81

CHAPTER-7GRAVITATION

FILL IN THE BLANKS1. The numerical value of the angular velocity of rotation of the earth shold be .... rad/s in order to make the effective

acceleration due to gravity at equator equal to zero. (1984, 2M)2. According to Kepler's second law, the radius vector to a planet from the sun sweeps out equal areas in equal

intervals of time. This law is a consequence of the conservation of ..... (1985, 2M)3. A geostationary satellite is orbiting the earth at a height of 6 R above the surface of the earth where R is the radius

of earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is .... hours.(1987, 2M)

4. The masses and radii of the Earth and the Moon are M1, R1 and M2, R2 respectively. Their centres are at distanced apart. The minimum speed with which a particle of mass m should be projected from a point midway between thetwo centres so as to escape to infinity is .... (1988, 2M)

5. The ratio of earth's orbital angular momentum (about the sun) to its mass is 4.4 1015 m2/s. The area enclosed byearth's orbit is approximately ....... m2. (1997C, 1M)

6. A particle is projected vertically upwards from the surface of earth (radius R) with a kinetic energy equal to halfof the minimum value needed for it to escape. The height to which it rises above the surface of earth is ....

(1997, 2M)TRUE/FALSE1. It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi.

(1984; 2M)

OBJECTIVE QUESTIONSOnly One option is correct :1. If the radius of the earth were to shrink by one per

cent, its mass remaining the same, the acceleration dueto gravity on the earth's surface would :(1981; 2M)(a) decrease (b) remain unchanged(c) increase (d) be zero

2. If g is the acceleration due to gravity on the earth'ssurface, the gain in the potential energy of an objectof mass m raised from the surface of the earth to aheight equal to the radius R of the earth, is :

(1983; 1M)

(a)12

mgR (b) 2 mgR

(c) mgR (d)14

mgR

3. Imagine a light planet revolving around a very massivestar in a circular orbit of radius R with a period ofrevolution T. If the gravitational force of attraction be-tween the planet and the star is proporional to R5/2,then : (1989; 2M)(a) T2 is proportional to R2(b) T2 is proportional to R7/2(c) T2 is proportional to R3/2(d) T2 is proportional to R3.75

4. If the distance between the earth and the sun were halfits present value, the number of days in a year wouldhave been : (1996; 2M)

(a) 64.5 (b) 129(c) 182.5 (d) 730

5. A statellite S is moving in an elliptical orbit around theearth. The mass of the satellite is very small comparedto the mass of the earth : (1998; 2M)(a) the acceleration of S always directed towards the

centre of the earth(b) the angular momentum of S about the centre of the

earth changes in direction, but its magnitude re-main constant

(c) the total mechanical energy of S varies periodi-cally with time

(d) the linear momentum of S remains constant inmagnitude

6. A simple pendulum has a time period T1 when on theearth's surface and T2 when taken to a height R abovethe earth' surface, where R is the radius of the earth.The value of T2/T1 is : (2001)

(a) 1 (b) 2(c) 4 (d) 2

7. A geostationary satellite orbits around the earth in acircular orbit of radius 36,000 km. Then, the time periodof a spy satellite orbiting a few hundred km above theearth's surface (Re = 6400 km) will approximately be :

(2002)(a) 1/2h (b) 1 h(c) 2 h (d) 4 h

82

8. A double star system consists of two stars A and Bwhich have time period TA and TB. Radius RA and RBand mass MA and MB. Choose the correct option.

(2006; 3M)(a) If TA > TB then RA > RB(b) If TA > TB then MA > MB

(c)

2 3A A

B B

T RT R

=

(d) TA = TB

9. A spherically symmetric gravitational system of par-

ticles has a mass density

>r

=rRrRr

for0for0

where r0 is

a constant. A test mass can undergo circular motionunder the influence of the gravitational field of par-ticles. Its speed v as a function of distance r from thecentre of the system is represented by (2008; 3M)

OBJECTIVE QUESTIONSMore than one options are correct?1. A solid sphere of uniform density and radius 4 units

is located with its centre at the origin O of coordinates.Two spheres of equal radii 1 unit, with their centres atA ( 2, 0, 0) and B (2, 0, 0) respectively, are taken outof the solid leaving behind spherical cavities as shownin figure. Then : (1993; 2M)

A Bm

O

z

y

x

(a) the gravitational field due to this object at theorigin is zero.

(b) the gravitational field at the point B (2, 0, 0) is zero

(c) the gravitational potential is the same at all pointsof circle y2 + z2 = 36

(d) the gravitational potential is the same at all pointson the circle y2 + z2 = 4

2. The magnitude of the gravitational field at distance r1and r2 from the centre of a uniform sphere of radius Rand mass M and F1 and F2 respectively. Then :

(1994; 2M)

(a) 1 1

2 2

F rF r

= if r1 < R and r2 < R

(b) 2

1 22

2 1

F rF r

= if r1 > R and r2 > R

(c) 3

1 13

2 2

F rF r

= if r1 < R and r2 < R

(d) 2

1 12

2 2

F rF r

= if r1 < R and r2 < R

SUBJECTIVE QUESTIONS1. Two satellite S1 and S2 revolve round a planet in

coplanar circular orbits in the same sense. Their peri-ods of revolution are 1 h and 8 h respectively. Theradius of the orbit of S1 is 10

4 km. When S2 is closestto S1. Find : (1986; 6M)(i) the speed of S2 relative to S1.(ii) the angular speed of S2 as actually observed by an

astronaut in S1.

2. Three particles, each of mass m, are situated at thevertices of an equilateral triangle of side length a. Theonly forces acting on the particles are their mutualgravitational forces. It is desired that each particlemoves in a circle while maintaining the original mutualseparation a. Find the initial velocity that should begiven to each particle and also the time period of thecircular motion. (1988; 5M)

3. An artificial satellite is moving in a circular orbitaround the earth with a speed equal to half the mag-nitude of escape velocity from the earth (1990; 8M)(i) Determine the height of the satellite above the

earth's surface.(ii) If the satellit is stopped suddenly in its orbit and

allowed to fall freely onto the earth, find the speedwith which it hits the surface of the earth.

4. Distance between the centre of two stars is 10a. Themasses of these stars are M and 16 M and their radiia and 2a respectively. A body of mass m is firedstraight from the surface of the larger star towards thesurface of the smaller star. What should be its mini-mum initial speed to reach the surface of the smallerstar? Obtain the expression in terms of G, M and a.

(1996; 5M)

83

ANSWERS

FILL IN THE BLANKS

1. 1.24 103 2. angular momentum 3. 8.48 4. 1 22 ( )Gv M Md

= +

5. 6.94 1022 6. h = R

TRUE/FALSE1. F 2

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (c) 2. (a) 3. (b) 4. (b) 5. (a) 6. (d) 7. (c)8. (d) 9. (c)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (a, c, d) 2. (a, b)

SUBJECTIV QUESTIONS

1. (i) p 104km/h (ii) 3 104 rad/s 2.3

, 23

Gm av Ta Gm

= = p

3. (i) 6400 km (ii) 7.9 km/s

4. 3 5

2Gm

a5. 99.5R

ASSERTION AND REASON1. (a)

5. There is a crater of depth 100R

on the surface of the

moon (radius R). A projectile is fired vertically upwardfrom the crater with velocity, which is equal to theescape velocity v from the surface of the moon. Findthe maximum height attained by the projectile.

ASSERATION AND REASON

This question contains, statement I (assertion) andstatement II (reasons).

1. Statement-I : An astronaut in an orbiting space stationabove the earth experiences weightlessness.

(2008; 3M)

Because :Statement-II : An object moving around the earthunder the influence of earth's gravitational force is ina state of 'free-fall'.(a) Statement-I is true, statement -II is true, statement-

II is a correct explanation for statement-I.(b) statement-I is true, statement-II is true; statement-

II is NOT a correct explanaion for statement-I.(c) statement-I is true, statement-II is false.(d) statement-I is false, statement-II is true.

(2003; 4M)

84

FILL IN THE BLANKS

1. g' = g Rw2 cos2fAt equator f = 0\ g' = g Rw2

0 = g Rw2

\ w =gR 3

9.8

6400 10=

= 1.24 10 3 rad/s

2. 2dA Ldt M

= = constant because angular momentum of

planet (L) about the centre of sun is constant.Thus, this law comes from law of conservation ofangular momentum.

3. T r3/2

\2

1

TT =

3 / 22

1

rr

or 2/3

1

2/3

1

22 7

5.3

=

=

RRT

rrT (24)h

= 8.48h

M1M2

m

d/2 d/2

4. Total mechanical energy of mass m at a point midwaybetween two centres is :

1 2 / 2 / 2

GM m GM mEd d

= 1 22 ( )Gm M M

d= +

Binding energy 1 22 ( )Gm M M

d= +

Kinetic energy required to escape the mass to infinityis,

212 e

mv = 1 22 ( )Gm M M

d+

\ ve =1 2( )2 G M Md+

5. Areal velocity of a planet round the sun is constantand is given by

dAdt = 2

Lm

where L = Angular momentum of planet (earth) aboutsun and m = mass of planet (earth).

Given : Lm = 4.4 10

15 m2/s

\ Area enclosed by earth in time T (365 = days) willbe

Area . .2dA LT Tdt m

= =

13 21 4.4 10 365 24 36002

m=

Area = 6.94 1022 m2.

6.

=

RGM

V2

21

From conservation of energy

)(42

21

hRGMm

RGMm

RGM

m+

-=-

)(4 hRGMm

RGMm

RGMm

+-=-

hRRR +-=-

1141

hRR +-=

- 14

41

RhR 4)(3 =+

3R

h =

TRUE FALSE1. New Delhi is not on the equatorial plane.

OBJECTIVE QUESTIONS (ONLY ONE OPTION)

1. g = 2GM

R

or g 21

Rg will increase if R decreases.

2. DU =1

mghhR

+

Given, h = R

DU =121

mgRmgRR

R

=+

SOLUTIONS

85

3.2mv

R R5/2

\ v R3/4

Now, T =2 R

vp

or T2 2R

v

or T2 2

3/4R

R

or T2 R7/2

4. From Kepler's third lawT2 r3 orT (r)3/2

\2

1

TT =

3 / 22

1

rr

or T2 = T1

3 / 22

1

rr

=3 / 21

(365)2

T2 = 129 days

5. Force on satellite is always towards earth, therefore,acceleration of satellite S is always directed towardscentre of the earth. Net torque of this gravitationalforce F about centre of earth is zero.Therefore, angularmomentum (both in magnitude and direction) of Sabout centre of earh is constant throughout. Since, theforce F is conservative in nature, therefore mechanicalenergy of satellite remains constant. Speed of S ismaximum when it is nearest to earth and minimumwhen it is forthest.

6. T 1g

i.e. 2

1

1

2

gg

TT =

where g1 = acceleration due to gravity at a height h =R from earth's surface = g/4

22

1Using ( ) 2

/ 41

Tg gg h

T ghR

= = = +

7. Time period of a satellite very close to earth's surfaceis 84.6 min. Time period increases as the distance ofthe satellite from the surface of earth increase. So, timeperiod of spy satellite orbiting a few 100 km above theearth's surface should be slightly greater than 84.6 min.Therefore, the most appropriate option is (c) or 2 h.

8. In case of binary star system angular velocity andhence the time period of both the stars are equal.

9. For Rr

rmv

rGmM 2

2= 3

3

3

3

3434

Rr

R

r

MM =

rp

rp=

3

2

3

32

RGMr

rRGMr

rGMv === 3

3

RMrM =

rR

GMv .3

=

rv For Rr >

rmv

rGMm 2

2=

or, vr

GM =

\ rv

1

So, (c) is correct.

OBJECTIVE QUESTIONS (ONLY ONE OPTION)1. (a) The gravitational field is zero at the centre of a solid

sphere. The small spheres can be considered as nega-tive mass m located at A and B. The gravitational forcedue to these masses on a mass at O is equal andopposite. Hence, the resultant force on mass at O iszero.(c and d) are correct because plane of these circlesis y-z, i.e., perpendicular to x-axis i.e., potential at anypoint on these two circles will be equal due to thepositive mass M and negative masses m and m.

2. For r < R, F 3

.GM

rR

= or F r

1 1

2 2

F rF r

= and r1 < R and r2 < R

and for r > R, F 2

GM

r= or F

21

r

i.e., 2

1 22

2 1

F rF r

= for r1 > R and r2 > R

86

SUBJECTIVE QUESTIONS1. T r3/2

or r T2/3

2

1

rr =

2 / 32

1

TT

( ) km 1041018 44

3/2

1

3/2

1

22 =

=

= r

TTr

v2

v1T2

T1

r1 r2

Now, v1 4

1

1

2 (2 )(10 )1

rTp p

= = = 2p 104km/h

v2 4

1

2

2 (2 )(4 10 )8

rTp p

= = = (p 104) km/h

(i) Seped of S2 relative to S1 = v2 v2 = p 104km/h

(ii) Angular speed of S2 as observed by S1

w =

4

1 27

2 1

510 m/s| | 18| | (3 10 m)

v vr r

p =

= 0.3 103 rad/s= 3.0 104 rad/s

2. Centre should be at O and radius r.We can calculate r by figure (b).

3030

(a)

a

rO

aa

30a/2

r

3030

FnetF

F

(b) (c)

/ 2ar

= cos30 3

2=

\ r =3

a

Further net force on any particle towards centreFnet = 2F cos 30

=2

23

22

Gm

a

=2

23Gm

a

This net force should be equal to 2mv

r

\2

23Gm

a=

2

/ 3mv

a

\ v =Gma

Time period of circular motion

T =2 2 ( / 3)

/r a

v Gm ap p=

=3

23aGm

p

3. (i) 2

2

)()( hRGMm

hRmv

+=

+

or, hRGMmmv

+=2

422

==+veGM

mvGMmhR

844..2

2

2

2

2===

vegR

veR

RGM

\ RVgRh

e

-=2

24

RRgR

gR=-=

24 2

(ii) From conservation of energy

2

21

2mv

RGMm

RGMm

+-=-

or, R

GMmR

GMmR

GMmmv222

2

=-=

RGMmmv

22

2

=

=2

4

87

gRR

GMV ==

310640010

8.9 = 4106498 =

km/s 9.7=

4. Le there are two stars 1 and 2 as shown below :

1 2

2a

C2C1a

M 16M

r1P r2

Let P is a point between C1 and C2, where gravita-tional field strength is zero. Or at P field strength dueto star 1 is equal and opposite to the field strength dueto star 2. Hence,

21

GM

r = 22

(16 )G M

r

or2

14

rr

= also r1 + r2 = 10 a

\ r2 =4

(10 ) 84 1

a a = + and r1 = 2aNow, the body of mass m is projected from the surfaceof larger star towards the smaller one. Between C2 andP it will be attracted towards. Therefore, the body shouldbe prjected to just cross point P because beyond that theparticle is attracted towards the smaller star itself.

From conservation of mechanical energy 2min21

mv

= Potential energy of the body at P Potential energy at the surface of large star

\ 21

min2

mv =1 2

16 GMm GMmr r

16

10 2 2GMm GMma a a

16 8

2 8 8GMm GMm GMm GMm

a a a a =

or 2min12

mV =458

GMma

\ Vmin =3 5

2GM

a

5. Speed of particle at A, vA= escape velocity on the surface of moon

2GMR

=

At highest B, vB = 0Applying conservation of mechanical energy, decreasein kinetic energy= increase in gravitational potential energy

AVA

B

h

R100

V = 0B

or 212 A

mv = UB UA = m (VB VA)

or2

2Av = VB VA

\

----

+--=

22

3 1003

2RRR

RGM

hRGM

RGM

or 21 1 3 1 99 1

.2 2 100R R h R R

= + + Solving this equation, we get

h = 99.5R

ASSERTION AND REASON1. Force acting on astronaut is utilised in providing

necessary centripetal force, thus he feels weightless-ness, as he is in a state of free fall.\ correct option is (a)

R=