GRAVITATION by ashok kumar

PHYSICS GRAVITATION PHYSICS : SHEET

GRAVITATION

SHEET LAWS OF MOTION

NEWTON’S LAW OF GRAVITATION Newton in 1687 proposed force law that we call Newton’s law of gravitation. According to him, “every particle of matter in the universe attracts every other particle with a gravitational force whose magnitude is directly proportional to the product of the masses of the particls and inversely proportional to the square of the distance between them”. Thus, m1m 2 r2 Here, m1 and m2 are the masses of the particles, r is the distance between them and G is the gravitational constant, with a value that is now known to be FG

G = 6.67 × 10–11 N–m2/kg2 G = 6.67 × 10–11 m3/kg–s2

1. 2. 3.

The direction of the force F is along the line joining the two particles. Regarding gravitational force, following points should be noted : The gravitational force between two particles is independent of the presence of other bodies or the properties of the intervening medium. Gravitational force is conservative force, therefore work done in displacing a body from one place to another is independent of path. It depends only on initial and final positions. The gravitational force obeys Newton’s third law i.e.,   F12   F21

C1 : Calculate the gravitational force of atraction between two spherical bodies, each of mass 1000 kg if the distance between their centre is 10 metre. (G = 6.67 × 10–11 Nm2 kg–2) Sol.

FG

m1m2 1000  1000  6.67  1011 2 r (10)2

F = 6.67 × 10–7 newton. Vector form of the law   Let us denote by F12 the force on m1 by m2. Similarly F21 denotes the force on m2 by m1. Then   F12   F 21 .

A m1 A

r12

F12 r1

F21

B That is, Newton’s law of gravitation obeys. Newton’s third law of motion m2 r2 (in strong form). The force on m1 is directed towards m2 and that on m2 is O directed towards m1. It is a central force.  Let be r12 be position vector of m1 with respect to m2. From vector addition law, we get,    r12  r 2  r1    r12  r1  r 2   Now F12 can be written as the product of magnitude and direction (unit vector). The direction of F12 is   r12   opposite to r12 . Hence  s the unit vector in the direction of F12 . | r12 |

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       r12    r 2  r1  F12  | F12 |     | F12 |      | r12 |   | r 2  r1 | 

  Gm1m 2 | F21 |  | F12 |  r2

  the distance between m1 and m2 may be written as r  | r1  r 2 | . This gives    Gm1m 2  r 2  r1  F12    2     | r1  r 2 |  | r 2  r1 | 

 Gm m   F12   1 23 r 2  r1 | r 2  r1 |



Thus,



    Gm m F12   1 2 3 r 2  r1  F21 | r 2  r1 |









This is the vector form of Newton’s law of gravitation. Principle of Superposition By using the principle of superposition, we can find the net (or result) gravitational force on any particle from others in a group of particles. This is a general principle that says a net effect is the sum of the individual effects. For n interacting particles, we can write the principle of superposition for gravitational force as       ...(i) F1,net  F12  F13  F14  F15  ......  F1, n  Here, F1,net is the net force on particle 1. We can express this equation more compactly as a vector sum. n   F1,net   F1i ...(ii) i2

For a particle which is from a real extended object the sum of Eq. (ii) becomes an integral and we have   F1   d F in which the integral is taken over the entire extended object and we drop the subscript “net”. y

Example 1 : Fig. shows an arrangement of three particles, particle 1 having mass m1 = 6.0 kg and particles 2 and 3 having mass m2 = m3 = 4.0 kg and with  distance a = 2.0 cm. What is the net gravitational force F1 that acts on particle 1 due to the other particles ?

m2 a m3

 Sol. The magnitude of the force F12 on particle 1 from particle 2 is F12 

Gm1m 2 a2

F12 

(6.67  1011 )(6.0)(4.0) (0.020)2

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F12 = 4.0 × 10–6 N  Similarly, the magnitude of force F13 on particle 1 from particle 3 is

F13 

Gm1m 2 (2a)2

F13 

(6.67  1011 )(6.0)(4.0) (2  0.020) 2

F12 = 1.0 × 10–6 N

  the force F12 is directed in the positive direction of y and has only the y-component F12. Similarly, F13 is directed in the negative direction of x and has only the x-component – F13. y

Thus, the net force on particle 1 is

F12

F1, net  (F12 ) 2  ( F13 )2 6 2

F13

6 2

F1, net  (4.0  10 )  (1  10 )

F1, net = 4.1 × 10–6 N C2 : Consider two particles of masses m and 4m separated by a distance of r. Find a point where the force exerted by these two particles on a third particle will be zero. Sol. Let the third particle of mass m´ be placed at a distance x from m on the line joining the particles. It must lie between the particles so that force may get balance. Then

G mm´ G 4 mm´  x2 (r  x)2

1 4 1 2    2 2 x (r  x) x rx r – x = 2x r = 3x x = r/3 The point is between the particles and at a distance of r/3 for m. C3 : Consider a particle moving in a circle of radius r due to gravitational force exerted by another particle of mass M at the centre. Calculate the speed of particle assuming the central particle at rest.

m M

Sol. Acceleration towards centre = v2/r GM m Force towards centre  r2 Using Newton’s second law,

GM mv 2 Mm v G 2 , r r r C4 : It is observed that two particles each of mass m are moving around a circle of radius R due to their mutual gravitational fore of attraction; then calculate the velocity of each particle. Sol. Here the force of attraction between them provides the necessary centripetal force. [as the two particles always ramain diametrically oppositive] hence.

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mV m G R 4R 2

V

F O

Gm 4R

R V

Example 2: Three particles are located at the corners of an equilatural triangle of side a. Mass of each particle is m. They are all moving with the same speed in the same sense along the circumstance of the triangle. Find the speed of each. Sol. Figure shows forces acting on one particle, say A. Net force is vector sum of the two forces (superposition principle is obeyed). V m

Gm 2 Net force towards centre = 2 cos 30º × 2 a = From figure,

r cos 30º =

a r

3 Gm 2 a2

30º

m a/2

a 2

r 3 a  , 2 2

r

a 3

v2 3 v2  Acceleration towards centre  r a Using Newton’s second law of motion m 3 v2 3 Gm 2  , a a2

v

GM a

Example 3 : Three particles each of mass ‘m’ are placed on the three corners of a square of side L metre. Calculate the gravitational force on the force on the fourth particle of mass m placed at the corner D.

Gm 2 Sol. Force due to the particle at C = 2 along DC. L

m A L B m

m D L

C m

Gm 2 Force due to the particle at A = 2 along DA. L Gm 2 Force due to the particle at B = along DB. ( 2 L)2  Resultant force on the particle m at D  Gm 2 Gm 2 Gm 2    2 cos 45º  2 cos 45º  , along DB L  L ( 2 L)2 



Gm 2  1 2   , along DB. 2  L  2

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Example 4 : A small mass and a thin uniform rod each of mass ‘m’ are positioned along the same straight line as shown. Find the force of gravitational attraction exerted by the rod on the small mass. Sol. The mass of considered element is m dm = dx 2L Gmdm Gm 2 dF = = dx  (L + x)2 2L(L + x)2 

F

Gm 2L

2 2L

 (L  x)

dx 

2L L m

dm m

F x

Gm  1  2L  (L  x)  0

F

Gm  1 1   2L  3L L 

F

Gm 2 2 Gm 2   2L2 3 3L2

Example 5 : Figure shows two sphere of mass M each (S1 and S2). A light rigid rod has two lead balls each of mass m. The rigid rod is suspended from vertical wire. S1 pulls A and S2 pulls B so that the rod turns by . The twist in suspension wire is producing a restoring torque. If AS1 = BS2 = d, determine the torisonal constant of suspension fibre, S. Find its order if G = 6.67 × 10–11 Nm2 kg–2, m M ~ 102 kg L ~ 1 m and d ~ 1 mm. Sol. Torque due to gravitational forces on A and on B is  

´ A d

d´ B

GM m L . This must be balanced by the restoring d2

torque ´ = c.  Now,

GM m L d2 d  (L / 2) c 



Gmm L d2

GMmL2 c 2d 3

6.67  10 4  1011  1 c ~ 10 4119 ~ 103 Nm rad 1 3 2  (10) GRAVITATIONAL FIELD Can force be exerted by one body on another without contact? Some people think, yes ! A magnet can pull iron nail from distance, earth pulls an apple from distance and so on. This view is known as action-at-adistance point of view.

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The question that haunted Newton also was : how is it that a body acts on another body from a distance ? He left the question unanswered but with the hope that there must be something that is mediating interaction; force cannot be exerted from distance. Today we know that this something is field. Faraday used to think in term of fields. Further investigations have given rise to certain particles mediating the interactions; we shall not go into such details but put here only the preliminary developments in the theory of gravitational field. Gravitational field is created by every mass around it and expends at speed of light. When this field reaches another mass, it exerts a foce on that mass. Thus one mass exerts force on the field and that field exerts force on this mass. The field plays intermediary role between two masses. Gravitational field intensity Let us put a test mass m0 at a point in a gravitational field created by some sources. The test mass experiences  a force F . This force is found to be proportional to the value of test mass m0.  F  m0  We can write equality sign by introducing proportionality constant, say g . Then   F  m0 g  Here g does not depend upon m0. Hence it is a vector property of the gravitational field. We call it ‘gravitational field intersity’. The strength of field at a point is described by force experienced by unit test mass placed at that point. The force per unit test mass is known as gravitational field intensity.  The value of F given above usually depends upon the location of observation the point and the mass distribution producting the field. Since the force is mutual, the test mass slightly disturbs the distribution of source masses  and g measured is not due to the given distribution but due to slightly different distribution of mases. This disturbance is negligible if m0  0. With this is mind the above definition is recast as    F g  lim  m0  0  m   0

 Unit and dimensions of g

S.I. unit — N kg–1 Dimensional formula : [LT–2] Gravitational Field intensity due to a point mass  Let us calculate as g at point due to a point mass M.

 For this we imagine a test mass m0 put at P. Then we write force using F using Newton’s law of gravitation.

 Mm ˆ F  G 2 0 (r) r  GM g   2 rˆ . The r  The minus sign shows that is g is directed towards the source mass M. Thus,

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Gravitational field intensity due to may point masses We can obtain net gravitational field intensity as the vector sum of field intensities due to each point mass.  If i-th source mass produces field intensity g i , then net field due to i = 1 to N point masses is given by     g  g1  g 2  .........  g N

 N  g   gi i 1

If the mass distribution be continuous, we have   g   dg where the integration covers the whole body. C5 : Three particles of equal mass are situated at the corners of an equilateral triangle of side ‘a’. How much is the gravitational field intensity at its centroid ? (2) m    Sol. Let us put a test mass m0 at the centroid. Forces acting on it are F1 , F2 and F3 . These are of equal magnitude and act at 120º with each other. Hence,    F1  F2  F3  0   g0

r F1

r r

F3 m (3)

(1)

C6 : In the previous example, find the gravitational field intensity at mid point of one side of the triangle.    Sol. The figure shows the situation. F1 is force on test mass m0 at P, due to M at A, etc. F2 and F3 are equal and opposite. AM Hence   F2  F3  0  The only force remains is F1 , whose F1 

GMm 0  a 3  2   

M B

F1 m0 P F3

M C

4 GM 3a 2 Thus the field intensity is 4 GM/3a2 in magnitude and directed towards the corner not jointed by the side containg observation point.



g

GRAVITATIONAL POTENTIAL Gravitational potentia at a point is defined as the work done by the field per unit test mass when the test mass is taken from given point to infinity or standard point. If W be the work done by the field when the test mass is taken to infinity the potential at the given point V

W m0 .

W is dependent upon the force exerted by the mass distribution and the mass distribution gets disturbed if m0 is not very small. Hence to get true gravitational potential m0 should be taken is very small. Thus

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W m0  0 m 0

V  lim

S.I. Unit : J kg–1 or, m2s–2 Dimensional Formula [M0 L2 T–2] Gravitational potential due to a point mass Let M be a point mass producing gravitational field. We want to find potential at point P. A test mass m0 is taken from P towards infinite. At some position in this course, force on it is given by

Fx 

GM m0 GM m0 . If the displacement be dx, the work done by Fx is given by dW = Fxdx = – dx. 2 x x2

Now the total work done in moving it from p to W is given by integration. That is, 

W  GM m 0  r 

dx x2

W  GM m0  x 2 dx r 



 x 21   GM m 0 W   GM m 0    r  1  r W  GM V  It is a negative scalar. m0 r ,

Gravitational potential due to many particles If there be many particles producing the field, the potential at a point is the scalar sum of potentials due to all the N

point masses. that is, V   Vi i 1

If the mass distribution be continuous we resort to integration V   dV where integration covers the whole system. Example 6 : Three particles, each of mass m, are located at the corners of an equilateral triangle of side a. Calculate potential at the centroid. a Sol. The distance r of centroid from a corner of the triangle in given by r cos 30º = . m 2 a/2

ra/ 3

30º

The potential at O due to mass m at the corner is equal to –Gm/( a / 3 ). Hence total potential at O due to three masses is given by    GM  Gm V     3  3 3 a .  a   3 

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O m

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GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy of a mass m placed in the field of mass M will be calculated. Using the definition of potential at site of m, work per unit mass = potential = –

GM . r

now the work done by field in taking m to infinite distance away is given by W

GMm r

This work is called potential energy U, of m in the field of M. Thus U = potential × m (Potential energy = mass × potential) Work done is assembling a system : Work done by field in taking m to  is the same as work done by agent bringing m from  to the given point without acceleration. Hence W = potential × mass. In bringing single mass m1, when no other mass is there, no work is done by agent  W1 = 0 ...(1) The second mass m2 is brought is presence of m1. Hence W2 = (potential due to m1) × m2

Similarly,

 Gm1  W2 =   r  m2  12 

...(2)

 m1 m 3  W3 = – G  r  r  m3  13 23 

...(3)

Summation gives total work done.  W = W1 + W2 + W3

 mm mm mm  W   G 0  1 2  2 3  3 1  r12 r23 r31   This called mutual potential energy of the assembly of masses. If summation notation be used, we get W

where factor of

G N  2 i 1

 j i

mi .m j ri j

m j .mi m i .m j 1 is to take into account the double terms like r and r . 2 ji ij

C7. A particle of mass 1 kg is kept on the surface of a uniform sphere of mass 20 kg and radius 1.0 m. Find the work to be done against the gravitational force between them to take the particle away from the sphere. Sol. Potential at the surface of sphere

GM (6.67  1011 )(20)  J / kg R 1 V = – 1.334 × 10–9 J/kg V

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i.e., 1.334 × 10–9 J work is obtained to bring a mass of 1 kg from infinity to the surface of sphere. Hence, the same amount of work will have to be done to take the particle away from the surface of sphere. Thus, W = 1.334 × 10–9 J Example 7 : Three particles each of mass m are placed at the corners of an equilateral triangle of side d. Calculate (a) the potential energy of the system, (b) work done on this system if the side of the triangle is changed from d to 2d. Sol. (a) As in case of two-particle system potential energy is given by (–Gm1m2/r), so UA = U12 + U23 + U31 So,

UA = –3

Gmm 3Gm 2 =– d d

(b) When d is changed to 2d,

UB  

60º

3Gm 2d

3Gm 2 So, work done = UB – UA = 2d

Example 8 : Four masses (each of m) are placed at the vertices of a regular pyramid m (triangular base) of side ‘a’. Find the work done by the system while taking them appart so that they form the pyramid of side ‘2a’.

m a

Sol. Ui = Initial gravitation potential Gm 2 energy  6  a Uf = Final gravitational potential Gm 2 energy  6  2a wext = U = Uf – Ui



6Gm 2  6Gm 2   2a  a 

3Gm 2 6Gm 2 6Gm 2 6Gm 2 w ext       a a 2a 2a Binding Energy The minimum energy needed to dismantle a system into its constituents far apart is known as binding every of the system. Let E1 be energy of bound system and E2 the energy of dismantled system. Then E2 is greater than E1 and (E2 – E1) is the binding energy. Example 9 : Write the BE of the binary star. m v

d m

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Sol. Let two identical stars move around common centre of mass at rest. They must have equal speeds and equal distance from centre of mass.

m2 The force on one star is F = G 2 , where d is separation between the two stars. This star has the speed v in d the circle of radius d/2, Newton’s second law of motion gives mv 2 Gm2  2 (d / 2) d Gm 2 2 mv  ...(1) 2d Using (1) we can write the kinetic energy K as 1 Gm 2 mv 2  ...(2) 2 4d The gravitational potential energy of the system is given by Gm 2 U ...(3) d The mechanical energy is bound state is K

Gm 2 Gm 2 3 Gm 2     4d d 4 d When the stars are far apart (dismantted), and at rest, E=0 The binding energy is now written as Eb = E´ – E E

...(4) ...(5)

 3 Gm2  Eb  0     4 d  3 Gm 2 . 4 d The binding energy of any bound system is positive. Eb 

Example 10 : A particle is projected from the surface of earth with an initial speed of 4.0 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km, and g = 9.8 m/s2. Sol. The maximum height attained by the particle is, h

v2 R Substituting the values, we have 2g 

h

(4.0  103 ) 2 (4.0  103 ) 2 2  9.8  6.4  106

h = 9.35 × 105 m or

h  935 km

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GRAVITATION

Example 11 : An object is projected vertically upward form the surface of the earth of mass M with a velocity such that the maximum height reached is eight times the radius R of the earth. Calculate : (i) the initial speed of projection (ii) the speed at half the maximum height. Sol. (a)At maximum height, velocity becomes zero. Applying conservation principle of mechanical energy. Ui + Ti = Uf + Tf GmM 1  GmM   mv 20  R 2 9R 1 GmM GmM mv 20   or 2 R 9R GmM 8 8GmM   = R 9 9R  (b) or or

4 GM 3 R Ui + Ti = Uf + Tf  GmM 1  GmM 1  mv02   mv 2 R 2 5R 2 GmM 1 16 GM  GmM 1  m   mv 2 R 2 9 R 5R 2



v0 =

2 3

2GM 5R

Example 12 : A projectile is thrown from the surface of the earth with a velocity v0. It loxses % of its mechanical energy in doing work against air friction. What height above the surface does it attain if (a) = 0 (b)   0. the radius of the earth is R and the value of gravitational field intensity is g at the surface. Sol. The initial energy of the projectile is the sum of kinetic energy

1 GmM mv20 and the potential energy  , i.e., 2 R

– mg R. Thus Ei 

1 mv 02  mg R 2

At a height h, the gravitational potential energy will be 

GMm  Rh

GMm mgR  h h  1 R 1    R R

The kinetic energy will be zero. Hence the mechanical energy at height h is mgR Ef   h   1   R According to question Ef = Ei –  % of Ei (a) For = 0, Ef = Ei .

v=0 v0

h R

(i)

(f)

 mg R 1  mv 20  mgR 1 h / R 2

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

v02 1 1  2gR 1  h / R  2gR  2gR  v02  h R  2  2gR  v0 

2 gR h  1 , 2 2 gR  v0 R

h

R  2Rg  ,  v 2  1

h

(b)

R n 1 2

ve where n  v 0

 mgR   1   1  mv02  mgR    h  100   2  1 R 1

h mgR h 2gR  1      R  R  2 2  1   mgR  mv 0 / 2 ,  1   mgR  v 0 100 100 









    2gR h  R  1  2gR  v 2 1     0    100  





Relationship between gravitational potential and field intensity  Let V be potential and g be gravitational field intensity at a point. If a test mass m0 be placed at this point, the potential energy U and force F, will be U = mass × potential U = m0V ...(1)   ...(2) F  m0 g    If m0 be displaced by dr , work done by gravitational force is m0 g . dr . This must be equal to the fall in potential energy, –dU. That is, –dU = dWcons   – d(m0 V) = m0 g . dr   – dV = g . dr ...(3) This is the relationship in differential form. In one dimension the above equation may be written as –dV = gx dx  dV gx  dx In two dimension, V is function of two variable. Then we use partial differentiations. We have y V y gx   x (x, y) V V x gy   y x

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GRAVITATION

SHEET LAWS OF MOTION

 means differentiate relative to x keeping other variables constant. x In three dimensions we have

Here

gy 

V V V ; gx   and g z  . y x z

       [In short g    ˆi  ˆj  kˆ   V   V ]  x y z  For a radial dependence V(r), we have gr = – dV/dr   Here gr is the component of g along r . Integral form : Integrating eqn. (3), we have V2



   dV   g . dr

r2   V1  V2   g . dr r1

This is the integral for of the relationship. Example 13 : Consider a gravitational potential given by V =   z – components of gravitational feild intensity g .

Sol. Here

k     where r  x i  y j  z k . Write down x, y and r

 | r |  x 2  y2  z2

k



V(x, y, z) =

x 2  y2  z 2

gx  

 V   k    x x  x 2  y 2  z 2 

gx  k

d(x 2  y 2  z 2 )1/ 2  2  (x  y 2  z 2 ) 2 2 2 d(x  y  z ) x

k 2 (x  y 2  z 2 ) 3/ 2 . 2x 2  kx kx gx  2  3 2 2 3/ 2 (x  y  z ) r gx 

Similarly,



ky r3 kz gz   3 r   k ˆ kr ˆ  g   3 (x i  y ˆj  z k) r r3 gy  

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

Gravitation potential and field intensity at a point on the axis of a uniform ring The figure shows a uniform ring of mass M and radius R. On its axis is situated a point P, at a distance x from the centre. Let us imagine the ring to be made up of elements of masses like dM. Potential at P due to dM is given by dM

dV = –G

x R

x2 + R2 R

Potential at P due to the whole ring is given by

V



G x R

Here

x2  R 2

ring

G

V

 dM   ring

GM x2  R2

x 2  R 2 is ‘slant’ distance of P from the ring.

Field intensity : As V = V(x), we use gx = 

dV . dx

Putting the expression for V, gx  

d   GM  dx  x 2  R 2 

g x  GM

d(x 2  R 2 )1/ 2 d(x 2  R 2 )  d(x 2  R 2 ) dx

 1  g x  GM   (x 2  R 2 ) 3 / 2  . 2x  2 

gx 

GMx ( x  R 2 )3 / 2 2

Plots : The graph of V(x) is shown in the figure. Its slope is

GMx which is zero at x = 0. (x  R 2 )3/ 2 2

V(x)

– GM R

x slope = zero

The plot of gx against x is as shown. We see that

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

(i)

x=0

gx = 0

(ii)

x << R

gx = –

GM x R3

(iii)

x >> R

gx = –

GM x2

(iv)

x 

gx = 0

Example 14 : A point P lies on the axis of a fixed ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed at C will be_______. Sol. Machenical energy consevation law Ui + Ti = Uf + Tf  GmM a2  a2

0

 GmM 1  mv 2 a 2

GmM  1  1 2 1    2 mv a  2

v2 



2GM  1  1   a  2

v

2GM  1  1   a  2

Example 15 : Find the gravitational field strength and potential at the centre of arc of linear mass density  subtending an angle 2at the centre. Sol.

2G  E sin R 2 2G 2 E sin R 2 E

2 R

2G sin  R

G m R  m = (2R) m = 2R G V 2 R v   2G   R Example 16 : Consider a uniform semicircular rod of mass m and length L. An equally massive particle is placed at the centre of curvature. Calculate the force acting on this particle. V

Sol. The length of the rod is half of perimetre of a circle.  r = L r = L/ ...(1) The mass per unit length of the rod is m/L.

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

Let us divide the rod into elementary masses, each of mass dm contained in angle d at position  (see figure). Then dm = m/Lr d. = /2 The force exerted by this ‘point mass’ on the mass m at centre  of currature is dF . It has two components –dF and dF sin . There is a symmetrical mass dm at – whose effect is to cancel out dF sin . The only contribution remains uncancelled is dF cos . Hence net force Fnet on the mass is givien by dm d

Fnet 



dF cos  =0

whole wire

dm Fnet  Gm  2 cos  r m cos  Fnet  Gm  d L r

Fnet 

Gm 2 rL

m dF dF sin

–

/2



cos  d

 / 2

= – /2

Gm 2 [sin ]/ 2/ 2 , rL r = L/ Fnet 

Putting

dFcos

dF sin

Fnet  2

Gm 2 rL

2 Gm 2 L2 This force is directed towards the mid point of the semicircular wire. We have

Fnet 

Gravitational potential and field intensity due to a uniform Spherical Shell Let us consider a uniform spherical shell of mass M. Its mass per unit area will be M/4R2 = . Let us consider a strip of width R d and radius R sin  (see figure). The whole shell is made up of such strips. Rd

d R sin x Uniform spherical shell

Potential at P due to the ring is given by dV = – G

dM z

2 R 2 sin  d z z  (x  R cos ) 2  (R sin ) 2 z2 = x2 + R2 – 2 xR cos  dV = – G

Now,

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...(1)

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

Differentiating, we get 2 xR sin = 2z dz z dz sin d= xR From (1) and (2) GM z dz . dV = – 2z xR GM dz dV = – 2x R

...(2)

...(3)

Case I : P is outside the shell. In this case minimum value of z is x – R and maximum value is x + R. Therefore, Potential

GM V 2x R

V

x R



xR

GM GM . 2R   2x R x

GM x The gravitational field intensity is obtained by using, dV d  GM  gx       dx dx  x  GM gx   2 x The shell is acting as if it were a point mass located at the centre.



V

Case II : Point P lies inside the shell. Here

Zmin = R – x



GM V 2x

V

GM R

x Z P

x R



Rx



gx = 0

Potential at all the inside points is equal to that at the surface. V(x)

x=R – GM R

not differentiable

The plot V(x) is shown here. At x = R, the plot has a corner where it is not differentiable. The plot of g(x) has a x finite discontinuity at x = R.

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PHYSICS GRAVITATION PHYSICS : SHEET

GRAVITATION

SHEET LAWS OF MOTION

g(x) gx = 0

discontinuity

x=R

x gx = – GM x2

C8. Two concentric spherical shells have masses m1 and m2 and radii r1 and r2(r2 > r1). What is the force exerted by this system on a particle of mass m3 if it is placed at a distance r (r1 < r < r2) from the centre. Sol. The outer shell will have no contribution in the gravitational field at point P. 

EP 

m2 m1

Gm1 r2

Thus, force on mass m3 placed at P is, F = m3EP

O r

Gm1m3 r2   The field EP and the force F both are towards centre O. or

F

Gravitational potential and field intensity at a point due to homogenous solid sphere The point where potential and field are to be evaluated may be (i) outside the sphere (ii) inside the sphere. Case I : Outside point. We assume the sphere to be made up of concentric spherical shells, one of which is shown. Its potential at P is given by dV = – GdM/r dm Integration gives G V = –  dm r GM r For observation at outside points, sphere like a point mass located at its centre. V

gr  

dV d  GM  GM     2  dr dr  r  r

GM r2 The solid homogenous sphere acts as a point mass at its centre. Case II : Inside point : Potential at P due to the whole sphere may be treated as the sum of two potentials : The potential due to gr  

(i)

Sphere of radius OP, V1 and

(ii) Hollow sphere of inner radius OP = r and outer radius R, V2.

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PHYSICS GRAVITATION PHYSICS : SHEET (i)

SHEET LAWS OF MOTION

GRAVITATION

Potential (V1) due to sphere of radius OP. Here P lies at the surface of this sphere. Hence r

P 4 / 3r 3 V   G  1 r R M M  where  = density = V 4 / 3R 3 (ii) Potential (V2) due to hollow sphere : The hollow sphere is assumed to be made up of spherical shells. One such shell, shown in the figure is producing potential at P given by

dV2   G

dM x

4x 2dx dV2  G x Due to the whole hollow sphere we have

P x dx

V2  4G  xdx r

4G 2 2 (R  r ) 2 Thus, using (i) and (ii) potential at P is given by V = V1 + V2 4 G r 3 4 2 2   3 V  G (R  r )  r 2   V2  

V

...(ii)

G 4r 2 G 4 2 2  (R  r ) 3 2

 r2 R 2  r2  V  G 4    2  3  3R 2  r 2  V   4 G  6   GM  3R 2  r 2  V  2 4 / 3R 3  3  GM (3R 2  r 2 ) 2R 3 This is as function of r. The gravitational field intensity along radial direction denoted by. It is given by dV  GM  gr      (0  2r)  3 dr  2R  GM gr   3 r R The graphs of V(r) and gr are shown below. Mark that | gr | is maximum at the surface of the sphere. V

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION V(r) r

r=R

outside sphere (hyperbola)

– 3GM 2R inside sphere (Parabola)

Table : Field and Potential Source

Potential (V) 

Point mass 

Ring

GM r



GM 2

x R



GM r2

GM x (x  R 2 )3/ 2 2

 GM  R , r  R   GM , r  R  r

rR  0,   GM  r 2 , r  R

 GM 2 2  2R 3 (3R  r ), r  R    GM , r  R  r

 GM  R 3 r, r  R   GM , r  R  r 2

Spherical Shell

Solid Sphere

Field intensity (gr)

C9 : A tunnel is made in the earth along its chord at a distance of R/2 from the centre. A mail is released from the mouth of the tunnel. there is no friction between the main and the tunnel wall. Calculate the normal force exerted by the tunnel on the mail if its mass is m. The mass of the earth is M and radius R.

N mg cos mg R/2

mg sin

Sol. Forces on the mail are (i) mg, towards the centre of the earth (ii) N, normal to tunnel Here g = GMr or, N = mg cos   GM  N = m  3 r cos  R

N=m N=

GM R R3 2

R   r cos    2

GM m 2R 2

Thus, N is independent of ; it depends on R.

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

C10 : A spherical mass of radius r = R/2 is taken out from a sphere of radius R and density . Calculate the force which this sphere having a cavity will exert on a mass m placed at a distance of x from its centre (x > R). Sol. For the whole sphere of radius R F1  G

Mm x2

4 / 3 R 3m x2 and for small sphere of radius r = R/2 M1m G4 / 3 (R / 2)3  m F2  G  (x  R / 2) 2 (x  R / 2)2 Resultant force = F1 – F2

F1  G



P r = R/2

1  4 1 R 3 Gm  2  2  3 2(2x  R)  x

KAPLER’S LAWS OF PLANETARY MOTION Before we actually put the kinematical laws of motion of planets given by Kepler, let us familiarise ourselves with an ellipse.

F2 F1

ellipse

Activity for an Idea of an ellipse : Take a drawing broad. Fix on it a graph paper. Fix two pins (F1, F2) on the paper. Take a thread whose length in larger than the distance between the two pins. Tie its ends to each of the pins. Take a pencil and pull the thread by it as in the figure. Move the pencil keeping the thread tight. You get a curve called an ellipse. The nails are two focii (plural world for focus) of the ellipse. Draw a line through F1 and F2. It cuts the ellipse at A and B. The length AB is called ‘major axis’ of the ellipse. Now draw a perpendicular bisector of the major axis. It is CD. It is called minor axis of the ellipse (because CD < AB as shown in the figure). The ratio OF1/OB or OF2/OA is known as eccentricity (e) of the ellipse. It is a measure of how much distorted the ellipse is releative to a circle (e = 0). Using the graph paper, you may find the area of the ellipse and compare with the standard formula, to judge how good your ellipse is : Area =  ab

A F2 C

2a D

F1 B 2b

The Kepler’s laws : Kepler (1571–1640) used an ellipse to describe his three laws of planentary motion. The three laws are :

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SHEET LAWS OF MOTION

GRAVITATION

(a) Law of orbits, (b) Law of areal velocity and (c) The law of periods. We shall now state and discuss these. The Law of orbits The closed path along which a planet moves around the sun is called an orbit. the first law gives its geometry and location of the sun relative to the orbit. Every plate moves around the sun in an elliptical orbit with the sun at one of the focii. P

In the figure sun is located at the focus F1, and F2 is vacant. The planet P is moving along the ellipse. Another planet will move in another ellipse but one focus of that ellipse will always be at the sun.

F2 S

The Law of Areal velocity or sector velocity If we join a planet (P) to the sun (S), the line (SP) turns as the planet moves. It sweeps some area in given time. The area swept per unit time is known as areal velocity (or sector velocity) of the planet. The Kepler’s second law tells us that this areal velocity remains uniform. The line joining a planet to the sun sweeps equal and in equal time interval. P´

Q (Perihelion) Fast

S F1

(Aphelion) slow

Q´

Let a planet move, from P to P´ in a time t, covering area SPP´. The planet goes from Q to Q´ in the same time interval t. Then, according to Kepler, Ar (SPP´) = Ar (SQQ´) If A be the area and t the time taken then, A = constant. t When the Earth, for example, is closest to the sun (at perihelion), its speed in larger than when it is farthest from the sun (apehelion).

The Law of Periods A planet takes some time to move once around the sun. This time is known as time period of the planet. If several planets are moving around the sun, they have their own periods. The time period of the earth for motion around the sun, for example, is one year. The law of periods connects this time period to size of ellipse. The law was formulated by Kepler’s in 1619. The square of period of revolution of planets around the sun are directly proportional to the cube of semimajor axes of respective orbits of the planets. Let T1 be the period of planet-1 moving in an ellipse of semimajor axis a1. Similarly T2, a2 are period and semimajor axis for planet-2, T3, a3 for planet-3, etc. Then

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T12 T22 T32   = ....... = constant. a13 a 32 a 33 In short, T2  a3. Notice the Table giving the value of constant. Table : Law of Periods Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

Semi-major axis (a)

Period (T)

Ratio T2/a3

(× 1010 m)

(years)

(× 10–34 year2/metre3)

5.79 10.8 15.0 22.8 77.8 143 287 450

0.24 0.615 1 1.88 11.9 29.5 94 165

2.95 3.00 2.96 2.98 3.01 2.98 2.98 2.99

The Physical basis  The sun exerts a force of pull ( F ) on a planet that passes through the centre of mass of the planet. Newton discovered the law that gives the force : It is central and attractive (always directed towards the fixed point, the sun), and obeys inverse square law of distance (the force is inversely proportional to square of distance between the sun and the planet). The torque of this force about the sun is zero. Hence angular momentum of the planet about the sun is conserved.    If r be the position vector and v be the velocity of the planet, the angular momentum L of the orbiting planet is given by v    m L L  r  mv d   r F planet Now the area of triangle formed by v dt and r can be represented by geometrical meaning of cross product of two vectors. Thus L p   1 r dA  | r  v dt | 2 Using this we have    | L |  | r  v| m

  dA  | L |  2 m  dt   The direction of L is constant and magnitude is also constant. Thus,  dA | L |  = constant dt 2m

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GRAVITATION

The law of areal velocity is, thus, consistent with conservation of angular momentum of the planet around the sun. In turn, it bases on the central nature of force between the planet and the sun. The other two laws are consistent with inverse square and attractive nature of gravitational force. Demonstrating these by calculations is straight forward but a bit cumbersome and not necessary at this level. Since a circle is also an ellipse with zero eccentricity, we have T2 = kr3 (Law of periods). 2

Also

 2  F  m r  m   r  T 2

4 2 m F .r kr 3 F

4 2 m kr 2

1 r2 Thus the period’s law is consistent with inverse square attractive force.

Thus

F

The value of ‘constant’ in Kepler’s law of periods for planetary motion The square of the period of revolution of a planet around the sun is proportional to the cube of semimajor axis of its orbit. T2  a3 T2 = ka3 ...(1) k is the constant of proportionality. To find k which is independent of the value of ‘a’, we take circle as a special case of an ellipse, where major axis equals minor axis Here we have P GmM s a S F= a2

mV 2 GmMs  a a2

v T

GMs a 2a v

T 2  4 2

 4 2  3 a2  a  GM s   GM s    a 

...(2)

From (1) and (2) k

4 2 GM s

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GRAVITATION

Example 17 : A saturn year is 29.5 times the earth year. Find the distance of saturn from the sun if the Earth is at 1.50 × 108 km from the sun. Sol. Using Kepler’s law of periods, 2

 TS   rS    T   r  E E T  rS   S  T 

2 /3

. rE

 29.5 y  rS    1 y 

2/3

. 150  108 km .

We need log table for exact evaluation. However we have only three significant calculation and can use binomial method. (29.5)

2/3

= (27 + 2.5)

2/3

= 27

 2.5  . 1    27 

2/3

 2 2.5   (3 × 3 × 3)2/3  1  3  27  = 9.5 (Exact evaluation gives it as 9.5413)

Using this

rS = 14.25 × 108 km rS = 1.42 × 1012 m

To show that Kepler’s laws are consistent with Newton’s laws of gravitation Let a planet be moving in an elliptical orbit around the sun. Since the mass of the sun is very large than that of the planet, the sun will exert a gravitational force (attractive) on the planet.

F MS

G mM s ...(1) R2 Since the eccentricities are small and sun is more massive than the planets, the path can be assumed as approximately a circle. The centripetal force, exerted by the sun on the planet, is given by F

mv 2 m  2R  Fc     R R T 

...(2)

From (1) & (2)

GmM m  4 2 R 2    R2 R  T 2  T 2 4 2  R 3 GM Here, 42, G and M are constants, therefore, T2 = constant. R3

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T2 = (const) R3 T2  R3 This is the law of periods. Areal velocity R 2 R 2   = constant. T  4 2 R 3   GM 

Thus the areal velocity in circular orbit is constant. C11 : The speed of a planet at perihelion is v0 and the distance from the sun is l. The major axis of elliptical path is 2a. Determine the speed at aphelion. If e be the eccentricity of the ellipse l = a(1 – e). Use it to expresses your answer.

ae a

S v0

Sol. At perihelion the angular momentum about the sun-axis normal to plane of the orbit, is mv0 l where m is the mass of the planet. When it travels to aphelion let its speed be v. Then distance from the sun is (2a – l). The angular momentum about the above axis is now mv(2a – l). Since action line of the force exerted by the sun on the planet is passing through the sun, angular momentum about the sun is conserved. Hence mv0l = mv (2a – l)

v

v0 l v  2a  l  2a / l  1

Using l = a (1 – e),

v0  2   1  1 e   1 e v  1  e  . Example 18 : A body is launched from the earth’s surface at an angle  = 30º to the horizontal at a speed v

1.5 GM . Neglecting air resistance and earth’s rotation. Find the height to which the body will rise. R Here M is mass of earth and R the radius of earth. Sol. Let velocity at highest point be v and R + h = r Applying conservation of angular momentum between P and Q, we have Q v mvr = mv0R cos 30º v0 

3 v0 R ...(1) 2r Applying conservation of mechanical energy between P and Q, we have : 1 GMm 1 GMm mv02   mv 2  2 R 2 r substituting the value of v from equation (1), we get

v

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=30 º

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GRAVITATION

1 GMm 1  3v 20 R 2  GMm 2 mv0   m  2 R 2  4r 2  r or

2GM 3v02 R 2 2GM v    R 4r 2 r

1.5GM 2GM 3  1.5GM  R 2 2GM      R R 4  R  r2 r

2 0

or or

1 9 R2 2  .  2R 8R r 2 r –4r2 = 9R2 – 16Rr 4r2 – 16Rr + 9R2 = 0

r

  1.5 GM is given   v0  R  



16R  256R 2  144R 2 8

16R  10.58R = 3.323 R and 0.677 R 8 r  R or h = r – R = 2.323 R

r= but Hence

r = 3.323 R

ACCELERATION DUE TO GRAVITY AND ITS VARIATION The force on a particle near the earth can be written as  F = mass × gravitational field intensity Newton’s second law of motion gives   F = mass × a grav where agrav is the acceleration due to gravity. This gives gravitational field intensity mathematically equal to the acceleration due to gravity.   a grav  g (gravitational field intensity) Variation in acceleration due to gravity g with altitude and depth If we assuming that the earth is a homogenous sphere (which she is not exactly) the gravitational field intensity may be written as g=

GM (outside) r2

GM r (inside) R3

where r = distance of observation point from the centre of the earth. The above expression shown that g decreases when we go away from the surface either inward (depth) outward (altitude). For small height h above the surface

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|g|=g

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g

Now,

GRAVITATION

GM  (R  h) 2

SHEET LAWS OF MOTION

GM h  R 1    R

GM  h 1  2  R  R

2

GM  g 0 (surface value R2 2

h  g  g0 1    [which h << R]  R  2h  g  g 0 1    R [Using binomial theorem (1 + x)n  1 + nx when | x | << 1.] If we go inside the earth at depth h g

GM (R  h) R3

h  g  g 0 1    R

The above discussion shows that g varies at faster rate is going upward than in going inward from the surface of the earth. As we go upward initially g falls at faster rate but at later distance its rate of fall becomes smaller-smaller whereas when we go inside the earth g goes on decreasing at constant rate becoming zero at the centre. C12 : The weight of a man on the surface of the earth is 50 kg, what would be his weight at the centre of earth ? Sol. Zero [Since ‘g’ at the centre of the earth is zero.] C13 : In above question, at what depth below the surface of the earth the weight of the man will be half of that on the surface of the earth ? Sol. Acceleration due to gravity at a given depth.  h g1  g 1    R

g1 h  1 g R

or,

1 h  1 2 R

h 1 1  1  R 2 2



R 6400 = = 3200 km. 2 2

C14 : In the above question, at what height the weight of the man will be half of that on the surface of the earth. Sol. The acceleration due to gravity at a given height ‘g’.  R  g1  g   R  h 



g1  R   g  R  h 

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PHYSICS GRAVITATION PHYSICS : SHEET or

GRAVITATION

1  R   2  R  h 

h



SHEET LAWS OF MOTION

Rh  2 R





2 1 R

h = 0.41 R = 0.41 × 6400 = 2624 km C15 : Calculate the value of acceleration due to gravity at a height R from the surface of the earth, where R is the radius of the earth. 2

g1  R  1  1     Sol.   g  R  h 4 1  h / R  g1 = g/4 C16 : Calculate the % change in acceleration due to gravity on the surface of the earth, if radius of the earth shirnks by 2%. (assuming density to be same). Sol. It will decrease by 2 %.

M 4 R3 4  G  2   GR 2 R 3 R 3 g  R; if  is constant. C17 : How high a man will be able to jump on a surface of a planet of radius 320 km, but having density same as that of the earth, if he jumps 5 metre on the surface of the earth (radius of the earth = 6400 km). gG

Sol. For the earth, For the planet;

g

g1 

GM 4  RG R2 3

GM1 4  R 1G R 12 3

g R 6400    20 g1 R 1 320

Let h and h1 be the distance upto which the man can jump on surface of the earth and planet, then mgh = mg1h1 

g h1 = g h = 20 × 5 = 100 m. 1

C18 : How high a man can jump on the surface of the moon if he jumps 5 metre on the surface of the earth, gm = 1/5 ge. Sol. gehe = gmhm 

 ge  hm =  g  × he = 25 meter..  m

C19 : Calculate the value of acceleration due to gravity on the surface of a planet whose mass is double of the earth and radius half of the earth. Sol. For the earth, g =

GM1 GM ; and for the planet, g = 1 R 12 R2 2

g1  M1   R   2M   R        = 2 × 4 = 8.  g  M   R1   M   R / 2  g1 = 8g = 78.4 m/sec2.

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C20 : Calculate the value of acceleration due to gravity on a planet whose radius is half that of the earth but the mean density is same as that of the earth. GM 4  RG R2 3 for same density, g  R

Sol. We have g 

g1 R 1 R g  or, g  g   4.9 m/sec2. 1 g R 2R 2 Example 19 : Calculate the distance from the surface of the earth at which above and below the surface acceleration due to gravity is the same. GM = Sol. Acceleration above the surface of the earth R + h 2 

(

)

d  Acceleration below the surface of earth = g  1   R According to question GM d   g 1   2  R  R  h GM

 R  h

h   g 1    R 2

GM  h 1  2  R  R

h   1   R

h  h  1   1   1    R  R

2

h   g 1    R

h   1    R

{Q d = h } or

or or or

2

h   g 1    R h   1   2  R h  1    R

 h 2 2h   h 1   1  2   1   R   R  R

h 2 2h h h 3 2h 2     R2 R R R3 R2 h h h2  h2 h h3 1   0  2   3  R  R R 2  = 0 R R R 1  1

h2 h  R2 R R2 – h2 = Rh h2 + Rh – R2 = 0 1

 R  R 2  4R 2 2 Physically possible value is 

h  g 1    R 1

h



R 2  h2 h  R2 R



h

R  5 R 2



5 1 R 2

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C21 : A mail is dropped along a smooth a tunnel along diameter of the Earth. assuming the Earth a homogenous sphere determine the type of motion executed by the mail. Sol. When the mail is at distance r from the centre, the force exerted on it is directed r

GMm r. towards the centre and is R3

In vector form,

 GMm  F r R3

This shows that the particle is pulled towards the centre, at the centre its velocity is maximum but force is zero. Due to inertia of motion it is deflected beyond the centre where it is again pulled back and decelerates. It stops at the mouth of the tunnel and is pulled back again. Thus the mail performs oscillations. Shape of the earth : The core of the earth is more massives and is responsible almost wholly for the value of g. If a point is nearer to this core (like polar points) the attraction is largter or g is larger. On the other hand at the equator the point is further away from the core where attraction is smaller. Thus g is smallest at the equator and largest at the pole.

near to centre oblate spheroid Low density High density

(farther from centre)

Variation in crust density : The density of crust is not uniform g is smaller if the density of curst below is smaller due to presence of low density material like oil and gas. Similarly g is larger if high density material like some iron ore is lying beneath the crust. Effect of rotation : If we use Earth frame, it is a rotating frame. An object at rest on it feels a centrifugal force directed normally away from rotation axis. It, in effect, reduces the value of g. At equator g´ = g – R. At the poles the effect of rotation on is zero. Weight the force with which a body acts on the support that prevents it from free fall in a reference frame is called to weight of the body in that reference frame. In a inertial reference frame, weight equals the gravitational pull by all the bodies in the universe. On a planet, it equals the pull of the planet. In interstellar space the pull becomes faint and it is called true weightlessness.  Apparent weightlessness is observed in a frame falling (rising) freely with acceleration g produced by all the gravitating bodies. Interial and Gravitational masses The mass due to which particle take part in gravitation in known as gravitational mass. The ‘mass’ due to which it opposes any change in its natural state of motion or rest (i.e., shows its inertia) is called inertial mass. Hence a body is likely to posses two kinds of masses – inertial and gravitational. Let m be inertial mass and m´ be gravitational mass. Then the gravitational force is decided by gravitational masses. GM´m´ r2 If ‘a’ be the acceleration produced in the body due to gravitational force, Newton’s second law of motion tells that it is equal to ma where m is inertial mass. Hence. F

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ma  a

SHEET LAWS OF MOTION

GM´m´ r2

GM´  m´   r2  m 

Near the earth, all the objects fall with the same acceleration. This was first notated by Galileo and is true. Hence m´/m= constant The difference between m and m´ were measured experimentally. Etvos measured equality of m and m´ upto 12th place after decimal. In the mean time Einstein floated theory of relativity. If we consider a freely falling lift, bodies becomes weightless in it. There is no compression or elongation in the spring balance

mg FS = 0 mg = m´g

g m´g

Gravitational force can be cancelled in lift only if m = m´. Einstein’s equivalence principle states that gravitational and inertial masses are the ‘same’. The question of two kinds of masses is solved – there is only one type of mass in a body playing both of the roles. ORBITAL SPEED OF A SATELLITE A body moving in a closed orbit around a planet is called its satellite. The Kepler’s laws are applicable to it. We shall consider approximately circular orbit for the satellite. The centripetal force is produced by gravitational force of a planet. Using dynamics of circular motion, we have

r m

mv GMm  r r2

m>>M

GM r This is called orbital velocity of a satellite. It is independent of mass of satellite only if m << M. If the satellite moves around the earth near the surface r  R = 6400 km 

v

GM GM  R  gR R R2

g  10 ms–2 –1 v  64  106 ms –1 = 8 kms

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PHYSICS GRAVITATION PHYSICS : SHEET

GRAVITATION

SHEET LAWS OF MOTION

Time period : The length 2r is covered with uniform speed v.

T

2r 2 r r 3/ 2   2 v GM GM r

Square of T is proportional to cube of r (Kepler’s law of periods)

2R 3/ 2 the smallest time period = GM

[  rmin  R ]

2R 3/ 2 2 3/ 21   R g GM 2 R R2  2

R . ( 84 min) g

The period of natural satellite of the earth, the moon, is 27.3 days. Example 20 : A satellite is revolving round the earth at a height of 6 × 105 m. Find : (a) the speed of satellite and (b) the time period of the satellite. Radius of earth = 6.4 × 106 m and mass of the earth = 6 × 1024 kg. Sol. The height of satellite from earth’s centre = 6.4 × 106 + 6 × 105 m = 7.0 × 106 m (a) The speed of satellite is

v

GM r

v

6.67  1011  6  10 24 7  106

v = 7.6 × 103 m/s (b) The time period is T=

2r v

2  7  106 7.6  103 T = 5.8 × 103 s T=

C22 : A satellite is rotating near the surface of the earth in a circular orbit; then calculate its speed. Sol. We have,

mV 2 mM G 2 R R

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GM R

V

GRAVITATION

but gR2 = GM

V  gR



 9.8  6400  103  8 km / sec. C23 : In the above Qus; calculate the time period of the satellite.

Sol. 1.

mV 2 = mg or, V2 = Rg R

w2R2 = Rg



 2    R  g T

 T  2

R g

2 R VT T = 2 or, V R C24 : Calculate the valocity of a satellite which is revolving round the earth in a circular orbit at distance of 6400 km above the surface of the earth. 2. T = 2

Sol. Since,

V

GM R



V2 R1 6400   V1 R2 12800

V1 8   4 2 km/sec. 2 2 C25 : In the above problem, calculate the time period of the satellite. V2 

Sol. We have,

GM mV 2 GmM 2  or, V  2 R R R  2R 2 

2 GM 4 2 3 GM 2  2  R . or,   R 2  or T  R GM  T R

Hence T  R 3/ 2 If T1 and T2 be the time period of the satellite near the surface of the earth and at a height 6400 km above the surface of the earth, then R 2 12800  2 R 1 6400



T2  R 2   T1  R1 

3/ 2

 23/ 2  2 2

T2 = 2 2 T1 = 2 2 × 84 = 238 minutes nearly.. GEOSTATIONARY SATELLITE If a satellite remains stationary relative to a place on the Earth, it is called geo-stationary satellite. For the satellite,

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

GMm  m 2e r r2

1/ 3

F  GM  h R 2   S  e  Putting the values, we get r  42400 km (from centre) h = 36000 km (above the surface). This satellite is used for communication. Hence it is also called communication satellite. Parking orbit is the Orbit in which the satellite is placed. A geostationary satellite is parked only in an equational orbit. If such a satellite is attempted in non-equatorial orbit the gravitational force by planet will have a component normal to the plane of the circle. This force will take the satellite in helical path. The orbit will be unstable. Hence the satellite will not be geostationary. A communication satellite will directly communicate with places with co-lititude not less than  as in the figure.

R r 6400 km 1 sin    42400 km 7  1   sin 1    8º   7  The lattitude will be     2 sin  

We have

N latitude

co-latitude

r S

Under direct communication

 1   cos 1    82º  7

C26 : Calculate the gain in the potential energy of an object of mass 2 kg. When it is raised from the surface of the earth to a height equal to the radius R of the earth. Sol. P.E. on the surface of the earth = –

GMm . R

P.E. at a ht. R above the surface of the earth = – 

gain in P.E. dU = –

GMm 2R

GMm  GMm  –    R  2R

GMm mgR   gR 2R 2 Example 21 : A satellite close to the earth is in orbit above the equator with a period of rotation of 1.5 hours. If it is above a point P on the equator at some time, it will be above P again after time______. Sol. v 0 = gR



R 





g R

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PHYSICS GRAVITATION PHYSICS : SHEET or

2  Ts

1 1  Ts 2

GRAVITATION

SHEET LAWS OF MOTION

g R g 1  R 2 R

{Q  2  g }

1 1  Ts 2 6400  103 1 1 1 = = Ts 1600 10 5060 1  0.000198 Ts 1 1  TE 24  60  60 For an observer at the surface of earth 2 2 T   s   E 2  2  Ts TE



1 1 1   T Ts TE

...(1) ...(2)

(when rotating from west to east)

= 0.000189

T = 5576 second = 1.6 hour Similarly, for rotating from east to west. 1 1 1 = + ...(3) T Ts TE Putting the values, from eqn (1) and (2) in eqn (3) T=

24 hour 17

Polar Satellites These pass through polar regions, are at attitudes of 500 ~ 800 km and scan the earth surface is each revolution. Information from there are used in remote sensing, meteorology and studies of earth’s environmental science. C27 : A polar satellite has a period of TP while the Earth turns around its axis at a period of TE. Neglecting orbital motion, calculate the angular shift of satellite orbit per revolution relative to earth. Sol. Relative to the earth the orbit is rotating at angular speed of the earth in opposite sense to that of the Earth. The angle turned – E   |  E |  TP 

2  TP TE

T   2  P   TE 

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SHEET LAWS OF MOTION

GRAVITATION

Example 22. If a satellite is revolving around a planet of mass M in an elliptical orbit of semi-major axis a, show that the orbital speed of the satellite when it is at a distance r from the focus will be given by : 2 1 v 2  GM    r a  GMm  Sol. As in case of elliptical orbit of a satellite mechanical energy E     remains constant, at any position 2a  of satellite in the orbit,

KE + PE = –

GMm 2a

...(i)

Now, if at position r, v is the orbital speed of satellite, KE =

1 GMm mv2 and PE = – 2 r

...(ii)

So from Eqs. (i) and (ii), we have 1 GMm GMm mv 2   2 r 2a

2 1 v 2  GM    r a Example 23 : Consider two satellites A and B of equal mass m, moving in the same circular orbit r of radius r around the earth E but in opposite sense of rotation and therefore on a B A collision course (see figure). Me (a) In terms of G, Me, m and r find the total mechanical energy EA + EB of the two satellite plus earth system before collision. (b) If the collision is completely inelastic so that wreckage remains as one piece of tangled material (mass = 2m), find the total mechanical energy immediately after collision. (c) Describe the subsequent motion of the wreckage.

i.e.,

Sol. (a)

 GmMe 1 2  Ei    mv 0   2 r 2  

GmM e  GmMe 1 GMe  Ei    m Ei    2, 2 r  r  r (b) According to conservation principle of momentum, Pi = Pf mv0 – mv0 = Pf  Pf = 0 It means just after collision, kinetic energy becomes zero.  Total mechanical energy after collision is Ef = P.E. + K.E. 2GmM e GmMe GmMe Ef    0  r r r (c) After collision, combined mass starts to move towards the centre of earth due to gravitational attraction of earth. Ef 

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Example 24 : Two planets of equal radius R are separated (centre-to-centre) by 8R. One has a mass M and the other a mass 4 M.

A small body is projected from the planet of mass 4M towards the centre of the other 8R u planet. What should be its minimum speed seas to reach the other planet ? Sol. The projectile should be able to reach the neutral point. As it crosses it, the other 4M planet will automatically pull it. Neutral point : Let its distance from 4M be x. Then a particle at this point will feel equal attraction by two planets. Thus, G(4M) m GM m  2 x (8R  x) 2 4(8R – x)2 = x2 2(8R – x) = ± x 16 R = 3x or x = 16 R 16 R or x = 16 R 3 At x = 16R the magnitude is equal but direction is not opposite. It is not a neutral point. Hence x = 16/3R. Now we consider energy conservation. Kinetic energyof the projectile at starting point is 1/2mv2. Potential energy is written as   G(4M) m   GM m  U   R    8R  R  Thus, total mechanical energy of the projectile at the surface of launch planet is x

Ei 

1  GM m  1   mv 2      4  2 R 7  

1 29  GMm  mv 2   . 2 7  R  When the projectile reaches neutral point, its velocity is negligible. Only then projection speed v will be a minimum. Thus,    G (4M) m GM m  Ef  0     16      16 R  8R  R   3     3  Ei 

GM m  3 3  3 GM m    R 4 8 4 R As the gravitational force is conservative, E f = Ei Ef  



3G Mm 1 29 GMm  mv 2  4R 2 7 R

1  29 3  GMm mv 2      7 4 R 2

v

95 GM . 14 R

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PHYSICS GRAVITATION PHYSICS : SHEET

GRAVITATION

SHEET LAWS OF MOTION m

Example 25 : A hypothetical planet of mass M has three moons each of equal mass ‘m’ each revolving in the same circular orbit of radius R. The masses are equally spaced and thus form an equailateral triangle. Find : (i) the total P.E. of the system m (ii) the orbital speed of each moon such that they maintain this configuration.  Gm 2  U  3  Sol. (i)  a 

a = 3R

}

º 30

30º

 3 Gm 2 3GmM UT   a R 3Gm  m  UT   M   R  3 (ii) Net force towards centre of planet is mv 02 F  2F1 cos 30º  F2  R 2 Gm 3 GmM mv02 2 2   or a 2 R2 R a R But 3

F2 R

G m  Gm GM v02 v0   M 3   or    R 3 3R 2 R2 R Example 26 : A satellite is moving in a circular orbit around the earth. The total energy of the satellite is E = – 2×105J. The amount of energy to be imparted to the satellite to transfer it to a circular orbit where its potential energy is U = –2 × 105J is equal to ________. GmM U  2  10 5 Sol. Here (R  h) GmM  2  105  ...(1) Rh 1 K.E. = Tf = mv 02 Also 2 2 GM  1  GM   Tf  m  Q v0    R  h  2  Rh  Tf 

1 GmM 2 (R  h)

1  2  105 (from eqn (1) 2 Tf = 105 Joule  Ef = Uf + Tf = –2 × 105 + 105 = –105 Joule  Required energy = Ef – Ei Tf 

= –105 – (–2×105)



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Required energy  1  105 Joule

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Example 27 : A satellite of mass m is orbiting the earth in a circular orbit of radius r. It starts losing energy due to small air resistance at the rate of C J/s. Then the time taken for the satellite to reach the earth is _____. Sol. As due to air drag some mechanical energy of sattelite with be converted into heat energy. Dur to this, radius of orbit will decrease Hence, the satellite will follow spiral path towards the earth. The total mechanical energy at orbital radius x is GmM 1 E  mv02 x 2 GmM 1 GM E  m x 2 x GmM E 2x (The path of satellite in dE GmM d 1 the presence of air drag)   x or dt 2 dt t R GmM dx GmM 2 dx  C dt  C    1 x or    2 x2 2 dt 0 r

 

Gm M  x  2  1   Ct    2  2  1  r



GmM  1  Ct  2  x  r

GmM  1 1  GmM  1 1  t      2C  R r  2 R r Example 28 : Two small dense stars rotate about their common centre of mass as a binary system with the period 1 1 year for each. One star is of double the mass of the other and the mass of the lighter one is of the mass of 3 the sun. Find the distance between the stars if distance between the earth & the sun is R. Gm  2m 2r  m 2 Sol. 2 r 3 3Gm 2   3 ...(1) r m C 2m GM s Orbital velocity of earth is 2r r R 3 3 GM s v0   R GM s R  or ...(2) R Time period of earth about Sun is one year.   = ’ From eqn (1) and (2) 3Gm GMs 3Gm 2  3  3  or r R3 r Ms m= But 3 G3m 3Gm  3 r R   R3 r

Ct 

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PHYSICS GRAVITATION PHYSICS : SHEET

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C28 : A body moving radially away from a planet of mass M, when at distance r from planet, explodes in such a way that two of its many fragments move in mutually perpendicular circular orbits around the planet. What will be (a) then velocity in circular orbits (b) maximum distance between the two fragments before collision and (c) magnitude of their relative velocity just befroe they collide. Sol. (a) Since, total momentum of system in perpendicular direction of radius is zero. So, after breaking, the momentum in perpendicular direction should be zero. This is possible only when these two fragments move with same speed in opposite direction. GmM mv 20  x Q ma r r r GM v0 =  r r (b)  l max = r 2 + r 2 = 2 r (c) Just before collision, their velocity are perpendicular to each other. 

v rel = v 02 + v02 v rel  2 v 0  v rel 

Gm r

2Gm r

ESCAPE VELOCITY The minimum speed needed for a body to get rid of gravitational attraction of a planet is known as escape speed of the body on that planet. It is also called first cosmic velocity.

ve R

GMm 1   mv 2e  0  0 r 2

ve 

V=0 F=0

2 GM R

ve  11.2 km s–1 for earth ve = 2 × orbital speed. Black Holes Black holes are astronomical bodies having very high gravitational field such that even light can’t escape from these bodies. Hence they are called black holes. Escape velocity on the surface of a black hole is larger than speed of light. Calculation of radius (Event horizon) : We use formula for escape speed

v

2GM R

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

Assuming the mass M distributed in the radius of R,

v R

4  2G  R 3   3  R c 4 G 3

Typical data are  = 1017 kgm–3 G = 10–10 Nm2 kg–2 We get 108

~ 105 m = 10 km.

10 Any signal of an event will come out only if it occurs outside the sphere of radius R.

c No news of events in the entrior to sphere of radius R = 8 G / 3 can be had. Hence R is called the radius of event horizon. Table : Some Speeds Orbital speed

GM 4 G  gr  r r 3

Escape speed

2GM 8 G  2gr  r r 3

Surface of Black hole

ve  c

Interstellar speed

v e n 2  1 , n  v / ve

Trajectory of a body pojected from point A in the direction AB with different initial velocities : Let a body be projected from point A with velocity v in the direction AB. For different values of v the paths are different. Here are the possible cases. A B (i) If v = 0, path is a straight line from A to O. v

(ii) If 0 < v < v0, path is an ellipse with centre O of the earth as a focus. (iii) If v = v0, path is a circule with O as the centre (iv) If v0 < v < ve, path is again an ellipse with O as a focus.

(v) If v = ve, body escapes from the gravitational pull of the earth and path is a parabola A

0 < v < v0

O Earth

v0 < v < ve

O O

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SHEET LAWS OF MOTION

 GM  (vi) If v > ve, body again escapes but now the path is a hyperbola. Here, v0 = orbital  r  at A and 

ve = escape velocity at A. Note : 1. From case (i) to (iv) total energy of the body is negative. Hence, these are the closed orbits. for case (v) total energy is zero and for case (vi) total energy is positive. In these two cases orbits are open. 2.

h R

If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.

C29 : Calculate the escape velocity from the moon. The mass of the moon = 7.4 × 1022 kg and radius of the moon = 1740 km. Sol. The escape velocity is

v

2GM R

v

2  6.67  10 11  7.4  10 22 1740  103

v = 2.4 km/s C30 : Calculate the escape velocity on the surface of a planet whose mass is 1025 kg. and radius is 6.67 × 109 metre. Sol.

2  6.67  1011  10 25 6.67  109

Ve 

2GM  R

Ve =

2  105 = 447.2 m/sec.

C31 : A body is thrown with a speed of 3ve where ve is escape speed from the surface of the Earth. Find its speed in interstellar space. Assume no other gravitational field than the earth, and neglect air. Sol. Conserving mechanical energy 

Now, 

GMm 1 1 2  m  3ve   0  mv 2 R 2 2

2 GM  v2e R

v 2e 1 1 2   3v e   v 2 2 2 2 2 2 2 v = (3ve) – ve 

v  2 2 ve Example 29 : A particle is fired vertically from the surface of the earth with a velocity ku e , where u e is the escape velocity and k < 1. Neglecting air resistance and assuming earth’s radius as Re. Calculate the height to which it will rise from the surface of the earth.

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GRAVITATION

Sol. Applying conservation principle of mechanical energy, Ui + Ti = Uf + Tf here Tf = 0 GmM 1 GmM  mk 2 ve2   Re 2 Re  h  GmM 1  GmM  mk 2  2gR   or Re 2 Re  h or

 GM  Q v e  2gR e and g  2  Re  

GmM 1 GM GmM  mk 2 2 2 R e  Re 2 Re Re  h

By solving,

Re 1 k2

R e  R e  R ek 2 1 k2



Re  h  h

Re  Re 1 k2 R ek 2 h 1 k2

h





Example 30 : Two stars each of radius R are at rest at a distance where the force between them is negligible. The mass of one star is M and that of the other is 2M. Find their speeds when their centres are a distance d > 2R apart. Sol. Let the stars of mass M be moving with velocity V1 and 2M with V2. Then MV1 + 2 MV2 = (M + 2M)Vcm Since their initial velocities are zero, Vcm = 0  V1 + 2V2 = 0 V1 = – 2 V2 M M Negative sign indicates that they are moving in opposite directions. When separated by distance d, the potential energy is given by V V/2 U

GM (2M) d2

The kinetic energy is V1   1 1 MV 2  (2M) (V / 2) 2   V2   2  2 2 3 K  MV 2 4 The sum of two energies, K and U, must not change. Hence Ki + Ui = Kf + Uf K

00 

3 2GM 2 MV 2  4 d

8 GM V 3 d The speed of star of mass M is

8GM and that of mass 2 M is 3d

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2 GM . 3d

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PHYSICS GRAVITATION PHYSICS : SHEET

SHEET LAWS OF MOTION

GRAVITATION

Example 31 : A spaceship of mass 1000 kg. is situated on the surface of a planet whose distance from the sun is 2.28 × 108 km. (a) How much energy is the spaceship having ? (b) How much energy must be expended on it to launch it to go to infinity ? The mass of the sun is 2 × 1030 kg, planet is 6.4 × 1023 kg, radius of planet is 3395 km and G = 6.67 ×10–11 Nm2 kg–2. Neglect motion of the planet. Sol. The spaceship at rest has only potential energy. It is given by

M M  U  Gm   S  R r´ 

m r´ R

But r´ >> R hence we take r´  r. Thus –11

U = – 6.67 × 10

 6.4  1023  2  1030  × 10 ×  3395  103 2.28  108  103    3

 6.4  1017  2 U  6.67  10    1019  J 2.28  3.395  8

200   6.4 U  6.67  109   J  3.395 2.28 

U = – 6.67 × 89.5 × 109 J U = – 597.62 × 109 U = – 6.0 × 1011 J When the spaceship will go to infinity, its total energy will be zero. Ef = 0

M M  Ei  Gm   S  R r  Energy supplied to spaceship = Ef – Ei

M M   GM   S  R r  = 6.0 × 1011 J [as calculated in part (a)]

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