12.1.
Model: Model the sun (s), the earth (e), and the moon (m) as spherical.
Solve: (b)
(a)
Fm on e =
Fs on e =
Gms me (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg)(5.98 × 10 24 kg) = 3.53 × 10 22 N = (1.50 × 1011 m ) 2 rs2− e
GMm Me (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(7.36 × 10 22 kg)(5.98 × 10 24 kg) = = 1.99 × 10 20 N (3.84 × 10 8 m ) 2 rm2− e
(c) The moon’s force on the earth as a percent of the sun’s force on the earth is
1.99 × 10 20 N × 100 = 0.56% 3.53 × 10 22 N
12.2. Solve:
Model: Assume the two lead balls are spherical masses. (a)
F1 on 2 = F2 on 1 =
Gm1 m2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(10 kg)(0.100 kg) = = 6.67 × 10 −9 N (0.10 m ) 2 r2
(b) The ratio of the above gravitational force to the weight of the 100 g ball is
6.67 × 10 −9 N = 6.81 × 10 −9 (0.100 kg)(9.8 m /s 2 ) Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated by 10 cm compared to the weight of the 100 g ball.
12.3. Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. GMs My GMe My and Fe on you = Solve: Fs on you = 2 rs− e re2 Dividing these two equations gives Fs on y Fe on y
2
2
M r 1.99 × 10 30 kg 6.37 × 10 6 m −4 = s e = = 6.00 × 10 5.98 × 10 24 kg 1.5 × 1011 m Me rs− e
12.4.
Model: Model the sun (s), the moon (m), and the earth (e) as spherical masses. GMs Mm GMe Mm and Fe on m = Solve: Fs on m = rs2− m re2− m Dividing the two equations and using the astronomical data from Table 12.2, 2
2
1.99 × 10 30 kg 3.84 × 10 8 m Fs on m Ms re − m = = 2.18 = Fe on m Me rs− m 5.98 × 10 24 kg 1.50 × 1011 m Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance.
12.5.
Solve:
Fsphere on particle =
GMs Mp rs2− p
and
Fearth on particle =
GMe Mp re2
Dividing the two equations,
Fsphere on particle Fearth on particle
2
2 M r 5900 kg 6.37 × 10 6 m = 1.60 × 10 −7 = s e = 5.98 × 10 24 kg 0.50 m Me rs− p
12.6. Solve:
Model: Model the woman (w) and the man (m) as spherical masses or particles.
Fw on m = Fm on w =
GMw Mm (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(50 kg)(70 kg) = = 2.3 × 10 −7 N (1.0 m ) 2 rm2− w
12.7. Model: Model the earth (e) as a sphere. Visualize:
The space shuttle or a 1.0 kg sphere (s) in the space shuttle is Re + rs = 6.37 × 10 6 m + 0.30 × 10 6 m = 6.67 × 10 6 m away from the center of the earth. Solve:
(a)
Fe on s =
GMe Ms (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg)(1.0 kg) = = 8.97 N ( Re + rs ) 2 (6.67 × 10 6 m ) 2
(b) Because the sphere and the shuttle are falling with the same acceleration, there cannot be any relative motion between them. That is why the sphere floats around inside the space shuttle.
12.8.
Model: Model the sun (s) as a spherical mass.
Solve: (b)
(a)
gsun surface =
gsun at earth =
GMs (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg) = = 274 m /s 2 (6.96 × 10 8 m ) 2 Rs2
GMs (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg) = = 5.90 × 10 −3 m /s 2 (1.50 × 1011 m ) 2 rs2− e
12.9.
Model: Model the moon (m) and Jupiter (J) as spherical masses. GMm (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(7.36 × 10 22 kg) Solve: (a) gmoon surface = = = 1.62 m /s 2 (1.74 × 10 6 m ) 2 Rm2 (b)
gJupiter surface =
GM J (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.90 × 10 27 kg) = = 25.9 m /s 2 (6.99 × 10 7 m ) 2 RJ2
12.10. Solve:
Model: Model the distant planet (P) as a spherical mass. The acceleration at the surface of the planet and at the altitude h are
gsurface =
GM P R2
and galtitude =
MP 1 1 GM P = gsurface ⇒ G 2 ( R + h) 2 2 R 2
⇒ ( R + h) 2 = 2 R 2 ⇒ R + h = 2 R ⇒ h = ( 2 − 1) R = 0.414 R That is, the starship is orbiting at an altitude of 0.414R.
12.11. Model: Model the earth (e) as a spherical mass. Visualize: The acceleration due to gravity at sea level is 9.83 m/s2 (see Table 12.1) and Re = 6.37 × 10 6 m (see Table 12.2).
GMe = ( Re + h) 2
GMe
gearth
2 2 = ( 9.83 − 0.0075) m /s h h Re2 1 + 1 + Re Re Here gearth = GMe Re2 is the acceleration due to gravity on a non-rotating earth, which is why we’ve used the value 9.83 m/s2. Solving for h,
Solve:
gobservatory =
2
=
9.83 h= − 1 Re = 2430 m 9 . 8225
12.12. Model: Model the earth (e) as a spherical mass. Solve:
Let the acceleration due to gravity be 3gsurface when the earth is shrunk to a radius of x. Then, GMe GM and 3gsurface = 2 e gsurface = Re2 x
GMe GMe R = 2 ⇒ x = e = 0.577 Re Re2 x 3 The earth’s radius would need to be 0.577 times its present value. ⇒3
12.13.
Model: Model the planet Z as a spherical mass. GM Z (6.67 × 10 −11 N ⋅ m 2 / kg 2 ) M Z ⇒ 8.0 m/s 2 = ⇒ M Z = 3.0 × 10 24 kg Solve: (a) gZ surface = 2 RZ (5.0 × 10 6 m ) 2 (b) Let h be the height above the north pole. Thus, gZ surface GM Z GM Z 8.0 m /s 2 2 gabove N pole = = = = 2 2 2 = 0.889 m /s 6 ( RZ + h) 2 10.0 × 10 m h h RZ2 1 + 1 + 1 + 5.0 × 10 6 m RZ RZ
12.14. Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as an isolated system, so mechanical energy is conserved. Visualize:
GMM (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(6.42 × 10 23 kg) = 3.77 m /s 2 = (3.37 × 10 6 m ) 2 RM2 (b) A height of 15 m is very small in comparison with the radius of Earth or Mars. We can use flat-earth gravitational potential energy to find the speed with which the astronaut can throw the ball. On Earth, with yi = 0 m and vf = 0 m/s, energy conservation is Solve:
(a)
gMar’s surface =
1 2
mvf2 + mgyf = 12 mvi2 + mgyi ⇒ vi = 2 ge yf = 2(9.80 m / s2 ) (15 m) = 17.15 m / s
Energy is also conserved on Mars, but the acceleration due to gravity is different. On Mars, with yi = 0 m and vf = 0 m/s, energy conservation is 1 2
mvf2 + mgyf = 12 mvi2 + mgyi ⇒ yf =
(17.15 m / s)2 vi2 = = 39.0 m 2 gm 2(3.77 m / s2 )
12.15. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize:
A pictorial representation of the before-and-after events is shown. Solve: After using v2 = 0 m/s, the energy conservation equation K 2 + U2 = K1 + U1 is
0J−
GMe mp Re + h
=
GMe mp 1 mp v12 − 2 Re
The projectile mass cancels. Solving for h, we find −1
1 v12 h= − − Re Re 2GMe = 4.18 × 10 5 m = 418 m
12.16. Model: Model the earth (e) as a spherical mass. Compared to the earth’s size and mass, the rocket (r) is modeled as a particle. This is an isolated system, so mechanical energy is conserved. Visualize:
Solve:
The energy conservation equation K 2 + U2 = K1 + U1 is
In the present case, r2 → ∞, so
GMe mr 1 GMe mr 1 mr v22 − = mr v12 − 2 2 r2 Re 2GMe GMe mr 1 1 mr v22 = mr v12 − ⇒ v22 = v12 − 2 2 Re Re
⇒ v2 = (1.50 × 10 4 m /s)2 − = 9.99 × 103 m /s
2(6.67 × 10 −11 N m 2 / kg2 )(5.98 × 10 24 kg) 6.37 × 10 6 m
12.17. Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize:
Solve:
The energy conservation equation K 2 + U2 = K1 + U1 is
GM J mo 1 GM J mo 1 mo v22 − = mo v12 − 2 2 r2 RJ where RJ and MJ are the radius and mass of Jupiter. Using the asymptotic condition v2 = 0 m /s as r2 → ∞,
0J=
GM J mo 1 mo v12 − ⇒ v1 = 2 RJ
2GM J = RJ
2(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.90 × 10 27 kg) = 6.02 × 10 4 m /s 6.99 × 10 7 m
Thus, the escape velocity from Jupiter is 60.2 km/s.
12.18. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize:
We denote by mp the mass of the probe. MS is the sun’s mass, and RS− p is the separation between the centers of the sun and the probe. Solve: The conservation of energy equation K 2 + U2 = K1 + U1 is
GMS mp 1 GMS mp 1 mp v22 − = mp v12 − 2 2 (r2 ) RS− p Using the condition v2 = 0 m/s asymptotically as r2 → ∞,
GMS mp 1 2 mp vescape = ⇒ vescape = 2 RS− p
2GMS = RS− p
2(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg) = 4.21 × 10 4 m /s (1.50 × 1011 m )
12.19. Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated, the mechanical energy of the object on both the planet and the earth is conserved. Visualize:
Let us denote the mass of the planet by Mp and that of the earth by Me. Your mass is m0. Also, acceleration due to gravity on the surface of the planet is gp and on the surface of the earth is ge. Rp and Re are the radii of the planet and the earth, respectively. Solve: (a) We are given that M P = 2 Me and gp = 14 ge . GM p GMe Since gp = and ge = , we have Rp2 Re2
GM p 2 p
R
=
1 GMe 1 G( M p / 2) = ⇒ Rp = 8 Re = 8 (6.37 × 10 6 m ) = 1.80 × 10 7 m 4 Re2 4 Re2
(b) The conservation of energy equation K 2 + U2 = K1 + U1 is
GM p mo 1 GM p mo 1 mo v22 − = mo v12 − 2 2 r2 Rp Using v2 = 0 m/s as r2 → ∞, we have
GM p mo 1 2 = mo vescape 2 Rp ⇒ vescape =
2GM p Rp
=
2 G( 2 M e ) = Rp
4(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg) = 9410 m /s 1.80 × 10 7 m
12.20. Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite, having mass ms and velocity vs, orbits the sun with a mass MS in a circle of radius rs. Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration for circular motion. Newton’s second law is GMS ms ms vs2 = rs2 rs Because vs = 2πrs / Ts where Ts is the period of the satellite, this equation simplifies to
GMS (2πrs ) 2 GMS Ts2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg)(24 × 3600 s) 2 3 = ⇒ = = ⇒ rs = 2.93 × 10 9 m r s 4π 2 4π 2 rs2 Ts2 rs
12.21. Model: Model the earth (e) as a spherical mass and the shuttle (s) as a point particle. Visualize: The shuttle, having mass ms and velocity vs orbits the earth in a circle of radius rs. We will denote the earth’s mass by Me. Solve: The gravitational force between the earth and the shuttle provides the necessary centripetal acceleration for circular motion. Newton’s second law is GMe ms ms vs2 GMe = ⇒ vs2 = ⇒ vs = rs2 rs rs
GMe rs
Because rs = Re + 250 miles = 6.37 × 106 m + 4.023 × 105 m = 6.77 × 106 m,
vs = Ts =
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg) = 7680 m /s (6.77 × 10 6 m ) 2πrs 2π (6.77 × 10 6 m ) = = 5542 s = 92.4 minutes 7.675 × 10 3 m /s vs
12.22. Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle. Visualize: The asteroid, having mass ma and velocity va, orbits the sun in a circle of radius ra. The asteroid’s time period is Ta = 5.0 earth years = 1.5779 × 108 s. Solve: The gravitational force between the sun (mass = Ms) and the asteroid provides the centripetal acceleration required for circular motion. 2
GMs Ta2 GMs ma ma va2 GMs 2πra = ⇒ = ⇒ ra = 2 ra ra ra 4π 2 Ta
1/ 3
Substituting G = 6.67 × 10 −11 N ⋅ m 2 / kg 2 , Ms = 1.99 × 10 30 kg, and the time period of the asteroid, we obtain ra = 4.37 × 1011 m. The velocity of the asteroid in its orbit will therefore be
va =
2πra (2π )( 4.37 × 1011 m ) = 1.74 × 10 4 m /s = 1.5779 × 10 8 s Ta
12.23. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs. The radius of the circular orbit is denoted by rs and the mass of the earth by Me. Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular motion: GMe ms ms vs2 GM (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg) = ⇒ rs = 2 e = = 1.32 × 10 7 m 2 (5500 m /s) 2 rs rs vs ⇒ Ts =
2πRs (2π )(1.32 × 10 7 m ) = = 1.51 × 10 4 s = 4.2 hr (5500 m /s) vs
12.24. Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle. Visualize: The geosynchronous satellite whose mass is ms and velocity is vs orbits in a circle of radius rs around Mars. Let us denote mass of Mars by Mm. Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion: GMm Ts2 GMm ms ms vs2 ms (2πrs ) 2 = = ⇒ rs = 2 2 rs rs rs (Ts ) 4π 2
1/ 3
Using G = 6.67 × 10 −11 N ⋅ m 2 / kg 2 , Mm = 6.42 × 1023 kg, and Ts = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we obtain rs = 2.05 × 10 7 m. Thus, altitude = rs − Rm = 1.72 × 10 7 m.
12.25. Model: Model the sun (s) and the earth (e) as spherical masses. Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs–e. The sun’s and the earth’s masses are denoted by Ms and me. Solve: The gravitational force provides the centripetal acceleration required for circular motion. GMs me me ve2 me (2πrs− e ) 2 = = rs2− e rs− e rs− e Te2 ⇒ Ms =
4π 2 rs3− e 4π 2 (1.50 × 1011 m ) 3 = = 2.01 × 10 30 kg −11 2 (6.67 × 10 N ⋅ m 2 / kg 2 )(365 × 24 × 3600 s) 2 GTe
Assess: The tabulated value is 1.99 × 10 30 kg. The slight difference can be ascribed to the fact that the earth’s orbit isn’t exactly circular.
12.26. Model: Model the planet and satellites as spherical masses. Visualize: Please refer to Figure Ex12.26. Solve: (a) The period of a satellite in a circular orbit is T = [(4π2/GM)r3]1/2. This is independent of the satellite’s mass, so we can find the ratio of the periods of two satellites a and b: r Ta = a Tb rb
3
Satellite 2 has r2 = r1, so T2 = T1 = 250 min. Satellite 3 has r3 = (3/2)r1, so T3 = (3/2)3/2T1 = 459 min. (b) The force on a satellite is F = GMm/r2. Thus the ratio of the forces on two satellites a and b is 2
Fa rb ma = Fb ra mb Satellite 2 has r2 = r1 and m2 = 2m1, so F2 = (1)2(2)F1 = 20,000 N. Similarly, satellite 3 has r3 = (3/2)r1 and m3 = m1, so F3 = (2/3)2(1)F1 = 4440 N. (c) The speed of a satellite in a circular orbit is v = (GM/r)2, so its kinetic energy is K = 12 mv2 = GMm/2r. Thus the ratio of the kinetic energy of two satellites a and b is
K a rb ma = K b ra mb Satellite 3 has r3 = (3/2)r1 and m3 = m1, so K1/K3 = (3/2)(1/1) = 3/2 = 1.50.
12.27. Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated system, so mechanical energy is conserved. Visualize: The initial position of the lunar lander (mass = ml) is at a distance r1 = Rm + 50 km from the center of the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is r2 = Rm + 300 km.
Solve:
The external work done by the thrusters is Wext = ∆Emech =
1 2
∆Ug
where we used Emech = Ug for a circular orbit. The change in potential energy is from the initial orbit at ri = Rm + 50 km to the final orbit rf = Rm + 300 km. Thus 1 2
− GMm m − GMm m GMm m 1 1 − Wext = 12 − = 2 ri rf rf ri =
(6.67 × 10 −11 N m 2 / kg 2 )(7.36 × 10 22 kg)(4000 kg) 1 1 − 1.79 × 10 6 m 2.04 × 10 6 m 2
= 6.72 × 10 8 J
12.28. Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an isolated system, so the mechanical energy is conserved. Visualize: The space shuttle (mass = ms) is at a distance of r1 = Re + 250 km.
Solve:
The external work done by the thrusters is Wext = ∆Emech =
1 2
∆Ug
where we used Emech = 12 Ug for a circular orbit. The change in potential energy is from the initial orbit at ri = Re + 250 km to the final orbit rf = Re + 610 km. Thus
− GMe m − GMe m GMe m 1 1 − Wext = 12 − = 2 ri rf rf ri =
(6.67 × 10 −11 N m 2 / kg 2 )(5.98 × 10 24 kg)(75,000 kg) 1 1 − 6 6.62 × 10 m 6.98 × 10 6 m 2
= 1.17 × 1011 J This much energy must be supplied by burning the on-board fuel.
12.29.
Solve:
We are given M1 + M2 = 150 kg which means M1 = 150 kg − M2 . We also have
GM1 M2 (8.00 × 10 −6 N )(0.20 m ) 2 −6 = 4798 kg 2 N ⇒ M1 M2 = 2 = 8.00 × 10 6.67 × 10 −11 N ⋅ m 2 / kg 2 (0.20 m) Thus, (150 kg − M2 ) M2 = 4798 kg 2 or M22 − (150 kg) M2 + ( 4798 kg 2 ) = 0. Solving this equation gives M2 = 103.75 kg and 46.25 kg. The two masses are 104 kg and 46 kg.
12.30. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will make an angle of θ with the vertical. The distance between the sphere centers is therefore going to be less than 1 m. The free-body diagram shows the forces acting on the lead sphere.
Solve: We can see from the diagram that the distance between the centers is d = 1.000 m – 2Lsinθ. Each sphere is in static equilibrium, so Newton’s second law is
∑F
x
= Fgrav − T sin θ = 0 ⇒ T sin θ = Fgrav
∑F
y
= T cosθ − mg = 0 ⇒ T cosθ = mg
Dividing these two equations to eliminate the tension T yields
Fgrav Gmm / d 2 Gm sin θ = tan θ = = = 2 cosθ mg mg d g We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to be enough to affect the value of Fgrav, the gravitational attraction between the two masses, so we’ll evaluate the right side of this equation by using 1.00 m for d. This gives
tan θ =
(6.67 × 10 −11 N m 2 / kg 2 ) (100 kg) = 6.81 × 10 −10 ⇒ θ = (3.90 × 10 −8 )° (1.00 m) 2 (9.80 m / s 2 )
This small angle causes the two spheres to move closer by 2Lsinθ = 1.4 × 10–7 m = 0.00000014 m. Consequently, the distance between their centers is d = 0.99999986 m.
12.31. Visualize:
The gravitational force between the lead sphere (Ml = 20 kg) and a smaller sphere (ms = 0.200 kg) leads to a deflection of the suspended rod when the heavier lead spheres are brought closer to the smaller spheres attached to both ends of the rod. GMl ms (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(20 kg)(0.200 kg) Solve: F = = = 1.1858 × 10 −8 N = 1.19 × 10 −8 N (0.15 m ) 2 d2 Because F = k∆θ,
∆θ =
F 1.1858 × 10 −8 N = = 1.19 × 10 −3 rad = 0.0679 degrees k 1.0 × 10 −5 N / rad
12.32. Model: Model the sun (s), the earth (e), and the moon (m) as spherical masses. Visualize:
In the case of a full moon, the net force is Fs on m + Fe on m. On the other hand, the net force in the case of a new moon is Fs on m – Fe on m. At a full moon, the separation between the sun’s center and the moon’s center is rs–m = Rs–e + Re–m, where Rs–e is the distance of the earth from the sun and Re–m is the distance of the moon from the earth. In the case of a new moon, rs–m = Rs–e – Re–m. Rs–e = 1.50 × 1011 m and Re–m = 3.84 × 108 m. This means rs–m is 1.50384 × 1011 m for a full moon and 1.49616 × 1011 m for a new moon. The masses of the moon, earth, and sun are represented by mm, Me, and Ms and have the values 7.36 × 1022 kg, 5.98 × 1024 kg, and 1.99 × 1030 kg. Solve: At a full moon:
Fs on m =
GMs mm GMs mm GMe mm = = 4.32 × 10 20 N and Fe on m = = 1.99 × 10 20 N 2 2 ( Rs− e + Re − m ) rs− m re2− m ⇒ Fnet = Fs on m + Fe on m = 6.31 × 10 20 N
At a new moon:
Fs on m =
GMs mm GMs mm = = 4.36 × 10 20 N 2 rs− m ( Rs− e − Re − m ) 2
⇒ Fnet = Fs on m − Fe on m = 2.37 × 10 20 N ⇒
( Fnet ) new moon 2.37 × 10 20 N = = 0.376 6.31 × 10 20 N ( Fnet ) full moon
Assess: This is reasonable because Fe on m is about (1/2)Fs on m and it can add and/or subtract from Fs on m to give the above result.
12.33. Visualize:
We placed the origin of the coordinate system on the 20 kg sphere (m1). The sphere (m2) with a mass of 10 kg is 20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses m1 and m2. This is because on such a point, the gravitational forces due to m1 and m2 are in opposite directions. As the gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of m1 or to the right of m2, will experience gravitational forces from m1 and m2 pointing in the same direction, thus always leading to a nonzero force. mm m2 m 20 10 Solve: Fm1 on m = Fm2 on m ⇒ G 12 = G ⇒ 2 = (0.20 − x ) 2 (0.20 − x ) 2 x x
⇒ 10 x 2 − 8 x + 0.8 = 0 ⇒ x = 0.683 m and 0.117 m The value x = 68.3 cm is unphysical in the current situation, since this point is not between m1 and m2. Thus, the point (x, y) = (11.7 cm, 0 cm) is where a small mass is to be placed for a zero gravitational force.
12.34. Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 5.0 kg mass (m3) is on the x-axis and the 10.0 kg mass (m2) is on the y-axis.
Solve:
(a) The forces acting on the 20 kg mass (m1) are r Gm1 m2 ˆ (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(20.0 kg)(10.0 kg) ˆ Fm2 on m1 = j= j = 3.335 × 10 −7 jˆ N (0.20 m ) 2 r122 r Gm1 m3 ˆ (6.67 × 10 −7 N ⋅ m 2 / kg 2 )(20.0 kg)(5.0 kg) ˆ Fm3 on m1 = i = i = 6.67 × 10 −7 iˆ N (0.10 m ) 2 r132 r Fon m1 = 6.67 × 10 −7 iˆ N + 3.335 × 10 −7 jˆ N ⇒ Fon m1 = 7.46 × 10 −7 N
Fm on m 6.67 × 10 −7 N θ = tan −1 3 1 = tan −1 = 63.4° 3.335 × 10 −7 N Fm2 on m1
r Thus the force is Fon m1 = 7.46 × 10 −7 N, 63.4° cw from the y -axis . (b) The forces acting on the 5 kg mass (m3) are r r Fm1 on m3 = − Fm3 on m1 = −6.67 × 10 −7 iˆ N
(
Fm2 on m3 =
)
Gm2 m3 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(10.0 kg)(5.0 kg) = = 6.67 × 10 −8 N [(0.20) 2 + (0.10) 2 ] m r232
r Fm2 on m3 = −(6.67 × 10 −8 N )cos φiˆ + (6.67 × 10 −8 N )sin φjˆ
20 cm ˆ 10 cm ˆ = −(6.67 × 10 −8 N ) i + (6.67 × 10 −8 N ) j 22.36 cm 22.36 cm = −(2.983 × 10 −8 N )iˆ + (5.966 × 10 −8 N ) jˆ
r Fon m3 = −6.968 × 10 −7 iˆ N + 5.966 × 10 −8 jˆ N
Fon m3 = ( −6.968 × 10 −7 N ) 2 + (5.966 × 10 −8 N ) 2 = 6.99 × 10 −7 N 6.968 × 10 −7 N θ ′ = tan −1 = 85.1° 5.966 × 10 −8 N
r Thus Fon m3 = 6.99 × 10 −7 N, 85.1° ccw from the y -axis .
(
)
12.35.
Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 10.0 kg mass (m2) is on the y-axis and the 5.0 kg mass (m3) has the coordinates (10 cm, 20 cm).
Solve:
(a) The forces on the 10.0 kg mass (m2): r Gm1 m2 ˆ G(20.0 kg)(10.0 kg) ˆ Fm1 on m2 = − j=− j = −3.33 × 10 −7 jˆ N (0.20 m ) 2 r122
Gm2 m3 ˆ G(10.0 kg)(5.0 kg) ˆ i = i = 3.33 × 10 −7 iˆ N (0.10 m ) 2 r232 r ⇒ Fon m2 = (3.33 × 10 −7 )iˆ N − (3.33 × 10 −7 ) jˆ N
Fm3 on m2 =
Fon m2 = (3.33 × 10 −7 ) 2 + ( −3.33 × 10 −7 ) 2 = 4.72 × 10 −7 N
3.33 × 10 −7 and θ = tan −1 = 45° 3.33 × 10 −7
r Thus Fon m2 = 4.72 × 10 −7 N, 45° ccw from the − y -axis . (b) The forces on the 20.0 kg mass (m1) are r Fm3 on m1 = +(1.33 × 10 −7 ) ⋅ cos φiˆ N + (1.33 × 10 −7 )sin φjˆ N
(
)
10 cm ˆ 20 cm ˆ i N + (1.33 × 10 −7 ) jN = (1.33 × 10 −7 ) 22.36 cm 22.36 cm = (0.60 × 10 −7 )iˆ N + (1.19 × 10 −7 ) jˆ N r Fm2 on m1 = 3.33 × 10 −7 jˆ N r r r ⇒ Fon m1 = Fm2 on m1 + Fm3 on m1 = (0.60 × 10 −7 )iˆ N + ( 4.52 × 10 −7 ) jˆ N
0.60 × 10 −7 Fon m1 = (894 G ) 2 + (6789 G ) 2 = 4.56 × 10 −7 N and θ ′ = tan −1 = 7.6° 4.52 × 10 −7 where the angle is clockwise of the positive y-axis. r Thus Fon m1 = 4.56 × 10 −7 N, 7.6° cw from the y -axis .
(
)
12.36. Model: The gravitational potential energy of two masses m1 and m2 , separated by a distance r12 , is defined as U = − Gm1m2/r12. Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 5.0 kg mass (m3) is on the x-axis and the 10.0 kg mass (m2) is on the y-axis
Solution: The gravitational potential energy is a scalar quantity. The gravitational potential energy of the 20.0 kg mass is the scalar sum of the potential energies from the 10.0 kg and 5.0 kg masses and the gravitational potential energy of the 5.0 kg mass is the scalar sum of potential energies from the 20.0 kg and 10 kg masses. Using G = 6.67 × 10 −11 N ⋅ m 2 / kg 2 , we obtain Gm1 m2 Gm1 m3 − (a) U m1 = U m1 , m2 + U m1 , m3 = − r12 r13
− G(20.0 kg)(10.0 kg) G(20.0 kg)(5.0 kg) − = −1.33 × 10 −7 J (0.20 m ) (0.10 m ) Gm2 m3 Gm1 m3 = U m3 , m2 + U m3 , m1 = − − r23 r13
= (b)
U m3
=
− G(10.0 kg)(5.0 kg) G(20.0 kg)(5.0 kg) − = −8.16 × 10 −8 J (0.2236 m ) (0.10 m )
12.37.
Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 10.0 kg mass (m2) is on the y-axis and the 5.0 kg mass (m3) has the coordinates (10 cm, 20 cm).
Solve:
(a) For the 10.0 kg mass (m2):
U m2 = U m2 , m1 + U m2 , m3 =
m m − Gm1 m2 Gm2 m3 − = − Gm2 1 + 3 r12 r23 r12 r23
20.0 kg 5.0 kg = −(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(10.0 kg) + = −10.0 × 10 −8 J 0.20 m 0.10 m (b) For the 20.0 kg mass (m1): m m U m1 = U m1 , m2 + U m1 , m3 = − Gm1 2 + 3 r13 r12 5.0 kg 10.0 kg = −(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(20.0 kg) + = −9.65 × 10 −8 J 0.20 m 0.2236 m Assess: Note that the gravitational potential energy is a scalar quantity.
12.38. Model: Model the planet as a spherical mass and the object as a point mass. Visualize:
Solve: The object is moving on the equator in a circular orbit of radius r = R, and therefore has a centripetal acceleration. Consequently, there is a net force on the object. We established an r-axis pointing toward the center of r the circle. The two forces acting on the object are the gravitational force Fgrav pointing toward the center of the earth and the normal force of the surface pointing away from the center. Newton’s second law along the r-axis is mv 2 ( Fnet ) r = Fgrav − n = mac = R
⇒ n = Fgrav −
GM v 2 mv 2 GMm mv 2 = − = m 2 − 2 R R R R R
But n = wapp = mgapp , where wapp is the apparent weight and gapp is the apparent acceleration due to gravity. If the planet didn’t rotate, so that v = 0 m/s, then gapp would be GM/R2 = gtrue. The apparent acceleration due to gravity is n GM v 2 gapp = = 2 − m R R An object moving in a circle of radius R with period T has speed v = 2πR/T, so we can write GM 4π 2 R 4π 2 R gapp = 2 − = gtrue − 2 R T T2 24 6 (b) Using earthly values M = Me = 5.98 × 10 kg, R = Re = 6.37 × 10 m, and T = 24 hours = 8.64 × 104 s, we can compute GMe 4π 2 Re gtrue = = 9.83 m /s 2 and = 0.03 m /s 2 2 Re T2
4π 2 R = 9.80 m /s 2 T2 (c) We have been using gapp. That is the actual acceleration measured for objects that “free fall” at the equator on a rotating earth. r (d) An r object on the north pole is not rotating about2 the earth’s center, so it has no centripetal acceleration. Thus, n and Fgrav are in equilibrium, and g = gtrue = 9.83 m/s . ⇒ gapp = gtrue −
12.39. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity (galtitude) is 10% of the surface value (gsurface). Solve: (a) Since galtitude = (0.10) gsurface, we have GMe GM = (0.10) 2 e ⇒ ( Re + h) 2 = 10 Re2 2 ( Re + h) Re ⇒ h = 2.162 Re ⇒ h = (2.162)(6.37 × 10 6 m ) = 1.377 × 10 7 m = 1.38 × 10 7 m (b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with velocity vs,
GMe ms ms vs2 = ⇒ vs = 2 ( Re + h) ( Re + h)
GMe = Re + h
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg) = 4450 m /s (6.37 × 10 6 m + 1.377 × 10 7 m )
12.40. Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize:
Solve:
(a) The conservation of energy equation K 2 + U g2 = K1 + U g1 is
GMe mo 1 GMe mo 1 mo v22 − = mo v12 − ( Re + y1 ) 2 2 Re
1 1 ⇒ v2 = 2GMe − Re Re + y1 1 1 = 3020 m /s = 2(6.67 × 10 −11 N ⋅ m 2 / kg2 )(5.98 × 10 24 kg) − 6.37 × 10 6 m 6.87 × 10 6 m (b) In the flat-earth approximation, Ug = mgy. The energy conservation equation thus becomes
1 1 mo v22 + mo gy2 = mo v12 + mo gy1 2 2 ⇒ v2 = v12 + 2 g( y1 − y2 ) = 2(9.80 m /s 2 )(5 × 10 5 m − 0 m ) = 3130 m /s (c) The percent error in the flat-earth calculation is
3130 m /s − 3020 m /s ≈ 3.6% 3020 m /s (d) No, because the hammer will have a forward motion such that while “falling” it will continue to circle the earth.
12.41. Model: Model the planet (p) as a spherical mass and the projectile as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The projectile of mass m was launched on the surface of the planet with an initial velocity v0.
Solve:
(a) The energy conservation equation K1 + U1 = K0 + U0 is
1 2 GM p m 1 2 GM p m = mv0 − mv1 − 2 Rp + y1 2 Rp −1
1 v02 6 ⇒ y1 = − − Re = 2.82 × 10 m 2 R GM p p (b) Using the energy conservation equation K1 + U1 = K 0 + U 0 with y1 = 1000 km = 100 × 10 6 m:
1 2 GM p m 1 2 GM p m mv1 − = mv0 − 2 Rp + y1 2 Rp 12
1 1 ⇒ v1 = v02 + 2GM p − Rp + y1 Rp 12 1 1 2 −11 2 2 24 = (5000 m /s) + 2(6.67 × 10 N ⋅ m / kg )(2.6 × 10 kg) − (5.0 × 10 6 m + 1.0 × 10 6 m ) 5.0 × 10 6 m = 3670 m /s
12.42. Model: Model the earth as a spherical mass and the meteoroids as point masses. Visualize:
Solve:
(a) The energy conservation equation K2 + U2 = K1 + U1 is 1 2
mv22 −
1 GMe m 1 2 GMe m 1 = 2 mv1 − ⇒ v2 = v12 + 2GMe − Re rm Re rm
1/ 2
= 1.13 × 10 4 m / s = 11.3 km / s The speed does not depend on the meteoroid’s mass. (b) This part differs in that r2 = Re + 5000 km = 1.137 × 10 7 m. The shape of the meteoroid’s trajectory is not important for using energy conservation. Thus 1 2
GMe m GMe m 1 1 mv − = 12 mv12 − ⇒ v2 = v12 + 2GMe − Re + 5000 km rm Re + 5000 km rm 2 2
= 8.94 × 10 3 m / s = 8.94 km / s
1/ 2
12.43.
Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize:
In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential energy. In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both kinetic energy and potential energy. Solve: The conservation of energy equation K f + U f = K i + U i is
1 2 GMm GMm mvf − − =0 J +0 J rf1 rf2 2 ⇒ vf =
4GM = rf
4(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg) = 32, 600 m /s (0.5 × 1012 m )
Assess: Note that the final velocity of 32.6 km/s does not depend on the mass of the comet.
12.44. Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The radius of the asteroid is Ma and its mass is Ra. Solve: The conservation of energy equation K f + U f = K i + U i for the asteroid is 1 2 GMa m 1 2 GMa m mvf − = mvi − 2 Ra + r 2 Ra The minimum speed for escape is the one that will cause you to stop only when the separation between you and the asteroid becomes very large. Noting that vf → 0 m/s as r → ∞, we have
vi2 =
2GMa 2(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.0 × 1014 kg) = ⇒ vi = 2.58 m /s (2.0 × 10 3 m ) Ra
That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on the earth. The conservation of energy equation is
0J−
1 2 mvi = − mg(0.50 m ) ⇒ vi = 2(9.8 m /s 2 )(0.50 m ) = 3.13 m /s 2
This means you can escape from the asteroid.
12.45. Model: Model the planets as spherical masses and the rocket as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: Please refer to Figure P12.45. The mass of the rocket is m. Solve: To calculate the escape speed from A, apply the energy conservation equation K f + U f = K i + U i : 1 2 GMm GMm 1 2 GMm GMm mvf − − = mvi − − r1 r2 R 2 2 7R The minimum speed for escape is the one that will cause the rocket to stop (vf = 0 m/s) only when r → ∞. This means
vi =
2GM 1 16GM 1+ = R 7 7R
We calculate the escape speed from B in the same way: 1 2 GMm GMm 1 2 GMm GMm mvf − − = mvi − − r1 r2 2 2 3R 3R Again, vf = 0 m/s as r1 → ∞ and r2 → ∞. Thus,
1 2 2GMm 4GM mvi = ⇒ vi = 2 3R 3R Assess: The ratio of the escape speed from point A to that from point B is
16 GM 7 R
4 GM 12 = = 1.31 3 R 7
That is, the rocket needs more energy to escape from point A then from point B.
12.46. Model: Model the two stars as spherical masses and the space capsule as a point mass. Visualize: We put the origin of the coordinate system on the left star. The right star is at x = 10 Rs, where Rs is the radius of the star. The mass of each star is Ms and the mass of the space capsule is m. Solve: (a) The potential energy of the space capsule along a line passing between the two stars is U=
GMs m − GMs m GMs m − GMs m − GMs m 1 1 − = − = + (10 − p) Rs x1 x2 pRs Rs p 10 − p
where p can vary from 1 through 9. Substituting into this equation gives
U=−
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg)(1 × 10 4 kg) 1 1 + 8 (6.96 × 10 m ) p 10 − p
1 1 = −(1.907 × 1015 J ) + p 10 − p A graph of this potential energy as a function of p is shown.
r r (b) This is a point of equilibrium because the two forces on the space capsule are equal but opposite, so Fnet = 0. However the equilibrium is unstable because a slight offset from the exact midpoint would mean more force from one star than from the other star, causing a motion toward the closer star. (c) Since the exact midpoint is 5Rs, the capsule will be 5 R + 0.5 m away from one star and 5 R − 0.5 m away from the other star. The m makes no numerical contribution. At the end, when colliding with the nearer star, one distance is Rs and the other is 9Rs. From the energy conservation equation K f + U f = K i + U i , with vi = 0, we have 1 2 GMs m GMs m 1 2 GMs m GMs m mvf − − = mvi − − 2 9 Rs 2 5 Rs 5 Rs Rs vf2 =
2GMs Rs
−11 30 2 2 1 + 1 − 1 − 1 = 2(6.67 × 10 N ⋅ m / kg )(1.99 × 10 kg) 32 ⇒ v = 521 km /s f 9 5 5 45 (6.96 × 10 8 m )
Assess: The speed is very large, but the sun is a very large attractor.
12.47. Model: Model the earth and moon as spheres and the spacecraft as a point mass. Visualize: Let Re–m be the earth-moon distance and r be the distance of the spacecraft from the center of the earth. This point is distance Re–m – r from the center of the moon. Then let r′ be the distance of the orbiting spacecraft from the center of the moon.
(a) The point at which there is no net force is the point at which the earth’s gravitational force on mass m equals the moon’s gravitational force on m. This is
GMe m GMm m M = ⇒ ( Re − m − r )2 = m r 2 = cr 2 r2 Me ( Re − m − r )2 where c = Mm/Me = 0.01231 is the moon-to-earth mass ratio. The resulting quadratic equation is
1 ± c (1 − c)r 2 − 2 Re − m r + Re2− m = 0 ⇒ r = Re − m 1− c The solution with r < Re–m (i.e., between the earth and the moon) is r = 0.900Re–m = 3.46 × 108 m. (b) The gravitational potential energy at distance r is GMe m GMm m U=− − r2 Re − m − r Since F = –dU/dr, the potential energy should be a maximum (dU/dr = 0) at r = 3.46 × 108 m where, as found in part (a), F = 0. Calculating the potential energy at several points from the earth to the moon produces the graph shown below. Note that the potential energy at the cross-over point, which will be needed in part (d), is Ucross = –1.28 × 1010 J.
(c) 300 km is very small in comparison with the distance from the earth to the moon, so we’ll assume that the spacecraft’s distance from the earth is always r = Re–m as it orbits. The mechanical energy of its orbit around the moon is Emech = K + Ug moon = 12 Ug moon. Including its gravitational potential energy with the earth, the spacecraft’s total mechanical energy is
Emech = 12 Ug moon + Ug earth = −
GMm m GMe m GMm m GMe m − =− − = −2.24 × 1010 J Re − m Re − m 2r ′ 2( Rm + 300 km)
(d) The minimum work would allow the spacecraft to reach the cross-over point with K = 0. Thus the work required is Wext = ∆Emech = Ucross – (Emech)moon orbit = –1.28 × 1010 J – (–2.24 × 1010 J) = 0.96 × 1010 J
(e) The conservation of energy equation is Kf + Uf = Ki + Ui. The initial point is the cross-over point, where Ki = 0 and Ui = –1.28 × 1010 J. The final point is rf = Re + 100 km = 6.47 × 106 m. Because Mm << Me and this point is very far from the moon, the moon’s contribution to the potential energy is very small and will be ignored. Thus 1 2
mvf2 −
2GMe 2U i GMe m = U i ⇒ vf = + rf m rf
1/ 2
= 11.0 × 10 4 m / s = 11.0 km / s
12.48. Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also use the law of conservation of momentum for our system. Visualize:
Solve:
The conservation of momentum equation pfx = pix is
M (vfx )1 + 2 M (vfx ) 2 = 0 kg m /s ⇒ (vfx )1 = −2(vfx ) 2 The equation for mechanical energy conservation K f + U f = K i + U i is
G( M )(2 M ) G( M )(2 M ) 1 1 1 0.8GM M (vfx )12 + (2 M )(vfx ) 22 − =− ⇒ [2(vfx ) 2 ]2 + (vfx ) 22 = R 2 2 2R 10 R 2 ⇒ (vfx ) 2 = −0.516
GM GM ⇒ (vfx )1 = −2(vfx ) 2 = 1.032 R R
The heavier asteroid has a speed of 0.516 (GM/R)1/2 and the lighter one a speed of 1.032 (GM/R)1/2.
12.49. Model: Gravity is a conservative force, so we can use conservation of energy. Visualize:
The planets will be pulled together by gravity and each will have speed v2 as they crash and the separation between their centers will be 2R. Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential and kinetic energy. Thus,
GMM GMM 1 1 2 2 K 2 + U2 = Mv2 + Mv2 − = K1 + U1 = 0 J − r1 r2 2 2 1 1 ⇒ v2 = GM − r2 r1 Because the planet is “Jupiter-size,” we’ll use M = MJupiter = 1.9 × 1027 kg and r2 = 2RJupiter = 1.4 × 108 m. The crash speed of each planet is v2 = 3.0 × 104 m/s. Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant acceleration motion. The formulas from constant-acceleration kinematics do not work for problems such as this.
12.50.
Solve:
(a) Using Equation 12.34 for a satellite in a circular orbit,
∆Emech = =
1 1 1 1 − ∆U g = ( − GMe m) 2 2 Re + 300 km Re + 500 km −1 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg)(5.00 × 10 4 kg) 2 1 1 = −4.35 × 1010 J × − 6.37 × 10 6 m + 3 × 10 5 m 6.37 × 10 6 m + 5 × 10 5 m
The negative sign indicates that 4.35 × 1010 J of energy is lost. (b)
The shuttle would fire retro-rockets to lose enough energy to go into an elliptical orbit. When halfway around, at rmin, the shuttle would fire retro-rockets again to lose the rest of the energy to stay at rmin.
Model: Model the earth (mass = Me) as a spherical mass and an object (mass = mo) as a point mass. Visualize:
12.51.
The figure shows the top view of the earth rotating at an angular velocity ω = v / Re . The forces on an object on the r r surface of the earth are the normal force n and the weight w. Solve: An object on the equator is in circular motion. Newton’s second law for the object is mv 2 Fr = w − n = Re As the speed increases, the apparent weight n decreases. The object will fly off when n becomes zero. At this speed
GmMe mv 2 GMe GMe mv 2 2πRe ⇒ = ⇒ v2 = ⇒ = T Re Re2 Re Re Re 2
w= ⇒T =
4π 2 Re3 = GMe
4π 2 (6.37 × 10 6 m ) 3 = 5, 058 s = 1.405 hrs (6.67 × 10 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg) −11
Assess: This time period is approximately 17 times smaller than the real time period, implying a much faster rotational motion of the earth.
12.52. Solve:
Model: Model Mars (M) and Phobos (P) as spherical masses. The period of a satellite orbiting a planet of mass Mm is
4π 3 T2 = r GMm Thus we can use Phobos’s orbit to find the mass of Mars:
Mm =
4π 2 r 3 GT
2
=
4π 2 (9.4 × 10 6 m ) 3 = 6.48 × 10 23 kg (6.67 × 10 N ⋅ m 2 / kg 2 )(2.7540 × 10 4 s) 2 −11
Assess: From Table 12.2, the mass of Mars is 6.42 × 1023 kg. The slight difference is likely due to Phobos’s orbit being somewhat non-circular.
12.53. Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated system, so mechanical energy is conserved. Visualize:
The orbital radius of the satellite is
r = Ra + h = 8, 800 m + 5, 000 m = 13,800 m Solve:
(a) The speed of a satellite in a circular orbit is 12
GM (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.0 × 1016 kg) = = 6.95 m /s r (13,800 m ) (b) The minimum launch speed for escape (vi) will cause the satellite to stop asymptotically (vf = 0 m/s) as rf → ∞. Using the energy conservation equation K 2 + U2 = K1 + U1 , we get v=
GMa ms 1 GMa ms GMa 1 1 2 ms vf2 − = ms vi2 − ⇒ 0 J − 0 J = vescape − 2 2 2 rf Ra Ra ⇒ vescape =
2GMa = Ra
2(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.0 × 1016 kg) = 12.3 m /s 8800 m
12.54. Solve:
Model: Model the star (s) and the planet (p) as spherical masses. A planet’s acceleration due to gravity is GM p gp = Rp2
⇒ Mp =
gp Rp2 G
=
(12.2 m /s 2 )(9.0 × 10 6 m ) 2 = 1.48 × 10 25 kg 6.67 × 10 −11 N ⋅ m 2 / kg 2
(b) A planet’s orbital period is
4π 2 3 T2 = r GMs ⇒ Ms =
4π 2 r 3 4π 2 (2.2 × 1011 ) 3 = = 5.22 × 10 30 kg GT 2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )( 402 × 24 × 3600 s) 2
12.55. Model: Model the moon as a spherical mass and the satellite as a point mass. Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own axis. This time happens to be 27.3 days. Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration necessary for circular motion around the moon. Therefore, GMm m 2π = mrω 2 = mr T r2 ⇒ r3 =
2
GMm T 2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(7.36 × 10 22 kg)(27.3 × 24 × 3600 s) 2 = 4π 2 4π 2 ⇒ r = 8.84 × 10 7 m
Since r = Rm + h , then h = r − Rm = 8.84 × 107 m – 1.74 × 106 m = 8.67 × 107 m.
12.56. Model: Model the earth as a spherical mass and the satellite as a point mass. Visualize: The satellite is directly over a point on the equator once every two days. Thus, T = 2Tearth = 2 × 24 × 3600 s = 1.728 × 10 5 s. Solve: A satellite’s period is
4π 2 3 T2 = r GMe GMe T 2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg)(1.728 × 10 5 ) 2 = 4π 2 4π 2 ⇒ r = 6.71 × 10 7 m Assess: The radius of the orbit is larger than the geosynchronous orbit. ⇒ r3 =
12.57.
Solve:
(a) Taking the logarithm of both sides of vp = Cuq gives
[log(v p ) = p log v] = [log(Cu q ) = log C + q log u] ⇒ log v =
q log C log u + p p
But x = log u and y = log v, so x and y are related by
q log C y = x + p p (b) The previous result shows there is a linear relationship between x and y, hence there is a linear relationship between log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u will be a straight line. (c) The slope of the straight line represented by the equation y = (q / p) x + log C / p is q/p. Thus, the slope of the log v-versus-log u graph will be q/p. (d) From Newton’s theory, the period T and radius r of an orbit around the sun are related by
4π 2 3 T2 = r GM This equation is of the form T p = Cr q, with p = 2, q = 3, and C = 4π2/GM. If the theory is correct, we expect a graph of log T-versus-log r to be a straight line with slope q/p = 3/2 = 1.500. The experimental measurements of actual planets yield a straight line graph whose slope is 1.500 to four significant figures. Note that the graph has nothing to do with theory—it is simply a graph of measured values. But the fact that the shape and slope of the graph agree precisely with the prediction of Newton’s theory is strong evidence for its correctness. (e) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is M = Msun.
1 1 4π 2 4π 2 1 = − ⇒ = 10 −18.528 = 18.528 log C = log . 9 264 GMsun 2 2 GMsun 10 4π 2 ⋅ 1018.528 = 1.996 × 10 30 kg G The tabulated value, to three significant figures, is Msun = 1.99 × 1030 kg. We have used the orbits of the planets to “weigh the sun!” ⇒ Msun =
12.58. Solve:
Visualize: Please refer to Figure P12.58. The gravitational force on one of the masses is due to the star and the other planet. Thus
Mm Gmm mv 2 m 2πr GM Gm 4π 2 r 2 + = = ⇒ + = r2 r r T r T2 4r (2 r ) 2 2
G
4π 2 r 3 G m 4π 2 r 2 1 M+ = ⇒T = 2 r T G M m 4 ( + / 4 )
12
12.59. Model: Model the earth and moon as spherical masses. Solve: The mechanical energy of the earth-moon system is Emech = 12 U g = −
GMe Mm (6.67 × 10 −11 N m 2 / kg 2 ) (5.98 × 10 24 kg)(7.36 × 10 22 kg) =− 2rm 2(3.84 × 10 8 m)
= −3.82 × 10 28 J Reducing the mechanical energy will make Emech more negative. A larger negative number requires a smaller value of rm, so the radius of the moon’s orbit will decrease. If all human energy is supplied by the moon for 100 years, the moon-earth energy change is
∆Emech = − P∆t = −(1.0 × 1013 J / s )(100 years × 3.16 × 10 7 s / year) = −3.16 × 10 22 J The energy loss is only about one part in a million. This is so small that it’s hard to calculate the radius change directly. Instead, we can make use of the fact that dE/dr ≈ ∆E/∆r for very small changes. The derivative of Emech is
dEmech GMe Mm = 2rm2 dr and thus ∆r ≈
2rm2 2(3.84 × 10 8 m) 2 ∆Emech = ( −3.16 × 10 22 J) 2 −11 (6.67 × 10 N m / kg 2 ) (5.98 × 10 24 kg)(7.36 × 10 22 kg) GMe Mm
= −317 m
We can relate the radius change to a period change by using T2 = (4π/GM)r3. The derivative is
4π 2 3 4π 3 3 2 2T dT = rm dr = T dr 3rm dr = GM r GM r e m e −317 m 3 ∆r 3 ⇒ ∆T ≈ T = (27.3 days × 86,400 s / day ) = −2.9 s 2 rm 2 3.84 × 10 8 m Thus the radius decreases by 317 m, or 8.26 × 10–5%, and the period decreases by 2.9 s, or 1.25 × 10–4%.
12.60.
Solve:
(a) Dividing the circumference of the orbit by the period,
v=
2πRs 2π (1.0 × 10 4 m ) = = 6.28 × 10 4 m /s 1.0 s T
(b) Using the formula for the acceleration at the surface,
gsurface =
GMs (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg) = = 1.33 × 1012 m /s 2 (1 × 10 4 m ) 2 Rs2
(c) The mass of an object on the earth will be the same as its mass on the star. The weight is
wstar = mg surface = 1.33 × 1012 N (d) The radius of the orbit of the satellite is r = 1 × 10 4 m + 1.0 × 10 3 m = 1.1 × 10 4 m. The period is
T2 =
4π 2 r 3 4π 2 (1.1 × 10 4 m ) 3 = ⇒ T = 6.29 × 10 −4 s GMs (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg)
This means there are 1589 revolutions per second or 95,360 orbits per minute. (e) Applying Equation 12.25 for a geosynchronous orbit,
r3 =
GMs 2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg)(1.0 s) T = ⇒ r = 1.50 × 10 6 m 4π 2 4π 2
12.61. Model: Assume the two stars are spherical masses. Visualize: The gravitational force between the two stars provides the centripetal acceleration required for circular motion about the center of mass.
Solve:
Newton’s second law is 1
GT 2 M 3 GMM 2π = = MRω 2 = MR ⇒ R = 2 T ( 2 R) 16π 2 2
Fgravitation
Using T = 90 days = 90 × 24 × 3600 s and M = 2Msun = 3.98 × 1030 kg, we get R = 4.667 × 1010 m. Thus the star separation is 2R = 9.33 × 1010 m.
12.62. Model: Assume the three stars are spherical masses. Visualize:
The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three stars. The gravitational force between any two stars is the same. On a given star the two forces from the other stars make an angle of 60°. Solve: The value of r can be found as follows:
L/2 L 1.0 × 1012 m = cos 30° ⇒ r = = = 0.577 × 1012 m r 2 cos 30° 2 cos 30° The gravitational force between any two stars is
Fg =
GM 2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 10 30 kg) 2 = = 2.64 × 10 26 N L2 (1.0 × 1012 m ) 2
The component of this force toward the center is
Fc = Fg cos 30° = (2.64 × 10 26 N )cos 30° = 2.29 × 10 26 N The net force on a star toward the center is twice this force, and that force equals MRω2. This means
2π 2 × 2.29 × 10 26 N = MRω 2 = MR T ⇒T =
4π 2 MR = 4.58 × 10 26 N
2
4π 2 (1.99 × 10 30 kg)(0.577 × 1012 m ) = 3.15 × 10 8 s = 9.98 years 4.58 × 10 26 N
12.63. Model: Angular momentum is conserved for a particle following a trajectory of any shape. Visualize:
For a particle in an elliptical orbit, the particle’s angular rmomentum is L = mrvt = mrv sin β , where v is the r velocity tangent to the trajectory and β is the angle between r and v. Solve: At the distance of closest approach (rmin) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved: mPluto v1rmin = mPluto v2 rmax
⇒ v2 = v1 (rmin / rmax ) = (6.12 × 10 3 m /s)( 4.43 × 1012 m / 7.30 × 1012 m ) = 3.71 km /s
12.64. Model: Angular momentum is conserved for a particle following a trajectory of any shape. Visualize:
For a particle in an elliptical orbit, the particle’s angular momentum is L = mrvt = mrv sin β , where v is the velocity r r tangent to the trajectory, and β is the angle between r and v. Solve: At the distance of closest approach (rmin) and also at the most distant point, β = 90°. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved: mMercury v1rmin = mMercury v2 rmax
⇒ rmin = rmax (v2 / v1 ) =
(6.99 × 10 9 m )(38.8 km /s) = 4.60 × 10 9 m 59.0 km /s
Model: For the sun + comet system, the mechanical energy is conserved. Visualize:
12.65.
Solve:
The conservation of energy equation K f + U f = K i + U i is
GMs Mc 1 GMs Mc 1 Mc v22 − = Mc v12 − 2 2 r2 r1 Using G = 6.67 × 10−11 Nm2/kg2, Ms = 1.99 × 1030 kg, r1 = 8.79 × 1010 m, r2 = 4.50 × 1012 m, and v1 = 54.6 km/s, we get v2 = 4.49 km/s.
12.66. Model: Model the planet (p) as a spherical mass and the spaceship (s) as a point mass. Visualize:
Solve:
(a) For the circular motion of the spaceship around the planet,
GM p ms 2 0
r
=
mv02 ⇒ v0 = r0
GM p r0
Immediately after the rockets were fired v1 = v0 / 2 and r1 = r0. Therefore,
v1 =
1 GM p 2 r0
(b) The spaceship’s maximum distance is rmax = r0 . Its minimum distance occurs at the other end of the ellipse. The energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is, GM p ms 1 GM p ms 1 ms v12 − = ms v22 − 2 2 r1 r2 Since the angular momentum at these two ends is conserved, we have
mv1r1 = mv2 r2 ⇒ v2 = v1 (r1 / r2 ) With this expression for v2, the energy equation simplifies to
GM p 1 2 GM p 1 2 v1 − = v1 (r1 / r2 ) 2 − 2 2 r1 r2 Using r1 = r0 and v1 = v0 / 2 =
1 GM p , 2 r0
r 1 1 GM p GM p 1 1 GM p r02 GM p 1 1 1 = − ⇒ − = 0 − − 2 4 r0 2 4 r0 r22 8r0 r0 8r22 r2 r0 r2 ⇒
7 r r 7 1 + 0 2 − = 0 ⇒ r22 − r2 + 0 = 0 8r0 8r2 8 r2 8r0
The solutions are r2 = r0 (the initial distance) and r2 = r0 / 7. Thus the minimum distance is rmin = r0 / 7.
12.67. Visualize:
Solve: We choose two equal time intervals tb − ta and tc − tb. A constant velocity and equal time intervals means that xb − xa = xc − xb. The area swept from t = 0 s to t = ta is Rx a / 2 and the area swept from t = 0 s to t = t b is Rx b / 2. Thus, the area swept between t = ta and t = t b is R( x b − x a ) / 2. In the same way, the area swept between t = t b and t = t c is R( x c − x b )/ 2. Since xc − xb = xb − xa, the area swept during time (tb − ta) is the same as the area swept during an equal time (tc − tb). Kepler’s second law is obeyed.
12.68. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on the surface of the earth? (b)
(c) The distance is 6.21 Ă&#x2014; 107 m. This is 1.06RSaturn.
12.69.
Solve:
(a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the orbit?
(b)
(c) The radius of the orbit is
r=
GME 6.67 × 10 −11 N ⋅ m 2 / kg 2 (5.98 × 10 24 kg) = = 1.00 × 10 8 m 2 (vpayload ) (1997 m/s) 2
12.70. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moonâ&#x20AC;&#x2122;s radius. At what speed does it impact the moonâ&#x20AC;&#x2122;s surface? (b)
(c) The speed is v2 = 1680 m /s.
Solve: (a) A 2.0 × 10 30 kg star and a 4.0 × 10 30 kg star, each 1.0 × 10 9 m in diameter, are at rest 1.0 × 10 m apart. What are their speeds as they crash together? (b)
12.71.
12
(c) The first equation is the conservation of momentum. It can be used in the conservation of energy equation to give vf1 = 596 km /s and vf2 = −298 km /s. That is, the speeds of the two stars are 596 km/s and 298 km/s.
12.72. Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the satellite’s distance from the earth is very small compared to the earth’s (and satellite’s) distance to the sun. Visualize:
Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational force toward the earth. This net force is responsible for circular motion around the sun. We want to chose the distance d to make the period T match the period Te with which the earth orbits the sun. The earth’s orbital period is given by Te2 = ( 4π 2 / GMs ) Re3 . Thus 2
Fnet =
GMs m GMe m GMs 4π 2 mv 2 m 2πr − = macentripetal = = = mr 2 = mr 3 2 2 r d r r Te Te Re
Using r = Re – d and canceling the Gm term gives
Ms M M − e = 3s ( Re − d ) Re ( Re − d )2 d 2 This equation can’t be solved exactly, but we can make use of the fact that d << Re to use the binomial approximation. Factor the Re out of the expressions (Re – d) to get
Ms M M − e = 2s (1 − d / Re ) Re2 (1 − d / Re ) 2 d 2 Re If we think of d/Re = x << 1, we can simplify the first term by using (1 ± x)n ≈ 1 ± nx. Here n = –2, so we get
Ms M M M M (1 − ( −2)d / Re ) − 2e = 2s (1 − d / Re ) ⇒ 2e = 3 2s Re2 d Re d Re Thus d = (Me/3Ms)1/3Re = 1.50 × 109 m. Assess: d/Re = 0.010, so our assumption that d/Re << 1 is justified.
12.73. Solve:
Model: Assume the solar system is a point particle. (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means
r = 25000 light years = 2500(3 × 10 8 )(365)(24)(3600) m = 2.36 × 10 20 m T=
2πr (2π )(2.36 × 10 20 m ) = = 6.64 × 1015 s = 2.05 × 10 8 years v (2.30 × 10 5 m /s)
5 × 10 9 years = 24.4 orbits. 2.05 × 10 8 years (c) Applying Newton’s second law yields (b) The number of orbits =
GMg center mss r2
=
mss v 2 v 2 r (2.30 × 10 5 m/s) 2 (2.36 × 10 20 m ) ⇒ Mg center = = = 1.87 × 10 41 kg 6.67 × 10 −11 N ⋅ m 2 / kg 2 r G
(d) The number of stars in the center of the galaxy is
1.87 × 10 41 kg = 9.4 × 1010 1.99 × 10 30 kg
12.74. Model: Model the earth as a spherical mass and the shuttle and payload as point masses. We’ll assume mpayload << mshuttle. Visualize:
Solve: (a) The payload, in steady state, is undergoing uniform circular motion. This means that the net force is directed toward the center of the earth. There is no tangential force component, since such a force would cause the payload to speed up or slow down and the motion wouldn’t be uniform. Since the net force is due to gravity and tension, and the gravitational force is radial, the tension force cannot have any tangential component. Thus the rope is radially outward at angle 0°. (b) To move with the shuttle, the payload’s period Tp in an orbit of radius rp must exactly match the period Ts of the shuttle in an orbit of radius rs. The shuttle’s period around the earth is given by T s2 = ( 4π 2 / GMe ) rs3 , where we’ve used the assumption mp << ms to infer that the rope’s tension will be too small to have any influence on the shuttle’s motion. Newton’s second law for the payload’s motion is
( Fnet ) r =
GMe mp rp 2
− T = m p ar =
mp v 2 rp
2
mp 2πrp 4π 2 = = mp rp 2 rp Tp Tp
Using Tp = Ts and the above expression for Ts, this becomes
GMe mp r2p ⇒T =
− T = mp rp
3 GMe GMe mp r p = rs3 r2p rs3
3 GMe mp rp GMe mp 1 − = r2p rs r2p
R + 290 km 3 1 − e = 4.04 N Re + 300 km
Assess: The fact that the tension is so small justifies our assumption that it will have no effect on the motion of the shuttle.
12.75. Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a spherical mass. Momentum is conserved during the inelastic collision of the two satellites. Visualize:
Solve:
For the given orbit, r0 = Re + 1 × 10 6 m = 7.37 × 10 6 m. The speed of a satellite in this orbit is
v0 =
GMe = 7357 m /s r0
The two satellites collide, stick together, and move with velocity v1. The equation for momentum conservation for the perfectly inelastic collision is
( 400 kg + 100 kg)v1 = ( 400 kg)(7357 m/s) − (100 kg)(7357 m/s) ⇒ v1 = 4, 414 m/s The new satellite’s radius immediately after the collision is still r1 = r0 = 7.37 × 10 6 m. Now it is moving in an elliptical orbit. We need to determine if the minimum distance r2 is larger or smaller than the earth’s radius Re = 6.37 × 10 6 m. The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is L = mrv sin β (see Equation 12.26) and is conserved in a trajectory of any shape. The angle β is 90° when v1 = 4414 m/s and when the satellite is at its closest approach to the earth. From the conservation of angular momentum, we have
r1v1 = r2 v2 = (7.37 × 10 6 )( 4414 m /s) = 3.253 × 1010 m 2 / s ⇒ r2 =
3.253 × 1010 m 2 / s v2
Using the conservation of energy equation at positions 1 and 2,
GMe (500 kg) 1 GMe (500 kg) 1 (500 kg)( 4414 m /s) 2 − = (500 kg)v22 − 6 2 7.37 × 10 m 2 r2 Using the above expression for r2, we can simplify the energy equation to v22 − (2.452 × 10 4 m/s)v2 + (8.876 × 10 7 m 2 / s 2 ) = 0 m 2 / s 2
⇒ v2 = 20,107 m/s and 4,414 m/s A velocity of 20,107 m/s for v2 yields
3.253 × 1010 m 2 / s = 1.62 × 10 6 m 20107 m /s Since r2 < Re = 6.37 × 106 m, the combined mass of the two satellites will crash into the earth. r2 =
12.76. Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated system, so mechanical energy is conserved. Also, the angular momentum of the satellite is conserved. Visualize: Please refer to Figure CP12.76. Solve: (a) Angular momentum is L = mrv sin β . The angle β = 90° at points 1 and 2, so conservation of angular momentum requires r mr1v1′ = mr2 v2′ ⇒ v1′ = 2 v2′ r1 The energy conservation equation is
1 1 GMm 1 GMm 1 m(v2′ ) 2 − = m(v1′ ) 2 − ⇒ (v2′ ) 2 − (v1′ ) 2 = 2GM − r2 r1 2 2 r2 r1 Using the angular momentum result for v1′ gives 2
r −r r −r r2 − r2 r (v2′ ) 2 − (v2′ ) 2 2 = 2GM 1 2 ⇒ (v2′ ) 2 1 2 2 = 2GM 1 2 r r1r2 r1r2 r1 1 v2′ =
2GM (r1 / r2 ) r1 + r2
r and v1′ = 2 v2′ = r1
2GM (r2 / r1 ) r1 + r2
(b) For the circular orbit,
v1 =
GM = r1
(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg) = 7730 m /s (6.37 × 10 6 m + 3 × 10 5 m )
For the elliptical orbit,
r1 = Re + 300 km = 6.37 × 10 6 m + 3 × 10 5 m = 6.67 × 10 6 m r2 = Re + 35900 km = 6.37 × 10 6 m + 3.59 × 10 7 m = 4.23 × 10 7 m v1′ =
2GM (r2 / r1 ) ⇒ v1′ = 10,160 m /s (r1 + r2 )
(c) From the work-kinetic energy theorem, W = ∆K =
1 1 mv1′ 2 − mv12 = 2.17 × 1010 J 2 2
2GM (r1 / r2 ) r1 + r2 Using the same values of r1 and r2 as in (b), v2′ = 1600 m /s. For the circular orbit, (d)
v2′ =
v2 =
GM = 3070 m /s r2
1 2 1 mv2 − mv2′ 2 = 3.43 × 10 9 J 2 2 (f) The total work done is 2.513 × 1010 J. This is the same as in Example 12.6, but here we’ve learned how the work has to be divided between the two burns. (e)
W=