INTRODUCTION

This book is based on our experience over the past few

years. This material covers extensively the fundamental
principles and concepts involved, solved problems which
highlight the application of these concepts, exercises and
assignments for practice by the students.

In order to get maximum benefit from this material, ‘word of

Advice’ given overleaf has to be carefully followed.

The book besides IITJEE will also prove useful to students

for other Engineering examinations as well as their school
curriculum.

Wishing you all success.

 A WORD OF ADVICE

 Try to do the solved problems and exercises given, after

completion of related topics in the chapter. Attempt the
assignments.

 The purpose of the assignments is to give you a practice in

solving various levels and varieties of problems. Each problem
has some important concept which it highlights. When you do a
problem from an assignment, make sure that you have completed
the study material, have committed the formulae to your memory
and have solved the solved problems (most of them on your own
before seeing the solution). Do not open the study material to
refer to formulae/theoretical concepts while doing the
assignment problems unless it is absolutely essential to do so.

 Do full justice to the exercises and assignment problems. Even

if you do not get the answer to a problem, keep trying on your
own and only approach your friends or teachers after making lot
of attempts.

 Do not look at the answer and try to work backwards. This would
defeat the purpose of doing the problem. Remember the purpose
of doing an assignment problem is not simply to get the answer
(it is only evidence that you solved it correctly) but to develop
your ability to think. Try to introduce twists and turns in given
problem to create similar problems.
 ABOUT THE CHAPTER

One of the most profound mysteries of nature is -why

objects fall down.

Why does the moon orbit the earth? Why does the earth
orbit the sun? Are these questions related to each other?

This was first solved by Sir Isaac Newton in the seventeenth

century  it is known as “the law of gravitation”.

Gravitation is one of the four fundamental forces of nature. It

acts between all material bodies in the universe  it is
universal.
 CONTENTS

JEE Syllabus .... …1

Newton’s Law of gravitation ........1
Gravitational Field and Intensity ........2
Variation of Acceleration Due to Gravity (g) ........5
Solved Problems ........6
Subjective ........6
Objective ........7
Exercise -1 ........8
Gravitational Potential Energy ........9
Gravitational Potential ........9
Binding Energy ......11
Escape Velocity ......11
Solved Problems ......12
Subjective ......12
Objective ......14
Exercise -2 ......15
Motion of Planets ......16
Motion of Satellites ......16
Kepler’s Second Law ......19
Solved Problems ......20
Subjective ......20
Objective ......23
Exercise -3 ......24
Answers to Exercise ......26
Concepts and formula a glance ......28
Chapter Practice Problem ......30
Assignments ......32
Section-I ......32
Section-II ......33
Section-III ......44
Answers to Chapter Practice Problem .. …46
Answers to Assignments .. …47
GRAVITATION
Syllabus
Universal law of gravitation, Gravitational field,
Gravitational potential energy and Gravitational potential,
Acceleration due to gravity, Motion of planets and
satellites in circular orbit.

NEWTON’S LAW OF GRAVITATION

Newton’s Law of Gravitation states that “Every particle in the universe attracts every other particle with a
force that is directly proportional to the product of their masses and inversely proportional to the square of
the distance between them”.
Consider two bodies A and B of masses mA and mB, mA mB
 
attracting each other with forces FAB (force on A due to B)  B
AF F
 AB BA

and FBA (force on B due to A), respectively.

 
Then, FAB   FBA
and

FAB  FAB = magnitude of the attractive force
m A mB
=G
d2
where d is the distance between them. G is a universal constant known as Universal Gravitational
constant. Its value was first measured by Cavendish and is now known to be:
G = 6.6726  1011 N-m2/kg2
  
If F1, F2 , F3 , … are the gravitational forces acting on the particle A due to particles P1, P2, P3, …,
respectively, then net gravitational force on a particle A due to particles P1, P2, …, Pn is given by:
   
F  F1  F2      Fn
This is called principle of superposition.

Important points
 The gravitational force is an attractive force.
 The gravitational force between two particles does not depend on the medium
 The gravitational force between two particles is along the straight line joining the particles (called
line of centers).
Illustration 1: Three identical particles, each of mass m, are placed at the three corners of an
equilateral triangle of side ‘a’. Find the force exerted by this system on another particle of
mass m placed at
(a) the mid-point of a side
(b) the centre of the triangle.

Solution: (a) Using the principle of superposition

   
F  FA  FB  FC
 
When the particle is placed at the mid point of a side (at P), FC   FB , and they cancel
each other.

IITJEE-2223-PHYSICS-GRAVITATION
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 
Hence, force experienced by the particle, F  FA A

 Gmm Gm2
| F |  | FA | = 
 AP 2  a sin602

4Gm2    FA
=  alongPA 
3a2   C  P
 B
FC FB

(b) If the particle is placed at the centre of the triangle, the net A
force on the particle P due to particles placed at the corners

A, B and C will be zero. FA
     P
Hence, F  FA  FB  FC  0 
 FC
FB
C B

Gravitational Field and Intensity

The space around a body where the gravitational force exerted by it can be experienced by any other
particle is known as the gravitational field of the body. The strength of this gravitational field is referred
to as intensity, and it varies from point to point.
Consider the gravitational field of a particle of mass m located at the origin (O).
Suppose that a test particle of mass m 0 is placed at the point P(x, y, z). The force of gravitational
attraction exerted on the test particle is given by,
 Gmm0
Fg   rˆ
r2
 
where the position vector OP  r ,
 
r = OP = | OP |  | r |

r
and the unit vector, r̂ 
r
The intensity of this gravitational field at a point (P) is given by the force per z

unit mass on a test particle kept at P, i.e. m0 P(x,y,z)

 Fg
E
m0 m
  O y
where E is the gravitational intensity and Fg is the gravitational force acting
x
on the mass m0. The gravitational field is, therefore, a vector field.

The gravitational

field at P due to a particle of mass m kept at the point O (origin) is given by
 Fg  Gmm0  1 G m
E   2
rˆ   =  2 r̂
m0  r  m0 r

where r  xiˆ  yjˆ  zkˆ represents the position vector of the point P with respect to the source at the

r
origin and r̂  represents the unit vector along the radial direction.
r
The superposition principle extends to gravitational fields (intensities) as well:
    
E  E1  E2  E3      En
  
where E1 , E2 , …, En are the gravitational field intensities at a point due to particles 1, 2, ..., n,
respectively.

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  
For a continuously distributed mass, the formula changes to E   dE , where dE = gravitational field

intensity due to an elementary mass dm.

The gravitational field of a ring on its axis

Let us consider a ring of mass M in the plane perpendicular to the d m

plane of the paper. We want to find the gravitational field on its axis Z
at a distance x. d E
Consider a differential length of the ring of mass dm. a

Gdm
dE = x
Z2
d E
The Y-components of the fields due to diametrically opposite
elements cancel each other. Thus, the X-components add up. d m

Gdm Gcos  GMcos  GMx

E 2
cos   2  dm  2

z z z (a  x 2 )3 / 2
2

Field due to a uniform thin spherical shell

Consider a thin spherical shell of radius ‘a’, mass M and of negligible
ad
thickness. Out of the spherical shell we consider a small ring of A Z
thickness (a d). The shaded ring has mass dm = (M/2) sin  d. The d
field at P due to this ring is  
O P
Gdm GM sin  d cos  a sin
dE  2 cos   
z 2 z2
From OAP,
r
z2 = a2 + r2  2ar cos 
or 2z dz = 2ar sin  d
or sin  d = z dz/ar
Also, from OAP,
z 2  r 2  a2
a2 = z2 + r2  2zrcos ; co s  
2zr
GM  a2  r 2 
Thus, dE  1   dz
4ar 2  z2 

GM  a2  r 2 
or  dE  4ar 2  z  z 


Case I: P is outside the shell, r > a
r a
GM  a2  r 2  GM
E 2  z    2
4ar  z  r
r a
We see that the shell may be treated as a point particle of the same mass placed at its centre to calculate
the gravitational field at an external point.
Case II: P is inside the shell, r < a
a r
GM  a2  r 2 
E  z   0
4ar 2  z 
a r
We see that the field inside a uniform spherical shell is zero.

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Gravitational field outside a solid sphere

The sphere can be thought of as composed of many shells from radius = 0 to radius = a.
The point P is at a distance r from the centre of all these concentric shells.
G
 E  2 [ M1  M2    ]
r
GM
E 2
r
Gravitational field inside a uniform solid sphere of radius ‘R’
To find the field at a point P inside the sphere at a distance r < R from the R
centre, let us consider a sphere of radius r.
Consider a point P on the surface of the shaded sphere. Since this point is
inside the shells having radii larger than r, they do not contribute to the field at
P. Shells that are less than radius ‘r’, contribute to the gravitational field at P. r P
The mass of the sphere of radius r is
4
M  r 3
3 Mr 3
M'   3
4 3 R
R
3
GM' GM r
 EP  2 
r R3
The adjacent graph shows the variation of E due to a solid sphere of radius R
with the distance r from its centre.
GM
E 2 (r  R) E Er 2
E  (1/r )
r
GM
E 3 r (r  R)
R R r

 This result holds good for the earth also, if it is assumed to be a uniform solid sphere.
Fg Fg
 As by definition, g = and also E = , so g = E, i.e. acceleration due to gravity and gravitational
m m
intensity E at a point are synonymous.

Illustration 2: On to a sphere of radius R/2 and density 2 with centre at C2 a second P m

solid sphere is moulded with density 1 radius R and centre C1. Find the y
force experienced by a point mass m at point P at a distance y from the
combination as shown.
1 R
C1
2 C
2

R/2

Solution: If we consider that a sphere of radius R is placed with centre at C1 of density 1 the force
on the mass at P is
(4 / 3)R31m
F1 = G towards the sphere.
(R  y)2
If we consider that a sphere of radius R/2 is placed with centre at C2 of density 1 the
force on the mass at P is
(4 / 3)(R / 2)3 1m
F2 = G towards the sphere.
(R / 2  R  y)2

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If we consider that a sphere of radius R/2 is placed with centre at C2 of density 2 the
force on the mass at P is
G(4 / 3)(R / 2)3 2m
F3 =
(R / 2  R  y)2
By the principle of superposition
F = F1 - F2 + F3
4 3  1 (2  1 ) / 8 
= R Gm  (R  y)2  ((3R / 2)  y)2 
3  

Variation of Acceleration due to Gravity (g)

(i) Due to altitude:

Consider a mass m at a height h from the surface of the earth. Now, the force acting on the mass due to
Mm
gravity is F = G , where M is the mass of the earth and R is the radius of the earth.
R  h 2
Mm
If the acceleration due to gravity at the given height is g, then mg = G ,
R  h 2
2
M GM  h
 g = G   1  
R2 1  h R 
2 2 R
R  
 g (1 – 2h/R) ( if h  R)
(Expanding binomially and neglecting the higher order terms).
(ii) Due to depth:
If a particle of mass m is kept at a depth ‘d’ from the surface of earth, d R
then gravitational force exerted on the particle of mass ‘m’. P
GMm
F=
R  d 2
where M = mass of earth within radius of (R – d)
M
 M = 3 (R  d)3
R
GM R  d m
F 
R3
GM  d  d
g = 3 R  1    g  1  
R  R  R

(iii) Due to rotation of the earth:

Consider a body at a point with latitude , on the surface of the earth. 
Pole mg
Let R = radius of the earth and  = angular velocity of the earth about its
m2Rcos
own axis.
mg
F.B.D. of the body with respect to the earth frame is shown in the equator 
adjacent figure. O

g = g  2R cos2
pole

Axis of rotation

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SOLVED PROBLEMS
SUBJECTIVE

Problem 1: Three particles, each of mass M, are located at the AM

vertices of an equilateral triangle of side 'a'. At what
30o
speed must they move if they all revolve under the
influence of their gravitational force of attraction in a
circular orbit circumscribing the triangle while still O r
preserving the equilateral triangle? C B
M M

    GM2   GM2 
Solution: FA  FAB  FAC  2  2  cos 30o   2  3 
 a   a 
a
r
3
Mv 2 Mv 2  3 GM2
Now, F   2  3
r a a
GM
or v 
a
Problem 2: The magnitudes of gravitational field at distances r1 and r2 from the centre of a uniform
sphere of radius R and mass M are I1 and I2 respectively. Find the ratio of (I1/I2) if (a) r1 >
R and r2 > R and (b) r1 < R and r2 < R.

Solution: In case of spherical volume distribution of mass,

GM GM
I  2 for r > R and I = 3 r for r < R
r R
(a) As both r1 and r2 are greater thatn R,
I1 (GM / r12 ) r22
 
I2 (GM / r22 ) r12
(b) As both r1 and r2 are lesser than R,
I1 (GM / R3 )r1 r1
 
I2 (GM / R3 )r2 r2

Problem 3: A uniform sphere has a mass M and radius R. Find the gravitational pressure P inside
the sphere, as a function of the distance r from its centre.

Solution: Consider a layer of thickness dr at a distance r from the

centre of the sphere.
dr
Now mass of the layer of thickness dr = 4r2dr r
4 P
G( r 3 )(4 r 2 dr)
force due to the layer dF = 3 R
r2
4
G( r 3 )(4r 2 dr )
 (dP)4r =
2 3
r2
(Where  is the mean density of sphere)

IITJEE-2223-PHYSICS-GRAVITATION
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4
or, dP = G  2 rdr
3
R
4
 P = G  2 rdr
r 3

2 2 2
or, P = G  (R  r 2 )
3
3 1  r / R  M
2 2 2

= G [   = M/ (4/3)R3]
8 R4

OBJECTIVE

Problem 1: A planet has twice the density of earth but the acceleration due to gravity on its surface is
exactly the same as that on the surface of earth. Then, its radius in terms of the radius of
earth (R) will be
(A) R/4 (B) R/2
(C) R/3 (D) R/8

Solution: (B)
We know that (for earth)
GM G 4 
g = 2 = 2  R3 d
R R  3 
where d = mean density of earth.
For the planet:
G 4 
g = 2
  R3 (2d)
(R)  3 
Given that g = g
G 4 3  G 4
 2 
R d = 2
 R3 (2d)
R 3  (R) 3
On solving, we get
R = (R/2).

Problem 2: The period of rotation of the earth so as to make any object weight-less on its equator is
(A) 84 min (B) 74 minutes
(C) 64 minutes (D) 54 minutes

Solution: (A)
Put ge = 0, in the expression
ge = g0  r2
g0
 =
r
r
or T = 2
g0
6
Putting r = 6.4  10 m
and g0 = 9.8 m/sec2, we obtain, T = 84 min

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EXERCISE -1
1. Newton’s apple fell towards the earth; didn’t the earth move towards the apple?

2. Somebody says “I weigh less than what I exactly weigh”. Is he right? Is there any place on earth
where he can weigh exactly?

3. Find the relation between the acceleration due to gravity g and the mean density  of earth in
terms of G (the gravitational constant) and Re (the radius of earth).

4. Two small particles of mass m each are placed at the vertices A and B of A

a right angle isosceles triangle. If AB = , find the gravitational field

strength at C. C B

5. Four particles of equal mass M move along a circle of radius R under the action of their mutual
gravitational attraction. Find the speed of each particle.

6. Two massive particles of masses M and m (M > m) are separated by a distance . They rotate
with equal angular velocity under their gravitational attraction. What is the linear speed of the
particle of mass m?

7. The weight of a body at the centre of the earth is

(A) zero (B) infinite
(C) same as on the surface of earth (D) none of these

8. Infinite number of bodies, each of mass 6 kg, are situated at distances 1 m, 2 m, 4 m, 8 m, ,

from the origin on the y-axis. The resultant gravitational field intensity at the origin is
(A) 4G (B) 8G
(C) 9G (D) 12G

9. The value of acceleration due to gravity at a point P inside the earth and at another point Q
outside the earth is g/2. (g being acceleration due to gravity at the surface of earth.) Maximum
possible distance in terms of radius of earth R between P and Q is
(A) 2R (B) 2R  2 1 
(C)
R
2
2 2 1  (D)
R
2
2 2 1 
10. Two particles, each of equal mass M, go round a circle of radius R under the action of their
mutual gravitational attraction. The speed of each particle is
GM 1 1
(A) v = (B) v =
2R 2R GM
1 GM 4GM
(C) v = (D) v =
2 R R
11. The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and 2.
The ratio of the acceleration due to gravity at their surfaces is
 
(A) g1 : g2 = 12 : 22 (B) g1 : g2 = R12 : R21
R1 R 2
(C) g1 : g2 = R12 : R21 (D) g1 : g2 = R11 : R22

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Gravitational Potential Energy

Change in gravitational potential energy of a system is defined as the ve of the work done by the
gravitational force as the configuration of the system is changed
f 

Uf  Ui   Wgr =   F  dr
i

Change in gravitational potential energy of two point masses m 1 and m2 as their separation is changed
from r1 to r2 is given by

Let a particle of mass m 1, be kept fixed at a point A and another particle of r
A B C
mass m2 is taken from a point B to a point C. Initially the distance between the
m1
particles is AB = r1 and finally it becomes AC = r2. We have to calculate the r1 dr
change in potential energy of the system of the two particles as the distance r2
changes from r1 to r2
Consider a small displacement when the distance between the particles changes from r to r + dr.
 
Gm1m2 r
F=   (force on m2 due to m1)
r2 r
The work done by the gravitational force in the displacement is
Gm1m2
dW =  dr
r2
The increase in potential energy of the two particles will be
Gm1m 2
dU = dW =
r2
U(r2 ) r2
Gm1m2
 
U(r1 )
dU  
r1 r2
1 1
U  r2   U  r1   Gm1m2   
 r1 r2 
If, at infinite separation, gravitational potential energy is assumed to be zero, then the gravitational
potential energy of the above two point mass system at separation r is
Gm1m2
U r   
r

Gravitational Potential

Gravitational field around a material body can be described not only by gravitational intensity vector E
but also by a scalar function, the gravitational potential V. The gravitational potential at any point may be
defined as the potential energy per unit mass of a test mass placed at that point.
U
V= (where U is the gravitational potential energy of the test mass m.)
m
Thus, if the reference point is taken at infinite distance, the potential at a point in the gravitational field is
equal to the amount of work done by the external agent per unit mass in bringing a test mass from infinite
distance to that point. The expression for the potential is given by
P 
V =  E  dr

With the above definition, the gravitational potential due to a point mass M at a distance r from it is
r GM  r GM GM
V = –  2 r̂  dr =  2 dr   , If reference point is taken as infinity
  r
r r

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Potential is a scalar quantity. Therefore, at a point in the gravitational field of a number of material
particles, the resultant potential is the arithmetic sum of the potentials due to all the particles at that point.
If masses m1, m2, …, mn are at distances r1, r2, r3, …, rn, then potential at the given point is
m m m 
V = G  1  2  3   
r
 1 r2 r3 
dV
The field and the potential are related as, E = 
dr
Gravitational potential due to a shell V
R r
GM
(i) at a point outside the shell is:  (r > R)
r
GM
(ii) at a point on the surface of the shell is: – GM/R
R
GM
(iii) at a point inside the shell is: –
R

Gravitational potential (V) due to a uniform solid sphere V R r

GM
(i) Outside of the sphere at a distance r from the centre, V = 
r
(ii) Inside the sphere at a distance r from the centre,
(GM/R)
3GM  R2 r2 
V 3   
R  2 6  3GM/2R


Illustration 1: Two spherical bodies of masses 2M and M and of radii 3R and 2M

R, respectively, are held at a distance 16R from each other in M
3R
free space. When they are released, they start approaching R
each other due to the gravitational force of attraction. Then,
find:
(a) the ratio of their accelerations during their motion. 16R
(b) their velocities at the time of impact.
Solution: (a) Due to the mutual attraction, the masses attract each 2M
other. 1 M
2
If the accelerations are a1 and a2, the net external
force on the system = 0

 aCM  0  m1a1  m2a2 = 0 v1 v2
a m 1
or 1  2 
a2 m1 2 4R

(b) Taking both the bodies as a system, from conserving momentum of the system,
m1 v 2
m1v1 – m2v2 = 0    2
m 2 v1
Now, conserving the total mechanical energy, we have
1 1 G  2M  M G  2M M
 2M v12  Mv 22   and solving it we get.
2 2 4R 16R
GM GM
v1 = and v2 = 2
8R 8R
Note: The velocities and accelerations are w.r.t. the inertial reference frame (i.e. the centre of mass of
the system).

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7
Illustration 2: At a point above the surface of earth, the gravitational potential is –5.12 x 10 J/kg and
the acceleration due to gravity is 6.4 m/s2. Assuming the mean radius of the earth to be
6400 km, calculate the height of this point above the earth’s surface.

Solution: Let r be the distance of the given point from the centre of the earth. Then,
GM 7
Gravitational potential =  = – 5.12  10 J/kg . . . (1)
r
and acceleration due to gravity,
GM 2
g = 2  6.4 m/s . . . (2)
r
Dividing (1) by (2), we get
5.12  107
r=  8  106 m  8000km
6.4
 Height of the point from earth’s surface = r – R = 8000 – 6400 = 1600 km

Binding Energy
Binding energy of a system of two bodies is the amount of minimum energy needed to separate the
bodies to a large distance.
If two particles of masses m1 and m2 are separated by a distance r, then the gravitational potential energy
of the system is given by
Gm1m2
U= …(i)
r
Let T amount of energy is given to the system to separate the bodies by a large distance. When the
bodies are separated by a large distance, gravitational potential energy of the system is zero. For
minimum T, conserving energy for initial and final positions,
T+U=0
Gm1m2 Gm1m2
 T  0 or T =
r r
Hence, binding energy of a system of two particles separated by a distance r is equal to
Gm1m2
T= , where m 1 and m 2 are the masses of the particles.
r

Escape Velocity
Escape velocity on the surface of earth is the minimum velocity given to a body to make it free from the
gravitational field, i.e. it can reach an infinite distance from the earth.
Let v e be the escape velocity of the body on the surface of earth and the mass of the body to be projected
be m. Now, conserving energy at the surface of the earth and infinity,
1 GMm 2GM
mv 2e   0  ve = .
2 R R

Illustration 3: The mass of Jupiter is 318 times that of earth, and its radius is 11.2 times the earth’s
radius. Estimate the escape velocity of a body from Jupiter‘s surface. [Given: The escape
velocity from the earth’s surface is 11.2 km/s.]

Solution: Hence, MJ = 318 Me; RJ = 11.2 Re; ve = 11.2 km/s

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2 GMJ 2 GMe
We know, v J = and ve =
RJ Re
vJ MJ Re
  
ve Me RJ

MJ Re
 v J = ve 
Me RJ
1 1
 318Me Re  2  318  2
v J = 11.2    = 11.2    59.7 km / s
 Me 11.2Re   11.2 

Illustration 4: Find the escape speed from a point at a height of R/2 above the surface of earth.
Assuming mass of earth as M and its radius as R.

Solution: Conserving mechanical energy of a point mass m which is to escape, and earth system
we have,
GMm 1
  mv 2  0
R  R / 2  2
4GM
 v
3R

SOLVED PROBLEMS
SUBJECTIVE
Problem 1: There are two fixed heavy masses of magnitude M of high density on x-axis at
(d, 0, 0) and (d, 0, 0). A small mass m moves in a circle of radius R about origin in the y-
z plane between the heavy masses. Find the speed of the small particle.

Solution: Force of attraction between M and m is m

GMm F F
F= 2
R  d2 R
By symmetry Fx components will cancel. M M
 The net force, which provides the centripetal
d d
force, is given by
GMm R GMm
2Fy = 2.  2 = 2R 2
(R  d ) (R  d )
2 2 2 1/ 2
(R  d2 )3 / 2
1/ 2
mv 2 GMm  2GMR 2 
  2R 2  v =  2 2 3/2 
.
R (R  d2 )3 / 2  (R  d ) 

Problem 2: What is the gravitational potential energy of a particle of mass m kept at a distance x from
the centre of a disc of mass M on its axis? The radius of the disc is R.

Solution: The gravitational potential of the differential ring at position

P is given as r
O m
dM dr P
dV = G
r x

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M 2Mrdr
where dM =  2rdr 
R 2 R2
& r 2  x2
r =
 2Mrdr 
 dV = G   r x
2 2
2
 R 
 The potential due to the entire disc at the point P is given as
2GM R rdr
V =  dV =  2 
R 0 r 2  x2
GM R 2rdr
=  
R2 0 r 2  x 2
2 2 2
Let, r + x = t
 2rdr = 2tdt
GM 2tdt 2GM
 V=  2   2 t
R t R
2GM 2 R
 V=  2 r x 2

R 0

2GM
=  2  R2  x 2  x 
R  
The gravitational potential energy of the system
U = mV = 
2GMm
R2

R2  x 2  x 
Problem 3: Distance between the centres of two rings is a. The M
masses of these rings are M and 4M. A body of mass m is 4M R a R
fired straight from the centre of the heavier ring. What
should be its minimum initial speed to cross the centre of
lighter ring? The radius of the both ring is R. (Assume
a>>R).

Solution: Let O be the point along O1O2 so that gravitational field intensities due to both the rings
balance each other. OO1 = x
G  4M x GM  a  x  M

 x2  R 2 
32 32
  a  x 2  R 2  4M R O R
  O1 O2
a >> R  x >> R x (ax)
4GM GM 4 1
  
a  x  (a  x)2
2 2 2
x x
 2a  2x = x  x = 2a/3
Now from conservation of energy between the point O1 and O
G  4M m GMm 1
 + mv2
R  a 
12
R 2 2 2

G  4M  m GMm
=  
R x R2   a  x 
2 2 2

G  4M m GMm 1
  + mv2
R a 2

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G  4M m GMm
= 
x a  x 
4GMm GMm 1 4GMm GMm
   mv 2   
R a 2 2a 3 a3
1 2 6GMm 3GMm GMm 4GMm
mv =   
2 a a a R
1 8GMm 4GMm   8 4
mv2 =   GMm   
2 a R  a R
 a >> R
GM
 v= 2 2
R

OBJECTIVE
Problem 1: Three particles, each of mass m, are placed at the corners of an equilateral triangle of
side d. The potential energy of the system is
3Gm2 Gm 2
(A) (B)
d d
3Gm 2
(C) (D) none of these
d

Solution: (C) m
For the system of two particles, gravitational energy is given
as 0
60
U = Gm1m2/r d
UA = U12 + U23 + U31
3Gm2 0
or UA = 
0
60 60
m m
d
[ ve sign indicates that the particles are bounded by their
mutual gravitational field]

Problem 2: The minimum energy required to remove a body of mass m from earth’s surface to far
away is equal to
(A) 2mgR (B) mgR
(C) –mgR (D) zero

Solution: (B)
GMm
The potential energy of the body on the surface of earth, U1 =  .
R
The potential energy of the body at infinity, U2 = 0
 U = U2  U1
GMm  GM 
=  mgR  g  2 
R  R 
3
Problem 3: The gravitational field due to a mass distribution at position x is given by I = (A/x ) in X-
direction. The gravitational potential at position x is equal to
(A) –A/x3 (B) –A/2x2
3
(C) +A/x (D) A/2x2.

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Solution: (D)
The potential at a distance x is
x x
A
V(x) =   I dx = –  3 dx
  x
x
 A  A
V(x) =  2  =
 2x   2x 2

EXERCISE -2
1. Does the escape velocity of a body from the earth depend upon
(a) the mass of the body, and (b) the direction of projection?

2. The masses and radii of earth and moon are M1, R1 and M2, R2, respectively. Their centres are at
a distance d apart. Find the minimum speed with which a particle of mass m should be projected
from a point midway between the two centres so as to escape to infinity.

3. Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released.
Assuming that only mutual gravitational forces are acting, find the speeds of the particles when
the separation decreases to 0.5 m.

4. A body of mass m is taken to a height kR from the surface of the earth very slowly, R being the
radius of the earth. Find the change in gravitational potential energy in this process. [Take m e the
mass of earth.]

5. A body starts from rest from a point at a distance r0(>Re) from the centre of earth. It reaches the
surface of earth. What is the velocity acquired by the body?

6. A space vehicle of mass m is in a circular orbit of radius 2Re about the earth (mass m e). What is
the work done by an external agent to transfer it to an orbit of radius 4Re?

7. Three stars each of mass M and radius R are initially at rest and the distance between centres of
any two stars is d and they form an equilateral triangle. They start moving towards the centroid
due to mutual force of attraction. What are the velocities of the stars just before the collision?
Radius of each star is R.

8. (i) What will be the escape velocity of a body if it is projected at an angle of 45 to the
horizontal?
(ii) Why are the lighter gases rare on the surface of earth?
(iii) If a projectile is fired straight up from the earth’s surface, what will happen if the total
mechanical energy is (a) less than zero, and (b) greater than zero? [Ignore the air
resistance and effects of other heavenly bodies.]

9. A particle of mass m1 lies inside a spherical shell of mass m2 and radius R at a distance r from
the centre. The gravitational potential energy of the system is
Gm1m2 Gm1m2
(A)  (B) 
r R
Gm1m2 Gm1m2
(C) + (D) +
r R

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10. If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a
body of mass m from the earth’s surface to a height R equal to the radius of the earth is
(A) mgR/2 (B) 2 mgR
(C) mgR (D) mgR/4

11. A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest but it
moves at a speed of 2 m/s as it reaches A. The work done by the person on the mass is 2J. The
potential at A is
(A) 2 J/kg (B) 2 J/kg
(C) 4 J/kg (D) none of these

Motion of Planets
One of the greatest ideas proposed in human history is the fact that the earth is a planet, among the other
planets, that orbits the sun. The precise determination of these planetary orbits was carried out by
Johannes Kepler, using the data compiled by his teacher, the astronomer Tycho Brahe. Johannes Kepler
discovered three empirical laws by using the data on planetary motion.
1. Each planet moves in an elliptical orbit, with the sun at one of focii of the ellipse.
2. A line from the sun to a given planet sweeps out equal areas in equal intervals of time.
3. The square of the periods of the planets are proportional to cube of their mean distance (or semi-
major axis) from the sun.
These laws go by the name ‘Kepler’s laws of planetary motion’. It was in order to explain the origin of
these laws, among other phenomena, that Newton proposed the theory of gravitation.
In our discussion, we are not going to derive the complete laws of planetary motion from Newton’s law of
gravitation. Since most of the planets actually revolve in near circular orbits, we’re going to assume that
the planets revolve in circular orbits.
Consider a planet of mass m rotating around the sun (mass M  m) in a circular orbit v
of radius r with velocity v. Then, by applying Newton’s law of gravitation and the P
r
second law of motion, we can write
Gravitational force = mass  centripetal acceleration S

GMm  v2 
i.e. =m   …(1)
r2  r 
 
GM
or, v2 = …(2)
r
As the moment of the gravitational force about S is zero, the angular momentum of the planet about the
sun remains constant. This is the meaning of Kepler’s 2nd law of motion, as will be shown later.
The time period of rotation, T, of the planet around the sun is given by,
2r 2r 2
T   = r3 / 2 …(3)
v GM / r GM
2  4 2  3
Squaring both sides, T = r …(4)
 GM 
 
rd
which is Kepler’s 3 law of motion.
Note: The constant of proportionality in the above equation depends only on the mass of the sun (M)
but not on the mass of the planet.
Kepler’s Laws are also valid for the motion of satellites around the earth.

Motion of Satellites
Here we discuss the motion of artificial earth satellites which moves in circular orbits.

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Orbital Speed
Suppose that the speed of an artificial earth satellite in its orbit of radius r be v 0. The satellite accelerates
GMm
towards the centre of earth due to the earth’s gravitational pull
r2

GMm
 FCP = V0
r2
mv 20 GMm M r
Fg m
=
r r2
GM
 v0 = …(1)
r
GM
Putting = g (acceleration due to gravity at the orbit), we obtain,
r2
 v0 = gr …(2)
When it orbits at an altitude h, putting r = (R+h)
GM
 v0 =
Rh
GM gR
= = ; where g is acceleration due to gravity at the earth-surface
R(1  h / R) (1  h / R)
If the height is very small compared to the radius of the earth, then (i.e., h  R)
v= gR

v
Angular Speed
The angular speed m
P
Earth Fg Satellite
v r
= 0
r M
R
GM h
Putting v0 = , we obtain,
r
GM GM
= = .
r3 (R  h)3

Time Period of Revolution

2
The period of revolution, T =

GM
Putting  = ,
r3
r3 (R  h)3
T = 2 = 2 .
GM GM

Potential Energy
The gravitational potential energy of interaction of any two masses m 1 and m2 kept at a separation r is
given by:
Gm1m2
PE = 
r
Gravitational potential energy of the planet satellite system

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GMm
PE =  .
r
ve sign indicates that the satellite is bounded by earth’s (planet's) gravitation. Instead of saying
gravitational potential energy of the system, we can say potential energy of the satellite for the sake of
simplicity.
Kinetic Energy
We know that the KE of a particle is equal to 1/2 mv 2. Therefore, KE of a satellite in the orbit can be given
2
2 GM  GM 
as KE = 1/2 mv , putting v0 = , we obtain, KE = 1/2 m  
0
r  r 
 
GMm
KE = .
2r

Total Energy
The total mechanical energy of the satellite is equal to the sum of its potential and kinetic energies.
 TE = PE + KE; putting the obtained values of PE and KE of the satellite we obtain,
 GMm   GMm  GMm
TE =    +   TE =  .
 r   2r  2r
Since, the total energy is negative, the satellite is bound by earth’s gravitational field.

Angular Momentum
The angular momentum of an earth satellite is given by
L = mvr.
GM
Putting v0 = , we obtain, L = m( GMr ).
r

Points to Remember
(a) The KE of the satellite is equal in magnitude to its total energy and half of its potential energy.
(b) If we want to remove the satellite from its orbit to infinity we have to impart an additional energy equal
GMm
to its total energy, that is , to increase its total energy to zero.
2r
(c) If the time period of the satellite is 24 hrs rotating in the same sense as the rotation of the earth and
the plane of the orbit is at right angle to the polar axis of the planet (earth), then the satellite will always be
above a certain place of the earth.
This kind of a satellite is called geostationary satellite.

Illustration 1: An artificial satellite of mass 100 kg is in circular orbit at 500 km above the earth’s
surface. Take the radius of the earth as 6.5  106 m.
(a) Find the acceleration due to gravity at any point on the satellite path.
(b) What is the centripetal acceleration of the satellite?

Solution: Here, h = 500 km = 0.5  106 m

R = 6.5  106 m
6 6 6
r = R + h = 6.5  10 + 0.5  10 = 7.0  10 m
2 2
 R   6.5  106  2
(a) g = g    9.8   = 8.45 m/s
R h  7.0  106
 
mv 2
(b) Centripetal force on the satellite, F =
r

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 
2
2 gR2 r
F v
 Centripetal acceleration, a =  
m r r
2 2
gR R
= 2 g  8.45 m / s 2
r  R  h  2

Kepler’s Second Law

Consider a planet P that moves in an elliptical orbit around the sun, P

and let P and P be the positions of the planet at time t and t + t  P

(where t is a very small time interval). If the angular displacement of S

the planet is , then the area swept out by the line joining the planet
and sun (SP) in time t is:
A = area of the section SPP
1
= r 2   ; where r = the length SP.
2
1 2
A 2 r  1 2
The areal velocity, v A =   r  = constant …(5)
t t 2
dA L
Areal velocity = = . . . (6)
dt 2m
This is the expression for the angular momentum of the planet,
L = I = mr2
 d 
= mr2   perpendicular to the plane of its orbit.
 dt 
The torque on the planet is zero,
    GMm  
i.e. r  F  r    2 rˆ   0 …(7)
 r 
Hence, the angular momentum of the planet does not vary, i.e. the areal velocity of the planet remains
constant. At its aphelion (farthest point from the sun, r is large), the planet moves slowly and at its
perihelion (nearest point from the sun, r is small) the planet moves fastest.

Illustration 2: Calculate the mass of the Sun from the following data; distance between the Sun and the
Earth = 1.49 x 1011 m, G = 6.67 x 1011 SI units and one year = 365 days.

mS mE
Solution: Force of attraction between the sun and the earth = G
d2SE
Considering the orbit of the earth as nearly circular, the centripetal force acting on the
2
earth is mE dSE .
m m
 mE dSE 2 = G S2 E
dSE
d3SE  2 42 d3SE
mS = 
G GT 2

 
3
4   3.14   1.49  1011
2
30
= = 1.97  10 kg.
  365  24  60  60 
11 2
6.67  10

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Illustration 3: A Saturn year is 29.5 times the earth year. How far is saturn from the sun (M) if the earth
is 1.5 x 108 km away from the sun?

Solution: It is given that

TS = 29.5 Te; Re = 1.5  1011 m
Now, according to Kepler’s third law
TS2 RS3

Te2 R3e
2 2
T  3 11  29.5Te 
3
RS = Re  S  = 1.5  10  
T
 e  Te 
12 9
= 1.43  10 m = 1.43  10 km

Illustration 4: A planet of mass m moves along an ellipse around the sun so that its maximum and
minimum distances from the sun are equal to R and r, respectively. Find the angular
momentum of this planet relative to the centre of the sun.

Solution: According to Kepler’s second law, the angular

v1
momentum of the planet is constant.
 mv1R = mv2r or v1R = v2r R S r
If the mass of the Sun is M, conserving total
mechanical energy of the system at two given v2
positions we have,
GMm 1 GMm 1
  mv12    mv 22
R 2 r 2
2 2 2 2 2
 1 1 v v  r  R  v1 v1 R
 GM     1  2 or GM    
R r  2 2  Rr  2 2r 2
2GM  R  r  r 2 2GMr
 v12  
 2
Rr R  r 2
 R R  r 

2GMRr
Now, angular momentum = mv1R = m
R  r 

SOLVED PROBLEMS
SUBJECTIVE
Problem 1: An artificial satellite is moving in a circular orbit around the earth with a speed equal to
half the magnitude of escape velocity from the earth.
(a) Determine the height of the satellite above the earth’s surface.
(b) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth,
find the speed with which it hits the surface of the earth.
2
(g = 10 m/s and RE = 6400 km.)

Solution: (a) We know that for satellite motion

GM g GM
vo = R [as g = 2 and r = R + h]
r  R  h  R
1 1
In this problem, vo = ve  2gR
2 2

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R2 g 1
So,  gR , i.e. 2R = h + R or h = R = 6400 km
Rh 2
(b) By conservation of ME,
 GMm  1 2  GMm 
0 +    mv    
 r  2  R 
1 1 
or v2 = 2GM    [as r = R + h = R + R = 2R]
 R 2R 
GM
or v =  gR  10  6.4  106 = 8 km/s.
R
Problem 2: Two satellites of same mass are launched in the same orbit around the earth so that they
rotate opposite to each other. If they collide in-elastically, obtain the total energy of the
system before and just after the collision. Describe the subsequent motion of the
wreckage.
GMm
Solution: Potential energy of the satellite in its orbit = 
r
| U | GMm
 K.E. = =
2 2r
where m is mass of satellite, M the mass of the earth and r the orbital radius.
GMm GMm GMm
Total energy of one satellite = K. E. + P.E. =  = 
2r r 2r
GMm
For two satellites, total energy E = 
r
Let v be the velocity after collision.
By conservation of momentum
 
mv1  mv 2  0  (m  m)v  v = 0
The wreckage of mass (2m) has no kinetic energy, but it has only potential energy. So,
GM(2m)
energy after collision =
r
The wreckage falls down under gravity.
Problem 3: A satellite of mass 2  103 kg has to be shifted from an orbit of radius 2R to another orbit
of radius 3R, where R is the radius of earth. Calculate the minimum energy required.
2
[R = 6400 km and g = 10 m/s .]

Solution: Total mechanical energy of a satellite in a circular orbit of radius r

1 GMm
=
2 r
1 GMm
E1 = energy in first orbit = 
2 2R
1 GMm
E2 = energy in second orbit = 
2 3R
1 GMm  1 1 
E, energy (minimum) required = E2  E1 =   
2 R  2 3
GMm mgR2 2
 E =  (GM = gR )
12R 12R
mgR 10
 E = = 1.067  10 J
12

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Problem 4: Two satellites A and B, of equal mass, move in the equatorial plane of earth close to the
earth's surface. Satellite A moves in the same direction as that of the rotation of the earth
while satellite B moves in the opposite direction. Determine the ratio of the kinetic energy
of B to that of A in the reference frame fixed to earth.

GM
Solution: The orbital speed of a satellite very close to earth = V0 =  g0 R
R
2
The peripheral speed of earth = Ve = Re = R
Te
 The velocities of the satellites relative to earth are given by
2 2
Vr  R  go R and Vr  goR  R
1
Te 2
Te
Positive and negative signs are for the satellites orbiting form east to west and west
to east respectively because earth rotates from west to east
2

1  2 
m Vr  Vr 
2  R  goR 
KE1  Te 
2

  2 
1
 1
=
KE 2 1  Vr   2 
2
m Vr  
  R  goR 
2
2
2 2

 Te 
6 2
Putting R = 6.4 x 10 m, g0 = 9.8 m/sec and Te = 86400 sec,
KE1
 1.265
KE 2

Problem 5: With what speed v 0 should a body be M

projected as shown in the figure, with respect
R
to a planet of mass M so that it would just be d
able to graze the planet and escape ? The m
radius of the planet is R. (Assume that the
A V0
planet is fixed).
D

Solution: By conservation of angular momentum we obtain v

mv0d = mvr
mv 0 d M r
B
 v= = v0 (d/r) m
mr R
By conservation of energy at A and B, we have d
1 GMm 1 GMm m
m v 20   mv 2 
2 D d
2 2 2 r A v0
When r  R we obtain, D
1 GMm 1 v 2 d2 GMm
mv 20   m. 0 2 
2 D2  d2 2 R R
1  d2  1 1 
 mv 02  2  1  GMm   
2 R  R D2  d2 
1 1 
2GM   
R D d 
2 2
 v0 =
 d2 
 2  1
 R 

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OBJECTIVE
Problem 1: A body of mass m starts approaching from far away towards the centre of a hypothetical
hollow planet of mass M and radius R. The speed of the body when it passes the centre
of the planet through its diametrical hole is
GM 2GM
(A) (B)
R R
(C) zero (D) none of these

Solution: (B)
At infinity, the total energy of the body is zero. Therefore, the total energy of the body just
before hitting the planet P will be zero according to the conservation of energy.
 Ep = E = 0
 Up + Kp = 0
GMm 1 2
   mv = 0
R 2
2GM
 v= .
R
Since the force imparted on a particle inside a spherical shell is zero, therefore the
velocity of the particle inside the spherical shell remains constant. Therefore, the body
2GM
passes the centre of the planet with same speed v = .
R

Problem 2: The energy required to shift a satellite from orbital radius r to orbital radius 2r is E. What
energy will be required to shift the satellite from orbital radius 2r to orbital radius 3r?
E
(A) E (B)
2
E E
(C) (D)
3 4

Solution: (C)
Energy required to shift the satellite from orbital radius r to orbital radius 2r,
 GMm   GMm 
E    
 4r   2r 
GMm  1 1 
or E  
r  2 4 
GMm
or E 
4r
Energy required to shift the satellite from orbital radius 2r to orbital radius 3r,
 GMm   GmM 
E'     
 6r   4r 
GmM  1 1 
or E'  
r  4 6 
GmM 1  GmM 
or E'   
12r 3  4r 
E
 E' 
3

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Problem 3: A satellite goes along an elliptical path around earth. The rate of change of arc length ‘a’
swept by the satellite is proportional to:
(A) r (B) r2
1/2
(C) r (D) r-1.

Solution: (D)
dA 1 2
 r   cons tan t = k
dt 2
1
 r(r)  k
2
 vr = 2k
 v  (1/r).

Problem 4: Imagine a light planet revolving around a very massive star in a circular orbit of radius R
with a period of revolution T. If the gravitational force of attraction between the planet and
-5/2 2
the star is proportional to R , then T is proportional to:
3
(A) R (B) R7/2
3/2
(C) R (D) R3.75

Solution: (B)
mR2 = Fgr
k
 mR2 =
R5 / 2
k
 2 =
mR7 / 2
 T2  R7/2

EXERCISE -3
1. Is it necessary for the plane of a satellite motion to pass through the centre of the earth?

2. A satellite is orbiting around earth. The centripetal force on the satellite is F. The gravitational
force of earth on the satellite is also F. What is the net force on the satellite?

3. If a graph is plotted between T2 and r3 for a planet, then what will be T

2

the value of the slope of the graph? (Letters have usual meanings.)

r3

4. Two satellites A and B revolve around a planet in coplanar circular orbit in the same direction with
period of revolutions 1 hour and 8 hours respectively. The radius of satellite A is 104 km then find
the angular speed of ‘B’ with respect to A ?

5. The radius of a planet is R1 and a satellite revolves round it in a circle of radius R2. The time
period of revolution is T. Find the acceleration due to the gravitation of the planet at its surface.

6. An artificial satellite, moving in a circular orbit around the earth, has a total energy (K.E. + P.E.)
E0. Find its potential energy.

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7. Two satellites are orbiting around the earth in circular orbits of same radii. One of them is 10
times greater in mass than the other. Their periods of revolution are in the ratio
(A) 100:1 (B) 1:100
(C) 10:1 (D) 1:1.

8. An earth satellite is moving round the earth in a circular orbit. For such a satellite, which of the
following statement is wrong?
(A) It is a freely falling body. (B) It is moving with a constant speed.
(C) Its acceleration is zero. (D) Its angular momentum remains constant.

9. A satellite of mass m is orbiting the earth in a circular path of radius r with velocity v. How much
energy is required to take the satellite from an orbit of radius r to 3r ? [M = mass of earth)
2GMm GMm
(A) (B)
3r 3r
GMm GMm
(C) (D)
r 2r

10. Two satellites of masses m1 and m2(m1 > m2) are revolving around the earth in a circular orbit of
radii r1 and r2(r1 > r2) respectively. Which of the following statements is true regarding their speed
v1 and v2?
(A) v1 = v2 (B) v1 > v2
(C) v1 < v2 (D) (v1/r1) = (v2/r2)

11. In an artificial satellite a space traveller tries to fill ink in pen by dipping it in ink. The amount of
ink filled in the pen as compared to the amount of ink filled on the earth’s surface will be
(A) less (B) more
(C) same (D) nil

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ANSWERS TO EXERCISE
Exercise -1
1. Yes, earth also moved towards the apple but with infinitesimally small
acceleration.
If mass of apple = m, m
GMe m
force of attraction = , and
Re2
Me
GMe m Gm
acceleration of earth =  0
MeRe 2 Re 2

2. Yes, he is right. Due to rotation of earth, he weighs less. Yes, at poles he can weigh exactly.

3g 2 2Gm
3.  4.
4GR e 2

GM  2 2  1 GM2
5.   6. v=
R  4  M  m  
7. A 8. B

9. D 10. C

11. D

Exercise -2

4G M1  M2 
1. (a) No. (b) No 2.
d

k  Gmme 
3. 4.2  105 m/s and 2.1  105 m/s 4.  
k 1  R 

 1 1 Gmme
5. v  2GM    6.
 R r0  8Re

 d  2R 
7. v= 2GM  .
 2dR 

2GMe
8. (i)
Re

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3RT
(ii) Vrms  .
M
2GMe
For lighter gases, Vrms is high while escape velocity is same for all gases, i.e. .
Re
GMe m
So, the lighter gas molecules may escape. Also, gravitational pull on the molecules is
R2e
near earth surface. Therefore, if m is small, gravitational force will be lesser.

(iii) If total mechanical energy is less than zero, projectile will come back to the earth’s surface.
If total mechanical energy is greater than zero, it means projectile has been given energy
more than binding energy and so it will escape and never return back.

9. B 10. A

11. C

Exercise -3
1. Yes, plane of satellite motion must pass through centre of earth; otherwise it will not be stable
in any other orbit because net force on it will not be directed towards centre of its orbit.

2. Net force on the satellite is F. In fact, the gravitational force of earth on the satellite provides it the
necessary centripetal force which makes it move in the circular orbit.

4 2
3. 4. /3 rad/hour
GM

4 2R32
5. 6. 2E0
T 2R12

7. D 8. C

9. B 10. C

11. B

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CONCEPTS AND FORMULAE AT A GLANCE

Gm
1. Potential due to a point mass m at a distance r is V = 
r

2. Gravitational Potential
GM M
(i) Outside the shell is : V =  (r > R)
r R
P
GM
(ii) On the surface of the shell is : V =  C r
R
GM
(iii) Inside the shell V = 
R

3. Potential due to the solid sphere

GM
(i) Outside the shell: V = 
r
3GM  R2 r 2 
(ii) Inside the shell: V =  3   
R  2 6
GM
(iii) On the surface: V = 
R

4. Potential due to a thin ring at a distance x. M

GM
V=  R
P
R2  x 2
C x

5. Gravitational potential energy of interaction of a system of two particles of masses m1 and m2

separated by a distance r
Gm1m2
U= 
r

GM
6. Field created by a point mass M at a distance r from the particle is | E | 
r2
 
7. A particle of mass m in a gravitation field EG experiences a force equal to F  mEG

 
GMx
8. Field at a distance x due to a circular ring, EG 
(R2  x 2 )3

9. Gravitational field due to a spherical shell M

GM R
EG = 2 (x > R) P
x
C X
EG = zero (x < R)

10. Gravitational field due to a solid sphere

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GM
EG = (x > R)
x2
GMx
EG = (x < R)
R3

11. The gravitational field intensity due to earth is called acceleration due to gravity.

12. Variation in the acceleration due to gravity: g

g0
x
(a) Inside the earth: g = g0 (x is distance from the centre of the earth) Inside
Outside
Re
2
R 
(b) Outside the earth: g = g0  e  Re x
 x 

13. Variation of gravity with latitude 

g = g0  R 2 cos2  ; g0 is the acceleration due to gravity at that place in absence of spinning.
2
 ; T=24 hrs
T

14. Motion of Planets and Satellites in Circular Orbits

v
GMm  v2 
F= = m   r
r2  r 
GM Mr
Speed of Satellite, v 
r

2r 2r 2
15. Time period of revolution; T   = r3 / 2
v GMe / r GMe
2GMe
Escape velocity: v e 
R
*In case of problems relating to elliptical orbits, conservation of angular momentum and
conservation of mechanical energy should be used.

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CHAPTER PRACTICE PROBLEM

1. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is
1.50 × 108 km away from the Sun?

2. The escape speed of a projectile on the Earth’s surface is 11.2 km s1. A body is projected out
with thrice this speed. What is the speed of the body far away from the Earth ? Ignore the
presence of the Sun and other planets.

3. If the Earth were a perfect sphere of radius 6.37 × 106m, rotating about its axis with a period of
4
one day (= 8.64 × 10 s), how much would the acceleration due to gravity (g) differ from the poles
to equator ?

4. Assuming the Earth to be a sphere of uniform mass density. How much would a body weigh half
way down to the centre of the Earth if it weighed 250 N on the surface ?

5. A rocket is fired vertically with a speed of 5 km s1 from the Earth’s surface. How far from the
Earth does the rocket go before returning to the Earth ? Mass of the earth = 6.0 × 1024 kg; mean
radius of the radius of the Earth = 6.4 × 106 m; G = 6.67 × 1011 N m2 kg2.

6. The ratio of KE of a planet at the point 1 and 2 is :

2 2
r  r  1 r1 r2 2
(A)  1  (B)  2 
 r2   r1  m m
r r2
(C) 1 (D)
r2 r1

7. A planet has twice the values of mass and radius of earth, acceleration due to gravity on the
surface of the planet is
(A) 9.8 m/s2 (B) 4.9 m/s2
2
(C) 980 m/s (D) 19.6 m/s2

8. If the change in the value of ‘g’ at a height h above the surface of the earth is the same as at a
depth x below it, then (both x and h being much smaller than the radius of the earth):
(A) x = h (B) x = 2 h
(C) x = (1/2)h (D) x = h2

9. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly
the same as on the surface of earth. Its radius in terms of radius of earth R will be
(A) R/4 (B) R/2
(C) R/3 (D) R/8

10. Inside a uniform spherical shell :

(A) Potential energy is zero (B) Field is non-zero
(C) Potential is constant (D) none of these

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11. The magnitude of the gravitational field at distance r1 and r2 from the centre of a uniform sphere
of radius R and mass M are f 1 and f 2 respectively then
f r2 f r
(A) 1  22 , if r1, r2 > R (B) 1  2 , if r1, r2 < R
f2 r1 f2 r1
f1 r12
(C)  , if r1, r2 > R (D) None of these
f2 r22

12. Two satellites A and B go round a planet in circular orbits having radii 4 R and R respectively. If
the speed of satellite A is 3V, the speed of satellite B will be
(A) 12 V (B) 6V
(C) 4V/3 (D) 3V/2

13. The distance from the surface of the earth at which the acceleration due to gravity is the same
below and above the surface of the earth.
 5  1
(A) 
5  1 Re  (B)   Re
 2 
 
5  1 Re
(C)
2
(D)  
5  1 Re

14. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential)
energy E0. Its potential energy is
(A) – E0 (B) 1.5 E0
(C) 2E0 (D) E0

15. Two identical particles each of mass M and distance d apart are released from rest and can move
forward to each other under the influence of their mutual gravitation forces. Speed of each
particle, when the separation reduces to half of initial value equals
GM 2GM
(A) (B)
d d
15.

GM
(C) (D) none of these
2d
16.

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ASSIGNMENTS
SECTION-I
1. Can an artificial satellite be put into orbit in such a way that it will always remain directly over New
Delhi?

2. How much quicker than at present must the earth revolve on its axis to make bodies at the
equator experience weightlessness? What will be the duration of day then?

3. How will weight of a body change at a height equal to earth’s radius?

 
4. The gravitational field in a region is given by E  (2 i  3 j ) N/kg. Show that no work is done by
the gravitational field when a particle is moved on the line 3y + 2x = 5.

5. A body is weighed by a spring balance to be 1.000 kg at the north pole. How much will it weigh at
the equator? Account for the earth’s rotation only.

6. Find the gravitational force of attraction between a particle of mass m and a uniform slender rod
of mass M and length L for the two orientations shown in the figure below.
m
M
m d
d
M
(a)
(b)

7. Two bodies of masses m1 and m2 are connected by a long

inextensible cord of length L. The combination is allowed to
m1 m2
fall freely towards the earth (mass me), the direction of the
O
cord being always radial as shown in the figure. Find: (a)
h L
the tension in the cord and (b) the accelerations of m1 and me
m2. Does the cord ever become slack? (ignore the
gravitational interaction between m 1 and m2) Earth

8. A body is thrown up (radially outward from the surface of the earth) with a velocity equal to one-
fourth of the escape velocity. Find the maximum height reached from the surface of the earth.
(Radius of earth is Re)

9. A satellite is put in an orbit just above the earth's atmosphere with a velocity 1.5 times the
velocity for a circular orbit at that height. The initial velocity imported is horizontal. What would be
the maximum distance of the satellite from the surface of the earth when it is in the orbit?

10. Two concentric shells of masses M1 and M2 are situated as shown in M1

B
Figure. Find the force on a particle of mass m when the particle is located M2
at (a) r = a (b) r = b and (c) r = c. O A
The distance r is measured from the centre of the shell. C r
Given that OA = a, OB = b, OC = c

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11. Does the escape speed of a body from the earth depend on: (a) the mass of the body, (b) the
location from where it is projected, (c) the direction of projection, and (d) the height of the location
from where the body is launched? Explain your answer.

12. A comet orbits the sun in highly elliptical orbit. Does the comet have a constant (a) linear speed,
(b) angular speed, (c) angular momentum, and (d) kinetic energy when it comes very close to the
sun?

13. The gravitational potential at an internal point inside a solid sphere is known to
GM
be Vr   3 (3a2  r 2 ) . Find the magnitude of gravitational field at that point.
2a
   
14. The gravitational field in a region is given by E  (10 N/ kg)( i  2 j  3k) . Find the work done by an
external agent to slowly shift a particle of mass 3 kg from the point O  (0, 0, 0) to a point
P  (3m, 4m, 5m).

15. The centres of two stars of masses M and 16M of 16M

M
radii R and 2R are separated by a distance of 10R.
Draw a graph between gravitational potential (V) and 2R
R
the distance (r) from the centre of the smaller star
along the line joining their centres for R  r  8R.
r
10Rr

SECTION-II
OBJECTIVE
(MULTI CHOICE SINGLE CORRECT)
1. Dimensional formula of ‘G’ is
(A) M1L3T2 (B) M2L1T2
(C) M1L2T2 (D) M1L2T1

2. The ratio of the radius of earth to that of the moon is 10. The ratio of the acceleration due to
gravity on the earth to that on the moon is 6. The ratio of the escape velocity from the earth’s
surface to that from the moon will be
(A) 4 (B) 6
(C) 7.75 (D) 12

3. The earth revolves round the sun in an elliptical orbit. Its speed
(A) goes on decreasing continuously (B) is greatest when it is closest to the sun
(C) is greatest when it is farthest from the sun (D) is constant at all the points on the orbit

4. An earth satellite has a mass M and angular momentum L. Its areal velocity is given by
(A) (L/M) m/s (B) (2L/M) m2/s
2
(C) (L/2M) m /s (D) (L/2M) m/s

5. Let V and E be the gravitational potential and gravitational field at a distance r from the centre of
a uniform spherical shell. Consider the following two statements:
(a) The plot of V against r is discontinuous, and (b) The plot of E against r is discontinuous.
Now, select the correct option from the following.
(A) Both (a) and (b) are correct. (B) (a) is correct but (b) is incorrect.
(C) (b) is correct but (a) is incorrect. (D) Both (a) and (b) are incorrect.

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6. If a body is projected with a speed less than the escape velocity,

(A) it must reach a certain height and may fall down following a straight path
(B) it must reach a certain height and may fall down following a parabolic path
(C) it may orbit the earth in a circular or elliptical orbit
(D) it must orbit the earth in a circular orbit

7. If the radius of the earth shrinks by 1%, its mass remaining same, the acceleration due to gravity
on the surface of the earth will
(A) decrease by 2% (B) decrease by 0.5%
(C) increase by 2% (D) increase by 0.5%

8. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is
launched into an orbit of radius 1.01R. The time period of the second satellite is larger than that of
the first one by approximately
(A) 0.5% (B) 1.5%
(C) 1% (D) 3.0%

9. The time period of revolution of a satellite is T. The kinetic energy of the satellite is proportional to
(A) T (B) T2
(C) T 3
(D) T2/3

10. A double star is a system of two stars (say having masses m1 and m2) moving around the centre
of inertia of the system due to gravitation. Then, ratio of sweeps of area of star of mass m 1 to the
star of mass m2 is
(A) m1 / m2 (B) m2/m1
(C) m12 / m22 (D) m22 / m12

11. R is the radius of earth,  is its angular velocity and gP is the value of g at poles. The effective
value of g at the latitude   60 will be equal to
1 3
(A) gP  R2 (B) gP  R2
4 4
1
(C) gP  R 2
(D) gP  R2
4

12. When a body is taken from equator to the poles, its weight
(A) remains the same
(B) increases
(C) decreases
(D) increases at north pole and decreases at south pole

13. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly
the same as on the surface of the earth. Its radius in terms of radius of the earth R will be
(A) R/4 (B) R/2
(C) R/3 (D) R/8

14. If a satellite be rotating about a planet, which of the following is true?

[U = Potential energy of planet satellite system, K = kinetic energy of satellite,
T = Total Mechanical energy]
(A) | U | = K = | T | (B) K = | T | = | 2U |
(C) 2K = | T | = | U | (D) K = | T | = |1/2 U |

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15. A person sitting in a chair in a satellite feels weightless because

(A) the earth does not attract the object in a satellite
(B) the normal force by the chair on the person balances the earth’s attraction.
(C) the normal force is zero
(D) the person in satellite is not accelerated.

MULTI CHOICE MULTI CORRECT

1. In case of earth:
(A) Potential is minimum at the centre of earth.
(B) Potential is same, both at centre and infinity but not zero.
(C) Potential is zero, both at centre and infinity
(D) Field is zero, both at centre and infinity

2. Two satellites of same mass of a planet in circular orbits have periods of revolutions 32 days and
256 days. If the orbital radius of the first is R, then
(A) the kinetic energy of the second is less than that of the first
(B) the total mechanical energy of the second is greater than that of the first
(C) Radius of the orbit of second is 4R
(D) Radius of the orbit of second is 8R

3. Gravitational field due to a point mass is

(A) A central field (B) always pointed towards the mass
(C) an inverse square field (D) a conservative field

4. Two satellite A and B move around the earth in a circular orbit. The mass of B is twice the mass
of A then
(A) speeds of A and B are equal (B) kinetic energy are equal
(C) kinetic energy of B is greater than A (D) kinetic energy of A is greater than B

5. Which of the following statement(s) is/are correct?

(A) An astronaut going from Earth to Moon will experience weightlessness once.
(B) When a thin uniform spherical shell gradually shrinks maintaining its shape, the gravitational
potential at the centre increase.
(C) In the case of spherical shell, the plot of potential versus distance from centre is continuous.
(D) In the case of spherical shell, the plot of gravitational field intensity I versus distance from
centre is continuous.

NUMERICAL BASED TYPE

1. Find the ratio of the kinetic energy required to be given to the satellite to escape earth’s
gravitational field to the KE required to be given so that the satellite moves in a circular orbit just
above earth’s atmosphere.

2. A mass equal to the mass of the earth is to be compressed in a sphere in such a way that the
8
escape velocity from its surface is 3  10 m/s. What should be the radius of the sphere in
13 2 -1
mm?(take GME = 40.5  10 Nm kg )

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NUMERICAL BASED WITH NON-NEGATIVE INTEGER TYPE

1. A thin spherical shell of total mass M and radius R is held fixed. m
There is a small hole in the shell. A particle of mass m is released
from rest at a distance R from the hole as shown in the figure. This
particle subsequently moves under gravitational force of the shell.
R
kR3
It travel from the hole to the point diametrically opposite is .
GM
Find the value of k. M

2. A satellite is moving around the sun in an elliptical orbit. ‘M’ is A

the mass of sun and ‘a’ is semi major axis of elliptical orbit.
kGM
The velocity of satellite when it reaches at A is . Find
a B D
the value of k. sun

NUMERICAL BASED DECIMAL TYPE

1. A solid sphere of mass M and radius R is fixed in free space, the
particle is released from the surface of sphere which is initially in rest M
as shown in the figure and surface of tunnel is frictionless. The
velocity of particle when it comes at point P in the tunnel is m C R
GM R/2
K m/s. Find the value of K.
R P

2. Assuming the Earth to be a sphere of uniform mass density. How much would a body weigh (in
newton) half way down to the centre of the Earth if it weighed 250 N on the surface?

NUMERICAL BASED (QUESTION STEM) DECIMAL TYPE

gRe
A body is thrown from the surface of earth with velocity at some angle from the vertical. If the
2
Re  X
. The angle of projection from the vertical is sin1 
 4 
maximum height reached by the body is .
4  
gRe
The velocity of the body at maximum height is v  . (Re = Radius of earth)
Y

1. The value of X is ………….

2. The value of Y is…………..

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LINKED COMPREHENSION TYPE

(I)
Artificial satellites orbiting the earth are a familiar fact of contemporary life. They have a wide area of
application, ranging from communication to weather forecast. So, study of artificial satellites is very
important. The questions we would like to probe are how do they stay in orbits and what determines the
properties of their orbits.
Launching a satellite is similar to the projection of a projectile from a height. The satellites are carried by
rocket to the pre calculated height and is projected with a definite velocity. The satellite then comes under
the effect of gravity which provides the centripetal force for the satellite to be in its orbit
A
The figure shows the trajectories of the orbits of satellites launched from a
h
height h with different initial velocities. Velocities increase from (1) to (6). 1
5
6
2
Trajectories (1) to (4) are closed orbits whereas (5) and (6) are open orbits.
Trajectory (3) is the simplest case. EARTH
RE 3
We would deal only with the circular trajectories. The only force acting on a 4
r
satellite is the earth’s gravitational force. We can find out the velocity of
projection for circular orbit frame.
GMEm mv 2

r2 r
GME
or v . . . (1)
r
2 R  h 
The time period of a satellite for a circular orbit is given by T  , where h is the height at which
v
the satellite is placed and v its velocity.
The mechanical energy E of the satellite always remains constant (since the satellite’s mass is much
smaller than earth’s mass. We assign U, K and E to represent potential energy, kinetic energy, and total
energy, respectively, for the earth–satellite system.) The potential energy of the system is given by
GMEm
U . . . (2)
r
1
and K  mV 2 . . . (3)
2
so, the total mechanical energy is given by
GMEm
E K U   . . . (4)
2r
where r = R + h
With the above theory, we can calculate the energies and time periods of satellites.
For all calculations in the questions given below take radius of earth RE = 6400 km, mass of earth
mE  5.98  1024 kg, (24 hours)2/3 = 1954 sec

1. Satellites move in closed orbits which means that

(A) the magnitude of potential energy is always greater than that of kinetic energy
(B) the magnitude of potential energy is always equal to that of kinetic energy
(C) the magnitudes of potential energy and total energy are equal
(D) the magnitudes of kinetic energy and the total energy are never equal

2. A ball of mass 7.2 kg is launched from the earth’s surface to a circular orbit at a height of 350 km
from the surface. In doing so, what is the change in mechanical energy of the ball?
(A) 250 MJ (B) 317 MJ
(C) 213 MJ (D) zero

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3. A satellite is orbiting close to the surface of earth. Then, its time period is
R 2R
(A) 2 (B) 2
g g
R 4R
(C) 2 (D) 2
2g g

(II)
Weight of a body depends directly upon acceleration due to gravity g. Value of g depends upon many
factors. It depends upon the shape of earth, rotation of earth etc. Weight of a body at a pole is more than
that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to
gravity g varies with latitude  as per relation given below:
2 2
grot = g  R cos  where R is radius of earth and  is angular velocity of earth. A body of mass m weighs
W r in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It
is observed that weight W m of the same body in the moving train is different from W r. Let ve be the velocity
of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the
relative velocity between earth and train will affect the weight of the body.
Now answer the following questions:
1. Difference between Weight W r and the gravitational attraction on the body can be given as
mv 2 1 mv 2
(A) (B)
R 2 R
1 mv 2e mv 2e
(C) (D)
2 R R

2. Weight W m of the body can be given as

(v e  v)2 (v e  v)2
(A) mg  m (B) mg  m
R R
m 2 (v  v)2
(C) [v e  (v e  v)2 ] (D) mg  m e
R R
(III)
Consider a satellite ‘S’ which is just above the Brazil and Brazil
rotating on equatorial plane of earth. The sense of rotation of
earth and satellite are similar means both are rotating from west
to east. The time period of satellite is 6 hours. At t = 0, satellite
is just above the Brazil. Now choose the correct option(s).
S
Earth
r0
1. The time after which the satellite will be above the Brazil again is
(A) 4 hours (B) 8 hours
(C) 12 hours (D) 16 hours

2. The height of geostationary satellite above the sea level is ……… (approximately).
(A) 15,800 km (B) 20,800 km
(C) 25,800 km (D) 35,800 km

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MATCH LIST TYPE

Answer questions 1 and 2 by appropriately matching the list based on the information given in the
paragraph.

A, B, C and D are four solid spheres. The mass density of solid spheres are , 2, 3 and 4
respectively. The radius of sphere A and C is R and for sphere B and D radius is 2R. List-I gives
the above for sphere while List-II lists the magnitude of some quantity.
List – I List – II

(I) Sphere A (P) 2

(II) Sphere B (Q) 1/2

(III) Sphere C (R) 16

(IV) Sphere D (S) 8

(T) 3/2

(U) 4

1. It is found that magnitude of gravitational field on the surface of sphere ‘A’ is E0. The correct
match for the magnitude of gravitational field at half of the radius in E0 units will be
Options
(A) I  Q, II  P, III  T, IV  U
(B) I  P, II  Q, III  T, IV  R
(C) I  R, II  S, III  U, IV  T
(D) I  S, II  Q, III  R, IV  T

2. It is found that gravitational potential on the surface of sphere A is V0. The correct match for
gravitational potential at a distance 2R from the centre of respective sphere in units of V0 will be
Options
(A) I  P, II  Q, III  S, IV  T
(B) I  Q, II  S, III  U, IV  P
(C) I  Q, II  S, III  T, IV  R
(D) I  U, II  P, III  Q, IV  R

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2R
3. Density of a sphere of radius R with concentric spherical cavity of radius r = is given by,
3
0  2R 
 (R-x), where x is distance from its centre and   x  R  , then
R  3 
List –I List-II

a0 GR 2
(P) If potential on its outer surface then a = (1) 11
81
b0 GR2
(Q) Potential at its inner surface then b = (2) 0
81
c0 GR
(R) Field on its surface , then c = (3) 12
81
R d0 GR
(S) Field at a distance from its centre , then d = (4) 14
2 81
Code:
P Q R S
(A) 1 4 1 2
(B) 2 3 3 4
(C) 1 3 1 2
(D) 2 3 3 1

4. Match the following.

List–I List-II
Separation from centre on axis of uniform ring radius R, having
(P (1) 2R
maximum gravitation field
Maximum height attained by the particle above earth surface
(Q)
 2GM  R
when projected up with velocity   where M and R are (2)
 3R  2
mass and radius of earth
(R)
Separation of centre of mass of one half of a uniform spherical R
(3)
shell of radius R from the centre of shell 2

Separation ‘S’ of a point on uniform rod of length R suspended

(S) freely from a fixed support at one end such that the total 2R
(4)
momentum of a particle hitting the rod normally at P and the rod 3
remains conserved.
Code:
P Q R S
(A) 3 1 2 4
(B) 1 4 2 3
(C) 4 1 2 3
(D) 2 3 3 4

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MATCH THE FOLLOWING TYPE

This question contains statements given in two columns which have to be matched. Statements (A, B, C,
D) in column I have to be matched with statements (p, q, r, s, t) in column II.

1. Match the following:

Column – I Column – II

Magnitude of gravitational field due to a r=R

(A) (p) r
spherical shell of mass M and radius R

Magnitude of gravitational field due to a solid r=R

(B) (q)
sphere of mass M and radius R r

(C) Gravitational potential V for a solid sphere (r) o

r=R r

(D) Gravitational potential V for a spherical shell (s) o

r=R r

(t) o
r=R r

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MATCHING TYPE WITH 3 COLUMNS & 4 ROWS

The following table has 3 columns and 4 rows. Based on table, there are THREE questions. Each
question has FOUR options (A), (B), (C), and (D). ONLY ONE of these four options is correct

Column 1 Column 2 Column 3

Uniform mass distribution of Represents potential Magnitude of gravitational
different geometry are shown in due to the given system field due to the given
figure at point ‘P’ system at the same point
‘P’

Point P lies on axis of ring.

GM GM
(I) P (i) (P)
R 2R 2 2R2

Ring (M, R)

GM GM
(II) (ii) (Q)
P R 2R2
Hemispherical
shell (M, R)

3GM 3GM
(III) (iii) (R)
P 2R 2R2
Solid hemisphere
(M, R)

GM GM
(IV) P (iv) (S)
2R 2R2
Quarter part of
spherical shell (M, R)

1. Which one of the following options is the correct combination?

(A) (IV) (i) (P) (B) (II) (ii) (S)
(C) (III) (iii) (P) (D) (IV) (ii) (S)

2. Which one of the following options is the correct combination?

(A) (I) (ii) (P) (B) (II) (ii) (Q)
(C) (III) (iii) (P) (D) (IV) (iv) (R)

3. Which one of the following options is the correct combination?

(A) (II) (i) (P) (B) (II) (ii) (Q)
(C) (III) (iii) (R) (D) (IV) (iv) (R)

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ASSERTION-REASONING TYPE
This question contains statement-1 (Assertion) and Statement-2 (Reason). Question has 4 choice (A),
(B), (C) and (D) out of which only one is correct.

1. STATEMENT -1
For an elliptical orbit of a planet around sun the total energy of the system is negative.
because
STATEMENT-2
Total energy is always positive.
(A) Statement-1 is true, Statement -2 is true, Statement -2 is a correct explanation for
statement-1.
(B) Statement-1 is true, Statement -2 is true, Statement -2 is not a correct explanation for
statement-1.
(C) Statement-1 is true, Statement -2 is false.
(D) Statement-1 is false, Statement -2 is true.

2. STATEMENT -1
A planet may orbit around a star either in orbit P or orbit Q. The speed of planet will not be same
for both orbits.
because Q

STATEMENT-2
Planets orbit around a star with uniform velocity.
P
(A) Statement-1 is true, Statement -2 is true, Statement -2 is a correct explanation for
statement-1.
(B) Statement-1 is true, Statement -2 is true, Statement -2 is not a correct explanation for
statement-1.
(C) Statement-1 is true, Statement -2 is false.
(D) Statement-1 is false, Statement -2 is true.

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SECTION-III
(MULTI CHOICE SINGLE CORRECT)

1. An artificial satellite of the earth release a package. If air resistance is neglected, the point where
the package will hit (w.r.t. the position at the time release) will be
(A) ahead (B) exactly below
(C) behind (D) it will never reach the earth
2. A system consists of n identical particles of mass m placed rigidly on the vertices of a regular
polygon with each side of length a. If K1 be the kinetic energy imparted to one of the particles so
that it just escapes the gravitational pull of the system and thereafter kinetic energy K2 is given to
the adjacent particle to escape, then the difference (K1  K2) is
nGm2 Gm2
(A) (B)
a na
 n  Gm
2
Gm2
(C)   (D)
 n  1 a a

3. A particle of mass M is projected from point P such that it grazes the P

surface of earth and escapes from earth’s gravitational field. The M1
total energy possessed by the particle at any instant is
GMm GMm Re
(A) (B) Me
Re Re
(C) zero (D) none of these

4. There is no atmosphere on the moon because:

(A) it is closer to the earth.
(B) it revolves round the earth.
(C) it gets light from the sun.
(D) the escape velocity of gas molecules is less than their root mean square velocity on the
moon.

5. A spherical mass is scooped out from a solid of radius R. The distance

between the cavity and centre of sphere is x. The gravitational field
inside the cavity is R
x
(A) zero
(B) uniform in direction and constant in magnitude
(C) non-uniform
(D) uniform in direction and variable in magnitude

6. Two isolated point masses m and M are separated by a distance . The moment of inertia of the
system about an axis passing through a point where gravitational field is zero and perpendicular
to the line joining the two masses, is
m2  M2 m M
(A) 2 (B) 2
( m  M) 2
( m  M)
mM
(C) 2 (D) none of the above
( m  M)2

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7. The mean radius of earth is R, its angular speed on its own axis is , and the acceleration due to
gravity at earth’s surface is g. The cube of the radius of the orbit of a geostationary satellite will
be
2 2
(A) g/ (B) R g/
2 2
(C) gR / (D) R22/g

NUMERICAL BASED TYPE

1. Inside an insulated fixed share of radius R and uniform density , there is a

R
R
spherical cavity of radius such that the surface of the cavity passes B
2 A
through the centre of the sphere as shown in the figure. A particle of mass R/2
m0 is released from rest at centre B of the cavity. Calculate velocity with
P
which particle strike the center A of the sphere. If your answer is GR2 ,
Q
find the value of P + Q. (neglect earth’s gravity)

2. The escape velocity of a body on an imaginary planet which is thrice the radius of the earth and
k
double the mass of the earth is v 0 . Find the value of k. (v0 is the escape velocity of earth)
3

NUMERICAL BASED DECIMAL TYPE

1. A solid sphere of uniform density and radius R applies a gravitational

force of attraction equal to F1 on a particle placed at a distance 2R from
R/2 A
the centre of the sphere. A spherical cavity of radius R/2 is now made in
the sphere as shown in the figure. The sphere with cavity now applies a R
F 70
gravitational force F2 on the same particle. The ratio 2  . Find the
F1 k 2R
value of k.

2. A small ball of mass ‘m’ is released at a height ‘R’ above the earth
surface, as shown in the given figure. The maximum depth of the
R
ball to which it goes is R/2 inside the earth through a narrow groove m R M
before coming to rest momentarily. The groove, contains an ideal K
spring of spring constant K and natural length R. The value of K, if
GMm
R is the radius of earth and M mass of earth is . Find the
10R3
value of .

3. Consider a planet whose density is half of the earth and its radius is double of the earth. What will
be the time period (in sec) of oscillation of the pendulum on this planet if it is a second pendulum
of the earth.

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ANSWERS TO CHAPTER PRACTICE PROBLEMS

1. 1.43 × 1012 m

2. 31.7 km/s

3. gpole – gequator = 3.37 × 102 ms2

4. 125 N

5. 8.0 × 106m from the earth’s centre

6. B

7. B

8. B

9. B

10. C

11. A

12. B

13. C

14. C

15. A

IITJEE-2223-PHYSICS-GRAVITATION
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ANSWERS TO ASSIGNMENTS
SECTION-I
1. No. 2. 17 times, 1.412 hrs

3. one-fourth 5. 0.997 kg

GmM 2GmM
6. (a) , (b) 
d(d  L) d L2  4d2

Gmem1m2  1 1 
7. Tension in the cord =   
m1  m2  (h  R) 2 2
(h  R  L) 

Gme  m1 m2 
Acceleration a1 = a2 =   
m1  m2  (h  R) 2 2
(h  R  L) 
towards the centre of the earth. The cord will always be in tension.

Re
8. 9. 2R
15

Gm GM2 m
2  1
10. (a) M  M2  (b) (c) zero
a b2

11. Depends on the height of location from where body is launched.

12. Only the angular momentum is constant.

GM
13. r toward the centre of solid sphere. 14.  780 J.
a3

15.
R 2R 8R

2 .5 G M

R O
25G M
 O
9R

8 .1 G M

R

IITJEE-2223-PHYSICS-GRAVITATION
48

SECTION-II
OBJECTIVE
(MULTI CHOICE SINGLE CORRECT)
1. A 2. C 3. B 4. C
5. C 6. C 7. C 8. B
9. D 10. D 11. A 12. B
13. B 14. D 15. C
MULTI CHOICE MULTI CORRECT
1. A, D 2. A, B, C 3. A, B, C, D 4. A, C
5. A, C
NUMERICAL BASED TYPE
1. 2 2. 9
NUMERICAL BASED WITH NON-NEGATIVE INTEGER TYPE
1. 4 2. 1
NUMERICAL BASED DECIMAL TYPE
1. 12.26 2. 0.75
NUMERICAL BASED (QUESTION STEM) DECIMAL TYPE
1. 5 2. 10
LINKED COMPREHENSION TYPE
(I) 1. A 2. D 3. A
(II) 1. D 2. A
(III) 1. B 2. D
MATCH LIST TYPE
1. A 2. C 3. A 4. D
MATCH THE FOLLOWING TYPE
1. (A) (s) (B) (r) (C) (p) (D) (q)
MATCHING TYPE WITH 3 COLUMNS & 4 ROWS
1. D 2. B 3. C
ASSERTION-REASONING TYPE
1. C 2. C
SECTION-III
(MULTI CHOICE SINGLE CORRECT)
1. D 2. D 3. C 4. D
5. B 6. A 7. C
NUMERICAL BASED TYPE
1. 5 2. 2
NUMERICAL BASED DECIMAL TYPE
1. 90.00 2. 70.00 3. 2.00

IITJEE-2223-PHYSICS-GRAVITATION