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SYLLABUS

8. Gravitation

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r

Newton’s law of universal gravitation

Gravitation

Kepler’s laws of planetary motion

Based on the regularities in the motion of the planets, Kepler formulated a set of three laws known as Kepler’s laws

of planetary motion.

I Law (Law of orbits)

All planets move round the Sun in elliptical orbits with Sun at one of the foci.

II Law (Law of areas)

A line joining any planet and the Sun sweeps out equal areas in equal intervals of time.

Areal velocity: The area swept by the radius vector of a planet around the sun, per unit time is called areal velocity

of the planet. Areal velocity of a planet remains constant.

III Law (Law of periods)

The square of the period of any planet about the Sun is proportional to the cube of the semi-major axis of its

orbit. T

2  a

3

where T is the period and a is the semi major axis.

If T 1 and T 2 are the periods of any two planets and r 1 and r 2 are their mean distances from the Sun, then

3 2

3 1 2 2

2 1

r

r

T

T

Nearer planets move faster. For example the orbital speed of Earth is about 30 km s

− 1

. The speed of Jupiter is about

13.2 km s

− 1 with a period of 11.86 years and that of Saturn is 9.7 km s

− 1 with a period of 29.46 years.



• Out of planets known before 18

th century, Saturn is the slowest. Infact, the Sanskrit name ‘shani’ refers to slowly

moving object. Saturn is seen for about 2½ years in each constellation

T 29. 2.5 years 12 12

   =^    

and passes

through 3 constellations in 7½ years, commonly known as ‘saade-sath’.

Newton’s Law of Gravitation: Every particle attracts every other particle with force that is proportional to the product of

the masses and inversely proportional to the square of their separation and acts along the straight line joining them.

2

1 2

r

mm F G

G is a universal constant, called the constant of gravitation.

11 2 2 G 6. 67 10 NmKg

− − = 

The dimensional formula for G is[M LT ]

− 1 3 − 2



• The gravitational force is the weakest known force of nature.
• The value of G is the same for two particles, two celestial objects and two terrestrial objects.
• A spherical shell of matter attracts a particle that is outside the shell as if all the shell’s mass were concentrated at

its centre.

• A uniform shell of matter exerts no gravitational force on a particle located inside it. It is a gravitational shield for

particles within it.

• The net gravitational force on a particle due to one or more particles is determined using the principle of

superposition.

• The gravitational force on a particle would first increase slightly, eventually reach a maximum and finally decrease

to zero at the centre of the earth as the particle is lowered down the centre. The reason for the initial increase is the

predominance of the effect of decrease in r over that of the shell of the earth’s crust that lies outside the radial

position of the particle. As the centre is approached, the effect of the outer shell predominates.

• If the earth were uniformly dense, the gravitational force would decrease to zero as the particle is lowered to the

centre of the earth.

Newton’s law in vector form

2 12

1 2 3 12

1 2 21 rˆ r

Gmm r r

Gmm F (^)  

→ →

where →

→ F 12 force exerted on particle of mass m 2 by particle of mass m 1

→ r 12 position vector of m 2 relative to m 1

rˆ 12 →^ unit vector in the direction of r 12

→

Gravity: It is the term used to describe the force on a body near the surface of a celestial body. The earth’s gravity is given

by

2 (R h )

GMm F

= where M → mass of the earth, R → average radius of the earth

h → height of a body of mass m above the surface of the earth.

Acceleration due to gravity (g): It is the acceleration of a body due to gravity. On the surface of the earth 2 R

GM

g =



• The value of g is independent of the mass of the body.
• In the absence of air resistance, heavy and light bodies released from the same height reach the ground

simultaneously.

• The average density of the earth is given by

3g

4 GR

 = 

.

• Acceleration due to gravity on the surface of the moon is about one-fifth of that on the surface of the earth.

Variation of g

(i) Due to altitude: Acceleration due to gravity at a height ‘h’ above surface of earth is

h (^2)

GM

g (R h)

h

2h g g 1 R

(for h << R) R → radius of earth

Thus g decreases with altitude.

(ii) Due to depth: Acceleration due to gravity at a depth ‘d’ below the surface of earth is

d

d g g 1 R

for (d << R)

at d = R, i.e., centre of earth

gd = 0

 g decreases with depth

(iii) Due to rotation of earth: Acceleration due to gravity at a latitude  is given by

g = g − R

2 cos

2   → angular velocity of earth

(a) At poles: 2

2 g (^) p g R cos 2

 = −  = gmax

Uf −Ui = − 

f

i

F. dr



2

1

r

r

2

1 2

r

Gmm dr = Gm 1 m (^2) 



1 r 2

r

We choose gravitational potential energy of this system equal to zero when the separation between the particles is

infinity. Then potential energy of the system for any separation r is U(r) = − r

Gm 1 m 2

For a system of n particles, the potential energy is the sum of the potential energy of every pair of particles in the

system. In other words, the potential energy is additive.

Gravitational potential

The gravitational potential at any point is defined the negative of the work done by the gravitational force to bring

unit mass from infinity to that point.

Or

The gravitational potential at any point is the negative of the work done by a force in displacing a unit mass from that

point to infinity.

If V is the gravitational potential at any point, the potential energy U of a mass m at that point is given by

U = mV.

Gravitational potential at a point due to a point mass

Let P be a point at a distance r, along the x-axis, from a point mass M kept at the origin O.

By definition of potential given earlier, the potential at the point P is

r

GM

Vp =− … ( 3 )

The following table gives potential due to various regular bodies.

Body Position Potential

1. Uniform ring of radius a

At a point on its axis at a

distance r from its

center

2 2 a r

GM

Uniform thin spherical shell of

radius a

At a distance r from its

center such that r 

a

− r

GM

At a distance r from its

center such that r <

a

− a

GM

3. Uniform solid sphere of radius a

At a distance r from its

center such that r 

a

− r

GM

At a distance r from its

center such that r <

a

− 3 2 a

GM

(3a^2 − r^2 )



Negative potential energy indicates that the gravitational force is attractive. Kinetic energy is always positive.

But potential energy can either be positive or negative. As long as the total energy (KE + PE) is

negative, the object is bound within the system.

Earth’s gravitational field

The gravitational force on a particle of mass m held at a distance h from the surface of the earth is

(R h)

GMm F 2

= towards the centre of the earth.

Hence gravitational field produced at a height h is G (^2) (R h)

GM

m

F

f

= = , towards the earth.

At the surface of the earth, G (^2) R

GM

f = , … (5)

towards the centre of the earth. We see that the gravitational field due to earth

is numerically equal to the acceleration due to gravity of the earth. The

variation of gravitational field due to earth with distance x from its centre is shown

in the figure.

Gravitational potential due to earth at a height h is R h

GM

V

At the surface  

2 R

GM

gr g R

GM

V  … (6)

For convenience in special cases potential at the surface is taken to be zero. Anyhow it is only the difference in

potential is that significant.

As the height above the surface increases potential increases and the gravitational field is in the direction of

decreasing potential, that is towards the surface.. dr

dV F =− … (7)



Gravitational field due to a spherical shell at any point inside it is zero. Potential is constant equal to R

GM

at all points on it. Hence gravitational field due to a shell is zero  

  

 = 0 dr

dV

. However, even inside

the spherical shell there will be gravitational field due to other objects. There is no gravitational

shielding.

The following table gives intensity of field due to various objects.

Object Position Field intensity

1. Point mass At a distance r from it

2 r

GM

2. Uniform ring of radius a At a point on its axis at a distance r from

center 2 2 3 / 2 (a r )

GMr

3. Uniform disc of radius a At a point on its axis at a distance r from

center



(^222) r x

r

a

2 GMr

4. Uniform thin spherical shell

of radius a

At a distance r outside the shell

2 r

GM

At a distance r inside the shell Zero

5. Uniform solid sphere of

radius a

At a distance r outside the sphere

2 r

GM

At a distance r inside the sphere

3 a

GMr

We have, the centripetal force 2

2

r

GMm

r

mv

 Kinetic energy is 2 r

GMm mv 2

K

2 = = … ( 1 )

Potential energy r

GMm U =− … ( 2 )

U = −2K

Total energy of the satellite E = K + U = 2 r

GMm 1 2

1

r

GMm =−  

  

 − … (3)

Close to the earth, 2

mgR

2 R

GMm E =− =−

As height increases, the kinetic energy decreases, potential energy increases.

(f) Rocket launching: It is a multistage process. Initially at the lift off, the launching rocket rises vertically to pass through

denser atmospheric layer with least fuel consumption. The first stage rocket falls off at about 60 km height, and the

second stage rocket is fired. The second stage rocket is gradually fitted by the guidance system. When it reaches the

desired height, the tracking system guides the rocket to move horizontally. At this stage, small rockets are fired to

separate the capsule from the second stage rocket and project it into space with the speed required to follow a

predetermined orbit.

Communication and Indian remote sensing (IRS) satellites

The artificial satellites are broadly classified into two types.

(i) Communication satellites and

(ii) Remote sensing satellites.

Communication satellites

1. Communication satellites are mainly used for communication.
2. They link remote areas of earth with telephone and television.
3. With a network of geostationary satellites a radio/TV program can be broadcasted all over the world

simultaneously.

4. These satellites are also used to take photographs of clouds around earth which help in weather forecasting.
5. INSAT series of satellites launched by India are communication satellites.

Remote sensing satellites

1. Remote sensing satellites are used to estimate the natural resources.
2. They are usually placed in low flying polar orbits.
3. These can take photograph of large areas of land. Hence they can be used to study resources such as forest, river,

water resources, mineral deposits, agricultural crops etc.

4. These are also used for military purposes.



• The geostationary circular orbit (also called the Clarke geosynchronous orbit or Clark Arc, after the famous science

fiction writer Arthur C. Clarke who proposed the idea of a communication satellite in 1945) is in the equatorial

plane of the earth. Its direction of angular motion coincides with that of the earth about its polar axis.

• All communication satellites are geostationary. TV programs are relayed live via these satellites
• The escape speed does not depend on the direction of projection
• It can be shown that the escape speed in terms of radius and density of a planet is given by

  

 = R v R 3

8 G v (^) e e

• An orbiting satellite has both potential energy U and kinetic energy K.

r

GMm U =− ; 2 r

GMm K =− K 2 r

GMm E =W+K=− =−

• It can be shown that 2 r

GMm E =− holds for an elliptical orbit of semi major axis ‘a’ if r is replaced by a. So, total

mechanical energy of a satellite remains the same if it is put in different elliptical orbits with the same a, but

different values of eccentricity e (including 2000)

• To change the speed and orbit of a satellite, a burn is executed in it. The values of K and E at P, the location of the

burn, are less than their corresponding values in the circular orbit.

W = −2K and E = −K for a circular orbit do not apply for the new elliptical orbit.

• Once the satellite is placed in the required orbit by burning small orbital engines, no engines are required to keep

it in the orbit since gravity takes care of that

• Physiological effects in weightlessness : (i) Astronaut’s face becomes puffy. (ii) Astronaut grows a little

temporarily (iii) Cardiovascular system of an astronaut does not need to work hard to pump blood around the body

(iv) There is no preferred direction, no upside down or right side up. All orientations of an astronaut are equally

comfortable.

• Physical effects in weightlessness : (i)A liquid column has no weight, no hydrostatic pressure, no buoyant effects,

no sedimentation. (ii) There is no convection. (iii) A column of air expands on heating but stays where it is. (iii)

Surface tension is much more evident. It is because of surface tension that near normal dinning is possible in space.

Space shuttle straws come with small clamps to pinch them closed and keep the drinks from climbing.

Illustrations

1. A uniform solid sphere of mass M and radius ‘a’ is surrounded symmetrically by a uniform thin spherical shell of

equal mass and radius 2a. Find the gravitational field at a distance

3

2

(^) a from the centre.

(A) 2

4

9

GM

a

(B) 2

4

16

GM

a

(C) 2

2

3

GM

a

(D) 2

21

25

GM

a

Ans (A)

The point P is at a distance

3

2

(^) a from the centre.

As P is inside the cavity of the thin spherical shell,

the field here due to the shell is zero.

The field due to the solid sphere is 2 2

4

3 9

2

= =      

GM GM g a a

2. A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown.

A particle of mass m 1 is placed on the line joining the two centres at a distance x from the point of contact of the

sphere and the shell [if r < x < 2 r ]. The magnitude of gravitational force due to sphere and shell on this particle is

(A)

1 3

Gmm x

r

(B)

1 3

Gmm ( x − r )

r

(C)

1 2 ( − )

Gmm

x r

(D)

1 2 ( − )

GMm

x r

Ans (B)

P (^) M

circular

Changing the orbit of a satellite

elliptical M

Three elliptical orbits with the same semimajor axis a, but e values (e = 0 for circular orbit of r = a)

2a 2a 2a

2 3 2 3

1 1 1

2 2 2

or 8

  =^     = 

    ^   

T R R

T R R

4 R 2 (^) = 4 R 1 = 4 10 km

1 4 1 1 1

2 4 1 2 2

2 2 10 km h

2 and 10 km h

R

T

R

T

−

−

 = =  

 = =  

v

v

In the given situation | v 1 − v 2 | =   10

4 km h

− 1

7. A body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity.

If the radius of the earth is R, the velocity of the body at a height h is

(A) gR (B)

2 2

3( + )

gR

R h

(C)

2

gR

R h

(D)

2 2

gR

R h

Ans (D): From conservation of energy

2 (^2 )

GMm m GM GMm m R R R h

v

2 1 2 2 2

2

• − =  =

GM GM GM GM gR

R h R R R h R h

v v =

8. A solid sphere of uniform density and radius 4 m is located with its centre at the origin ‘O’ of

coordinates. Two spheres of equal radius 1 m with their cavities at A (−2, 0, 0) and B (2, 0, 0)

respectively are taken out, leaving behind spherical cavities. The mass of each sphere taken

out is M. The gravitational field at B is

(A) GM (B)

21

5

GM

(C)

31

16

GM (D)

31

8

GM

Ans (C)

Mass of whole sphere of radius 4 M (without cavities) is

3

0 3

4 (4) 3 64 4 (1) 3

 

= =



M

M M

The gravitational field at B = field due to whole sphere − field due to sphere A

0 = 3 − 2

GM r GM

R AB

3 2

(64 ) 2 31

4 4 16

G M GM GM

 = − =

9. Three uniform spheres each having a mass m and radius R are kept in such a way that each touches the other two.

Find the magnitude of the gravitational force on any of the spheres due to the other two.

(A)

2

2

Gm

R

(B)

2

2

3

2 2

Gm

R

(C)

2

2

3

4 2

Gm

R

(D)

2

2 2 2

Gm

R

Ans (A)

 = 60 ,  PQS is an equilateral triangle of side 2 R.

Forces on S:

2 2

2 2 (2 ) 4

Gm Gm F R R

= = and F  = F

2 2 2 Fnet = F + F + 2 F cos60 = 3 F

2

2

net

Gm F R

10. The distance between two bodies A and B is r. Taking the gravitational force according to the law of inverse square

of r , the acceleration of body A is a. If the gravitatinal force follows an inverse fourth power law, then what will be

the acceleration of the body A?

(A) 3

a

r

(B)

a

r

(C)

a

r

(D) 2

a

r

Ans (D)

2 2

A B B A A

Gm m F Gm F a a r m r

=  = = = ...(i)

4 4

A B B A A

Gm m F Gm F a a r m r

 (^) =  = = = ...(ii)

From (i) and (ii) 2 2

a a a a r r

11. Value of g on the surface of earth is 9.8 ms

− 2

. Find its value on the surface of a planet whose mass and radius both

are two times that of earth.

(A) 9.8 ms

− 2 (B) 19.6 ms

− 2 (C) 4.9 ms

− 2 (D) 10.32 ms

− 2

Ans (C)

2 2

9.8 ms ;

GM

g R

− = =

2 2 2

. 4.9 ms (2 ) 2

G M GM

g R R

−  = = =

12. Calculate the change in the value of g at latitude 45 (when compared with g at equator). The radius of earth = 6.

 10

3 km.

(A) 0.021 ms

− 2 (B) 0.0013 ms

− 2 (C) 0.043 ms

− 2 (D) 0.017 ms

− 2

Ans (D)

2 2 2 g (^) equator = g − R  cos 0= g − R 

2 2 2 45 cos 45^ 2

R g (^)  g R g

 = −   = −

2 2 2 ( ) 2 2

R R g g g R

     = (^)  − (^) − −  =  

6 2 6.37 10 2

2 24 60 60

 (^)    =     ^  

= 0.017 ms

− 2

13. A body is weighed by a spring balance, 1000 N at the north pole. If only the rotation of earth is accounted for, how

much will it weigh at the equator?

(A) 1002 N (B) 997 N (C) 995 N (D) 999 N

Ans (B)

2 2 g (^) pole = g − R  cos 90  = g  mg = 1000

2 2 2 g  = gequator = g − R  cos 0 = g − R 

2 (^1000 ) ( ) 1000 1 997 N

R mg g R g g

    = −  = (^)  − (^) =

 

14. A particle of mass 20 g experiences a gravitational force of 4N along + ve X-direction. Find the gravitational field at

that point (magnitude)

(A) 50 N kg

− 1 (B) 100 N kg

− 1 (C) 200 N kg

− 1 (D) 150 N kg

− 1

Ans (C) 3 m 20 g 20 10 kg; F 4 N

− = =  =

3 1 3

0.2 10 200 N kg 20 10

F

E

m

− −

15. Two spheres one of mass m has radius r. Another sphere has mass 4m and radius 2 r. The centre to centre distnace

betwen them is 12 r. Find the distance from the centre of smaller sphere where net graviational field is zero.

(A) 2 r (B) 4 r (C) 5 r (D) 3.5 r

We have F = 2

1 2

r

Gmm

2

1

F

F

2

1 2

2

1 2

( 3 r )

Gmm

r

Gmm

= 9 F 2 =

F 1

 % decrease in force = 100 F

F F

1

1 2 

20. The ratio between masses of two planets is 2 : 3 and the ratio between their radii is 3 : 2. The ratio between

acceleration due to gravity on these two planets is

(A) 4 : 9 (B) 8 : 27 (C) 9 : 4 (D) 27 : 8

Ans (B)

We know that 2 R

GM

g = 

2

1

g

g

2 2

2

2 1

1

R

GM

R

GM

2

1

M

M

2 1

2 2

R

R

2

21. If the change in the value of g at a height h above the surface of the earth is same as at a depth d below it, then (both

d and h are much smaller than the radius of the earth)

(A) d = 2

h (B) d = h (C) d = 2h (D) d = h^2

Ans (C)

The acceleration due to gravity at height h is g

/ = g  

R

2 h 1

The acceleration due to gravity at a depth d is g

// = g  

R

d 1

Given that, g

/ = g

//

g

/  

R

2 h 1 = g

//  

R

d 1 2h = d

22. If R is the radius of the earth and g is acceleration due to gravity on the earth’s surface, the mean density of earth is

(A)

3 gR

4 G

(B)

4 gR

3 G

(C)

4 RG

3 g

(D)

12 G

Rg

Ans (C)

We have, g = 2 R

GM

R

G

2

 R

3    = 4 RG

3 g

1. The escape speed of a body from the earth depends on

(1) Mass of the body

(2) The direction of projection

(3) The height of location from where the body is launched

(4) All of these

2. A planet of mass m revolved around the sun of mass Min an elliptical orbit. The maximum

and minimum distance of the planet form the Sun are r and 3r respectively. The time period of

the planet is proportional to

(1) r

3

3 2

(2r) (3) 4r (4)

2 3

(4r)

NCERT LINE BY LINE QUESTIONS

3. Two point masses m and 9m are separated by a distance don a line. A third point mass of 1 kg

is to be placed at a point on the line such that the net gravitational force on it is zero.

The distance of 1 kg mass from mass m is

d

d

d

d

4. The force of gravitation between two masses is 10 mN in vacuum. If both the masses are placed

in a liquid at the same distance, then new force of gravitation will be

(1) 10mN (2)

mN

mN

(4) Can’t say

5. Three equal masses of 3 kg each are fixed at the vertices of an equilateral triangle ABC. The

gravitational force acting on mass 2 kg placed at the centroid of triangle is

(1) Zero (2)

3

6.67 10 N

−

9

9 10 N

−

 (4) Data is insufficient

6. An object is projected from earth’s surface, with speed half of the escape speed of earth, then

maximum height attained by it is

RE

RE

3) RE 4) 2RE

7. The change in gravitational potential energy when a body of mass m is raised to height

4RE from the earth surface is 4RE is radius of earth)

(1) E

mgR

(2) mgR

E

mgRE

(4) E

mgR

8. The potential energy of a system of four particles each of mass m, placed at vertices of a square

of side a is

2

Gm

a

2

Gm

a

2

Gm

a

4Gm

a

9. A satellite of mass m is in a circular orbit of radius 2R

E

around the earth. The energy required to

transfer it to a circular orbit of radius 4RE is

mgRE

(2) E

mgR

mgRE

mgRE

10. If the gravitational potential at the surface of earth is V

o

, then potential at a point at height equal

to radius of earth is

(1) V 0 (2)

V 0

V 0

V 0

11. A satellite revolving around earth has potential energy - 2 MJ, then the binding energy of the

satellite is

(1) 1 MJ (2) 2MJ (3) – 1 MJ (4) 8 MJ

12. Starting from the centre of earth having radius R E , the variation is acceleration due to gravity

is best represented by the curve

(4) Escape out with speed some interstellar speed

NCERT BASED PRACTICE QUESTIONS

1. “All planets move i elliptical orbits with the sun situated at one of the foci”. This law

was given by

(a) Newton (b) keplar (c) Archimedesc (d) Pascal

2. Planet revolving round sun swept out equal area in equal time because

(a) Its linear momentum remain constant

(b) It angular momentum remain constent

(c) Its (linear + angular) momentum remain constant

(d) None of these

3. Planet revolving around the sun is shown in figure its speed will be maximum at

(a) A B

(b) B C A

(c) C

(d) same at all points

4. A planet revolving round the sun has time perid T and semi major axis R then which

of the following relation is correct

(a) T^2 R^3 (b) T R^2 (c) T R^3 (d) T^2 R

5. A planet revolving around the sun has angular momentum L and mass m then areal

velocity of the planet is

(a)

m

L

(b)

m

L

(c)

L

2 m

(d) 2 mL

6. Which of the following is not correct for gravitation force?

(a) It is a conservation force

(b) It is a central force

(c) It depends on the medium between two particles

(d) all of the above

7. Two particles of mass m 1 and m 2 are laced at distance r has a force of attraction F if a

third mass m 3 is placed near these two particles then force of attraction between them

will be F’ then

(a) F’ > F (b) F’ < F (c) F’ = F (d) can not be said

8. If earth pall moon with force F the moon pull the earth with force

(a) greater than F (b) less than F

(c) equal to F (d) none of these

9. If a body is at a height h from the surface of earth than the value of acceleration due

to gravity at that height if acceleration due to gravity at earth surface is g is

(a) g 

R

2 h

1 (b) g 

R

h

(c) g 

R

h

1 (d) None of these

10. A planet has radius half the radius of earth and mass double than that of earth then

value of acceleration due to gravity at the surface of that planet is (g = 10 m/s^2 )

(a) 40 m/s^2 (b) 160 m/s^2

(c) 80 m/s^2 (d) 90 m/s^2

11. The value of acceleration due to gravity at a depth of d from the earth surface is

(a) g 

R

2 d

1 (b) g 

R

d

sum

(c) g 

R

d

1 (d) None of these

12. If value of acceleration due to gravity at a height h from the earth surface is same as

at a depth d the which of the following is correct

(a) h = 2d (b) d = 2h (c) d =

h

(d) d = 4h

13. The value of acceleration due to gravity ----------when one move from equator to pole

(a) increases (b) decreases

(c) remains same (d) can not be said

14. Acceleration due to gravity at the centre of earth is

(a) ∞ (b) 0 (c) g (d) None of these

15. Two particles of mass m 1 and m 2 are placed at distance & then gravitational potential

energy of to particle system is

(a)

r

Gm 1 m 2

− (b)

r

Gm 1 m 2

(c) 2

1 2

r

Gmm

− (d)

2

1 2

r

Gmm

16. Potential energy of a system of four particles placed at the vertices of a square of side

l is

(a)

l

Gm

2

− (b) 

2

l

Gm

(c) (^2 2 )

2

l

Gm

(d) 

2

l

Gm

17. Escape velocity of a body at the earth surface is

(a)

gRE

(b)

2 gRE

(c)

gR E

(d) E

E

R

GM

18. Kinetic energy of a satellite of mass M revolving in orbit of radius R is

(a) 2

R

GM Em

(b)

R

GM Em

(c)

R

GM Em

(d)

R

GM Em

19. If kinetic energy of a satellite revolving around the sun is k then total energy of the

satellite is

(a ) 2k (b) – k (c)

k

(d) – 2k

20. Weight of a person at the surface of earth is w then weight of the person in a satellite

revolving around the earth is

(a ) 0 (b) w (c)

w

(d) 2w

21. Which of the following symptoms is not likely to afflict an astronaut in space?

(a) swollen feet (b) swollen face

(c) headache (d) orientational problem

22. The gravitational intensity at the centre of hemispherical shell of uniform mass density

has the direction