GRAVITATION

5.1. INTRODUCTION: motion which accurately described the motions of

From the study of lower classes we are aware planets around the sun. These laws were formed on
of the action of gravitation. For example, activities like the basis of the famous Newton’s law of Universal
walking on a road, throwing an object etc, are gravitation.
influenced by the very interesting phenomenon called The first systematic study of gravitation was made
gravitation.Gravitation is one of the fundamental forces by Galileo, who was unaware of the basic laws
of nature. The motion of the earth, other planets and governing that phenomenon.
the moon has been a subject of greater interest for a In the year 1665 Issac Newton focussed his
long time. Though weakest among the fundamental attention on the motion of the moon about the earth.
forces, it has its major role on astronomical scales e.g Newton made several assumptions which later proved
the birth of a star and evolution of entire universe etc. to be very interesting turning points in physics. Netwon
Aryabhatta, a famous indian astronomer suggested that discovered the universal law of gravitation which is
the earth is a solid sphere and it spins around itself. applicable to all material bodies in the universe.
The early theory i.e. Geo centric theory (which In this topic we discuss gravitation at the terrestrial
was universally accepted for more than 14 centuries) level by considering Newton's law of gravitation. A very
was proposed by a Greek astronormer Cladius brief idea of Einstien's general theory of relativity, which
Ptolemy. In this theory the earth was assumed to be at is the modern theory of gravitation is discussed. The
the centre of the universe and the Sun , all the planets gravitational force is also responsible for one of nature's
along with the Moon revolve around the earth in circular strongest objects called blackhole. We discuss about a
orbits. black hole in brief. Then we discuss about the reference
frames and principle of equivalnce. We conclude this
In the 16 th century, an astronomer Nikolaus
topic with the explanations about escape,and orbital
Copernicus suggested that all the planets move around
velocities and geo stationary satellit es and
the sun in circular orbits and proposed Heliocentric
weightlessness.
theory.
5.2. BASIC FORCES IN NATURE :
Copernicus also believed that the earth rotates
about its axis once everyday.Copernicus theory was We know that there exits a gravitational field
not readily accepted by the scientists of the period. around mass. Similarly an electric field exists around a
Tycho brahe a Danish astronomer made very careful charge and magnetic field around a magnetic pole. It is
and accurate measurements of the motion and H.Kepler in these fields, that, there develops an interaction (force)
(15th century AD) studied the planetary motion in great between two bodies. Thus we have the gravitational
detail. force, electromagnetic force etc. in nature. These forces
Kepler formulated three laws of planetary come into play due to the exchange of field particles
electrons to huge planets and stars. position vector r12 In the vector form
5.4 The Law of Universal Gravitation
Newton’s law of universal gravitation can be stated as r12
follows. F12 F21
m1 m2

Direction of gravitational force

 Gm1m2 
F12  2
r12
r12
where r12 is the unit vector in the direction of r12 and
Every particle in the universe attracts every r12 is the distance between the two bodies.
other particle with a force that is directly   Gm m  
proportional to the product of their masses and is  F12   1 2
 r12 ...........(12.1)
inversely proportional to the square of distance  r12 3 
between them and this force acts along the line By Newton’s third law of motion, the force of attraction
joining the two particles. of the second body on the first body is equal and
If m1,m2, are the masses of the particles separated by a opposite to the force of attraction of the first body on
distance r, the magnitude of the gravitational force is  
given by the second body i.e., F12  F21 .
Hence, gravitational force is a mutual force.
F  m1m2
5.1. Note :
1
F i) In case a system consists of many masses m1,
r2 m2, ..... the resultant force on any one of the masses m1
Gm1m2 due to remaining masses, m2, m3, ..... can be found
F from superposition principle.
r2    
Here G is constant called the universal gravitational  F1  F12  F13  F14 
constant. The value of G = 6.67 x 10-11 N m2/kg2. 5.5 Characteristics of Gravitational Force
Unit of G: SI unit: N m2/kg2, CGS unit: dyne cm2/g2. 1. This force depends upon the product of masses
Dimensions of G: 2. This force obeys inverse square law
 MLT 2 L2  3. F - d2 graph isForce
a rectangular hyperbola as shown.
Fr 2    M 1L3T 2 
G ;[G ]  
m1m2    
 MM 
f
Universal gravitational constant (G) does not
depend upon the medium between the particles or
any other factors. It is constant every where in the Distance2
universe. So it is called universal gravitational constant. d2
4. The gravitational force between two bodies
While calculating the gravitational force of attraction forms an action - reaction pair which are equal in
between two bodies we have to take ‘r’ as the distance magnitude and opposite in direction. This force is mutual
between their centres of mass. and internal.
The direction of gravitational force is along the line joining 5. Gravitational force acts along the line joining the
the particles or the centres of mass of the bodies. The two interacting bodies. It is a central force. This force
force of attraction on the first body of mass m1 due to exhibits no angular dependence. Its magnitude depends
the second body of mass m2 is in the direction of the
on r . We say that gravitational force posseses spherical
 B
G  m 2m 4Gm 2
FB  
FA FB

30 0
30 0
 a 
2
a 2 along OB
 
 2 
FR
600 600 G  m  4m 8Gm 2
A C FD 
 a 
2

a 2 along OD.
 
   2 
solution : FA = FB , F = 2FAcos  
2
Their resultant force,
Gm 3 2
Gm 2
 4Gm 2
=2 2
 3 2 and    300 FDB = FD – FB = along OD.
a 2 a 2 a2
* Problem – 5.5 : Four particles of masses m, 2m, The resultant of FAC and FDB is of magnitude
3m and 4m are placed at the corners of a square
FR =  FAC    FDB  as these two forces are
2 2
of side length a. Find the gravitational force on a
particle of mass ‘m’ placed at the centre of the perpendicular to each other.
square.
Gm 2
Solution : Let m, 2m, 3m and 4m be at A, B, C and D FR = 4 2
a2
vertices of a square ABCD respectively. At the centre
‘O’ a mass ‘m’ is placed. Force due to the mass at A FAC
Direction; tan   F  1 ;
on the mass at the centre is DB

G  m m 2Gm 2  = 450 i.e., parallel to AD.

FA  
 a 
2
a 2 along OA. *Problem – 5.6 : Find the point at which the gravi-
 
 2  tational force acting on any mass is zero due to
Force due to the mass at C on the mass at the the earth and the moon system. The mass of the
centre is earth is approximately 81 times the mass of the
G  m 3m 6Gm 2
FC   moon and the distance between the earth and the
 a 
2
a 2 along OC.
  moon is 3,85,000 km. (such a point is called neu-
 2  tral point).
B C C
2m 3m Solution : The force due to the mass m1 on the mass
FB
x d
FC FR
FA ‘m’ is m1
FD O m m2
a O   450

2 Gmm1
A D F1  towards m1.
m 4m D x2
The resultant fo rce of F A and F C is The force due to the mass m2 on the mass ‘m’ is
4Gm2 Gmm2
FAC  FC  FA  F2 
a2
along OC.
d  x2  towars m2.
Similarly the forces due to the masses at the op- If the resultant force on the mass ‘m’ is to be
posite corners B and D are zero, F1 must be equal to F2 in magnitude as these two
 M and length L as shown in fig. Find the
d G 
i.e.,  2 mM  m2   0 i.e., M–2m = 0 magnitude of the gravitational force between
dm  r 
them:
[as G / r 2  0 ] x
or (m/M) = (1/2), i.e., the force will be maximum m
d L
when the parts are equal.
Solution :
Problem :5.10 Two particles of equal mass move Consider an element of mass 'dm' and length 'dx'
in a circle of radius r under the action of their at a distance 'x' from the point mass.
mutual gravitational attraction.Find the M
Mass of the element dm  dx .
speed of each particle if its mass is m. L
Solution : The particles will always remain dia- Gravitational force due to the element is
metrically opposite so that the force on each par- M 
Gm  dx
ticle will be directed along the radius. Consider-  L 
dF 
ing the circulation of one particle we have x2

mv 2 Gmm  M 
 Gm d  L Gm 
 L 
i.e.,   dx
2 r  ,
2
r 4r F  x2
d
Problem :5.11 Imagine a light planet revolving
around a very massive star in a circular or- d  L 
GmM
bit of radius 'r' with a period of revolution F
L  x2 dx
d
'T' On what power of 'r' will the square of
time period depend if the gravitational force GmM  x1 d  L
F  
L  1 
of attraction between the planet and the star  d
is proportional to r–5/2
 1 
d  L
GmM
Solution : As gravitation provides centripetal F  
L  x  d
force
1 
mv2 k k
F
GmM   1 
 5 / 2 , i.e., v  3 / 2
2
 d d  L
r r mr L  

2r mr3 / 2 GmM  d  Ld  GmM

F  
so that T   2r  d  L d   d d  L
v k L  
Problem :5.13 Two metal spheres of same mate-
2r mr3 / 2
T  2r ; so T2  r 7 / 2 rial and each of radius 'r' are in contact with
v k each other. The gravitational force of attrac-
Note:- If gravitational force is proportional to r–n then tion between the spheres is proportional to
n 1 –––
time period of a planet T r 2
Problem :5.12 A particle of mass m is is situated
at a distance d from one end of a rod of mass Solution: r r
 gravitation the gravitational force on the body is
Gm 2 Gm 2
FR  2  2 GMm
a2 2a F ....................................(2)
R2
Gm 2  1
mr 2   2  From Eqs. 1 and 2, we have
a2  2 
GMm GM
mg   g  2 .......................(3)
Gm  2 2  1
2
a R R
2 
2 a 2  2  This relation is valid for any planet including the
earth.
2 a) In terms of Radius (R) and mean density (  )
Find  and use T 
 GM
we know that g 
R2
a  3
2 
T  2 4 3
Gm  2 2  1 But M   R  
3 
12.6 Relation Between Universal Gravitational
Constant (G) and Acceleration due to Gravity (g): 4 
G  R 3 4 
 3   g   G R
g  3 
The uniform acceleration produced in a R2
freely falling body due to the gravitational force b) In terms of Mass(M) and density (  )
of a planet is known as acceleration due to gravity
on that planet. GM
As g =
It is denoted by ‘g’ and it varies from place to R2
place where as universal gravitational constant (G) is 4
same everywhere in the universe. But M = R 3 
3
m
 3M 2 / 3
 R 2   
F=mg  4 
M
GM
g 2/3
 3  M2 / 3
Consider a body of mass m on the surface of a   .
 4  2 / 3
planet, or very near to it so that the distance ‘r’ of the
body from the centre of the planet becomes the radius
R of the planet. Here we assume that the planet to be a  4 2 / 3
g    G. M1/ 32 / 3
homogeneous sphere of a uniform density (  ) whose  3 
entire mass “M” is supposed to be concentrated at the
* Problem – 5.18 : If the acceleration due to grav-
centre of it.
ity on earth is 9.81 m/s2 and the radius of the earth
If m, g, F represent the mass of the body,
is 6370 km find the mass of the earth ?
acceleration due to gravity and the force acting on the
body due to gravitational pull of the planet then (G = 6.67 x 10–11Nm2/kg2)
F = mg. ................... (1) Solution :
According to Newton’s law of universal g = 9.81m/s2, R = 6370 km = 6.370 x 106m;
 4 g
 R 3 gd  (r) ( R  d  r)  gd  r
4 R
 g  G 3 2   GR  ................... (1)
R 3  gd versus r graph is a straight line passing through
Now, consider a body of mass m at a depth d, the origin as shown in fig.
from the surface of the earth. The outer shell of thickness Above the earth :
d which is exterior to the body exerts no resultant gR 2
gravitational force on the body. The force on the body gh 
will only be due to the mass of the earth confined in the (R  h) 2
inner solid sphere of radius (R-d). Let the acceleration
gR 2 1
due to gravity at this depth d be gd, gh  2
( R  h  r)  gh  2
r r
R is replaced by (R-d) in Eq.(1)
 gh versus r graph is a curve as shown in the fig.
4
 gd   G (R  d) .....(2) 12.7.4 Variation of g with Latitude :
3
From Eqs. 1 and 2 we get When the earth rotates about the axis passing
through its north and south poles, a frame attached to
gd R  d the earth is a non inertial frame of reference. In such a

g R frame so as to apply Newton’s laws we have to assume
 d a centrifugal force on the objects at rest
g d  g 1   ........(3)
 R Consider an object of mass m at latitude  . The radius
From the above equation it is clear that the value
of the circle on which it is rotating about the axis of
of ‘g’ decreases with the depth.
earth is r.
Eq. 3 gives the variation of ‘g’ with depth.  r 
i) ‘g’ decreases as depth d increases. r = Rcos  .  cos   
 R
ii) ‘g’ becomes zero at the centre of the earth.
iii) ‘g’ is maximum on the surface of the earth
and it decreases with height and depth.
iv) Decrease in ‘g’ at small heights is more than
the decrease in ‘g’ at small depths.
v) at large heights decrease in ‘g’ is less than the
decrease in g at large depths.
12.7.3 Graphical representation of variation of g
with height and depth :
The variation of g with the distance r of a place
from the centre of the earth is shown below

g=9.8ms-2
due to gravity

1
acceleration

gr g The forces acting on the object are (i) gravitational force
r2 (ii) centrifugal force.
inside
the earth above the surface GMm  GM 
Gravitational force FG   mg
x
0   g 
R  R 2 
distance from centre of earth (r) R2
g Centrifugal force FC  mr2
Inside the earth: g d  (R  d) The resultant force due to these two forces acting at an
R
angle of (180°-  ) can be found by using parallelogram
law of vectors. Let the magnitude of the resultant be F.
Case – 1: When an object is in a freely falling lift. Its  2 
 where  0  rad / sec is the present value 
apparent weight is zero. This phenomenon is known as  20 60 60 
weightlessness.
As,   0
a=g
N
R  0.034m / s2 
2
0
a=g  ge  g  R 02
using free body diagram Where, a  R 20 , is the centripetal accelaration
w|=m(g-a)=0
mg If a = g,
mg –N = ma ge = g–a = g–g = 0, Hence object appears weight
ligtless ness
 N  m g  a   m g  g
This is possible for an angular velicity  such
N0 that
 Wapp  0 g  R 2

g
Case – II : Consider a person in a satellite orbiting 
R
around earth close to the surface of Earth. Let V0 be
If the present value of angular velocity is  0
GM
the orbital velocity of the satellite then V0  g R 2 9.8
Then,    289
R
N
a R 0 0.034
2

N
 2 
ac   0  
 24 6060 
V0 ac
mg
M
mg   2  289 20
V02 GM   17 0
Centripetal acceleration Vc   g
R R2 Hence, present angular velocity of earth has to
Using free body diagram increase to 17 times for which object becomes
mg –N = mac weightless at equator
 N  mg  ma c Note : i) If R 2  g  object appears weightless
 N  m g  a c  ii) R 2  g  object fliesoff from surface of
N  0 planet.
 Wapp  0 iii) R 2  g  object will remain stuck
It means every thing inside the satellite is in a state of with the surface of planet.
free fall. This is just like a body or a person falling Problem–5.19: What is the time period of ratation
towards the centre of the earth from a height. So a of the earth around its axis so that the objects at
person inside a satellite experiences weightlessness. the equator becomes weightless? (g=9.8m/s2, Ra-
12.7.5 Condition for weightlessness of an object dius of earth = 6400km)
at the equator of the earth Solution: g at eh equator is
We know that at equator g0 = g- g 0  g  R 2
g   g  R  Cos 
2 2
0
If bodies are to become weightless at the equa-
  2h  astronomical distances, influence one another
Sol: i) Using gh = g 1  instantaneously without any intervening signal that can
 r 
move faster than light or that can move with an infinite
with h = 100km and R= 6400km velocity. In other words, the transmission of the
gh = 9.494m/s2
gravitational force of one body to another body can
g not be instantaneous, as given by Newton’s third law.
ii) using gh = 1  h  with h= R g = 2.45ms–2
2
This indicates a limtations of Newton’s third law when
 R  applied to transimission of gravitational force.
a) Action at a distance
* Problem : 5.24 b) Field concept
If ‘g’ on the surface of the earth is 9.8 ms–2, Action at a distance concept:
find its value at a depth of 3200km (radius of the Newton’s law of universal gravitation assume
earth = 6400km) that gravitational force acts directly and instantaneously
Sol: with d= 3200km even when the bodies are not in physical contact or
 d when they are separated by vast distance. This point of
gd  g 1  
 R  view is known as ‘action at a distance’ or ‘ force at a
 gd  4.9ms2 .
distance’ concept.
However this theory can not explain how the
12.8 Limitations of Newton’s Third Law
gravitational force can be transmitted instantaneously
Newton third law has certain limitations and is
over large distances without any intervening medium
not valid strictly in all situations.
between them.
The limitations are:
Field concept :
1) The law does not apply strictly when the
Gravitational field : It is the space around a mass in
objects move with speed close to that of light
which its influence is felt
2) It does not hold good where the gravitational
The field concept assumes that the presence of a body
fields are very strong
(mass) modifies the space around in someway and thus
3) It also fails to explain how one body exerts
a gravitational field is developed
gravitational force on the other though they are not in
A body placed in this field inturn exerts a gravitational
physical contact
force of attraction on the other. This gravitational field
4) The law is not strictly applicable when the
has both momentum and energy. Thus the field is the
gravitational interaction between two bodies involves
agent through which two bodies exert forces on each
vast distances
other
5) The transmission of gravitational force of one
body on another can’t be instantaneous as stated by 12.9 GRAVITATIONAL INTENSITY OR
this law STRENGTH OF GRAVITATIONAL FIELD
Newtons’s third law implies that mutual Gravitational intensity, at any point, in a
gravitational forces of one body on the other constitute gravitational field is defined as the gravitational
an action and reaction forces which are instantaneous. force experienced by a unit mass placed at that
These instantaneous transimissions of mutual point.
gravitational forces from one body to the other that are 
separated by vast distances give rise to some If F is the force experienced by a body of mass

conceptual difficulties. Here the question arises as to m, gravitational intensity E is given by

how does two non contacting bodies separated by  F
E
m
 It is not necessary that the gravitational lines of
that point.
force may be straight. In the case of two or more masses,
6) Gravitational fields are supposed to propagate
the gravitational lines of force will be curved.
through particles known as gravitons.
(i) Field due to a point mass :
7) Gravitational fields are supposed to propagate Suppose, a point mass M is placed at point O. We
with velocity of light. 
want to find the intensity of gravitational field E at a
12.9.4 REPRESENTATION OF point P, a distance r from O. Magnitude of force F acting
GRAVITATIONAL FIELD on a particle of mass m placed at P is,
Gravitational field intensity is a vector quantity.
GMm
The magnitude of the field at any point is the force acting F
r2
on a unit mass kept at that point. The direction of the
field is the direction in which a freely placed unit mass F GM
 E  2
moves on its own m r
In order to represent the gravitational field , we GM
make use of the fact that the gravitational field intensity or E
r2
is a vector quantity. A vector can be associated with The direction of the force F and hence of E is from P to
every point in the gravitational field. The magnitude of a O as shown in fig.
vector at any point in the field is equal to the magnitude
of the force acting on a unit mass placed at that point.
M r
The direction of this vector will be the direction in which O  P
the test mass placed at that point will move if free to do E
so. So, a gravitational field can be represented completely (ii) Gravitational field due to a uniform solid sphere
with the help of vectors. The gravitational field is a vector Field at an external point
field. A uniform sphere may be treated as a single
particle of same mass placed at its centre for calculating
the gravitational field at an external point. Thus,
GM
E(r)  for r  R
r2
1
or E(r) 
r2
Here r is the distance of the point from the centre
Lines of force of the sphere and R the radius of sphere.
Field at an internal point
An important way to represent gravitational field The gravitational field due to a uniform sphere at
is by gravitational lines of force.A gravitational line of force an internal point is proportional to the distance of the
is the path traced out by a unit mass when allowed to point from the centre of the phere. At the centre itself, it
move freely in the gravitational field. The direction of the GM
gravitational field, at a point, is given by the tangent to the is zero and at surface it is , where R is the radius
R2
line of force at that point. of the sphere. Thus,
In fig, the gravitational field of an isolated spherical
GM
body is represented with the help of gravitational lines of E(r)  r for r  R
force. The lines of force are directed towards the centre R3
of the body. or E(r)  r
   GM GM
W   Fdr   2
.dr 
r r r r
Hence, the work done in bringing unit mass from R
P
GM r
infinity to P will be  . Thus, the gravatational
r
potential at P will be,
GM
V 
r At some internal point P, potential at a distance
Conclusions
r from the centre is given by,
(i) Gravitational potential at any point is a
scalar quantity
(ii) Its unit, in C.G.S. system, is erg g–1 and in S.I. GM 3R r  r r 
Vr      forr  R
J Kg–1. 
R3 2 
 
(iii) It is always negative since gravitational force
is always attractive in nature for a point on the surface of sphere r = R hence
(iv) With an increase in’ r’,it becomes less negative GM
potential on the surface is Vr   
i.e. it increases as we move away from the source of R
gravitational field. At the centre of sphere r = 0 hence potential at
(v) Maximum value of gravitational potential is
3 GM
zero and this happens only at infinite distance from the the centre of sphere is V =  .
2 R
source
* Potential due to a uniform solid sphere
a) Potential at an external point GM
1.5
R
To calculate gravitational potential due to a solid
sphere at an external point total mass of the sphere 
GM
R

is to be taken at its centre and distance of the r

R
external point is to be measured from the centre
of the sphere.
V

i.e., at the centre of the sphere the potential is 1.5

M
r times the potential at surface. The variation of V
versus r graph is as shown in Fig.
GM (iii). Potential due to a uniform thin spherical shell
V r   rR
r a) Potential at an external point
GM
At the surface, r  R and V   To calculate gravitational potential due to a
R
spherical shell at an external point total mass of
b) Potential at internal point
the shell is to be taken at its centre and distance of
the external point is to be measured from the cen-
tre of the shell.
 Let P be the position of ‘m2 ‘ ( distant x from
 dv    Ex dx  E y dy  E z dz 
m1) at any instance. Gravitational force F acting upon it
 ^ ^ ^
is
Here, d r  dx i  dy j  dz k and
 ^ ^ ^ Gm1m2
E  Ex i  E y j  E z k F
x2
Note : a) If E is given V can be calculated by the Let dw is the small work done to displace it from P to
r r
Q through a distance dx towards the mass m1 .
formula V   dv   E.dr  
  or dW  F .dx  F dx cos 
b) The negative of the slope of V - r curve gives
E or dW = F dx   0
12.11. GRAVITATIONAL POTENTIAL
Gm1m2
ENERGY OF A SYSTEM dW  dx .... (i)
x2
When two or more bodies interact with each other r
G m1m 2
due to gravitatinal forces some work has to be done in
 W   dW   x2
dx
assembling them together in their respective places. 

Total work done in assembling the bodies G m1m 2

W   ....... 5.19
together, in their respective places, is called potential r
energy of the system.
We know that work done by conservative
Case (i) A system of two masses
force is equal to negative of change in potential
Potential energy of a system of two masses energy
is defined as the amount of work done in bringing
W = – ( Uf – Ui)
these two masses from infinity to their respective
W = Ui – Uf
places.
Since PE at infinity is zero
A B
  W = Ui
F
r P
m2 Q
m1 dx Gm1m2
r  U 
r
x
Conclusions.
(i) Negative sign of the potential energy is due to
Consider a spaces where there is no gravitatinal
the fact that the gravitational forces are attractive in
feeld. In order to bring a body of mass m1 from infinity
nature.
to a point A in the space no work is done because
there is no gravitatinal force on m1. Now m1 produceses (ii) The above equation indicates that with an
a gravitatinal field in the space. To bring another body increase in r the potential energy of the system becomes
of mass m2 from infinity to a point B at a distance r lesser negative i.e. it increases.
from m1 some work is required. This work is done by (iii) When distance between two bodies is infinite,
the gravitatinal force of atraction of m1 acting on m2. gravitational potential energy is zero. This is the maximum
value of gravitational potential energy of the system
 r2
 
r2

 dv   E.dr
V
2 gh
r1 r1 2GM  h   
  r2 V   or 1  h 
V2 V1   E.dr R Rh   R 
r1

dU 12.12. ESCAPE VELOCITY FROM THE

b) F   SURFACE OF A PLANET:
dr
The minimum velocity with which a body must
be projected from the surface of a planet so as to
dU  Fdr
eccape from gravitational influence of planet is
  r2
called escape velocity of that planet.
U2 U1  mE.dr
r1
Consider an object of mass ‘m’ at rest on the
surface of a planet of mass M and radius R.
U 2 U 1 W
c) V2 V1    field The gravitational potential energy of the body.
m m
r2 ve
Wext agent m
  E.dr  R
r1
m
M
Note: 2) If a body is moving only under the influence of
gravitational force, law of conservation of mechanical Negative sign indicates that the body is attracted by
energy is applicable the planet and the body is bound to the planet. Now
 U 1  K1  U 2  K 2 the body can be sent out of gravitational field by making
Note 5.2 : The velocity required to send an object its total energy either zero or positive.
to height ‘h’ above the earth surface. Supposse the body is projected with a velocity
Ve such that kinetic energy imported to the body must
m v=0 be equal and oposite to the potential energy. Then a
u=v h total energy of the body becomes zero and a body
m
R escapes.
R
M M  PE  KE  0 (or ) PE  KE.
1  GMm 
 mve2  
When the body is on the surface of earth.
2  R 
GMm 1
PE1  , KE1  mV 2 1 GMm
R 2 mve2 
When the body is at its maximum hight 2 R

GMm 2GM
PE2  , KE2  0 ve  ................ 5.24
Rh R
Using law of conservationof energy TE1 = TE2 In terms of g, the expression for ve is
 PE1  KE1  PE2  KE2
2gR 2  
  g  GM 
R  R 
2
early period of moon’s formation, its temperature was
1 1 
so high and due to thermal agitations, oxygen and  V2  2GM   
nitrogen molecules acquired speeds greater than 2.38  R R  h 
km/s resulting in their escape from the moon’s
surroundings. 2gh
 V2 
h
4  1
Note 5.4 : Ve  2 gR  2.  RG  R g
3 

4
 g   RG 2gR 2gh
3 
Ve 2  
if V  then n 1  h 
8 n  R 
Ve  GR 2  Ve  R 
3
 = mean density of the planet. R
 h
BEHAVIOUR OF A BODY PROJECTED n 2 1
VERTICALLY UP WITH DIFFERENT
VELOCITIES FROM THE SURFACE OF A Case II: If the velocity of projection V = Ve then total
PLANET: energy of the body just becomes zero so that the body
just esacpes from the planet and goes to infinity
r E=0
V=0 E=-ve r E=+ve
R
h
E=-ve n 2 1
E=0 E=+ve
h
R   V  Ve 
E=-ve E=0 E=+ve
  
2 
1 1   n  1

V<Ve V=Ve V>Ve
1 GMm
K.E < P.E
(Numerically) 2
mV 2 
r
K.E > P.E
 h 
As total energy of the body is zero at infinity
(a) V < Ve (b) V = Ve (c) V > Ve
GMm 1 2
T.E =  mV 0
Consider a body of mass ‘m’ projected with a  2
velocity ‘V’ from the surface of a planet of mass
V  0
‘M’ and radius ‘R’ then
Case I: If the velocity of projection V<Ve then body i.e., body posses zero velocity at infinity
fails to escape. It goes to a certain max. Height and Case III: If the velocity of projection V>Ve then total
then falls back using law of conservation of energy T.E energy fo the body becomes positive. Hence the body
surface = T.Emax. height escapes from the planet and goes to infinity let V be
GMm 1 2 GMm the velocity of the body at in intersteller space then using
  mv  
R 2 Rh law of conservation of energy
T.Esurface of planet = T.Einter steller space
 E = P.E + K.E
4 2
T  2
.r 3 GMm
4 3
G. R .  E 
3 2r
KE
Note 5.9 :
 32  3 PE
T2   Total energy, TE  KE 
r
r
 GR 3 .  2 TE
PE
so, PE : KE : TE = –2 : 1 : –1
ii) For surface satellite (i.e., h < < R)
r = R Note 5.13 : When a satellite falls from an higher
orbit to a lower orbit its
3 3 R i) PE decreases
 T2   T  T  2
G G g ii) TE decreases
iii) KE increases
a) In case of earth, the time period of surface
iv) Speed increases
satellite is v) Linear momentum increases
T = 84.6 min vi) Angular momentum decreases
b) If two planets have same mean density then Note 5.13 : When a satellite rises from lower orbit
satellites revolving close to their surfaces will have same to higher orbit
time periods. i) PE increases
Note 5.7 : ANGULAR MOMENTUM OF A ii) TE increases
SATELLITE iii) KE decreases
iv) Speed decreases
GM
L  mv0 r  m r  GMm 2 r v) Linear momentum decreases
r vi) Angular momentum increases
Trajectaries of a body projected at with different
L  GMm 2 R  h
velocities
for the surface satelite (h <<R)
1) A body revolves around a planet only when it
 L  GMm 2 R  gR 3m 2 is projected with sufficient velocity in a direction
Note 5.8 : ENERGIES OF A SATELLITE pepenrdicular to the gravitational force of attrac-
Consider a satillite of mass m revolving in an tion of planet on the body
orbit of radius r, around the earth of mass M, then
GMm
(a) P.E of satellite = 
r
1 2
(b) Its kinetic energy KE  mv
2
GM
But v 
r
1  GM 
2

 KE  m  
2  r 
GMm
KE 
2r
(c) Therefore total energy of satellite

If v  gr body falls on the surface of earth

 in terms of G, M and a. face of earth
Soluton: The distance x (from the smaller planet)  GMm  GMm
E S  KE  PE  0     
where the gravitational pull of two planet's bal-  R  R
ance each other will be given by And the energy of the sky lab in an orbit of radius
r
GMm G 16M  m

10a  x 
2
x2
1  GMm  GMm
E  mv20   
i.e., So the body will reach the smaller planet 2  r  2r
due to planet's gravitational field if it has suffi-  
cient energy to cross the point B (x = 2a), i.e., asv  GM 
 0
r 

1
m 2  m  VB  VS  (a) So the energy required to place the lab from
2
the surface of earth to the orbit of radius 2R,
16GM GM 
V     
65GM
but S  2a 
10  2a 
 8a GMm  GMm  3 GMm
E1  E S     
10a 2(2R)  R  4 R
i.e.,
3m 3
a B S E   gR 2  mgR
2a 4R 4
 GM 
M
16M asg  
 R 2 
i.e.,
16GM GM  20GM
and VB      
 E 
3
 
3
 
2 103 106.4106  12.81010  9.6 1010 J
 8a 2a  8a 4 4
(b) As for II orbit r= 3R,
1  65GM 20GM 
so m  m
2
  
2  8a 8a 
GMm GMm
E11   
3 5GM 2(3R) 6R
i.e., min 
2 a GMm  GMm  1 GMm
Problem :5.29 A sky lab of mass 2 x 103 kg is first  E11  E1   6R   4R 
 12 R
launched from the surface of earth in a cir-
cular orbit of radius 2R (from the centre of But as g  (GM / R 2 ),
earth) and then it is shifted from this circu- i.e., GM = gR2
lar orbit to another circular orbit of radius or
3R. Calculate the minimum energy required
(a) to place the lab in the first orbit (b) to E 
1
12
mgR 
1
12

12.81010  1.11010 J 
shift the lab from first orbit to the second
orbit. Given, R = 6400 Km and Problem :5.30 The masses and radii of the earth
g = 10m/s2. and moon are M1, R1 and M2, R2 respec-
Solution : The energy of the sky lab on the sur- tively. Their centres are at distance d apart.
 earth and the other around Mars close to their
Ve1  6.01024  1740  4.695
    surface. (mass of the earth = 6 x 1024kg, Mass of
Ve 2  7.41022   6400  1 the Mars = 6.4 x 1023 kg, Radius of the earth =
* Problem 5.36: Show that the velocity of an ob- 6400km and Radius of Mars = 3400km)
Solution
ject released from rest at infinity reaches the earth
The orbital speed of a satellite close to the surface of a
with a velocity equal to the escape velocity.
Solution: GM
By the principle of conservation of energy for the ob- planet, V0 
R
ject,
TE at infinity = TE at the surfece of the earth V01  M 1   R2 
   
(KE +PE) at infiniy = (KE+PE) at the surface of the The ratio of orbital speeds, V  M   R 
02 2 1
earth.
As the body is released from rest at infinity, its KE at
 61024   3400 
inifinity is zero and its PE at infinity is also zero. If v is its V01
      6034  2.232
velocity on reaching the earth its KE becomes equal to V02  6.410   6400 
23
6.4 64 1
1 2 GMm * Problem 5.39: A satellite is at a height of 25,
mv and PE 
2 R 600km from the surface of the earth. If its orbital
1 2 GMm 1 2 Gm speed is 3.536km/s find its time pariod. (Radius
 00  mv  ; v   of the earth = 6400km)
2 R 2 R
Solution:
2GM R = 6400km = 6.4 x 106m, h = 25, 600km = 25.6 x
The velocity of the object, V   Ve . 106 m, v0 = 3.536 x 103 m/s
R
The time period
* Problem 5.37: Find the ratio of the orbital speeds
ot two satellites of the earth if the satellites are at 2   R  h  2  25.6  6.4 10 6
T 
height 6400km and 19200km. (Radius of the earth V0 3.536 10 3
= 6400km.)
2 32
Solution:  103  56,870s
3.536
GM * Problem 5.40: What is the work done in taking
Orbital speed, V0 
Rh an object of mass 1kg from the surface of the earth
The ratio of the orbital speeds of two satellites to a height equal to the radius of the earth?
around the earth is (G = 6.67 x 10–11 Nm2/kg2, Radius of the earth =
6400km, Mass of the earth = 6 x 1024kg.)
V01 R  h2 Solution:

V02 R  h1 The work done in taking an object to a height
Where h1 and h2 are the heights of the satellites GMm

In this problem h 1 = 6400km, h2 = 19200km, R  R  h
R=6400km.
Here h = R
V01 6400  19200 25600 2
   6.67 1011  61024 1
V02 6400  6400 12800 1 
2 6.4106
* Problem 5.38: Find the ratio of the orbital speeds
of two satellites one of which is rotating round the
Use the known value of G. GMm 4GMm
Sol. Here, mass of each star, M = 2 × 1030 kg. 2

r (6 R  r )2
Initial distance between two stars,
r = 109 km = 1012 m. (6 R  r )2  4r 2
GMm  r  2 R or  6 R
Initial potential energy of the system  
r we discard the negative value. It is sufficient to
1 1 project the body with a speed which would enable it to
Total K.E.of the stars  Mv2  Mv2  Mv2
2 2
reach the null point. After that the greater gravitational
where v is the speed of stars with which they pull of 4M would act. At the neutral point, the speed of
collide. When the stars are about to collide, the d i s - approachis zero.
tance between their centres, r|  2R. From conservation of mechanical energy
GMM
 Final potential energy of two stars   2R . 1 2 GMm 4GMm GMm 4GMm
mv     +0
2 R 5R 2R 4R
Since gain in K.E. is at the cost of loss in P.E.
3GM
2 GMM  GMM   GMM GMM or v 2  2GM  4  1   v 
 Mv       R  5 2 5R
r  2R  r 2R
Here one must note that speed of projectile is
11
 (2  1030 ) 2 zero at neutral point but it is non-zero when it strikes
or 2  1030 v2   6.67  10
1012 the other sphere.
6.67  1011  (2  1030 )2 Problem: 5.47
 A rocket is fired vertically from the surface of
2  107
43
Mars with a speed of 2 kms–1 . If 20% of its
 2.668  1038  1.334  10 43  1.334  10 J
initial kinetic energy is lost due to Martian at-
1.334  10 43 mospheric resistance, how far will the rocket go
 v  2.583 ×106 ms 1 .
2  1030 from the surface of Mars before returning to it.
Problem : 5.46 Mass of Mars = 6.4 × 1023 kg; radius of Mars =
Two uniform solid spheres of equal radii R, but 3395 km; G = 6.67 × 10 –11 Nm2 /kg.
mass M and 4M have seperation 6R between their Sol. Let m  mass of the rocket, M  mass of the
centres. The two spheres are held fixed. A Mars and R  radius of Mars. Let v be the initialv e -
projectile of mass m is projected from the surface locity of rocket.
of the sphere of mass M directly towards the 1 GMm
Initial K.E.  mv2 ; initial P.E.  
centre of the second sphere. Obtain an expression 2 R
for the minimum speed v of the projectile so that 1 GMm
Total initial energy  mv2 
it reaches the surface of the second sphere? 2 R
Sol: The body projected is acted upon by the Since 20% of K.E. is lost, only 80% is left be-
gravitational forces of those two spheres in opposite hind to reach the height. Therefore,
directions. These two fores will be equal in magnitude Total initial energy available
at null point between them. If that null point is at a 80 1 2 GMm
 mv   0.4 mv 2 
GMm
distance r from the centre of the first sphere, 100 2 R R
If the rocket reaches the highest point which is at
a height h from the surface of Mars, its K.E. is
GMm
zero and P.E. 
(R  h)
Using principle of conservation of energy, we
expended to rocket the satellite out of the 12.14.1 Uses of Geostationary Satellites
gravititional influence of earth? Mass of the satel- 1) Study the upper layers of the atmosphere.
lite is 200 kg , mass of the earth = 6.0 × 1024 kg, 2) Forecast the changes in the atmosphere.
radius of the earth = 6.4 × 106 m, G = 6.67 × 10–11 3) Know the shape and size of the earth.
Nm2 kg–2. 4) Identify the minerals and natural resources
Ans. Total energy of orbiting satellite at a height h present inside and on the surface of the earth.
GMm 5) Transmit the T. V, programmes to distant
 places.
2(R  h)
Energy expended to rocket the satellite out of 6) Study the properties of radio waves in the
the earth’s gravititional field.   (total energy of the upper layers of the atmosphere.
satellite) 7) Under take space research i.e., to know about
the planets, satellites and comets etc.
GMm (6.67  1011 )  (6  1024 )  200 8) These are widely used for telecomunication
 =  5.9  109 J .
2(R  h ) 2(6.4  106  4  105 ) purpose
12.14. Geostationary Satellite : Problem–5.51 : An astronaut orbiting in a
The satellite which revolves round the earth and spaceship round the earth, has centripetal accelera-
which appears to be stationary with respect to an tion 2.45m/s 2. Find the height of the spaceship. (Take
observer on the earth is called geostationary satellite. R = 6400 km).
“If the period of revolution of an artificial Sol : As centripetal acceleration equals to acceleration
satellite is equal to the period of rotation of earth, due to gravity at that height, then
then such a satellite is called geostationary
satellite”. GM GM gR 2
a  gh   
When an earth satellite satisfies the following r2  R  h  2  R  h 2
conditions then it becomes geostationary satellite.
(1) Height of the orbit above the earth’s surface gR 2
a
is about 35,800 km  R  h2
(2) The orbit must lie in the equatorial plane of
2
the earth.  R  a 2.45 1
(3) The sense of revolution must be same as that =    
 R  h g 9.8 4
of the rotation of the earth about its own
axis (from west to east in the equitorial plane) R 1
 
(4) Time period of revolution must be same as Rh 2
that of the earth’s rotation (24 hours)  R  h  2R
(5) The orbital angular velocity of Geostationary R = h = 6400 km
satellite is same as that of the earth  spaceship is at a height 6400 km
(6) Speed of revolution is 3.1 km/sec. Problem–5.52: A satellite is revolving round the earth
(7) Relative angular velocity of geostationary in a circular orbit with a velocity of 8 km/s. at a height
satellite w.r.t earth about the centre of the earth is zero where acceleration due to gravity is 8 m/s2. How high
* The orbit of geostationary satellite is called is the satellite from the earth ? (Take R=6000km)
parking orbit. Sol : As centripetal acceleration equals to acceleration
* Orbital radius of Geostationary satellite is due to gravity at that height, then
42,250 km (approximately 7R here R is the radius of
the earth)
 m1 F m2
O r
v fixed point
M A central force is that force which is always
v
m directed towards or away from a fixed point. i.e; along
r II
m I
the position vector of the point of application of the
force with respect to the origin or fixed point. The
magnitude of central force depends on r.
So the total energy of the system before collision
a) For the motion of a body under a central force the
GMm angular momentum of the body about centre of force is
E i  E1  E 2  2E  
r always conserved because torque acting on the body
As the satellites of equal mass are moving in op- by the central force is zero.
posite direction and collide inelastically, the ve- b) The motion of a particle under the central forceis
locity of wreckage just after collision, by conser- always confined in a plane
vation of linear momentum will be c) The postion vector of the particle in circular or
mv  mv  2mv, i.e., v= 0 elleptical orbit with respect to the centre of force (i,e;
i.e., just after collision wreckage comes to rest in da
fixed points) has constant areal velocity  0 .
the orbit. So energy of the wreckage just after dt
collision will be totally potential and will be

GM(2m) 2GMm
EF   
r r
And as after collision the wreckage comes to
rest in the orbit, it will move along the radius to-
wards the earth under its gravity.
12.17PROPAGATION OF GRAVITATIONAL FIELD :
Problem : 5.57
A geostationary satellite orbits the earth at a Gravitational field as described by Einstein is dis-
height of nearly 36,000 km from the surface of tortion of space due to the presence of matter. Ac-
earth. What is the potential due to earth’s gravity cording to Einstein's theory of relativity, whenever the
at the site of this satellite? (Take the potential mass particles are accelerated, the gravitational fields
energy at infinity to be zero). Mass of the earth = around them undergo rapid changes. In Plank's mod-
6.0 × 1024 kg, radius = 6400 km, G = 6.67 × 10–11 ern theory, all fields are considered to have the quan-
Nm2/kg2. tum nature. The minimum energy packet of electromag-
Ans. Gravititional potential at height h from the sur- netic radiation is called photon and it travels with the
GM velocity of light. Similarly, graviton is a hypothetical quan-
face of earth is V   (R  h)  tum of gravitational energy and propagates in the form
of massless particles which travels with velocity of
6.67  10 11  (6  1024 )
 9.4  106 J kg . light.Thus gravitational fields are supposed to propa-
(6.4  106  36  106 )
gate with velocity of light.
12.16 Central force : 12.18 GRAVITATIONAL WAVES :
According to Einstein's theory of special relativity,
nothing can travel faster than speed of light. This
collapse ( contraction) since the gravitational attraction dwarf having the mass of the sun has approcimatly the
of matter towards the centre of the star is balanced by size of the earth. If the mass is greater than 1.4Ms , the
the out ward radiation presure. A star will remain stable degeneracy pressure between electrons will not be able
like this for millions of years, until it runs out of nuclear to half further gravitational collapse. The star then
fuel such as H2 and He. The more massive a sta is , collapse a neutron star or a black hole.
faster will be the rate at which it will use its fuel because Neutron Star
grater energy is required to balance the greater For a star become a Neutron star, its initial mass must
gravitational attraction owing to great mass i.e., massive be grater than the solar masses. (M > 10Ms).As a star
stars burn out quickly. When the nuclear fuel is over, with initial mass M > 10Ms cools off the large mass of
i.e., when the star cools off, the radiation pressure is the star causes it to contract abruptly, and when it runs
not sufficient to hall the gravitational collapse. The star out of fuel it spring back and explodes violently. The
then begins to shrink with tremendous increase in the explosion flings most of the star matter into space and
density. The star eventually settles into a white dwarf, such a state of a star is called a Supernova. A supernova
Neutron star or black hole depending upon its initial explosion is very bright and outshines the light of an
mass. entire galaxy. The mass of the matter left behind is greater
As the star collapse after cooling off, the radiation than 1.4 Ms. If the mass of the left over matter is between
emitted due to fusion of remaining H2 nuclei at the outer 1.4 Ms and3 Ms Neutron stars evolve. At this stage the
edge of the core, cause the lighter outer mantle of the repulsion between electrons will not be able to half
star to expand to several times its original diameter. further gravitational collapse. Under such conditions,
Such expanding star is called a Red Giant. After several the protons and electrons present in the star combine
millions of years, H2 is exhausted and the material from to form neutrons. After the formation of neutrons, the
the outer edge mantle of the Red Giant is blown off and outward left over is called the Neutron star. Neutron
remaining core left over, which is very dim is called star has a density much larger than a white dwarf and
white dwarf. A white dwarf is barely visible and has a has a radius of about 20 km. Neutrons stars are also
mass of less then 1.4 Ms where Ms is the mass of the called pulsars, because they emit regular pulses
sun. For a star to become a white dwarf, initial mass of radio waves. Stars turn into balckholes, when the
must be less than tn solar masses ( M>10Ms) mass of the remaining matter after a supennova explosion
Chandrasehar Limit is greater than 3Ms, the initial mass being greater than
The maximum mass that a white dwarf can have 10Ms. When the mass of the remaining star is grater
is called Chandrasekhar limit, which is 1.4 Ms. A white than 3Ms even the degeneracy pressure between
dwarf cannot have a mass greater than 1.4M. The neutrons cannot prevent the gravitational collapse and
volume of a white dwarf is about 10–6 times the volume Black hole is formed.
of the original star. In the white dwarf stage further Problem 5.58
gravitational collapse is halted due to the balance What is the schwarzchild radius for a star with a mass
between repulsion of electrons and gravitational 10 times that of the sun. ?
attraction. The repulsion between electrons causes 2GM
degenaracy pressure. The degeneracy pressure arises Sol: Schwarzchild radius, R =
C2
because all the lower available quantum energy states,
is filled up by electrons.The pauli Exclusion Principle 2  6.67 1011 10 2 1030
R 
prevents further filling up of these energystates. This (3108 ) 2
causes the remaining electrons to fill up higher structures.  R  30,000 m
Matter in a white dwarf has a very high density. A white
 According to the law of universal gravitation, the Intertial M ass G ravitational M ass
gravitational force on a body is proportional to its mass. 1. The ratio of weight of
1 . T h e ra tio o f fo rc e
t h e b o d y a n d t h e
The gravitational force experienced by the body is acting on the body and
acceleration due to
called its weight. Gravitational mass is denoted by mg. acceleration gained by
gravity at that place is the
it is the inertial m ass
gravitational m ass.
W = mg g 2 . W h ile d eterm in in g
2 . T h e b o d y w ill b e a t
W th e in e rtia l m a ss, th e
re st w h ile d e te rm in in g
 mg  body shall be in the
g state of m otion.
gravitational m ass.
The mass calculated using this formula is called 3. It is very difficult to
gravitational mass. Gravitational mass is a coefficient determ ine this m ass
which determines the force of attraction experienced b e c a u s e a ll k i n d s o f 3. Gravitational m ass can
external forces on the b e d e t e r m i n e d v e r y
by a body in a gravitational field. It is equal to the weight b ody like friction, air easily.
divided by the acceleration due to gravity. resistance etc m ust be
12.21.3 Relation between Inertial and accurately assessed.
4 . I n e rt ia l m a s s o f a
Gravitational Mass b o d y in c r e a s e s w it h
Every object on the surface of the earth is pulled v e l o c i t y w h e n 4. G ravitational m ass of
v e l o c i t i e s a r e a body is constant.
towards its center even when it is at rest.
co m p arab le to that of
Magnitude of this force is known as weight and it is light.
proportional to the mass of the object. 12.22. PRINCIPLE OF EQUIVALENCE:
W  m g or W  m g g
If this body is dropped from certain height it falls towards Statement : It states that a uniformly accelerated
earth with an acceleration ‘a’. As the body is in motion reference frame in the absence of any
under the action of force W, we can use gravitational effect is completly equivalent to a
homogeneous gravitational field.
W
 mi or W  mi a OR
a
mi a  mg g For all practical purposes the inertial and
It has been experimentally observed that the ratio non inertial frames are equivalent as experiments
(mi /mg ) is equal to one and hence inertial mass (mi) conducted in identical conditions in both the
and its gravitational mass (mg ) of an object are equal. frames give identical results.
We generally use gravitational mass since it is easy to OR
determine it. It is taken as the mass m of a body while No experiment performed inside an
solving any problem. accelerated , closed system can distinguish
12.21.4 DISTINCTION BETWEEN INERTIAL between the effects of gravitational field and the
AND GRAVITATIONAL MASS: effects of an accelerated motion.
Note : This principle established that the inertial mass
and gravitational mass are completely equivalent.
Illustration:
The Principle of Equivalence can also be stated as
follows.
“A uniformly accelerated reference frame in the
absence of any gravitational effect is completely
equivalent to a homogeneous gravitational field”.
This principle established that the inertial mass and
 SP = rP and SA = rA . If m is mass of the planet,
from conservation of angular momentum, mvArA =
mvPrP
or VArA  Vprp

It means that
 vP  vA as rA  rP
3. Law of periods:
As shown in fig., If a planet moves from p1 to p2
or from p3 to p4 in the same duration of the time. Then, The square of the time period of revolution
the areas A1 and A2 are equal. From this law it of a planet is directly proportional to the cube of
can be observed that planets move slower when they the semi major axis of the elliptical path or the
are farther from the sun than when they are nearer. This average distance between the sun and the planet
law is a direct consequence of law of conservation of If T is the time period of revolution of the planet
angular momentum. and 2a is the length of major axis of the elliptical path
Note: I) In case of planetory motion areal velocity traced by the planet, then T 2 a 3
dA 1 (r)( dt) 1
  r and as L = mvr  T1 
2
 a1 
3
dt 2 dt 2
i.e.,     
 T2   a2 

Semi major axis = average distance between the

sun and the planet.
dA L dA
so  vdt Problem : 5.59
dt 2m r
A Saturn year is 29.5 times the earth year. How
far is the Saturn from the sun if the earth is 1.5 ×
108 km away from the sun.
The gravitational force is a central force . Ans. Given that Ts = 29.5 Te and Re = 1.5 × 108 km
Therefore this force acting on the planet due to the sun According to Kepler’s law we can write, T2  R3
does not create any torque on it. In the absence of Ts2 R3s
 
torque angular momentum is conserved Te2 R3e
 L = mvr = Constant for a planet  T 2 / 3
 Rs  Re  s 
But as L = constt., areal velocity (dA/dt) =  Te 
constant which is Kepler’s II law, i.e., kepler’s II law  29.5Te 2 /3
 Rs  1.5108  
or constancy of areal velocity is a consequence of  Te 
law of conservation of angular momentum.
 Rs  1.43109 km .
ii) Let vp and vA be the magnitudes of the velocities of
the planet at the perihelion and aphelion. S denotes the
position of the sun.

VA
P S
A
VP
2. What is a Black hole ? Describe the different for it.
stages in the formation of ‘Black holes’. Can any 18. What is escape velocity ? Obtain an expression
material object escape from a black hole ? If it for it.
can escape what velocity should it have ? 19. What is a geo stationary satellite ? State its uses.
SAQ : 20. There is no atmosphere on the moon’s surface.
1. What is the acceleration due to gravity (g) and Explain.
universal gravitational constant (G). Derive the VSAQ :
relation between them ? 1. Which is the weakest force of all the basic forces
2. State the different types of basic forces in nature ? in nature ?
3. Among the following, pick out the basic force 2. Which is the strongest force of all the basic forces
involved. in nature ?
(a) Force of friction (b) Force between 3. Among the following, which of the forces are long
two neutrons inside a nucleus (c) β - decay range ?
(d) Muscular strength (e) Moon revolving i) Force between two nucleons ii) Force be-
round the earth (f) Surface tension tween two electric charges
(g) Tension on a string (h) Tides iii) Force between the earth and the moon
4. State and exlain Newton’s law of gravitation and 4. State the units and dimensions of universal gravi-
on what factor ‘G’ depend ? tational constant.
5. Obtain an expression for the acceleration due to 5. State the vector forms of Newton’s law of gravi-
gravity at a height ‘h’ from the surface of a planet. tation.
6. Derive an expression for the acceleration due to 6. The gravitational force of the earth on the moon
gravity at a depth d. is F. What is the gravitational force of the moon
7. Plot a graph between the acceleration due to on the earth ? Do these forces form an action -
gravity and the distance from the centre of the reaction pair ?
earth. State the nature of the obtained curve. 7. State the relation betwen g and G in terms of (i)
8. How does ‘g’ change for the same values of mass and radius (ii) density and radius of a planet.
height h and depth d. 8. Keeping the mass of the earth constant, if the
9. Derive an expression for the variation of ‘g’ with radius of the earth decreases by 2%. What would
latitude. be the change in ‘g’ ?
10. Deduce a relation between ‘g’ at poles and ‘g’ 9. Keeping the density of the earth constant, if the
at equator. radius of the earth decreases by 2%/ What woild
11. Discuss the effect of the shape of the earth on ‘g’. be the change in ‘g ?
12. Define gravitational field and gravitational field 10. As we go from on planet to another planet, how
strength. State the units and dimensions of gravi- will (a) the mass and (b) weight of a body
tational field strength. change.
13. State the properties of gravitational field. 11. Keeping the length of a simple pendulum the
14. Distinguish between inertial and non - inertial same, will the time period be the same on all plan-
frames of reference. ets ? Give also the reason.
15. Distinguish between inertial and gravitational 12. Give the equation for the value of g at a height h
masses. above the surface of the earth. How does it
16. State and explain the principle of equivalence. change at small heights ?
17. What is orbital velocity ? Obtain an expression 13. Give the equation for the value of g at a depth
2. Arrange the basic forces in the ascending order same, does ‘g’ change at a point which is at a
of their strengths and mention which of the forces, height h from its surface before shrinking ?
weak nuclear and gravitational is weaker ? Ans. No, Since r & M does not change
Ans. Gravitational force < weak nuclear force < elec- 10. According to Newton’s third law of motion, if
tromagnetic force < strong nuclear force. Gravi- the earth pulls the stone, stone must also pull the
tational force is weaker than weak nuclear force. earth. Then why doesn’t the earth fall towards
3. Why can’t two persons each of say 100 kg stand- the stone ?
ing 1 m part do not experience any gravitational Ans. Stone also pulls the earth with the same force as
attraction between them ? the earth pulls the stone, but due to the large mass
Ans. According to Newton’s law of gravitation the of the earth its acceleration towards the stone is
force of attraction between the two persons turns quite negligible and we just can’t notice it.
out to be 10–7N, which is very very small. Hence 11. Can the sun become a black hole ?
the persons in question do not experience the Ans. No, since a star can become a black hole only
minute force that exists between them. when its initial mass is 10 times the mass of the
4. What causes the tides in the oceans ? sun.
Ans. The gravitational force of attraction between the 12. Why does a tea spoon full of a neutron star weighs
moon and the earth causes the tides. almost thousands of tonnes ?
5. If heavier bodies are attracted more strongly by Ans. As density of a neutron star is very very high of
the earth, why don’t they fall faster than the lighter the order of 1017 kg/m3
bodies ? 13. What stops the collapse of a white dwarf and a
Ans. As acceleration due to gravity (g) is same for Neutron star ?
both heavier and lighter bodies, both fall with the Ans. Electron degeneracy pressure stops the collapse
same rate. of white dwarf whereas Neutron degeneracy
6. Cavendish is credited with weighing the earth. pressure stops the collapse of Neutron stars.
How did he do that ? 14. At what stage of the star do most of the ele-
Ans. Cavendish determined experimentally the value ments originate ?
GM Ans. supernova explosion
of G. Then he used the equation g  2 to de- 15. Which type of frame of reference does a bus
R
starting from rest represent ?
termine the mass, M of the earth. Ans. Non inertial frame of reference, since it is accel-
7. How does the acceleration due to gravity change erating at that moment
inside the mines ? 16. Which mass gravitational or inertial does the mass
Ans. Decreases of a celestial bodies refer to ?
8. Consider two points. One above the surface of Ans. Inertial mass
the earth and other inside the surface towards 17. Does principle of equivalence applicable to elec-
the centre at the same distance of 1 km. Where tromagnetic phenomena ?
is ‘g’ greater ? Ans. Yes
Ans. Below the surface of t he earth since 18. What is the escape velocity of an object from
 d  2h  the surface of a black hole ?
g d  g  1   and g h  g  1   , If
 R  R Ans. Greater than or equal to the velocity of light,since
d = h  g d  gh
9. If the earth suddenly shrinks, keeping its mass