus, this results in the plots shown in Figure 9 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 10.

18

Figure 9: Venus’s Distance and Speed (Ephemeris Comparison)

19

Figure 10: Venus’s SMA, eccentricity, and inclination (Ephemeris Comparison)

20

Problem 14 Repeat Problem 13 for Mercury.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Mercury, this results in the plots shown in Figure 11 where the Earth is also shown for comparison. Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 12.

21

Figure 11: Mercury’s Distance and Speed (Ephemeris Comparison)

22

Figure 12: Mercury’s SMA, eccentricity, and inclination (Ephemeris Comparison)

23

Problem 15 Repeat Problem 13 for Jupiter.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Jupiter, this results in the plots shown in Figure 13 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 14.

24

Figure 13: Jupiter’s Distance and Speed (Ephemeris Comparison)

25

Figure 14: Jupiter’s SMA, eccentricity, and inclination (Ephemeris Comparison)

26

Problem 16 Repeat Problem 13 for Saturn.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Saturn, this results in the plots shown in Figure 15 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 16.

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Figure 15: Saturn’s Distance and Speed (Ephemeris Comparison)

28

Figure 16: Saturn’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Problem 17 Repeat Problem 13 for Uranus.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Uranus, this results in the plots shown in Figure 17 where the Earth is also shown for comparison. Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 18.

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Figure 17: Uranus’s Distance and Speed (Ephemeris Comparison)

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Figure 18: Uranus’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Problem 18 Repeat Problem 13 for Neptune.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Neptune, this results in the plots shown in Figure 19 where the Earth is also shown for comparison. Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 20.

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Figure 19: Neptune’s Distance and Speed (Ephemeris Comparison)

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Figure 20: Neptune’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Problem 19 Repeat Problem 13 for Pluto.

Solution Refer to the solution approach given in Problem 13.

For this problem, we will use the initial date of Jan 1, 2025 and final date of Jan 1, 2035 (10 years later).

For Pluto, this results in the plots shown in Figure 21 where the Earth is also shown for comparison.

Although not requested by the problem statement, the semimajor axis, eccentricity, and inclination differences are shown in Figure 22.

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Figure 21: Pluto’s Distance and Speed (Ephemeris Comparison)

37

Figure 22: Pluto’s SMA, eccentricity, and inclination (Ephemeris Comparison)

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Problem 20 Use Standish and Williams’ method to estimate the date on which the next conjunction between Earth and Mars occurs, i.e. when Earth and Mars are on the opposite side of the Sun. Solution For this problem, we will use Jan 1, 2025 as the starting date, i.e. we want to estimate when the next Earth-Mars conjunction happens from this given date. To do this, we will look at dates that go from this given date for a full Earth-Mars synodic period (780 days), i.e. until Feb 20, 2027.

Two possible ways to estimate this are to 1. find the maximum distance between Earth and Mars during the given synodic period; or 2. find when the angle between the positions of Earth and Mars is close to 180˝ ; or 3. plot the orbits of Earth and Mars and visually estimate when the planets are in conjunction (we won’t show this method) For the first method, we compute the ephemeris data and search for the maximum distance, i.e. ` ˘ 𝑑𝑚𝑎𝑥 “ max |rD{@ ´ rC{@ | for the given dates. This also means that the one-way light time (OWLT) is maximized. Referring to Figure 7, we can estimate this date to be around Dec 1, 2025.

For the second method, we compute the ephemeris data and search for the value of the angle between the Earth and Mars position vectors (Δ𝜃) that is closest to 180˝ , i.e. we compute Δ𝜃 using the definition of the dot product ˙ ˆ rD{@ ¨ rC{@ Δ𝜃 “ arccos |rD{@ ||rC{@ | Plotting Δ𝜃 as a function of time for the given dates results in Figure 23, from which we can estimate the conjunction date to be around Jan 9, 2026.

While these are estimates, if the orbits of the planets were exactly circular and coplanar, these methods would provide the same (exact) answer.

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Figure 23: Angle between Earth’s and Mars’s position vectors

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Interplanetary Astrodynamics Chapter 5 Problem Solutions

Problem 1 A spacecraft is launched into a circular orbit around the Earth at an altitude of 450 km. The upper stage of the launch vehicle initiates a Hohmann transfer to GEO through a GTO. How much Δ𝑣 is the launch vehicle expected to perform (Δ𝑣1 ) and how much Δ𝑣 is the spacecraft’s propulsion system expected to deliver in order to perform the adequate orbit insertion at GEO (Δ𝑣2 )? How long does the transfer take?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Earth are 𝜇C “ 398, 600 km3 {s2 𝑅C “ 6, 371 km Therefore, the spacecraft starts in a LEO having radius and velocity 𝑟1 “ 𝑅C ` 450 km “ 6, 821 km c 𝜇C 𝑣1 “ “ 7.64442 km/s 𝑟1 The spacecraft needs to transfer to GEO, i.e. an orbit which has period equal to the length of a sidereal day, or 23 hours 56 minutes and 4.091 seconds, which gives 𝑇𝐺𝐸𝑂 “ 23 ˆ 3600 ` 56 ˆ 60 ` 4.091 “ 86, 164.091 s Thus, the semimajor axis corresponding to this orbital period is ˆ ˙2{3 𝑇𝐺𝐸𝑂 1{3 𝑎𝐺𝐸𝑂 “ 𝜇C “ 42, 164.15 km 2𝜋 Using Equations (5.4) – (5.10) with 𝑟1 “ 6, 821 km and 𝑟2 “ 𝑟𝐺𝐸𝑂 gives Δ𝑣1 “ 2.38553 km/s Δ𝑣2 “ 1.452085 km/s Δ𝑣𝑡𝑜𝑡 “ 3.83762 km/s And the time of flight is d 𝑇 𝑂𝐹 “ 𝜋

p𝑟1 ` 𝑟2 q3 “ 5.29820 hours 8𝜇C 1

Problem 2 A space vehicle in a circular orbit at an altitude of 300 km above Mars’s surface executes a Hohmann transfer. The total Δ𝑣 available is 0.35 km/s. This means that this Δ𝑣 needs to be split between Δ𝑣1 (to initiate the transfer) and Δ𝑣2 (to complete the transfer). What is the maximum altitude of the final orbit that this spacecraft can reach with the allotted Δ𝑣? How long does this transfer take?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Mars are 𝜇D “ 43, 050 km3 {s2 𝑅D “ 3, 389.5 km so that the spacecraft starts at a radius and orbital speed 𝑟1 “ 𝑅D ` 300 km “ 3, 689.5 km c 𝜇D 𝑣1 “ “ 3.40706 km/s 𝑟1 We then use Equation (5.10) Δ𝑣𝑡𝑜𝑡 “ 𝑣𝑐1

ˆ

1 1´ 𝑅

˙c

2𝑅 1 ` ? ´1 𝑅`1 𝑅

with Δ𝑣𝑣𝑐𝑡𝑜𝑡 “ 0.102728 to numerically solve for 𝑅, which leads to 𝑅 “ 1.24293. This means that the 1 final altitude 𝑧2 that the spacecraft reaches is 𝑧2 “ 𝑟2 ´ 𝑅D “ 𝑅𝑟1 ´ 𝑅D “ 1, 196.27 km

2

Problem 3 A spacecraft completes an interplanetary Hohmann transfer from Earth to Venus. The crew wants to come back to Earth. (a) How long do they have to wait before Earth and Venus align themselves to allow for a Hohmann transfer from Venus to Earth? (b) What is the angle between Earth and Venus at departure? (c) What is the angle between Earth and Venus at arrival?

Solution Using Appendix A, we can find the gravitational parameter of the Sun, the Earth’s and Venus’s distances from the Sun 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝑟C{@ “ 149, 597, 800 km 𝑟B{@ “ 108, 204, 088.7 km We then compute the orbital periods of Earth and Venus, along with their mean motions d 3 𝑟C{@ “ 365.25 days 𝑇C “ 2𝜋 𝜇@ d 3 𝑟B{@ 𝑇B “ 2𝜋 “ 224.65 days 𝜇@ 360˝ 𝑛C “ “ 0.985626 ˝ {day 𝑇C 360˝ “ 1.602493 ˝ {day 𝑛B “ 𝑇B Computing the semimajor axis and time of flight of an Earth-Venus Hohmann transfer gives 1 𝑎𝐻 “ p𝑟C{@ ` 𝑟B{@ q “ 128, 900, 944 km 2d 𝑎3𝐻 𝑇 𝑂𝐹𝐻 “ 𝜋 “ 146 days 𝜇@ Figure 1 shows the positions of Earth and Venus at departure from Earth and arrival at Venus for the outbound (Earth-Venus) trajectory, while Figure 2 shows the positions of Earth and Venus at departure from Venus and arrival at Earth for the inbound, or return, (Venus-Earth) trajectory.

For the outbound transfer (Earth-Venus) we arbitrarily set Earth’s angle at 𝛼C,𝑑𝑒𝑝 “ 0˝ at departure (or launch) from Earth. Therefore, for the outbound trajectory we can compute the Earth and Venus angles when the spacecraft leaves Earth at departure ‘dep’ and at Venus upon arrival ‘arr’

3

as follows 𝛼C,𝑑𝑒𝑝 “ 0˝ 𝛼B,𝑎𝑟𝑟 “ 180˝ 𝛼C,𝑎𝑟𝑟 “ 𝑛C 𝑇 𝑂𝐹𝐻 “ 143.9˝ 𝛼B,𝑑𝑒𝑝 “ 𝛼B,𝑎𝑟𝑟 ´ 𝑛B 𝑇 𝑂𝐹𝐻 “ 234˝ ´ 180˝ “ ´54˝ This means that at departure Venus is 54˝ behind Earth and at arrival Venus is 𝛼B,𝑎𝑟𝑟 ´𝛼C,𝑎𝑟𝑟 “ 36.1˝ ahead of Earth.

For the return trip from Venus, we must wait until Earth and Venus are once again aligned appropriately to perform a Hohmann transfer – this time from Venus to Earth. Mathematically, this means that the angle between Earth and Venus must grow by 360˝ ´ 2p𝛼B,𝑎𝑟𝑟 ´ 𝛼C,𝑎𝑟𝑟 q “ 287.8˝ which corresponds to a wait time of Δ𝑛 “ 𝑛B ´ 𝑛C “ 0.616866 ˝ {day 287.8˝ 𝑇𝑤𝑎𝑖𝑡 “ “ 466.55 days Δ𝑛 To summarize a the crew must wait approximately 467 days b For the outbound trajectory, at departure Venus is 54˝ behind Earth, while at arrival Earth is 36.1˝ ahead of Venus c For the inbound trajectory, at departure Earth is 36.1˝ behind Venus, while at arrival Venus is 54˝ ahead of Earth

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Figure 1: Earth-Venus Hohmann Transfer (outbound)

5

Figure 2: Earth-Venus Hohmann Transfer (outbound)

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Problem 4 For a Hohmann transfer from Earth to Venus, plot C3 vs. time for a window that begins 15 days before the optimal Hohmann geometry and ends 15 days after the Hohmann alignment. Increment launch date by 1 day, and use the time of flight of 146 days. Keep the date of arrival constant.

Solution In order to compute 𝐶3 , we need to first compute the solution to Lambert’s problem (refer to Sections 5.3.1 or 5.3.2) using r1 “ 𝑟C{@ rcos 𝜃 sin 𝜃 0s r2 “ 𝑟B{@ r´1 0 0s c 𝜇@ vC{@ “ r´ sin 𝜃 cos 𝜃 0s 𝑟C{@ such that 𝜃 changes following Earth’s mean motion, or 𝑛C “

360˝ 𝑇C

which gives 𝜃 “ 𝑛C Δ𝑇 with Δ𝑇 P r´15, 15s days. Then, solving Lambert’s problem with 𝑇 𝑂𝐹 “ 146 days gives the required initial velocity for the transfer to take place, or v1 . We can therefore compute the velocity of the spacecraft at Earth’s sphere of influence as 𝑣8{C “ |v1 ´ vC{@ | From which we compute 𝐶3 2 𝐶3 “ 𝑣8{C

This results in the plot shown in Figure 3.

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Figure 3: Earth-Venus 𝐶3 as a function of time

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Problem 5 Starting with Equations (5.12) – (5.15) and the definitions of R and S, prove Equation (5.16). Hint: an equation that you will find useful in this derivation is that of angular momentum ℎ written as ? b 𝑟𝑎 𝑟𝑝 a function of apoapse radius ra and periapse radius rp only, i.e. ℎ “ 2𝜇 𝑟𝑎 `𝑟𝑝 Solution For this derivation, we will use the given definitions of 𝑅 and 𝑆 such that 𝑟2 𝑟1 𝑟𝑖 𝑆“ 𝑟2 𝑟𝑖 𝑅𝑆 “ 𝑟1 𝑅“

Referring to Figure 5.2 in Chapter 5, our goal is to write the various velocities shown based on the initial circular speed and the above definitions. We therefore start by writing the initial and final circular speeds c 𝜇 𝑣𝑐1 “ 𝑟1 c 𝜇 𝑣𝑐2 “ 𝑟2 For the bi-elliptic transfer, two transfer trajectories are used having periapse and apoapse radii as given in the following table Transfer Periapsis

Using the suggested equation, ℎ “

Apoapsis

1

𝑟𝑝1 “ 𝑟1

𝑟𝑎1 “ 𝑟𝑖

2

𝑟𝑝2 “ 𝑟2

𝑟𝑎2 “ 𝑟𝑖

? b 𝑟𝑎 𝑟𝑝 2𝜇 𝑟𝑎 `𝑟𝑝 and definitions of 𝑅 and 𝑆, we compute the angular

momenta of these transfer trajectories as c

𝑅𝑆 ? 𝑟1 𝑅𝑆 ` 1 c a 𝑆 ? ℎ2 “ 2𝜇 𝑟2 𝑆`1

ℎ1 “

a

2𝜇

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and we compute the velocities on these trajectories at periapsis and apoapsis c c ℎ1 a 𝑅𝑆 2𝑅𝑆 1 𝑣𝑝1 “ “ 2𝜇 𝑣𝑐 ? “ 𝑟1 𝑅𝑆 ` 1 𝑟1 𝑅𝑆 ` 1 1 c c 𝑅𝑆 1 1 2𝑅𝑆 1 ℎ1 a “ 2𝜇 𝑣𝑐 𝑣𝑎1 “ ? “ 𝑟𝑖 𝑅𝑆 ` 1 𝑅𝑆 𝑟1 𝑅𝑆 ` 1 𝑅𝑆 1 c c ℎ2 a 𝑆 1 2𝑆 ? 𝑣𝑝2 “ “ 2𝜇 “ 𝑣𝑐 𝑟2 𝑆 ` 1 𝑟1 2 𝑆`1 2 c c ℎ2 a 𝑆 1 1 2𝑆 1 “ 2𝜇 𝑣𝑐 𝑣𝑎2 “ ? “ 𝑟𝑖 𝑆 ` 1 𝑆 𝑟2 𝑆 `1𝑆 2 Thus, each Δ𝑣 maneuver is the difference of the required vs. current velocities, such that ¸ ˜c 2𝑅𝑆 Δ𝑣1 “ 𝑣𝑝1 ´ 𝑣𝑐1 “ ´ 1 𝑣𝑐1 𝑅𝑆 ` 1 ¸ ˜c c 2𝑆 1 1 2𝑅𝑆 1 ? ´ Δ𝑣2 “ 𝑣𝑎2 ´ 𝑣𝑎1 “ 𝑣𝑐1 𝑆 `1𝑆 𝑅 𝑅𝑆 ` 1 𝑅𝑆 ¸ ˜c 2𝑆 1 1 ? ´? 𝑣𝑐1 Δ𝑣3 “ 𝑣𝑝2 ´ 𝑣𝑐2 “ 𝑆`1 𝑅 𝑅 where we used the fact that 𝑟2 “ 𝑅𝑟1 . Combining all Δ𝑣 maneuvers together and dividing by 𝑣𝑐1 gives Δ𝑣𝑡𝑜𝑡 “ Δ𝑣1 ` Δ𝑣2 ` Δ𝑣3 Δ𝑣𝑡𝑜𝑡 Δ𝑣1 Δ𝑣2 Δ𝑣3 “ ` ` 𝑣𝑐1 𝑣𝑐1 𝑣𝑐1 𝑣𝑐1 which, combining all the previous Δ𝑣 expressions in terms of 𝑅 and 𝑆 can be simplified to d c ˆ ˙ 2 2p1 ` 𝑆q 1 Δ𝑣𝑡𝑜𝑡 “ p𝑅𝑆 ´ 1q ` ´ 1` ? 𝑣𝑐1 p𝑅𝑆 ` 1q𝑅𝑆 𝑅𝑆 𝑅 which is equivalent to Equation (5.16), thus completing the proof.

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Problem 6 Compute the Δ𝑣 and TOF for a bi-elliptic heliocentric transfer between Earth (1 AU) and Uranus (19.19 AU) using S = 2. How do Δ𝑣 and TOF for this transfer compare with those of a Hohmann transfer?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Earth are 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝑟1 “ 𝑟C{@ “ 149, 597, 800 km 𝑟2 “ 𝑟G{@ “ 2, 870, 991, 219 km The ratio of 𝑟2 to 𝑟1 is 𝑅 “ 𝑟𝑟21 “ 19.1914, which is larger than the critical R value discussed in Chapter 5, so we expect the Δ𝑣 for a bi-elliptic transfer to be lower than a Hohmann transfer. Also, since 𝑆 “ 2, then 𝑟𝑖 “ 𝑆𝑟2 “ 5, 741, 982, 438 km.

We start in a heliocentric orbit having radius 𝑟1 so the orbital speed is c 𝜇@ 𝑣𝑐1 “ “ 29.7833 km/s 𝑟C{@ Computing the total Δ𝑣 for the bi-elliptic transfer using Equation (5.16) and the Hohmann transfer using Equation (5.10) gives ˙ Δ𝑣𝑡𝑜𝑡 “ 0.535175 𝑣𝑐1 𝐻 𝑜ℎ𝑚𝑎𝑛𝑛 ˙ Δ𝑣𝑡𝑜𝑡 “ 0.526874 𝑣𝑐1 𝐵𝑖´𝑒𝑙𝑙𝑖𝑝𝑡𝑖𝑐 so that the total Δ𝑣 for each transfer type is Δ𝑣𝑡𝑜𝑡 q𝐻 𝑜ℎ𝑚𝑎𝑛𝑛 “ 15.9393 km/s Δ𝑣𝑡𝑜𝑡 q𝐵𝑖´𝑒𝑙𝑙𝑖𝑝𝑡𝑖𝑐 “ 15.6920 km/s This means that this type of bi-elliptic transfer (with 𝑆 “ 2) has a Δ𝑣 saving of 0.247259 km/s, or approximately 1.55% when compared to a Hohmann transfer. However, the difference in time of flight is much more significant ˘ 1` 𝑎𝐻 “ 𝑟C{@ ` 𝑟G{@ “ 1, 510, 294, 510 km 2d 𝑎3𝐻 𝑇 𝑂𝐹𝐻 𝑜ℎ𝑚𝑎𝑛𝑛 “ 𝜋 “ 16.0403 years 𝜇@ d ˜d ¸ 3 p𝑟1 ` 𝑟𝑖 q p𝑟𝑖 ` 𝑟2 q3 𝑇 𝑂𝐹𝐵𝑖´𝑒𝑙𝑙𝑖𝑝𝑡𝑖𝑐 “ 𝜋 ` “ 120.927 years 8𝜇@ 8𝜇@ which means that this bi-elliptic transfer takes approximately 7.53897 times longer than a Hohmann transfer.

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Problem 7 A resupply of potatoes for Mark Watney is currently orbiting Mars in a circular orbit with altitude of 2,000 km (green orbit in Figure 5.71). The resupply vehicle performs a retrograde Δ𝑣 that puts it on an elliptical orbit (blue orbit) intersecting Mars at an angle 𝛼 “ 30˝ from the line of apsides (LOA) of the new orbit, as shown in Figure 5.71. Calculate the Δ𝑣 (in km/s) required to perform this orbital maneuver.

Figure 5.71: Mark Watney, Visibly Angry Solution Let’s define point A as the starting spacecraft location at apoareion, and point B as the point where the spacecraft impacts Mars. Then, in polar coordinates, the p𝑟, 𝜃q coordinates of points A and B are p𝑟𝐴 “ 𝑅D ` 2, 000 km, 190˝ q and p𝑅D , ´30˝ q, respectively, where 𝑟𝐴 “ 5, 389.5 km. Let’s denote with subscripts ‘1’ and ‘2’ the initial and final orbits of the spacecraft, respectively. Then, the orbital speed and angular momentum of the spacecraft at point A on orbit 1 are: c 𝜇D 𝑣𝐴 q1 “ “ 2.81896 km/s 𝑟𝐴 ℎ1 “ 𝑟𝐴 𝑣𝐴 q1 “ 15, 192.8 km2 {s Evaluating the orbit equation at points A and B on orbit 2 gives us the following system of

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equations in terms of the angular momentum and eccentricity of orbit 2, i.e. ℎ2 and 𝑒2 𝑟𝐴 “ 𝑅D “

ℎ22 {𝜇D : ´1

1 ` 𝑒2 cosp180 ˝ q ℎ22 {𝜇D

?

3

: 2

cosp´30 1 ` 𝑒2 ˝ q

which simplifies to ℎ22 “ 𝑟𝐴 p1 ´ 𝑒2 q 𝜇D ? ˙ ˆ ℎ22 3 “ 𝑅D 1 ` 𝑒2 𝜇D 2 Equating these two expressions and solving first for 𝑒2 and then for ℎ2 gives 𝑒2 “ 0.240243 ℎ2 “ 13, 242.7 km2 {s Thus, the orbital speed of the spacecraft on orbit 2 at point A can be computed 𝑣𝐴 q2 “

ℎ2 “ 2.45713 km/s 𝑟𝐴

So the required Δ𝑣 to transfer from orbit 1 to orbit 2 is Δ𝑣 “ 𝑣𝐴 q1 ´ 𝑣𝐴 q2 “ 0.361831 km/s

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Problem 8 A spacecraft is orbiting Earth with semimajor axis 𝑎1 = 15,000 km, and eccentricity 𝑒1 = 0.45. (a) Calculate the flight-path angle 𝜙1 when the spacecraft reaches a true anomaly of 𝜃1 “ 258˝ . (b) Calculate the Δ𝑣 needed to circularize the orbit at 𝜃1 “ 258˝ where the radius is 𝑟1 . (c) How much has the line of apsides rotated from the initial elliptical orbit to the final circular orbit?

Solution As provided by Appendix A, the gravitational parameter of the Earth is 𝜇C “ 398, 600 km3 {s2 . Let ‘1’ denote the initial orbit and ‘2’ denote the final orbit. Then, 𝑎1 “ 15, 000 km 𝑒1 “ 0.45 (a) We are given that 𝜃1 “ 258˝ . We start by computing the semilatus rectum, energy, radius, speed, and angular momentum of the spacecraft on orbit 1 using the information that we are given 𝑝1 “ 𝑎1 p1 ´ 𝑒12 q “ 11, 362.5 km 𝜇C 𝐸1 “ “ ´13.2867 km2 {s2 2𝑎1 𝑝1 𝑟1 “ “ 13, 197.2 km 1 ` 𝑒1 cos 𝜃1 d ˆ ˙ 𝜇C “ 5.81664 km/s 𝑣1 “ 2 𝐸1 ` 𝑟1 ? ℎ1 “ 𝜇C 𝑝1 “ 69, 052.5 km2 {s Thus, the flight path angle 𝜙 can be computed as ˆ ˙ ℎ1 𝜙1 “ ˘ arccos “ ˘0.452061 rad 𝑟1 𝑣1 To solve the sign ambiguity, we refer to where on the orbit the spacecraft is located, which can be determined by the value of 𝜃1 . Since 180˝ ă 𝜃1 ă 360˝ , then 𝜙1 must be negative, i.e. 𝜙1 “ ´0.452061 rad or ´25.9012˝ . (b) To circularize an orbit it means to transfer the spacecraft from orbit 1 to a circular orbit (orbit 2). Thus, we must have that 𝜙2 “ 0 c 𝑣2 “ 𝑣𝑐 “

𝜇C “ 5.49575 km/s 𝑟1

Thus, the change in flight path angle is Δ𝜙 “ |𝜙2 ´ 𝜙1 | “ 0.452061 rad 14

Using the law of cosines, we can compute the Δ𝑣 needed to perform this circularization maneuver a Δ𝑣 “ 𝑣12 ` 𝑣22 ´ 2𝑣1 𝑣2 cos Δ𝜙 “ 2.55444 km/s (c) Since the line of apsides (LOA) for a circular orbit is undefined, we cannot determine the LOA rotation from orbit 1 to orbit 2.

15

Problem 9 Calculate the Δ𝑣 required to change the inclination of an Earth orbit at apoapsis, using the following data: 𝑎 = 13,952 km, 𝑒 = 0.47, 𝑖1 “ 28.5˝ , 𝑖2 “ 63.4˝ , 𝜔 “ 0˝ . Note that these orbital elements define an orbit with periapsis at the ascending node.

Solution As provided by Appendix A, the gravitational parameter of the Earth is 𝜇C “ 398, 600 km3 {s2 . We can summarize the given orbital parameters as follows 𝑎 “ 13, 952 km 𝑒 “ 0.47 𝑖1 “ 28.5˝ 𝑖2 “ 63.4˝ 𝜔 “ 0˝ The inclination change maneuver needs to perform an inclination change of Δ𝑖 “ 𝑖2 ´ 𝑖1 “ 34.9˝ .

Since 𝜔 “ 0˝ , the inclination change maneuver must be performed at the ascending or descending node (here, periapsis or apoapsis) in order to be a pure inclination change, i.e. the other orbital parameters are left unchanged by the maneuver. Since Δ𝑣 for an inclination change is directly proportional to the transverse speed of the spacecraft, we would perform this maneuver at apoapsis, where the spacecraft is going slower. Let’s compute the apoapse speed of the spacecraft 𝜇C “ ´14.2847 km2 {s2 2𝑎 𝑟𝑎 “ 𝑎p1 ` 𝑒q “ 20, 509.4 km c 𝜇C q “ 3.20944 km/s 𝑣𝑎 “ 2p𝐸 ` 𝑟𝑎 𝐸“

Thus, the Δ𝑣 required to perform the inclination change is ˆ ˙ Δ𝑖 Δ𝑣 “ 2𝑣𝜃 sin 2 ˆ ˙ Δ𝑖 “ 2𝑣𝑎 sin “ 1.92485 km/s 2

16

Problem 10 Consider a spacecraft in GEO. It is desired to shift its longitude by 12˝ westward using three revolutions of its phasing orbit. Calculate the orbital period of the phasing orbit, the total TOF and Δ𝑣 for such phasing maneuver.

Solution As provided by Appendix A, the gravitational parameter of the Earth is 𝜇C “ 398, 600 km3 {s2 .

Furthermore, from the solution of Problem 1, we know that 𝑇𝐺𝐸𝑂 “ 86, 164 s 𝑟𝐺𝐸𝑂 “ 𝑎𝐺𝐸𝑂 “ 42, 164.15 km 𝑣𝐺𝐸𝑂 “ 3.074659 km/s from which we can compute the mean motion of a spacecraft in GEO 𝑛𝐺𝐸𝑂 “

2𝜋 “ 7.29212 ˆ 10´5 rad/s 𝑇𝐺𝐸𝑂

Shifting spacecraft position means performing a phasing maneuver in 𝑁 “ 3 revolutions. We start by computing the orbital period of the phasing orbit 𝑇𝑝ℎ for the required phasing angle 𝛼 “ 12˝ (note that for westward phasing 𝛼 ą 0, while for eastward phasing 𝛼 ă 0) ` ˘ 𝑛𝐺𝐸𝑂 𝑁 𝑇𝑝ℎ “ 2𝜋𝑁 ` 𝛼 from which 2𝜋𝑁 ` 𝛼 𝑛𝐺𝐸𝑂 𝑁 𝑇𝐺𝐸𝑂 2𝜋 𝛼 “ ` 𝑛𝐺𝐸𝑂 𝑛𝐺𝐸𝑂 𝑁 “ 86, 164 ` 957.4 “ 87, 121.4 s

𝑇𝑝ℎ “

Thus, the time of flight for this maneuver to occur is 𝑇 𝑂𝐹 “ 𝑁 𝑇𝑝ℎ “ 261, 364 s or approximately 3.02505 days. We then compute the semimajor axis, apoapse radius, angular momentum, and periapse velocity associated with the phasing orbit in order to quantify Δ𝑣 ˆ ˙2{3 𝑇𝑝ℎ 1{3 𝑎𝑝ℎ “ 𝜇C “ 42, 475.9 km 2𝜋 𝑟𝑎,𝑝ℎ “ 2𝑎𝑝ℎ ´ 𝑟𝐺𝐸𝑂 “ 42, 787.6 km c a 𝑟𝑎,𝑝ℎ 𝑟𝐺𝐸𝑂 ℎ𝑝ℎ “ 2𝜇C “ 130, 115 km2 {s 𝑟𝑎,𝑝ℎ ` 𝑟𝐺𝐸𝑂 ℎ𝑝ℎ 𝑣𝑝,𝑝ℎ “ “ 3.08592 km/s 𝑟𝐺𝐸𝑂 17

Therefore, the required Δ𝑣 to initiate the phasing maneuver is Δ𝑣1 “ 𝑣𝑝,𝑝ℎ ´ 𝑣𝐺𝐸𝑂 “ 11.2626 m/s and the Δ𝑣2 required to complete the maneuver is equal to Δ𝑣1 , so that Δ𝑣2 “ Δ𝑣1 “ 11.2626 m/s Δ𝑣𝑡𝑜𝑡 “ Δ𝑣1 ` Δ𝑣2 “ 22.5252 m/s

18

Problem 11 In this problem use canonical units such that 𝜇d “ 4𝜋 2 LU3 /TU2 , where 1 LU = 1 AU and 1 TU = 1 year. Planet Express has been tasked with delivering a shipment of buggalo milk from Mars to a colony of Amphibiosans orbiting Jupiter. Assume all transfers in this problem are prograde and that Mars and Jupiter are in circular orbits around the Sun. Also, ignore the gravity of Mars and Jupiter. (a) Determine the type of orbit (i.e., conic section) for the heliocentric cruise required to transfer from Mars to Jupiter provided that the transfer angle Δ𝜃 is 192.558˝ and the time of flight (TOF) is 1.4 years. (b) Treating the transfer angle Δ𝜃 as a variable, determine the range of values for the speed 𝑣1 (in LU/TU) that the Planet Express Ship (“Old Bessie”) would require for a minimum-energy transfer ellipse as it enters the transfer orbit at Mars.

Solution We use the constants given in the problem statement so that 𝜇@ “ 4𝜋 LU3 {TU2 𝑟1 “ 𝑟D{@ “ 1.523679 LU 𝑟2 “ 𝑟E{@ “ 5.202887 LU where here 1 LU is equivalent to 1 AU and 1 TU is equivalent to 1 year. We are also given the transfer angle and time of flight Δ𝜃 “ 192.558˝ “ 3.3608 rad 𝑇 𝑂𝐹 “ 1.4 TU (a) In order to determine the orbit type for this transfer, we start by computing the chord and semiperimeter a 𝑐 “ 𝑟12 ` 𝑟22 ´ 2𝑟1 𝑟2 cos Δ𝜃 “ 6.6983 LU 1 𝑠 “ p𝑟1 ` 𝑟2 ` 𝑐q “ 6.7124 LU 2 ˝ ˝ Then, since 180 ă Δ𝜃 ă 360 we compute the parabolic time of flight as follows ? ‰ 2 “ 3{2 𝑡𝑝 “ ? 𝑠 ` p𝑠 ´ 𝑐q3{2 “ 1.304897 TU 3 𝜇@ Since 𝑇 𝑂𝐹 “ 1.4 TU ą 𝑡𝑝 , the orbit must be elliptical. (b) Since we want to compute the range of 𝑣1 for the minimum energy case, we start by computing the minimum energy semimajor axis as a function of Δ𝜃, which is the only unspecified parameter 𝑠 1 1a 2 𝑎𝑚𝑖𝑛 “ “ p𝑟1 ` 𝑟2 q ` 𝑟 ` 𝑟22 ´ 2𝑟1 𝑟2 cos Δ𝜃 2 4 4 1 1? “ 1.6816 ` 29.3916 ´ 15.8551 cos Δ𝜃 4 19

We then compute the minimum and maximum 𝑎𝑚𝑖𝑛 values, which occur for Δ𝜃 “ 0 and Δ𝜃 “ 𝜋, respectively (one can more rigorously prove this by taking a derivative of 𝑎𝑚𝑖𝑛 with respect to Δ𝜃) # 2.6014 LU Δ𝜃 “ 0 𝑎𝑚𝑖𝑛 “ 3.3633 LU Δ𝜃 “ 𝜋 Using the vis-viva equation we compute the energy and velocity associated with each Δ𝜃 value # ´7.5878 LU2 {TU2 Δ𝜃 “ 0 ´𝜇@ 𝐸𝑚𝑖𝑛 “ “ 2𝑎𝑚𝑖𝑛 ´5.8690 LU2 {TU2 Δ𝜃 “ 𝜋 d ˆ ˙ # 6.053452 LU{TU Δ𝜃 “ 0 𝜇@ “ 𝑣1 “ 2 𝐸𝑚𝑖𝑛 ` 𝑟1 6.331019 LU{TU Δ𝜃 “ 𝜋 Thus, 𝑣1 P r6.053452, 6.331019s LU/TU. Notice that the minimum energy ellipse has a size (semimajor axis) based on the value of transfer angle Δ𝜃. So one should be mindful of the fact that the minimum energy ellipse for a transfer is not unique if only time of flight and the magnitudes of the initial and final positions are given.

The most notable minimum energy ellipse is the Hohmann transfer, which has transfer angle of Δ𝜃 “ 180˝ along with the assumptions that the starting and final orbits are coplanar and circular.

20

Problem 12 A spacecraft is transferring from Earth to Mars. Assume a change in true anomaly of 120˝ (true anomaly from Earth departure to Mars arrival). Find the heliocentric transfer orbit characteristics (𝑎 and 𝑒) assuming a 300-, 200-, 100-, 50-, and 25-day transfer. Be sure to state all of your assumptions.

Make sure to compute the minimum transfer time you can have and still have an elliptical transfer to en-sure that you are solving Lambert’s equation using the correct conic type.

Solution In this problem, we use the physical constants in canonical units, as follows 𝜇@ “ 4𝜋 LU3 {TU2 𝑟1 “ 𝑟C{@ “ 1.00000 LU 𝑟2 “ 𝑟D{@ “ 1.523679 LU where here 1 LU is equivalent to 1 AU and 1 TU is equivalent to 1 year. We are also given the transfer angle Δ𝜃 “ 120˝ “ 2.09439 rad We start by computing the chord and semiperimeter a 𝑐 “ 𝑟12 ` 𝑟22 ´ 2𝑟1 𝑟2 cos Δ𝜃 “ 2.20149 LU 1 𝑠 “ p𝑟1 ` 𝑟2 ` 𝑐q “ 2.36200 LU 2 Then, since Δ𝜃 ă 180˝ we compute the parabolic time of flight as follows ? ‰ 2 “ 3{2 𝑡𝑝 “ ? 𝑠 ´ p𝑠 ´ 𝑐q3{2 “ 97.695 days 3 𝜇@ which means that transfer times below 97.695 days will result in hyperbolic orbits. We solve Lambert’s problem following the solution method given in Section 5.3, making use of the elliptical or hyperbolic case, as appropriate. This gives the results summarized in the following table.

TOF (days)

SMA (AU)

Eccentricity Conic Type

300

1.22027

0.4768

Elliptical

200

1.20035

0.2720

Elliptical

100

83.1556

0.9452

Elliptical

50

–0.224615

4.1741

Hyperbolic

25

–0.0418759

16.1385

Hyperbolic

Notice how the 100-day transfer has an extremely large semimajor axis and an eccentricity close to 1 since this transfer time is only about 3 days more than that for a parabolic transfer orbit.

21

Problem 13 Consider Lambert’s problem. Determine the expression for the parabolic transfer time 𝑡𝑝 , i.e. the transfer time between 𝑃1 and 𝑃2 assuming that the orbit is a parabola, or ? ‰ 2 “ 𝑡𝑝 “ ? 𝑠 3{2 ´ sgnpsin Δ𝜃qp𝑠 ´ 𝑐q3{2 3 𝜇 Start with Lambert’s equation for an elliptical orbit, ? 𝜇p𝑡2 ´ 𝑡1 q “ 𝑎3{2 r𝛼 ´ 𝛽 ´ psin 𝛼 ´ sin 𝛽qs and proceed to the limit as 𝑎 Ñ 8. Be sure to account for the two cases, Δ𝜃 ď 𝜋 and Δ𝜃 ą 𝜋.

Solution We start with Lambert’s equation ? 𝜇p𝑡2 ´ 𝑡1 q “ 𝑎3{2 r𝛼 ´ 𝛽 ´ psin 𝛼 ´ sin 𝛽qs As 𝑎 Ñ 8, then p𝑡2 ´ 𝑡1 q Ñ 𝑡𝑝 . Let’s start the proof by considering the following term of Lambert’s equation 𝑎3{2 p𝛼 ´ sin 𝛼q Using Equations (5.59) and (5.61) with the fact that the orbit is parabolic, we can rewrite 𝛼 as c 𝑠 𝛼 “ 2 arcsin 2𝑎 so we clearly see that as 𝑎 Ñ 8, then 𝛼 Ñ 0. Furthermore, recall the Mclaurin series expansion for the sine and inverse sine functions for a generic function of x 8 ÿ 𝑥3 𝑥5 p´1q𝑛 2𝑛`1 𝑥 “𝑥´ ` ´ ...` sin “ p2𝑛 ` 1q!

3!

5! 𝑛“0

arcsin 𝑥 “

8 ÿ

p2𝑛q! 𝑥 3 3𝑥 5 2𝑛`1 𝑥 “ 𝑥 ` ` ` ...

4𝑛 p𝑛!q2 p2𝑛 ` 1q 6 40 𝑛“0

Using these expansions, we can rewrite 𝛼 ´ sin 𝛼 as ˆ ? ˙ 2 2 𝛼 ´ sin 𝛼 “ ` HOTs 3!𝑎3{2 where the first order terms in 𝛼 sum to zero, so we keep the next term in the series, while HOTs represent Higher Order Terms of the series expansion which approach zero as 𝑎 Ñ 8. Combining this term with the 𝑎3{2 term from Lambert’s equation and taking the limit as 𝑎 Ñ 8, we get ˆ ? ˙ 2 2 3{2 3{2 3{2 𝑎 p𝛼 ´ sin 𝛼q “ 𝑎 𝑠 ` HOTs 3!𝑎3{2 ?

2 3{2 3{2 𝑎 p𝛼 ´ sin 𝛼q Ñ 𝑠 3 22

as 𝑎 Ñ 8. Let’s now look at the term involving 𝛽 𝑎3{2 p𝛽 ´ sin 𝛽q Similarly to before, we can rewrite 𝛽 as c 𝛽 “ 2 arcsin

𝑠´𝑐 2𝑎

which approaches zero as 𝑎 Ñ 8. Expanding and combining the 𝛽 and sin 𝛽 terms, along with using the McLaurin series expansions stated above, gives „ ȷ p𝑠 ´ 𝑐q3{2 3{2 3{2 𝑎 p𝛽 ´ sin 𝛽q “ 𝑎 8 ? 3{2 ` HOTs 2 2𝑎 ?

2 p𝑠 ´ 𝑐q3{2 𝑎3{2 p𝛽 ´ sin 𝛽q Ñ 3 as 𝑎 Ñ 8. Notice, however, that the sign of 𝛽 is determined by the value ofΔ𝜃 such that # `𝛽 if Δ𝜃 P r0, 𝜋q 𝛽“ ´𝛽 if Δ𝜃 P r𝜋, 2𝜋q Combining all terms gives ? ‰ 2 “ 𝑡𝑝 “ ? 𝑠 3{2 ´ sgnpsin Δ𝜃qp𝑠 ´ 𝑐q3{2 3 𝜇 which is the equation for 𝑡𝑝 that we needed to prove.

23

Problem 14 Show that 𝑝𝑚 “

𝑟1 𝑟2 p1 ´ cos Δ𝜃q 2p𝑠 ´ 𝑟1 qp𝑠 ´ 𝑟2 q “ 𝑐 𝑐

Solution We can simplify the denominator in the problem statement so that 2p𝑠 ´ 𝑟1 qp𝑠 ´ 𝑟2 q “ 𝑟1 𝑟2 p1 ´ cos Δ𝜃q Defining the left-hand side of this equation ‘LHS’ and the right-hand side as ‘RHS’, we thus need to prove that LHS = RHS. Seeing that the RHS has a cos Δ𝜃 term, we start with the definition of the chord using the law of cosines and simplify until we obtain the RHS of the above equation 𝑐 2 “ 𝑟12 ` 𝑟22 ´ 2𝑟1 𝑟2 cos Δ𝜃 2𝑟1 𝑟2 cos Δ𝜃 “ 𝑟12 ` 𝑟22 ´ 𝑐2 ´𝑟 2 ´ 𝑟22 ` 𝑐2 𝑟1 𝑟2 ´ 𝑟1 𝑟2 cos Δ𝜃 “ 1 ` 𝑟1 𝑟2 2 ´𝑟 2 ´ 𝑟22 ` 2𝑟1 𝑟2 ` 𝑐 2 𝑟1 𝑟2 p1 ´ cos Δ𝜃q “ 1 2 2 𝑐 ´ p𝑟2 ´ 𝑟1 q2 𝑟1 𝑟2 p1 ´ cos Δ𝜃q “ 2 𝑐2 ´ p𝑟2 ´ 𝑟1 q2 RHS “ 2 2

2

Thus, it is now sufficient to show that LHS is equal to 𝑐 ´p𝑟22 ´𝑟1 q “∶ 𝐴. Since LHS has terms involving the semiperimeter 𝑠, we rewrite those terms using Equation (5.50) 𝑟1 ` 𝑟2 ` 𝑐 𝑐 ` p𝑟2 ´ 𝑟1 q ´ 𝑟1 “ 2 2 𝑟1 ` 𝑟2 ` 𝑐 𝑐 ´ p𝑟2 ´ 𝑟1 q p𝑠 ´ 𝑟2 q “ ´ 𝑟2 “ 2 2

p𝑠 ´ 𝑟1 q “

so that LHS can be rewritten as 1 LHS “ 2p𝑠 ´ 𝑟1 qp𝑠 ´ 𝑟2 q “ r𝑐 ` p𝑟2 ´ 𝑟1 qsr𝑐 ´ p𝑟2 ´ 𝑟1 qs 2 𝑐2 ´ p𝑟2 ´ 𝑟1 q2 “ 2 which is equal to 𝐴, thus concluding the proof – we used the difference of squares formula p𝑎 ´ 𝑏qp𝑎 ` 𝑏q “ 𝑎2 ´ 𝑏2 in the last step.

24

Problem 15 At time 𝑡0 “ 0, the International Space Station (ISS) is ejecting a CubeSat through its CubeSat deployer mechanism. Assume the ISS is in a geocentric circular orbit with period of 92 minutes.

With respect to the Cartesian Hill-Clohessy-Wiltshire (HCW) frame of reference, the CubeSat is ejected with a relative velocity of 𝛿vp𝑡𝑜 q “ 5i ` 10j ` 0k m/s (a) Using the HCW relative motion equations, compute how far away (in km) the satellite is from the ISS at t = 46 minutes. (b) What happens to the relative distance as time increases? Does this make physical sense?

Briefly explain why or why not.

Solution (a) The CubeSat is ejected from the ISS at 𝑡 “ 𝑡0 “ 0 so 𝛿r0 “ 0𝑖̂ ` 0𝑗̂ ` 0𝑘̂ km 𝛿v0 “ 5 ˆ 10´3 𝑖̂ ` 10 ˆ 10´3 𝑗̂ ` 0𝑘̂ km/s Noting that the final time 𝑡 “ 46 min “ 𝑇2 2𝜋 “𝜋 𝑇 𝑇2 2𝜋 “ 1.13826 ˆ 10´3 rad/s 𝑛“ 𝑇 Using Equation (5.125), which is the state transition matrix definition of the HWC equations, we can write " * „ ȷ " * 𝛿rp𝑡q Φ𝑟𝑟 p𝑡q Φ𝑟𝑣 p𝑡q 𝛿r0 “ “ 𝛿vp𝑡q Φ𝑣𝑟 p𝑡q Φ𝑣𝑣 p𝑡q 𝛿v0 𝑛𝑡 “

so that :0

𝛿rp𝑡q “ Φ𝑟𝑟 p𝑡q𝛿r 0 ` Φ𝑟𝑣 p𝑡q𝛿v0 “ Φ𝑟𝑣 p𝑡q𝛿v0

In Cartesian coordinates »

fi 0 0 fl Φ𝑟𝑣 p𝑡q “ –´ 𝑛2 p1 ´ cos 𝑛𝑡q 𝑛1 p4 sin 𝑛𝑡 ´ 3𝑛𝑡q 1 0 0 sin 𝑛𝑡 𝑛 » fi 4 0 0 𝑛 4 3𝜋 – “ ´ 𝑛 ´ 𝑛 0fl 0 0 0 2 p1 ´ cos 𝑛𝑡q 𝑛

1 sin 𝑛𝑡 𝑛

so that , $ , fi $ 4 0 0 &5 ˆ 10´3 . & 35.1414 . 𝑛 10´4 𝛿rp46 minq “ –´ 𝑛4 ´ 3𝜋𝑛 0fl “ ´100.371 km % - % 0 0 0 0 0 »

25

(b) As 𝑡 Ñ 8, then |𝛿r| Ñ 8, which does not make physical sense since the spacecraft is not on an escape trajectory from Earth. However, this happens because the HWC equations are only valid for “small" |𝛿r|.

26

Problem 16 During the Apollo program during the lunar orbit rendezvous, the lunar module launched from the Moon and would rendezvous with the command module. Had the lunar module been able to get into orbit, but was not able to get to the command module, the command module could have maneuvered to rendezvous with the lunar module. (a) Assume that the command module is in a 100 km altitude circular orbit about the Moon.

The lunar module is a circular orbit 100 km behind and 10 km below the command module.

What is the Δ𝑣 necessary for the lunar module to rendezvous with the command in 1 hour? (b) If the command module had to go after the lunar module in its orbit, what Δ𝑣 would be needed to rendezvous in 1 hour (assuming the same relative positions in part (a)?

Solution From Appendix A, we can find the gravitational parameter and physical radius of the Moon 𝜇K “ 4, 903 km3 {s2 𝑅K “ 1, 732 km (a) For part (a) we assume that the target is the Command Module (CM), which orbits the Moon in a circular orbit with radius 𝑟0 “ 𝑅K ` 100 km “ 1, 838 km, and the chaser is the Lunar Module (LM) which has relative position and velocity given as follows 𝛿r0 “ ´10𝑒̂𝑟 ` ´100𝑒̂𝜃 ` 0𝑒̂ℎ km 𝛿v0 “ 0𝑒̂𝑟 ` 0𝑒̂𝜃 ` 0𝑒̂ℎ km/s Using the HCW definitions in cylindrical coordinates as given by Equations (5.131) - (5.136) along with the provided time of flight 𝑡 “ 1 hour, we can compute the necessary Δ𝑣 maneuvers to initiate and complete the rendezvous between the LM and CM Δ𝑣1 “ 19.7023 m/s Δ𝑣2 “ 16.3806 m/s Δ𝑣𝑡𝑜𝑡 “ 36.0899 m/s This rendezvous maneuver is shown in Figure 4. (b) For part (b) we switch the target and chaser roles, so that the target is now the LM and the chaser is the CM. This means that the relative motion of the (new) chaser with respect to the target is 𝛿r0 “ 10𝑒̂𝑟 ` 100𝑒̂𝜃 ` 0𝑒̂ℎ km 𝛿v0 “ 0𝑒̂𝑟 ` 0𝑒̂𝜃 ` 0𝑒̂ℎ km/s We again use the HCW definitions in cylindrical coordinates as given by Equations (5.131) - (5.136) along with the provided time of flight 𝑡 “ 1 hour, to compute the necessary Δ𝑣 maneuvers to 27

initiate and complete the rendezvous between the LM and CM Δ𝑣1 “ 19.6377 m/s Δ𝑣2 “ 16.1805 m/s Δ𝑣𝑡𝑜𝑡 “ 35.8182 m/s This rendezvous maneuver is shown in Figure 5.

The table below summarizes the results from parts (a) and (b).

Figure 4: Lunar Rendezvous between the CM (target) and LM (chaser)

28

Figure 5: Lunar Rendezvous between the LM (target) and CM (chaser) Part (a) (b) (a) (b) (a) (b) (a) (b) (a) (b)

Parameter Value (m/s) |𝛿v` 0 | |𝛿v` 𝑓 |

Δ𝑣1

Δ𝑣2

Δ𝑣𝑡𝑜𝑡

22.4770 22.4189 16.3806 16.1805 19.7023 19.6377 16.3806 16.1805 36.0899 35.8182

29

Reference Equation (5.133) Equation (5.135) Equation (5.134) Equation (5.136)

Δ𝑣1 ` Δ𝑣2

Problem 17 Compute the Δ𝑣1 , Δ𝑣2 , and Δ𝑣𝑡𝑜𝑡𝑎𝑙 (in km/s) and TOF (in years) required to perform an interplanetary Hohmann transfer from Earth to Jupiter. Assume that you start from a circular LEO having altitude of 300 km and you are targeting a circular orbit around Jupiter having altitude of 4,300 km. How could one lower the Δ𝑣 requirements for a mission targeting Jupiter? Explain, referring to maneuvers discussed in this chapter.

Solution Using Appendix A, we can find that the gravitational parameters and physical constants for this problem 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝜇C “ 398, 600 km3 {s2 𝜇E “ 1.268 ˆ 108 km3 {s2 𝑟C{@ “ 149, 597, 800 km 𝑟E{@ “ 778, 327, 433.8 km 𝑅C “ 6, 378.140 km 𝑅E “ 71, 490 km We assume that the orbits of Earth and Jupiter are circular and coplanar. We start our analysis by computing the orbital speeds of the planets with respect to the Sun c 𝜇@ 𝑣C{@ “ “ 29.7831 km/s 𝑟C{@ c 𝜇@ 𝑣E{@ “ “ 13.0563 km/s 𝑟E{@ We then compute the semimajor axis and energy of the Hohmann transfer ellipse between Earth and Jupiter 1 𝑎𝐻 “ p𝑟C{@ ` 𝑟E{@ q “ 464, 059, 200 km 2 ´𝜇@ 𝐸𝐻 “ “ ´142.965 km2 s2 2𝑎𝐻 We can therefore compute the required spacecraft orbital speeds at perhelion (Earth departure) and aphelion (Jupiter arrival) as follows d ˆ ˙ 𝜇@ 𝑣𝑝{@ “ 2 𝐸𝐻 ` “ 38.5763 km/s 𝑟C{@ d ˆ ˙ 𝜇@ 𝑣𝑎{@ “ 2 𝐸𝐻 ` “ 7.41205 km/s 𝑟E{@ 30

Then, we compute the speeds that the spacecraft will have at the boundary of sphere of influence with respect to Earth at departure and Jupiter at arrival 𝑣8{C “ 𝑣𝑝{@ ´ 𝑣C{@ “ 8.79325 km/s 𝑣8{E “ 𝑣E{@ ´ 𝑣𝑎{@ “ 5.64299 km/s Now we have enough information to compute the Δ𝑣 maneuvers required to leave Earth and be captured as Jupiter. We start by computing the radii and velocities of the starting and arrival orbits at Earth and Jupiter, respectively 𝑟1 “ 𝑅C ` 300 km “ 6, 678.140 km c 𝜇C 𝑣1 “ “ 7.72584 km/s 𝑟1 𝑟2 “ 𝑅E ` 4, 300 km “ 75, 490 km c 𝜇E “ 40.8867 km/s 𝑣2 “ 𝑟2 In order to compute Δ𝑣1 , we perform an energy balance using the vis-viva equation, equating the energy immediately after Δ𝑣1 is performed and at the boundary of Earth’s sphere of influence 2

𝑣8{C p𝑣1 ` Δ𝑣1 q2 𝜇C ´ “ 2 𝑟1 2 which gives Δ𝑣1 “ 6.29908 km/s (considering only the positive root of the quadratic equation).

2 Here, 𝐶3 “ 𝑣8{C “ 77.3227 km2 /s2 . Although we do not use 𝐶3 in these calculations, 𝐶3 is generally used when determining launch vehicles for interplanetary missions – in fact, launch vehicle performance is often cited using 𝐶3 . Furthermore, a few simple algebraic manipulations can be used to show that Δ𝑣1 can be computed as b 2 Δ𝑣1 “ 𝑣𝑒𝑠𝑐{C ` 𝐶3 ´ 𝑣1 ? where 𝑣𝑒𝑠𝑐{C “ 2𝑣1 is Earth’s escape velocity at 𝑟1 . At arrival at Jupiter, we can perform a similar energy balance to that used for Earth departure. Alternatively, we can compute the velocity that the spacecraft has at perijove c 2𝜇E 2 𝑣𝑝{E “ 𝑣8{E ` “ 58.0973 km/s 𝑟2 b ?

2𝜇E where the second term under the square root can be recognized to be 𝑣𝑒𝑠𝑐{E “ “ 2𝑣2 , 𝑟2 i.e. Jupiter’s escape velocity at 𝑟2 . The required Δ𝑣2 to be captured into a Jovian circular orbit at altitude 4,300 km can be computed as Δ𝑣2 “ 𝑣𝑝{E ´ 𝑣2 “ 17.2106 km/s which, when summed to Δ𝑣1 gives a total Δ𝑣 of Δ𝑣𝑡𝑜𝑡 “ Δ𝑣1 ` Δ𝑣2 “ 23.50968 km/s 31

Time of flight can be computed as d 𝑇 𝑂𝐹 “ 𝜋

𝑎3𝐻 “ 998.0 days or 2.734 years 𝜇@

In order to reduce the mission Δ𝑣, one could opt to enter an elliptical orbit at Jupiter instead of a circular orbit. Assuming a fixed perijove at an altitude of 4,300 km, the capture speed can be computed as a function of final orbital eccentricity d 𝜇E p1 ` 𝑒q 𝑣𝑐𝑎𝑝𝑡𝑢𝑟𝑒 “ 𝑟2 so the minimum capture speed the spacecraft must achieve can be computed by setting 𝑒 “ 1 (parabolic orbit), which gives 𝑣𝑐𝑎𝑝𝑡𝑢𝑟𝑒 “ 57.8226 km/s. This means that the minimum Δ𝑣2 to be captured must be Δ𝑣2 ą 𝑣𝑝{E ´ 𝑣𝑐𝑎𝑝𝑡𝑢𝑟𝑒, 𝑒“1 “ 0.274723 km/s in order to be captured into an elliptical orbit. Another option would be to perform planetary flybys of other planets, increasing TOF but reducing total Δ𝑣.

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Problem 18 Compute the Δ𝑣1 , Δ𝑣2 , and Δ𝑣𝑡𝑜𝑡𝑎𝑙 (in km/s) and TOF (in years) required to perform an interplanetary Hohmann transfer from Earth to Venus. Assume that you start from a circular LEO having altitude of 300 km and you are targeting a circular orbit around Venus having altitude of 500 km.

Solution Using Appendix A, we can find that the gravitational parameters and physical constants for this problem 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝜇C “ 398, 600 km3 {s2 𝜇B “ 325, 700 km3 {s2 𝑟C{@ “ 149, 597, 800 km 𝑟B{@ “ 108, 204, 088.7 km 𝑅C “ 6, 378.140 km 𝑅B “ 6, 051.9 km We assume that the orbits of Earth and Venus are circular and coplanar. We start our analysis by computing the orbital speeds of the planets with respect to the Sun c 𝜇@ 𝑣C{@ “ “ 29.7831 km/s 𝑟C{@ c 𝜇@ 𝑣B{@ “ “ 35.0214 km/s 𝑟B{@ We then compute the semimajor axis and energy of the Hohmann transfer ellipse between Earth and Venus 1 𝑎𝐻 “ p𝑟C{@ ` 𝑟B{@ q “ 128, 900, 944 km or 0.861650 AU 2 ´𝜇@ 𝐸𝐻 “ “ ´514.7829 km2 s2 2𝑎𝐻 We can therefore compute the required spacecraft orbital speeds at aphelion (Earth departure) and perihelion (Venus arrival) as follows d ˆ ˙ 𝜇@ 𝑣𝑎{@ “ 2 𝐸𝐻 ` “ 27.28891 km/s 𝑟C{@ d ˆ ˙ 𝜇@ 𝑣𝑝{@ “ 2 𝐸𝐻 ` “ 37.72835 km/s 𝑟B{@

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Then, we compute the speeds that the spacecraft will have at the boundary of sphere of influence with respect to Earth at departure and Venus at arrival 𝑣8{C “ 𝑣C{@ ´ 𝑣𝑎{@ “ 2.495734 km/s 𝑣8{B “ 𝑣𝑝{@ ´ 𝑣B{@ “ 2.706970 km/s noting that since we are analyzing a transfer from an outer planet to an inner planet, we would need to have the spacecraft depart Earth in the opposite direction of Earth’s motion with respect to the Sun. Now we have enough information to compute the Δ𝑣 maneuvers required to leave Earth and be captured as Jupiter. We start by computing the radii and velocities of the starting and arrival orbits at Earth and Venus, respectively 𝑟1 “ 𝑅C ` 300 km “ 6, 678.140 km c 𝜇C 𝑣1 “ “ 7.72584 km/s 𝑟1 𝑟2 “ 𝑅B ` 500 km “ 6, 551.9 km c 𝜇B 𝑣2 “ “ 7.050587 km/s 𝑟2 In order to compute Δ𝑣1 , we perform an energy balance using the vis-viva equation, equating the energy immediately after Δ𝑣1 is performed and at the boundary of Earth’s sphere of influence 2

𝑣8{C p𝑣1 ` Δ𝑣1 q2 𝜇C ´ “ 2 𝑟1 2 which gives Δ𝑣1 “ 3.481531 km/s (considering only the positive root of the quadratic equation).

2 Here, 𝐶3 “ 𝑣8{C “ 6.228687 km2 /s2 . Although we do not use 𝐶3 in these calculations, 𝐶3 is generally used when determining launch vehicles for interplanetary missions – in fact, launch vehicle performance is often cited using 𝐶3 . Furthermore, a few simple algebraic manipulations can be used to show that Δ𝑣1 can be computed as b 2 Δ𝑣1 “ 𝑣𝑒𝑠𝑐{C ` 𝐶3 ´ 𝑣1 ? where 𝑣𝑒𝑠𝑐{C “ 2𝑣1 is Earth’s escape velocity at 𝑟1 . At arrival at Venus, we can perform a similar energy balance to that used for Earth departure. Alternatively, we can compute the velocity that the spacecraft has at pericytherion c 2𝜇B 2 𝑣𝑝{B “ 𝑣8{B ` “ 10.331952 km/s 𝑟2 b ? where the second term under the square root can be recognized to be 𝑣𝑒𝑠𝑐{B “ 2𝜇𝑟2B “ 2𝑣2 , i.e.

Venus’s escape velocity at 𝑟2 . The required Δ𝑣2 to be captured into a Venusian circular orbit at altitude 500 km can be computed as Δ𝑣2 “ 𝑣𝑝{B ´ 𝑣2 “ 3.281365 km/s 34

which, when summed to Δ𝑣1 gives a total Δ𝑣 of Δ𝑣𝑡𝑜𝑡 “ Δ𝑣1 ` Δ𝑣2 “ 6.762896 km/s Time of flight can be computed as d 𝑇 𝑂𝐹 “ 𝜋

𝑎3𝐻 “ 146.07 days or 0.399922 years 𝜇@

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Problem 19 Create a porkchop plot for Earth-Venus transfers for the following sets of dates • Departure dates: 01 Sep 2052 through 01 Mar 2054 • Arrival dates: 01 Sep 2053 through 01 Jul 2054 using 1-day time steps. Use planetary ephemerides provided either by JPL Horizons or MATLAB’s planetaryEphemeris function. Assuming departure and arrival planetary altitudes of 400 km and 500 km, respectively, compute and plot a porkchop plot for total mission Δ𝑣. What is the minimum total Δ𝑣? For which departure and arrival dates does it occur? Is the transfer type I or type II? Solution Figure 6 shows the porkchop plot for Earth-Venus transfers during the desired time frame.

Figure 6: Earth-Venus Hohmann Transfer (outbound) The following table summarizes the minimum total Δ𝑣 case for the given time frame. A type I transfer is the optimal transfer for this time frame.

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Parameter

Value

Δ𝑣𝑡𝑜𝑡

7.5630 km/s

Δ𝑣1

3.9533 km/s

Δ𝑣2

3.6096 km/s

𝐶3 at launch

17.4250 km2 {s2

𝑣8 at arrival

3.7791 km/s

Departure date

30 Oct 2053

Arrival date

14 Feb 2054

TOF (interplanetary)

107 days

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Problem 20 Recreate the porkchop plot shown in Figure 5.22. Use similar contour levels for C3 at launch, arrival 𝑣8 , and TOF. Solution In order to recreate the porkchop plot in Figure 5.22, the classical or unversal variable solutions for Lambert’s problem can be used. Refer to Section 5.3 and 5.8 for additional details.

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Problem 21 Assume that a spacecraft performs an Earth-Venus interplanetary Hohmann transfer – assume circular coplanar orbits for simplicity. However, at arrival, instead of performing an orbit insertion, the spacecraft does a leading edge flyby of Venus. Compute and plot the turn angle, hyperbolic flyby eccentricity, impact parameter, and post-flyby aphelion as a function of flyby altitude, from a minimum altitude of 200 km to a maximum altitude of 10,000 km.

Solution Using Appendix A, we can find that the gravitational parameters and physical constants for this problem 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝜇C “ 398, 600 km3 {s2 𝜇B “ 325, 700 km3 {s2 𝑟C{@ “ 149, 597, 800 km 𝑟B{@ “ 108, 204, 088.7 km 𝑅C “ 6, 378.140 km 𝑅B “ 6, 051.9 km Refer to the solution of Problem 18 to obtain the required 𝑣8{B “ 2.706970 km/s.

For each flyby trajectory, we have a pericytherion ranging from 𝑟𝑝 “ 𝑅B ` 200 km “ 6, 251.9 km to 𝑟𝑝 “ 𝑅B ` 10, 000 km “ 16, 051.9 km, which we use in combination with 𝑣8{B to compute flyby eccentricity, turn angle, and impact parameter as follows (for simplicity, we dropped the subscript {B) 2 𝑟𝑝 𝑣8 𝜇 ˆ ˙ 1 𝛿 “ 2 arcsin 𝑒𝐹 𝐵 d 2𝜇 Δ “ 𝑟𝑝 1 ` 2 𝑟𝑝 𝑣8

𝑒𝐹 𝐵 “ 1 `

To compute aphelion, we follow the steps given in Section 5.12.1.1 to compute the post flyby heliocentric trajectory. Then, we compute aphelion as follows (all the following orbital parameters

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are with respect to the post-flyby heliocentric trajectory) 𝑎“

𝜇@ @ ` 2 p𝑣 q ´ 𝑟2𝜇 B{@

ℎ “ 𝑟B{@ 𝑣𝜃` ℎ2 𝑝“ 𝜇 c@ 𝑝 𝑒 “ 1` 𝑎 𝑟𝑎{@ “ 𝑎p1 ` 𝑒q Figure 7 shows the turn angle, eccentricity, impact parameter, and aphelion as a function of flyby altitude at Venus.

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Figure 7: Turn angle, Eccentricity, Impact Parameter, and Aphelion as a Function of Flyby Altitude

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Problem 22 For a hyperbolic gravity assist (GA) of a target planet “b2 𝑤𝑖𝑡ℎ𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝜇b and physical radius 𝑅b , the periapsis of the hyperbolic GA trajectory has eccentricity 𝑒, periapse radius ` 𝑟𝑝 , and the inbound and outbound velocities at the sphere of influence of the planet are v´ 8 and v8 , respectively. Determine the values of 𝑟𝑝 and magnitude of v´ 8 that yield the maximum possible ´ magnitude of the gravity assist Δ𝑣𝐺𝐴 . Express your answers for 𝑟𝑝 and 𝑣8 in terms of 𝜇b and 𝑅b (noting that 𝑟𝑝 ě 𝑅b ). Then, determine the maximum value of Δ𝑣𝐺𝐴 in terms of 𝜇b and 𝑅b . Lastly, determine the numerical values for the corresponding turn angle 𝛿 and eccentricity 𝑒.

Solution ` We use the law of cosines to compute Δ𝑣𝐺𝐴 , using the fact that |v´ 8 | “ |v8 | “ 𝑣8 a 2 ` 𝑣 2 ´ 2𝑣 𝑣 cos 𝛿 Δ𝑣𝐺𝐴 “ 𝑣8 8 8 8 a “ 𝑣8 2p1 ´ cos 𝛿q 𝛿 “ 2𝑣8 sin 2

where we used a trig identity in the last step. Turn angle 𝛿 and eccentricity can be written as ˆ ˙ 1 𝛿 “ 2 arcsin 𝑒 2 𝑟𝑝 𝑣 𝑒 “1` 8 𝜇b Combining the above equations with the expression for Δ𝑣𝐺𝐴 we get Δ𝑣𝐺𝐴 “

2𝑣8 𝜇b 2 𝑟 𝜇b ` 𝑣8 𝑝

from which we can see that Δ𝑣𝐺𝐴 is maximized when 𝑟𝑝 is minimized, i.e. when 𝑟𝑝 “ 𝑅b . To find the value of 𝑣8 that maximizes Δ𝑣𝐺𝐴 , we take a partial derivative of Δ𝑣𝐺𝐴 with respect to 𝑣8 2 2𝜇2 ´ 2𝜇b 𝑟𝑝 𝑣8 BΔ𝑣𝐺𝐴 “ b 2 𝑟 q2 B𝑣8 p𝜇b ` 𝑣8 𝑝

Setting the above derivate equal to zero and solving for 𝑣8 gives c 𝜇b 𝑣8 “ 𝑟𝑝 We can easily show that the second derivative is negative at this value of 𝑣8 , thus confirming that this value is a maximum. Since we know that 𝑟𝑝 “ 𝑅b maximizes Δ𝑣𝐺𝐴 , we can rewrite 𝑣8 as c 𝜇b 𝑣8 “ 𝑅b 42

Using the first equation we derived for Δ𝑣𝐺𝐴 and plugging in this value results in the maximum value of Δ𝑣𝐺𝐴 c 𝜇b Δ𝑣𝐺𝐴, 𝑚𝑎𝑥 “ 𝑅b Interestingly, this value, which is the same as that computed for 𝑣8 , is equivalent to the circular orbital speed a spacecraft would need to have to orbit the targeted planet at a theoretical altitude of zero. Using these computed values, we can find the turn angle and eccentricity that maximize Δ𝑣𝐺𝐴 ´b ¯2 𝜇b 𝑅b 2 𝑅b 𝑟𝑝 𝑣 8 𝑒 “1` “1` “2 𝜇b 𝜇b ˆ ˙ ˆ ˙ 1 1 “ 2 arcsin “ 60˝ 𝛿 “ 2 arcsin 𝑒 2

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Problem 23 A spacecraft is in orbit around Mars and is beginning to aerobrake. Initially, it starts out in an orbit with periapse and apoapse altitudes of 140 and 1000 km, respectively. Use the following atmospheric and spacecraft physical characteristics Density Model

Values

Reference Density

30 kg/km3

Reference Altitude

140 km

Scale Height

10 km

Drag Coefficient

2.2

Projected Area

23 m2

Mass

1100 kg

a Propagate the orbit and estimate how long it takes to reduce the apoapse to 140 km altitude. b If a propulsive maneuver of 1 m/s is added at apoapse on the first pass, how long does it take to reduce the apoapse to 140 km? c If a propulsive maneuver of 10 m/s is added at apoapse on the first pass, how long does it take to reduce the apoapse to 140 km?

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Mars are 𝜇D “ 43, 050 km3 {s2 𝑅D “ 3, 389.5 km In order to determine when the apoapse altitude reaches 140 km, we numerically integrate Equations (5.200) using the spacecraft parameters provided and the atmospheric model given by Equation (5.213). (a) When no propulsive maneuver is implemented, the time it takes to lower apoareion altitude to 140 km is 50.3512 hours. Figures 8, 9, and 10 show the aerobraking orbits along with the change of various orbital parameters over time. Note that while apoapse altitude decreases significantly, periapse altitude also decreases, although not as rapidly. (b) When a 1 m/s propulsive maneuver is implemented, the time it takes to lower apoareion altitude to 140 km is 32.7142 hours. Figures 11, 12, and 13 show the aerobraking orbits along with the change of various orbital parameters over time. Maneuvers at apoapsis lower periapsis, thus increasing the effects of drag and speeding up the aerobraking maneuvers. However, this is expected to increase the flux and temperature experienced by the spacecraft. (c) When a 10 m/s propulsive maneuver is implemented, the time it takes to lower apoareion altitude to 140 km is only 1.05319 hours. Figures 14, 15, and 16 show the aerobraking orbits along with the change of various orbital parameters over time.

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Figure 8: Aerobraking Orbits (no Propulsive Maneuver)

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Figure 9: Altitude, Velocity, Energy, and Dynamic Pressure as functions of time (no Propulsive Maneuver)

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Figure 10: Semimajor Axis, Eccentricity, Apoapse Altitude, and Periapse Altitude as functions of time (no Propulsive Maneuver)

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Figure 11: Aerobraking Orbits (1 m/s Propulsive Maneuver)

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Figure 12: Altitude, Velocity, Energy, and Dynamic Pressure as functions of time (1 m/s Propulsive Maneuver)

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Figure 13: Semimajor Axis, Eccentricity, Apoapse Altitude, and Periapse Altitude as functions of time (1 m/s Propulsive Maneuver)

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Figure 14: Aerobraking Orbits (10 m/s Propulsive Maneuver)

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Figure 15: Altitude, Velocity, Energy, and Dynamic Pressure as functions of time (10 m/s Propulsive Maneuver)

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Figure 16: Semimajor Axis, Eccentricity, Apoapse Altitude, and Periapse Altitude as functions of time (10 m/s Propulsive Maneuver)

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Problem 24 A spacecraft is entering the Martian atmosphere, on its way to land. At 112,000 m altitude, it is traveling at 5588 m/s. a Using the atmospheric drag parameters in the table below, integrate the equations of motion and estimate the velocity 7 minutes later. Be sure to state your assumptions.

Density Model

Range

Values

Reference Density

0–25 km

0.0525 kg/m3

25–125 km

0.0159 kg/m3

0–25 km

11.049 km

25–125 km

7.295 km

Scale Height

Ballistic Coefficient N/A

65 kg/m2

b Provide three plots: altitude vs. time, heating vs. time, and dynamic pressure vs. time.

Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Mars are 𝜇D “ 43, 050 km3 {s2 𝑅D “ 3, 389.5 km We numerically integrate Equations (5.200) using the spacecraft parameters provided and the atmospheric model given by Equation (5.213), noting that here we have a piece-wise distribution for the Martian atmospheric density (which is what causes discontinuities in the final results).

For this problem, we simulated the entry, descent, and landing (EDL) for entry flight path angles (EFPA) ranging from ´10˝ to ´5˝ . Figure 17 shows the results of altitude, heat flux, and dynamic pressure as functions of time. It can be seen that if EFPA ą ´7˝ the spacecraft does not fly through the atmosphere at a low enough altitude to land in one pass.

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Figure 17: Altitude, Heat Flux, and Dynamic Pressure as Functions of Time for Various EFPAs

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Problem 25 Design a roundtrip trajectory for an Earth-Mars crewed mission during the 2050s, such that a crew is given a stay time between 50 and 100 days. What types of outbound and inbound transfers would you choose? Explain why, considering the time the crew spends in space and ensuring that your total mission Δ𝑣 is within a reasonable value. Use JPL’s planetary ephemeris data to obtain the position and velocities of Earth and Mars. (Just like real mission design, there is no unique answer for this problem) Solution Design problem.

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Problem 26 Compute the semimajor axis, eccentricity, radii of perihelion and aphelion for the Aldrin cycler discussed in Section 5.15.3. For this problem, assume that the orbits of Earth and Mars are coplanar and circular. Furthermore, assume that Earth and Mars orbit the Sun in exactly 1 year and 1.875 years, respectively. This means that the cycler would encounter Earth on successive passes that are 1/7 of an orbit (or 51.4˝ ) apart. Lastly, assume that Mars flybys do not provide any changes in the heliocentric orbit of the cycler, so that only Earth flybys can provide the necessary orbital changes to maintain the cycler’s orbit periodicity.

Hint: you will need to solve a multi-rev Lambert’s problem to find the semimajor axis of the cycler orbit.

Solution The problem statement tells us that 𝑇C “ 1 year 𝑇D “ 1.875 years So we can compute the synodic period of Earth and Mars as 𝑇C 𝑇D 𝑇𝑠𝑦𝑛 “ “ 2.1429 years |𝑇C ´ 𝑇D | which is equivalent to 2 and 1/7 years. Since the cycler needs to rotate its heliocentric orbit by 51.4˝ at every Earth flyby, it means that the heliocentric pre-flyby and post-flyby true anomalies at every Earth flyby are ´25.7˝ and 25.7˝ , i.e. half of 51.4˝ each. Thus, in order to compute the semimajor axis of the cycler orbit, we need implement multi-rev Lambert’s problem, or Equations (5.96) and (5.97) ? 𝜇p𝑡2 ´ 𝑡1 q “ 𝑎3{2 r2𝑁 𝜋 ` 𝛼 ´ 𝛽 ´ psin 𝛼 ´ sin 𝛽qs where 𝑁 “ 1 since the spacecraft re-encounters Earth after one full revolution around the Sun. Therefore, the boundary conditions are 𝜇@ “ 4𝜋 2 km3 {s2 𝑟1 “ 𝑟C{@ “ 1 AU 𝑟2 “ 𝑟C{@ “ 1 AU Δ𝜃 “ 51.4˝ 𝑡2 ´ 𝑡1 “ 𝑇𝑠𝑦𝑛 “ 2.1429 years where we used canonical units for simplicity. Solving Lambert’s equation using one of the methods described in Sections 5.3.1 or 5.3.2 gives 𝑎 “ 1.59959 AU, which corresponds to an orbital period of 2.024506 years. We use the orbit equation to solve for eccentricity 𝑎p1 ´ 𝑒 2 q 1 ` 𝑒 cos 𝜃 𝑟 ` 𝑒𝑟 cos 𝜃 “ 𝑎 ´ 𝑎𝑒 2 𝑟 cos 𝜃 𝑟 ´𝑎 𝑒2 ` 𝑒` “0 𝑎 𝑎 𝑟“

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which gives a quadratic equation for 𝑒. We use 𝑎 “ 1.59959 AU, 𝑟 “ 1 AU, and either 𝜃 “ ´25.7˝ or 𝜃 “ 25.7˝ to get 𝑒 “ 0.392283 by solving the quadratic equation and consider only the positive root of the equation. We can now compute perihelion and aphelion of the cycler orbit as follows 𝑟𝑝{@ “ 𝑎p1 ´ 𝑒q “ 0.972097 AU 𝑟𝑎{@ “ 𝑎p1 ` 𝑒q “ 2.227083 AU

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Problem 27 Compute the minimum radial-only acceleration needed to transfer a spacecraft from a circular LEO with 400 km altitude to GEO. Then, using the acceleration value you computed, find the time at which the spacecraft would reach GEO. Solution Using Appendix A, we can find that the gravitational parameter and physical radius of Earth are 𝜇C “ 398, 600 km3 {s2 𝑅C “ 6, 378.140 km Therefore, the spacecraft starts in a LEO having radius and velocity 𝑟0 “ 𝑅C ` 400 km “ 6, 778.140 km c 𝜇C 𝑣0 “ “ 7.66855 km/s 𝑟1 and the minimum radial-only acceleration required to achieve GEO (or to escape) must be at least 𝑎𝐿𝑇 ,𝑟, 𝑚𝑖𝑛 “

𝜇C “ 1.0844917 ˆ 10´3 km/s2 8𝑟0

Recall from Problem 1 that GEO has a radius of 𝑟𝐺𝐸𝑂 “ 42, 164.15 km. In order to find the time the spacecraft takes to reach GEO, we need to numerically integrate the equations of motion, Equation (5.247), using the radial-only acceleration that we computed. The spacecraft takes approximately 1.332057 days to reach GEO with the above constant acceleration. Figure 18 shows a plot of the spacecraft’s low-thrust orbit along with its starting orbit in LEO and the final targeted orbit, GEO.

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Figure 18: LEO, GEO, and the Radial-only Low-thrust Orbit Connecting Them

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Problem 28 Assume that a spacecraft starts from a circular heliocentric orbit equal to that of the Earth (𝑟0 “ 1 AU). The vehicle turns on its thrusters, producing a constant tangential acceleration of 1{32 𝜇{𝑟0 .

How long will the spacecraft take to escape the Solar System? Compute this value by integrating the equations of motion and also by using the approximations given in Section 5.16.3.2. How do the values compare to each other?

Solution Using Appendix A, we can find the gravitational parameter of the Sun and the Earth’s distance from the Sun 𝜇@ “ 1.327 ˆ 1011 km3 {s2 𝑟C{@ “ 149, 597, 800 km or 1 AU The tangential-only acceleration the spacecraft receives is 𝑎𝐿𝑇 ,𝑇 “

𝜇@ “ 1.85315 ˆ 10´7 km/s2 32𝑟0

With the analytical approximation discussed in Section 5.16.3.2 we get the following escape radius and escape time (we arbitrarily set 𝑡0 “ 0 here) 𝑟0 𝑣0 “ 2.674961 AU p20𝑎2𝐿𝑇 ,𝑇 𝑟02 q1{4 « ˆ 2 2 ˙1{8 ff 20𝑎𝐿𝑇 ,𝑇 𝑟0 𝑣0 1´ 𝑡𝑒𝑠𝑐 “ “ 1.979047 years 𝑎𝐿𝑇 ,𝑇 𝑣04

𝑟𝑒𝑠𝑐 “

If we numerically integrate the equations of motion, Equation (5.229), using the tangential-only acceleration computed above, we get that the escape radius and escape time are 4.762696 AU and 3.455814 years, respectively. As mentioned in Section 5.16.3.2, the analytic approximation is valid only when 𝑎𝐿𝑇 ,𝑇 ăă 𝑟𝜇2 , which explains the discrepancy in the results. A plot of the numerically 0 integrated trajectory can be seen in Figure 19.

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Figure 19: Tangential-only Low-thrust Trajectory from Earth’s Orbit to Escape

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Interplanetary Astrodynamics Chapter 6 Problem Solutions

Problem 1 Consider a Mars-orbiting spacecraft in a circular orbit at an altitude of 500 km. The spacecraft initiates a maneuver to land on Mars, targeting a periapsis equal to the radius of Mars at 180˝ from the location of the maneuver. However, the propulsion system malfunctions, causing the spacecraft to deliver a higher retrograde Δ𝑣 than expected. It is believed that the extra Δ𝑣 that the propulsion system delivered is somewhere between 1% and 5% of the nominal Δ𝑣. Based on this, what is the range of new landing location angles? What is the new range of times of flight for the spacecraft to hit the Martian surface? For this problem, ignore the atmospheric effects of Mars.

Solution As provided in Appendix A, the symbol for Mars is D, and Mars’s gravitational parameter and physical radius are 𝜇D “ 43, 050 km3 /s2 𝑅D “ 3, 397 km Thus,b the starting circular orbit has radius 𝑟1 “ 𝑅D ` 500km “ 3,897 km, and an orbital speed of 𝑣1 “ 𝜇𝑟D1 “ 3.32369 km/s. Figure 1 shows a diagram of the starting orbit (green), the nominal deorbit trajectory (blue), and an example of a deorbit trajectory whose Δ𝑣 was larger than the nominal Δ𝑣 (purple).

The nominal Δ𝑣 to deorbit the spacecraft is equivalent to the Δ𝑣1 of a Hohmann transfer whose apoapsis is 𝑟1 and peripasis is 𝑅D ˆc ˙ 2𝑅D Δ𝑣 “ 𝑣1 ´ 1 “ ´0.115941 km/s 𝑟1 ` 𝑅 D where the negative indicates that the Δ𝑣 is performed in the retrograde direction, as expected.

The deorbit trajectory has the following semimajor axis and TOF 1 𝑎 “ p𝑟1 ` 𝑅D q “ 3, 647 km 2d 𝑎3 𝑇 𝑂𝐹 “ 𝜋 “ 55.58 min 𝜇D

1

Figure 1: Diagram of the Deorbit Trajectory (not to scale) We can therefore compute the 1% and 5% extra Δ𝑣 values to obtain a range of minimum and maximum apoapse velocities, 𝑣𝑎 that the spacecraft would have as a result of the additional velocity change Δ𝑣𝑒𝑥𝑡𝑟𝑎, 𝑚𝑖𝑛 “ 1.01Δ𝑣 “ 0.117100 km/s Δ𝑣𝑒𝑥𝑡𝑟𝑎, 𝑚𝑎𝑥 “ 1.05Δ𝑣 “ 0.121738 km/s so that 𝑣𝑎1 “ 𝑣1 ´ Δ𝑣𝑒𝑥𝑡𝑟𝑎, 𝑚𝑖𝑛 “ 3.20659 km/s 𝑣𝑎2 “ 𝑣1 ´ Δ𝑣𝑒𝑥𝑡𝑟𝑎, 𝑚𝑎𝑥 “ 3.20195 km/s where we used the subscript ‘a’ to denote apoapsis, ’1’ for the maximum, and ’2’ for the minimum values.

We can now compute various orbital parameters of the deorbit trajectories that received more Δ𝑣 than expected. We only need to compute these for the minimum and maximum Δ𝑣 to obtain a range of values. The following table summarizes these computed parameters.

2

Parameter

1

2

Reference

𝑎

3,644.7 km

3,635.5 km

Equation (2.61)

𝑟𝑝

3,392.4 km

3,374.1 km

Equation (2.33)

𝑒

0.069224

0.071915

Equation (2.39)

𝑝

3,627.2 km

3,616.7 km

Equation (2.30)

𝜃 (deg)

348.218˝

334.069˝

Equation (2.40)

𝜃 (rad)

6.07756

5.83060

Equation (2.40)

Thus, the new landing location angles (true anomaly, 𝜃, as shown in Figure 1) are given by the range 𝜃 “ p5.83060, 6.07756q. Since the nominal landing location is at periapsis (𝜃 “ 2𝜋), then the difference in landing location angles is Δ𝜃 “ p0.205626, 0.452581q. Converting these values into actual distances, 𝑑, on the surface can be done using the fact that 𝑑 “ Δ𝜃𝑅D which gives values between 698.5 km and 1,537 km. This means that the spacecraft is predicted to be off target between 698.5 km and 1,537 km.

To compute TOF, we need to use Kepler’s time equation. However, we must first convert true anomaly into eccentric anomaly using Equation (2.91) keeping in mind that all calculations must be done using radians, not degrees, ˆ ˙ȷ „c 𝜃 1´𝑒 tan 𝐸 “ 2 arctan 1`𝑒 2 where no quadrant check is needed since 𝐸2 and 𝜃2 always reside in the same quadrant. This gives 𝐸1 “ ´0.191939 rad or 6.09125 rad, and 𝐸2 “ ´0.422087 rad or 5.86110 rad. Then, from Kepler’s Time Equation, or Equations (2.96) and (2.97), we compute TOF d 𝑎3 p𝐸 ´ 𝜋 ´ 𝑒 sin 𝐸q 𝑇 𝑂𝐹 “ 𝜇D where the ´𝜋 comes from the fact that we are computing time from apoapsis to the landing location, not from periapsis. This gives 𝑇 𝑂𝐹1 “ 52.37 min and 𝑇 𝑂𝐹2 “ 48.40 min, which correspond to a range of difference in TOF of Δ𝑇 𝑂𝐹 “ p3.21, 7.18q min. While the difference in TOF may not be significant, we earlier computed that even these relatively small time differences can lead to the spacecraft being off target by hundreds of km, or more.

3

Problem 2 Consider Problem 1 from Chapter 5. Assume that the upper stage has a velocity deviation which can be modeled as a normal distribution with zero mean and 0.01% of the spacecraft speed in LEO in all x-, y-, and z-directions. Create a Monte Carlo simulation with these parameters, and propagate the orbit until apoapsis. How does the orbit differ from the nominal GTO? Create scatter plots similar to Figure 6.15 showing the variation of inclination, semimajor axis, and eccentricity.

Solution For reference, Problem 1 in Chapter 5 states: A spacecraft is launched into a circular orbit around the Earth at an altitude of 450 km. The upper stage of the launch vehicle initiates a Hohmann transfer to GEO through a GTO. How much Δ𝑣 is the launch vehicle expected to perform (Δ𝑣1 ) and how much Δ𝑣 is the spacecraft’s propulsion system expected to deliver in order to perform the adequate orbit insertion at GEO (Δ𝑣2 )? How long does the transfer take?

Using the given parameters and the Earth’s radius and gravitational parameter from Appendix A, we get that the departing LEO has radius and speed 𝑟1 “ 𝑅‘ ` 450km “ 6821km c 𝜇‘ 𝑣1 “ “ 7.64442km/s 𝑟1 The spacecraft needs to transfer to GEO, i.e. at a radius of 𝑟𝐺𝐸𝑂 “ 42, 164 km (see Chapter 5 for details). In order to do so, the nominal Δ𝑣 required is computed using Equation (5.12) ˙ ˆc 2𝑟𝐺𝐸𝑂 ´ 1 “ 2.38553km/s Δ𝑣1 “ 𝑣1 𝑟1 ` 𝑟𝐺𝐸𝑂 resulting in a post-Δ𝑣 speed of 𝑣2 “ 𝑣1 ` Δ𝑣1 “ 10.02995 km/s. The semimajor axis and TOF for this transfer are 1 𝑎 “ p𝑟1 ` 𝑟𝐺𝐸𝑂 q “ 24, 493km 2d 𝑎3 𝑇 𝑂𝐹 “ 𝜋 “ 5.29820hrs 𝜇‘ Using these nominal parameters as initial conditions, we then perturb 𝑣2 in all Cartesian direction as given by the problem statement. Propagating 100 such perturbed orbits gives the variations of eccentricity, inclination, semimajor axis, and apoapsis that are shown in Figure 2. While these variations may not seem particularly significant, they would lead to the need of trajectory correction and stationkeeping maneuvers to maintain the spacecraft within its predefined allocated GEO location.

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Figure 2: Variation of Orbital Parameters from Monte Carlo Simulation of GTO Trajectories

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Problem 3 A spacecraft is arriving in the sphere of influence of Jupiter, where it is expected to perform a gravity assist of the planet at an altitude of 1 million km. The spacecraft has a nominal hyperbolic eccentricity of 2.5. Create a Monte Carlo simulation assuming that the spacecraft position and velocity at the sphere of influence can be modeled as provided in Equation (6.72) and that the nominal 𝑣8 has a deviation that can be represented as a normal distribution with zero mean and 1% of the nominal 𝑣8 (in all directions). Propagate 1,000 orbits, plot them and create a B-plane analysis similar to Figure 6.14, showing the 3𝜎 uncertainty ellipse resulting from your simulation.

How likely is it that the trajectory would impact Jupiter?

Solution As provided in Appendix A, the symbol for Jupiter is E, and Jupiter’s gravitational parameter and physical radius are 𝜇E “ 1.268 ˆ 108 km3 /s2 𝑅E “ 71, 492 km The following table summarizes the computed parameters for this problem, including the equations used for each computation.

Parameter

Value

Reference

𝑒

2.5

Given

𝑟𝑝{E

1,071,492 km

𝑅E + 1,000,000 km

𝜇E

1.268ˆ108 km3 /s2

Appendix A

𝑎

-714,328 km

Equation (2.37)

𝑣8{E

13.3233 km/s

Equation (2.50)

Δ

1,636,725 km

Equation (4.15)

𝑟𝑆𝑂𝐼

48,223,800 km

Equation (5.148)

𝛿

47.1564˝

Equations (5.176) and (5.177)

If 𝑣8{E is not computed using 𝑟𝑆𝑂𝐼 Ñ 8, but rather using the value indicated in the table above, then 𝑣8{E would be slightly larger, i.e. 13.5192 km/s. Similarly, Δ would instead be approximately 1,589,600 km.

Thus, the initial conditions of the spacecraft at the boundary of the sphere of influence of Jupiter are a 2 r8{E “ ´ 𝑟𝑆𝑂𝐼 ´ Δ2 𝑆̂ ` Δ𝐵̂ ` 0𝑅̂ km v8{E “ 𝑣8{E 𝑆̂ ` 0𝐵̂ ` 0𝑅̂ km/s Applying the deviations on the above v8{E with zero mean and 1% standard deviation means 6

perturbing v8{E in all directions by 𝛿v8{E “ 𝛿𝑣𝑥 𝑆̂ ` 𝛿𝑣𝑦 𝐵̂ ` 𝛿𝑣𝑧 𝑅̂ Running a Monte Carlo simulation with 1,000 orbits gives the orbits shown in Figure 3, where the black orbit is the nominal trajectory while the blue orbits are the perturbed trajectories. The 3𝜎 ellipse shown in this figure has semimajor axis of approximately 709,200 km and semiminor axis of approximately 517,200 km. Numerically, we get the results shown in the following table.

Parameter Minimum Value Maximum Value 𝑒

1.86363

3.10975

𝑎

-729,261 km

-699,915

𝛿

37.5157˝

64.9032˝

𝑟𝑝{E

617,339 km

1,504,493

Due to the inherent randomness of the 𝛿𝑣𝑥 , 𝛿𝑣𝑦 , and 𝛿𝑣𝑧 deviations, one would unlikely get these exact numerical results, though they should be similar in value. Using these results, we can create the plot requested for the B-plane, as shown in Figure 4. In Figure 4, the black dot represents the nominal trajectory, the blue crosses are the Monte Carlo simulation trajectories and the blue dot is their average, and the blue ellipse represents the 3𝜎 ellipse based on this particular simulation.

Although it is nearly impossible for any of these trajectories to impact Jupiter, some get very close to the planet, varying greatly from the nominal trajectory.

Although not requested by the problem statement, we are also reporting the variation on various orbital elements, as shown in Figure 5.

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Figure 3: Monte Carlo Simulation of Jupiter Flyby Trajectories (SB plane)

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Figure 4: Monte Carlo Simulation of Jupiter Flyby Trajectories (B-plane)

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Figure 5: Variation of Orbital Parameters from Monte Carlo Simulation of Jupiter Flyby Trajectories (B-plane)

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