Gravitation

The Discovery of the Law of Gravitation

The way the law of universal gravitation was discovered is often considered the paradigm of modern scientific
technique. The major steps involved were
• The hypothesis about planetary motion given by Nicolaus Copernicus (1473–1543).
• The careful experimental measurements of the positions of the planets and the Sun by Tycho Brahe
(1546–1601).
• Analysis of the data and the formulation of empirical laws by Johannes Kepler (1571–1630).
• The development of a general theory by Isaac Newton (1642–1727).

Universal Law of Gravitation: Newton's Law

According to this law "Each particle attracts every other particle. The force of attraction between them is directly
proportional to the product of their masses and inversely proportional to square of the distance between them".

m1m2 m1m2
F or F =G
r2 r2
where G = 6.67 × 10–11 Nm2 kg–2 is the universal gravitational constant.

Dimensional Formula of G:
Fr 2  MLT −2  L2 
G= = =  M −1 L3T −2 
m1m2 M2 

Vector Form of Newton's Law of Gravitation

Let r12 = Displacement vector from 𝑚1 to 𝑚2
Y m1 F12
r21 = Displacement vector from 𝑚2 to 𝑚1 r12 F21
r21
r1 m2
F21 = Gravitational force exerted on 𝑚2 by 𝑚1
r2
F12 = Gravitational force exerted on 𝑚1 by 𝑚2
O X
Gm m Gm m Z
F12 = − 12 2 rˆ21 = − 13 2 r21
r21 r21

Negative sign shows that :

(i) The direction of F12 is opposite to that r̂21

(ii) The gravitational force is attractive in nature

Gm1m2 Gm m
Similarly, F21 = − rˆ12 or F21 = − 13 2 r12  F12 = −F21
r122 r12

The gravitational force between two bodies are equal in magnitude and opposite in direction.

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Important characteristics of gravitational force:

(i) Gravitational force between two bodies form an action and reaction pair i.e. the forces are equal in
magnitude but opposite in direction.

(ii) Gravitational force is a central force i.e. it acts along the line joining the centers of the two interacting bodies.

(iii) Gravitational force between two bodies is independent of the nature of the medium, in which they lie.

(iv) Gravitational force between two bodies does not depend upon the presence of other bodies.

(v) Gravitational force is negligible in case of light bodies but becomes appreciable in case of massive bodies
like stars and planets.

(vi) Gravitational force is long range-force i.e., gravitational force between two bodies is effective even if their
separation is very large.

For Illustration gravitational force between the Sun and the Earth is of the order of 1027 N although distance
between them is 1.5 × 107 km

Illustration 1:

Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm. Assuming that the only forces acting on
the particles are their mutual gravitation, find the initial acceleration of heavier particle.
Solution:

Gm1m2 6.67  10−11  1  2

Force exerted by one particle on another F = = = 5.3  10−10 N
r2 (0.5)2

F 5.3  10−10
Acceleration of heavier particle = = = 2.65  10−10 ms −2
m2 2

This Illustration shows that gravitation is very weak but only this force keep bind our solar system and also this
universe, all galaxies and other interstellar system.

Illustration 2:

Two stationary particles of masses M1 and M2 are at a distance 'd' apart. A third particle lying on the line joining the
particles, experiences no resultant gravitational forces. What is the distance of this particle from M1.
Solution:

GM1 m M1 m M2
The force on m towards M1 is F1 = r
r2 d
GM2m
The force on m towards M2 is F2 =
(d − r )2

According to question net force on m is zero i.e. F1 = F2

2
 GM1m GM2m  d −r  M2
=  r  =M
r2 (d − r )
2
  1

d M2  M1 
 −1 =  r = d 
r M1  M1 + M2 

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Principle of superposition

The force exerted by a particle on other particle remains unaffected by the presence of other nearby particles in
space. Total force acting on a particle is the vector sum of all the forces acted upon by the individual masses when
they are taken alone.

F = F1 + F2 + F3 + .......

m1
F1
m F2 m2

F3
m3

Illustration 3:

Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side 'a'. The only forces
acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while
maintaining their original separation 'a'. Determine the initial velocity that should be given to each particle and time
period of circular motion.

m m

Solution:

The resultant force on particle at A due to other two particles is

Gm2  Gm2 
FA = FAB
2
+ FAC
2
+ 2FAB FAC cos60 = 3 ...(i)  FAB = FAC = 2 
a2  a 

2 a
Radius of the circle r =  a sin60 =
3 3

If each particle is given a tangential velocity v,

so that F acts as the centripetal force,

mv 2 mv 2
Now = 3 ...(ii)
r a

mv2 Gm2 3 Gm
From (i) and (ii) 3 =  v=
a a2 a

Time period, T = 2r = 2a a

= 2
a3
v 3 Gm 3Gm

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Illustration 4:
Four point masses each of mass 'm' are placed on the corner of square of side 'a'. Calculate magnitude of
gravitational force experienced by each particle.
a
m m

m m
Solution:
F
m m
45°
45° F
F 1

m m

Fr = resultant force on each particle = 2F cos 45°+ F1

2G.m2 1 Gm2 G.m2

=  + = ( 2 2 + 1)
a2 2 ( 2a )
2
2a

Illustration 5:

Find gravitational force exerted by point mass ‘m’ on uniform rod (mass ‘M’ and length ‘’).

Rod (M,ℓ)
dM
m
a
dx ℓ
x
Solution:
G.dM .m
dF = force on element in horizontal direction =
( x + a )2

M
where dM = dx.

G.Mmdx G.Mm dx G.Mm  1 1 GMm

 F =  dF =  =  =  − + =
0 ( x + a)2
0 ( x + a)
2
 ( + a ) a  ( + a)a

Gravitational Field
The space surrounding the body within which its gravitational force of attraction is experienced by other bodies is
called gravitational field. Gravitational field is very similar to electric field in electrostatics where charge 'q' is replaced
by mass 'm' and electric constant 'K' is replaced by gravitational constant 'G'. The intensity of gravitational field at a
point is defined as the force experienced by a unit mass placed at that point.

F
E=
m
The unit of the intensity of gravitational field is N kg–1.

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Intensity of Gravitational Field due to Point Mass

The force due to mass m on test mass m0 placed at point P is given by :
GMm0
F=
r2

F GM
Hence E =  E= m 𝑟Ƹ P
m0 r2
r
In vector form
GM
E =− rˆ
r2
F  MLT −2   0 −2 
Dimensional formula of intensity of gravitational field = = = M LT
m M 
Illustration 6:
Find the distance of a point from the Earth’s centre where the resultant gravitational field due to the Earth and the
moon is zero. The mass of the Earth is 6.0 × 1024 kg and that of the moon is 7.4 × 1022 kg. The distance between the
Earth and the moon is 4.0 × 105 km.
Solution:
The point must be on the line joining the centres of the Earth and the moon and in between them. If the distance of
the point from the Earth is x, the distance from the moon is (4.0 × 105 km-x). The magnitude of the gravitational field
due to the Earth is

GMe G  6  1024 kg
E1 = =
x2 x2
and magnitude of the gravitational field due to the moon is
GMm G  7.4  1022 kg
E2 = =
( 4.0  105 km − x ) 2
( 4.0  105 km − x )2

These fields are in opposite directions. For the resultant field to be zero 𝐸1 = 𝐸2 .

or, 6  1024 kg 7.4  1022 kg

=
x 2
( 4.0  105 km − x )2

x 6  1024
or, = =9
4.0  10 km − x
5
7.4  1022

or, x = 3.6 × 105 km.

Illustration 7:
Calculate gravitational field intensity due to a uniform ring of mass M and radius R at a distance x on the axis from
center of ring.
Solution:
A dm

R
θ P
O x

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Consider any particle of mass dm. Gravitational field at point P due to dm

Gdm
dE = along PA
r2
Gdm
Component along PO is dE cos  = cos
r2
Net gravitational field at point P is

Gdm G cos 
E=
r2 
2
cos  = dm
r
GMx
= towards the center of ring
( R2 + x 2 )3/2

Illustration 8:
Calculate gravitational field intensity at a distance x on the axis from centre of a uniform disc of mass M and radius R.
Solution:
Consider a elemental ring of radius r and thickness dr on surface of disc as shown in figure

Gravitational field due to elemental ring

GdMx
dE = Here Disc (M, R)
( x + r 2 )3/2
2

M 2M dr
dM = 2 .2rdr = 2 rdr
R R r
G.2Mxrdx P
 dE =
dE
R2 ( x 2 + r 2 )
3/2
x
R
  2GMx  rdr
E =   2
0 R  ( x + r 2 )3/2
2

2GMx  1 1 
 E= −
R2  x 
x +R 
2 2

Illustration 9:
For a given uniform spherical shell of mass M and radius R, find gravitational field at a distance r from centre in
following two cases:

(a) r  R (b) r < R

Solution: Ring of radius = Rsin 
Rd 
GdM
dE = 2
.cos  r R d ℓ

M  
dM =  2R sin Rd P
4R2 r

M
dM = sin  d
2

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GM sin  cos  d
 dE =
22
Now  2 = R2 + r 2 – 2rR cos .....(i)

R 2 = 2 + r2 – 2r cos .....(ii)

2
+ r 2 − R2
 cos  =
2r
R2 + r 2 − 2
cos  =
2rR
Differentiating (1)
 2  d  = 2rR sin d

GM d 2
+ r 2 − R2 GM  r 2 − R2 
 dE =    dE = 1 + d 
2 2 Rr (2 r ) 4Rr 2  2

GM  r −R r −R d  GM
 E =  dE = 2  r +R
d + ( r 2 − R2 )    E = 2 ,r R
4Rr  r +R 2
 r
If point is inside the shell limit changes to [(R – r) to R + r]
E = 0 when r < R.

Illustration 10:
Find the relation between the gravitational field on the surface of two planets A and B of masses mA, mB and radius RA
and RB respectively if
(i) they have equal mass
(ii) they have equal (uniform) density
Solution:
Let EA and EB be the gravitational field intensities on the surface of planets A and B.
4
G R3A A
GmA 3 4G
then, EA = 2 = =  A RA
RA RA2 3
GMB 4G
Similarly, E A = =  B RB
RB2 3

E A RB2
(i) for mA = mB =
E B RA2
E A RA
(ii) For and  A = B =
E B RB

Gravitational Potential
The gravitational potential at a point in the gravitational field of a body is defined as the amount of work done by an
external agent in bringing a body of unit mass from infinity to that point, slowly (no change in kinetic energy).
Gravitational potential is very similar to electric potential in electrostatics.

r
Q P
M dr

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Gravitational Potential Due to a Point Mass

Let the unit mass be displaced through a distance 𝑑𝑟 towards mass M, then work done is given by

GM
dW = F dr = dr
r2

Total work done in displacing the particle from infinity to point P is

r
GM −GM
W =  dw =  2
dr =
 r
r

Thus, gravitational potential, V = − GM .

r

The unit of gravitational potential is J kg–1. Dimensional Formula of gravitational potential

Work  ML2T −2   0 2 −2 
= = = M LT
mass M 

Illustration 11:

Find out potential at P and Q due to the two point mass system. Find out work done by external agent in bringing
unit mass from P to Q. Also find work done by gravitational force.

m
1
ℓ
ℓ/2
ℓ Q P
ℓ/2 ℓ
2
m
Solution:

Gm
(i) VP1 = potential at P due to mass ‘m’ at ‘1’ = −

Gm
VP 2 = −

2Gm
 VP = VP1 + VP2 = −

Gm Gm
(ii) VQ1 = −  VQ 2 = −
/2 /2

Gm Gm 4Gm
 VQ = VQ1 + VQ2 = − − =−
/2 /2

Force at point Q = 0

2GM
(iii) work done by external agent = (VQ – VP) × 1 = −

2GM
(iv) work done by gravitational force = VP – VQ =

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Illustration 12:

Find potential at a point ‘P’ at a distance ‘x’ on the axis away from centre of a uniform ring of mass M and radius R.

Solution:
Ring R,M

ඥ𝑅2 + 𝑥2
R
P
x

Ring can be considered to be made of large number of point masses (m1, m2 ..........etc)

Gm1 Gm2
VP = − − − .......
R +x
2 2
R2 + x 2

G GM
=− ( m1 + m2 ..........) = ,
R +x
2 2
R2 + x 2

where M = m1 + m2 + m3 + ............

G .M
Potential at centre of ring = −
R

Relation between Gravitational Field and Potential

The work done by an external agent to move unit mass from a point to another point in the direction of the field E, slowly
through an infinitesimal distance dr = Force by external agent × distance moved = – Edr.

Thus dV = – Edr

dV
 E =− .
dr

Therefore, gravitational field at any point is equal to the negative gradient at that point.

Illustration 13:

The gravitational field in a region is given by E = 20(iˆ + ˆj ) N / kg Find the gravitational potential at the origin
(0, 0) (in J/kg)

(A) zero (B) 20 2 (C) −20 2 (D) can not be defined

Ans. (A)

Solution:

V = − E .dr =   Ex .dx +  Ey.dy  = 20x + 20 y

At origin V = 0.

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Illustration 14:
In above problem, find the gravitational potential at a point whose co-ordinates are (5, 4) : (in J/kg)
(A) – 180 (B) 180 (C) – 90 (D) zero
Ans. (B)
Solution:
V = 20 × 5 + 20 × 4 = 180 J/kg

Illustration 15:
In the above problem, find the work done in shifting a particle of mass 1 kg from origin (0, 0) to a point (5, 4): (In J)
(A) – 180 (B) 180 (C) – 90 (D) zero
Ans. (B)
Solution:
W = m (Vf – Vi) = 1 (180 – 0) = 180 J

Illustration 16:
Find gravitational field at a point (x, y, z).
v = 2x2 + 3y2 + zx,
Solution:
−v
Ex = = – 4x – z
x
E y = –by
Ez = – x

 Field = E = – [(4x +z) iˆ + by ĵ + x k̂ ].

E = −V

Analogy between Electrostatics and Gravitation

(1) Point Charge Point Mass
kQ GM
(a) E= g=
r2 r2
kQ −GM
(b) V= VG =
r r
(2) Uniform charged ring Ring of uniform mass distribution
kQx GMx
(a) E= on axis g= on axis
(r 2 + x 2 )3/2 (r 2 + x 2 )3/2

r r
E is max. when x = g is max. when x =
2 2

kQ kQ −GM −GM
(b) V= on axis, at center VG = on axis, at center
r2 + x2 r r2 + x2 r

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(3) Uniform linear charge Uniform linear mass

Gλ
 ሺcosβ − cosαሻ
r  𝑘𝜆 r
r 
 sinሺα + βሻ
𝑟 Gλ
ሺsinα − sinβሻ
𝑘𝜆 r
ሺcosβ − cosαሻ
𝑟 (=mass per unit length)
(4) Infinite Linear charge Infinite linear mass
∞ ∞
r2 r2
r1 r1

A B A B
∞ ∞

2K  2G
(a) E= g=
r r
r2
(b) VB–VA= –2K ln VB–VA= 2G ln  r2 
r1  r1 

(5) Infinite Sheet of charge Infinite Sheet of mass

 
E= g =  4G = 2G
20 2

= 2K  ( = mass per unit area)

• Notice gravitational force is always attractive and hence gravitational potential is always –ve. (for a
repulsive force potential is positive). This can be explained from the sign of Wext in moving the test
charge from  to the point under consideration.

• Since g points from B towards A potential increases as we move from A to B. Just like electric potential
gravitational potential also increases opposite to field direction.

(6) Uniformly charged hollow sphere Hollow sphere of uniform mass

Charge Q, radius R Mass M , radius R

distance of field point from center r distance of field point from center r

Case I. r>R
Q
r P
R GM
g=
r2
kQ GM
E= VG = −
r2 r
kQ
V=
r

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Case II. r<R

Q
R P
r
g=0

GM
E=0 VG = −
R
kQ
V=
r
(7) Electrostatics self energy of uniformly Gravitational self energy of uniform

charged thin spherical shell. thin spherical shell.

KQ2 GM 2
U= U=
2R 2R
(8) Uniformly charged solid sphere Uniformly solid sphere

mass M, radius R mass M, radius R

kQ GM
E= ,r>R g= ,r>R
r2 r2
kQr GM
,r<R r,r<R
R3 R3
kQ GM
V= ,r>R Va = − ,r>R
r r
kQ −GM
2R3
( )
3R3 − r 2 , r > R
2R3
( )
3R3 − r 2 , r > R

(9) Electrostatics self energy of uniformly Gravitational self energy of uniform

charged solid sphere. solid sphere.

3 KQ 2 3 GM 2
U= U=
5 R 5 R

Illustration 17:

A uniform solid sphere of density  and radius R has a spherical cavity of radius r inside it as shown. Find gravitation
field at

(a) O (b) C (c) P (prove that field inside cavity is uniform)

P
O C

4 
k  r 3  
 3  CO 4GOC 4
Ans. (a) (b) (c) GO1O2
(OC )2 3 3

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Solution:
g = gravitational field at any point inside sphere

GM G 4
g= 3
r = 3 R3r
R R 3
4
g = G r
3
Let the sphere with cavity is formed by superimposing it with a small sphere of density (–) as shown


P P g1 P
+ g2
O1 O2 O1 – O2

Resultant field g = g1 + g2
4  4 
=  G  O1 P +  G  PO2
 3   3 
4
= GO1 O2 O1 O2 = OC 
3  
It is independent of position of point inside cavity

At O g = g1 + g2 = 0 +
GM
(CO )
(CO )
2

4
G r 2 
= 3 CO
(CO )
2

4 2 
 3 Gr  
=   CO
(CO )
2

Illustration 18:
(a) Find where is gravitational field intensity is zero.
(b) Find gravitational potential at P.
M 16M

Solution:
GM G16M
(a) =
x2 (D − x )2
P
D−x D
y=  x= x (D – x)
x 5
−GM  −G(16M )  25GM
(b) vP = +  =−
D 4D D
 
5  5 

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Illustration 19:

GFI at P = ?

R/2 P
R

Solution:
 43 R = M
3

 R
3
– 43  R2 = M'

GM GM '
EP = +
(2R ) 2
(3R / 2)2

Illustration 20:

A particle is released at a distance x from centre of Earth as shown in figure. Show that it performs SHM and find T.

Me

Solution:
F = m × GFI

 M x
ma = m  G e3 
 R 

GMe x
a=
R3

2

R3
T = 2
GMe

Illustration 21:

A particle is released in a tunnel across Earth as shown in figure. Show that it performs SHM and find T.

x
r 

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Solution:

 GM r 
Fx =  3e sin   m = ma
 R 

 GM 
a =  3e  x
 R 
2

R3
T = 2 (as x = r sin)
GMe

Gravitational Potential Energy

Gravitational potential energy of two mass system is equal to the work done by an
external agent in assembling them, while their initial separation was infinity. Consider a
body of mass m placed at a distance r from another body of mass M. The gravitational
GMm
force of attraction between them is given by, F =
r2
Now, Let the body of mass m is displaced from point. C to B through a distance 'dr'
towards the mass M, then work done by internal conservative force (gravitational) is
given by,

GMm r
GMm
dW = F dr = dr   dW =  dr
r2  r
2

 Gravitational potential energy, U = − GMm

r

Illustration 22:

Mass 𝑀1 is fixed and mass, 𝑚 is released from rest at infinity, find its velocity, when separation between them is ′𝑟′.
M1 
fix m released from rest
Solution:
WG + Wext = KE f − KEi
 
0 0

Ui – Uf = KEf – 0

GM1m  GM1m  1 2
− −−  = Mv − 0
  r  2

GM1m 1 2
= Mv
r 2

2GM1
v=
r

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Illustration 23:
Particle is projected vertically with speed 0, particle reaches maximum height of 3R, find 0 (where 𝑅 is radius of
Earth)
µ0

R
M2
Earth

Solution:
Height always measured from surface.
WG = KE f − KE i

0

Ui – Uf = KEf – 0
GMe m  GMe m  1
− −−  = − Mµ0
2

R  4R  2

GMe m GMe m 1
− + = − Mµ02
R 4R 2
1 2 3GMe m
Mµ0 =
2 4R
3GMe
µ0 =
2R

Illustration 24:
Distance between centres of two stars is 10 a. The masses of these stars are M and 16 M and their radii are a and 2a
respectively. A body is fired straight from the surface of the larger star towards the smaller star. What should be its
minimum initial speed to reach the surface of the smaller star?
Solution:
Let P be the point on the line joining the centres of the two planets s.t. the net field at it is zero
GM G.16M
Then, − =0  (10 a–r)2 = 16 r2
r 2
(10a − r )2

 10a – r = 4r  r = 2a
Potential at point P,
−GM G.16M −GM 2GM −5GM
VP = − = − =
r (10a − r )2
2a a 2a

Now if the particle projected from the larger planet has enough energy to cross this point, it will reach the smaller
planet.
For this, the K.E. imparted to the body must be just enough to raise its total mechanical energy to a value which is
equal to P.E. at point P.
1 2 G (16M ) m −GMm
i.e., mv − = mVP
2 2a 8a

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v 2 −8GM −GM −5GMm

or =
2 a 8a 2a

45GM
or v2 =
4a

3 5GM
or vmin =
2 a

Gravitational Self-Energy
The gravitational self-energy of a body (or a system of particles) is defined as the work done by an external agent in
assembling the body (or system of particles) from infinitesimal elements (or particles) that are initially an infinite
distance apart.

Gravitational Self Energy of a System of n Particles

Potential energy of n particles at an average distance 'r' due to their mutual gravitational attraction is equal to the
sum of the potential energy of all pairs of particle, i.e.,

mi m j
Us =– G 
all pairs rij
j i

This expression can be written as

1 i =n j =n mi mj
Us = − G 
2 i =1 j =1 rij
j i

If consider a system of 'n' particles, each of same mass 'm' and separated from each other by the same average
distance 'r', then self energy

1 n n
 m2 
or Us = − G   
2 i =1 j =1 r ij
j i

1 m2
Thus, on the right hand side 'i' comes 'n' times while 'j' comes (n – 1) times. Thus U s = − Gn( n − 1) .
2 r

Gravitational Self Energy of a Uniform Sphere (Star)

4 3 
 r   ( 4r dr )
2

 3  M 1
= − G ( 4 ) r 4dr
2
U sphere = −G where  =
r 4
  3 3
  R
3
R 2
2 r 
R
1 1 5
3  4  1
U star = − G ( 4 )  r 4dr = − G ( 4 )   = − G  R3 
2

3 0
3  5  0 5  3  R

3 GM 2
 U star = −
5 R

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Acceleration Due to Gravity

It is the acceleration, a freely falling body near the Earth’s surface acquires due to the Earth’s gravitational pull. The
property by virtue of which a body experiences or exerts a gravitational pull on another body is called gravitational
mass mG, and the property by virtue of which a body opposes any change in its state of rest or uniform motion is
called its inertial mass mI thus if E is the gravitational field intensity due to the Earth at a point P, and g is
acceleration due to gravity at the same point, then mI g = mG E .

Now the value of inertial and gravitational mass happen to be exactly same to a great degree of accuracy for all
bodies. Hence, g = E

The gravitational field intensity on the surface of Earth is therefore numerically equal to the acceleration due to
gravity ሺ𝑔ሻ, there. Thus, we get,

where, Me = Mass of Earth

Re = Radius of Earth

Note:
Here the distribution of mass in the Earth is taken to be spherical symmetrical so that its entire mass can be assumed
to be concentrated at its center for the purpose of calculation of g.

Variation of Acceleration Due to Gravity

(a) Effect of Altitude:
GMe
Acceleration due to gravity on the surface of the Earth is given by, g = .
Re2

Now, consider the body at a height 'h' above the surface of the Earth, then the acceleration due to gravity at
height 'h' given by
−2
GMe  h  2h 
gh = = g1 +  g  1 −  when h << R.
( Re + h)
2
 Re   Re 

2 gh
The decrease in the value of 'g' with height h = g – gh = . Then percentage decrease in the value of
Re
g − gh 2h
'g' =  100 =  100% .
g Re

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(b) Effect of Depth:

GMe m
The gravitational pull on the surface is equal to its weight i.e. mg =
Re2
4
G  Re3m
3 4
 mg = 2 or g = GRe  ...(i)
Re 3

When the body is taken to a depth d, the mass of the sphere of radius (Re – d) will only be effective for the
gravitational pull and the outward shall will have no resultant effect on the mass. If the acceleration due to
gravity on the surface of the solid sphere is gd, then
4
gd = G ( Re − d )  ...(ii)
3
By dividing equation (2) by equation (1)

 gd = g  1 − d 
 Re 

Important Points:

(i) At the center of the Earth, d = Re, so gcentre = g  1 − R e  = 0 .

R e 
Thus weight (mg) of the body at the centre of the Earth is zero.

g − gd 
Percentage decrease in the value of 'g' with the depth = 
d
(ii)   100 =  100 .
 g  Re

(c) Effect of the Surface of Earth

The equatorial radius is about 21 km longer than its polar radius.
GMe
We know, g = Hence gpole > gequator. The weight of the body increases as the body taken from the
Re2
equator to the pole.

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(d) Effect of Rotation of the Earth

The Earth rotates around its axis with angular velocity . Consider a particle of mass m at latitude . The
angular velocity of the particle is also .

According to parallelogram law of vector addition, the resultant force acting on mass m along PQ is
F = [(mg)2 + (m2 Re cos)2 + {2mg × m2 Re cos} cos (180 – )]1/2
= [(mg)2 + (m2 Re cos)2 – (2m2 g2 Re cos) cos]1/2

  R 2 2 R 2 
= mg 1 +  e  cos2  − 2 e cos2  
  g  g 

 
At pole  = 90°  gpole = g, At equator  = 0  gequator = g 1 − Re   .
2

 g 

Hence gpole > gequator

 
If the body is taken from pole to the equator, then g ' = g 1 − Re   .
2

 g 

 R 2 
mg − mg  1 − e 
 g  mRe 2 R 2
Hence % change in weight =  100 =  100 = e  100
mg mg g

Key Points:
1. Apparent weight is equal to normal reaction is equal to reading of weighing machine is equal to
m × gapparent.
[Apparent weight = Normal reaction = Reading of weighing machine = m × gapparant]
2. Motion of an object is referred as free fall if its acceleration is same as acceleration due to gravity.

Illustration 25:
At same planet gapp at equator is equal to half of gapp of at poles.
(a) Find  in terms of g and R.
Solution:
1 1 g g
(a) ge = gp  g − R2 = g  2 R =  =
2 2 2 2R

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Weightlessness
 GM 
State of the free fall  a = − 2 r  is called state of weightlessness. If a body is in a satellite (which does not produce
 r 
its own gravity) orbiting the Earth at a height h above its surface then
mGM mg
True weight = mgh = = ; Apparent weight = m(gh – a)
(R + h)2  h
2

1 + 
 R
v02 GM GM
But, a = = 2 = = gh ; Apparent weight = m(gh – gh) = 0
r r (R + h)2

Note:
The condition of weightlessness can be overcome by creating artificial gravity by rotating the satellite in addition to
its revolution.

Illustration 26:
Find ratio of gravitational field on the surface of two planets which are of uniform mass density and have radius R1
and R2 if
(a) They are of same mass
(b) They are of same density
g1 R22 g1 R
Ans. (a) = (b) = 1
g2 R12 g2 R2

Solution:
4
G R3
GM  4GR 
g= 2 = 3 2 = 
R R  3 

Escape Speed
The minimum speed required to send a body out of the gravity field of a planet (send it to r → ).

Escape Speed at Earth's Surface

Suppose a particle of mass m is on Earth's surface. We project it with a velocity V from the Earth's surface, so that it
just reaches r →  (at r → , its velocity become zero). Applying energy conservation between initial position (when
the particle was at Earth's surface) and find positions (when the particle just reaches to r → )

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Ki + Ui = Kf + Uf

1 2  Gm   GMe  2GM0
mv + m0  − e  = 0 + m0  −
( )  v=
2  R   r →   R

2GMe
Escape speed from Earth is surface ve =
R

If we put the values of G, Me, R the we get Ve = 11.2 km/s.

Escape Speed Depends on:

(i) Mass (Me) and size (R) of the planet
(ii) Position from where the particle is projected.

Escape speed does not depend on:

(i) Mass of the body which is projected (m0)
(ii) Angle of projection.
If a body is thrown from Earth's surface with escape speed, it goes out of Earth's gravitational field and never returns
to the Earth's surface. But it starts revolving around the sun.

Illustration 27:
A very small groove is made in the Earth, and a particle of mass m0 is placed at R/2 distance from the centre. Find the
escape speed of the particle from that place.

Solution:
Suppose we project the particle with speed v, so that it just reaches at (r → ).
Applying energy conservation Ki + Ui = Kf + Uf

1  GM  2  R 2  
m0v + m0  − 3e  3R −     = 0 + 0
2

2  2R   2  

11GMe
v= = ve at that position.
4R

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Illustration 28:
Find radius of such planet on which the man escapes through jumping. The capacity of jumping of person on Earth is
1.5 m. Density of planet is same as that of Earth.
Solution:
GMP m GMP m
For a planet : 1 mv 2 − 1
= 0  mv 2 −
2 Rp 2 Rp

On Earth → 1 mv 2 = m  GM2E  h
2  RE 

GM p m GME .m Mp M h
 = h  = E2
Rp RE2 Rp RE

4 / 3Rp3 4 / 3RE3   RP = RE h
 Density () is same  =
Rp RE2

Kepler’s Law for Planetary Motion

Suppose a planet is revolving around the sun, or a satellite is revolving around the Earth, then the planetary motion
can be studied with help of Kepler’s three laws.

Kepler’s Law of Orbit

Each planet moves around the Sun in a circular path or elliptical path with the Sun at its focus. (In fact circular path is
a subset of elliptical path)

Planet

SUN

Law of Areal Velocity

To understand this law, lets understand the angular momentum conservation for the planet.
If a planet moves in an elliptical orbit, the gravitation force acting on it always passes through the centre of the sun.
So, torque of this gravitation force about the centre of the Sun will be zero. Hence, we can say that angular
momentum of the planet about the centre of the Sun will remain conserved (constant)
 about the Sun = 0
dL
 =0  Lplanet / Sun = constant  mvr sin  = constant
dt
Now we can easily study the Kepler's law of aerial velocity.

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If a planet moves around the sun, the radius vector ( r ) also rotates are sweeps area as shown in figure. Now let's find
rate of area swept by the radius vector ( r ).

Suppose a planet is revolving around the Sun and at any instant its velocity is v, and angle between radius vector ( r )
and velocity ( v ). In dt time, it moves by a distance vdt, during this dt time, area swept by the radius vector will be OAB
which can be assumed to be a triangle

dA = 1/2 (Base) (Perpendicular height)

dA = 1/2 (r) (vdt sin )
dA 1
So, rate of area swept = vr sin 
dt 2
dA 1 mvr sin 
We can write =
dt 2 m
Where mvr sin  = angular momentum of the planet about the sun, which remains conserved (constant)
dA Lplanet /sun
 = = constant
dt 2m
So, rate of area swept by the radius vector is constant.

Illustration 29:
x 2 y2
Suppose a planet is revolving around the Sun in an elliptical path given by + = 1 . Find time period of
a 2 b2
revolution. Angular momentum of the planet about the Sun is L.

Solution:
dA L
Rate of area swept = = constant
dt 2m
L A=ab t =T
L L 2mab
 dA = dt ;  dA =  2mdt  ab = TT=
2m A=0 t =0
2m L

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Kepler’s Law of Time Period

m0v2 GMS m0
Suppose a planet is revolving around the Sun in circular orbit then =
r r2
GMs
v=
r

Time period of revolution is T = 2r = 2r r

v GM s

 42  3  T 2 = r 3
T2 =  r
 GM s 

For all the planet of a Sun, T  r

2 3

Illustration 30:
The Earth and Jupiter are two planets of the sun. The orbital radius of the Earth is 107 m and that of Jupiter is
4 × 107 m. If the time period of revolution of Earth is T = 365 days, find time period of revolution of the Jupiter.

Jupiter
<

2 7
r = 4 × 10 m
<

T2 = ?
SUN

7
r1 = 10 m
< Earth

T1 = 365 days

<
Solution:
For both the planets
T 2  r3
2 3 2 3
 TJupiter   TJupiter   TJupiter   4  107 
  =     = 7 
 TEarth   rEarth   365 days   10 
Tjupiter = 8 × 365 days
Graph of T vs r :
Graph of log T v/s log R :
T

Tr
3/2

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 42  3  2log T = log  42  + 3log R

T2 =  R  
 GM s   GM s 

1  42  3
T= log   + log R
2  GMs  2

log T

3
slope =
2
1  4 2 

C= log
2  GMs 

log R

• If planets are moving in elliptical orbit, then T2  a3 where a = semi major axis of the elliptical path.

Circular Motion of a Satellite Around a Planet

Suppose at satellite of mass m0 is at a distance r from a planet. If the
satellite does not revolve, then due to the gravitational attraction, it
may collide to the planet.
To avoid the collision, the satellite revolve around the planet, for
circular motion of satellite.

GMe m0 m0v2
 = ....(i)
r2 r
GMe
 v= this velocity is called orbital velocity ሺ𝑣0 ሻ
r

GMe
v0 =
r

Total Energy of the Satellite Moving in Circular Orbit

1
(i) KE = m0v 2 and from equation (1)
2
m0v2 GMe m0 GMe m0 1 GMe m0
=  m0v 2 =  KE = m0v 2 =
r r2 r 2 2r
GMe m0
(ii) Potential energy U = −
r
 GMe m0   −GMe m0 
Total energy = KE + PE =  + 
 2r   r 

GMe m0
TE = −
2r
Total energy is negative. It shows that the satellite is still bounded with the planet.

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Essential Condition's for Satellite Motion

• Centre of satellite's orbit coincide with centre of Earth.
• Plane of orbit of satellite is passing through centre of Earth.

m
v0
M

Geo-Stationary Satellite
We know that the Earth rotates about its axis with angular velocity Earth and time period TEarth = 24 hours.
Suppose a satellite is set in an orbit which is in the plane of the equator, whose  is equal to Earth,
(or its T is equal to TEarth = 24 hours) and direction is also same as that of Earth. Then as seen from Earth, it will appear
to be stationery. This type of satellite is called geo- stationary satellite. For a geo-stationary satellite,

wsatellite = wEarth
 Tsatellite = TEarth = 24 hr.
So, time period of a geo-stationery satellite must be 24 hours. To achieve T = 24 hour, the orbital radius
geo-stationary satellite :
 42  3
T2 =  r
 GMe 

Putting the values, we get orbital radius of geo stationary satellite r = 6.6 Re (here Re = radius of the Earth)
height from the surface h = 5.6 Re.

Special Points about Geo–Stationary Satellite

• All three essential conditions for satellite motion should be followed.
• It rotates in equatorial plane.
• Its height from Earth surface is 36000 km. ሺ~6𝑅𝑒 ሻ
• Its angular velocity and time period should be same as that of Earth.

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• Its rotating direction should be same as that of Earth (West to East).

• Its orbit is called parking orbit and its orbital velocity is 3.1 km./sec.

• Maximum latitude at which message can be received by geostationary satellite is  = cos−1  Re 


R
 e + h 

2hRe2
• The area of Earth's surface covered by geostationary satellite is S = Re2 =
Re + h

• The area of Earth's surface escaped by geostationary satellite is

2 = 90 – 
R
R 
cos 2 = 2
R+h 
R h
4R2 − 2R2(1 − cos 2 ) = 2r 2 + 2r 2 cos(90 − )

= 2R2(1 + sin )

Polar Satellite (Sun–Synchronous Satellite)

It is that satellite which revolves in polar orbit around Earth. A polar orbit is that orbit whose angle of inclination with
equatorial plane of Earth is 90° and a satellite in polar orbit will pass over both the north and south geographic poles
once per orbit. Polar satellites are Sun–synchronous satellites. Every location on Earth lies within the observation of
polar satellite twice each day. The polar satellites are used for getting the cloud images, atmospheric data, ozone
layer in the atmosphere and to detect the ozone hole over Antarctica.
Polar orbit

Equatorial
plane
Equator

Equatorial
orbit

Only the equatorial orbits are stable for a satellite. For any satellite to orbit around in a stable orbit, it must move in
such an orbit so that the centre of Earth lies at the centre of the orbit.

Binding Energy
Total mechanical energy (potential + kinetic) of a closed system is negative. The modulus of this total mechanical
energy is known as the binding energy of the system. This is the energy due to which system is closed or different
parts of the system are bound to each other.

Binding Energy of Satellite (System):

1 2 GMm L2 − P.E.
B.E. = – T.E. B.E. = mv0 = = Hence B.E. = K.E. = – T.E. =
2 2r 2mr 2
2

Work Done in Changing the Orbit of Satellite:

− GMm
W = Change in mechanical energy of system but E = , so W = E2 – E1 = GMm  1 – 1 
2r 2  r1 r2 

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Illustration 31:
A satellite moves eastwards very near the surface of the Earth in equatorial plane with speed (v0). Another satellite
moves at the same height with the same speed in the equatorial plane but westwards. If R = radius of the Earth and
 be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period
as observed on the Earth.
Solution:
2R 2R
Twest = and Teast =  T = Teast – Twest = 2R  2R 
2 2 
4R2
= 2
v0 + R v0 − R  v − R   v0 − R 
2
0
2 2

Illustration 32:
Find Wext to change orbital radius of satellite from r1 to r2.
Solution:
Wext + WG = KEf – KEi
Wext + Ui – If = KEf – KEi
Wext = (Kf + Uf) + (Ki + Ui)
= TEf – TEi
−GMp MS  −GMp MS 
Wext = − 
2r2  2r1 

Illustration 33:
A satellite of mass m, initially at rest on the Earth, is launched into a circular orbit at a height equal to the radius of
the Earth. The minimum energy required is
3 1 1 3
(A) mgR (B) mgR (C) mgR (D) mgR
4 2 4 4
Ans. (D)
Illustration 34:
An artificial satellite (mass m) of a planet (mass M) revolves in a circular orbit whose radius is n times the radius R of
the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming the
force of resistance on satellite to depend on velocity as F = av2 where 'a' is a constant, calculate how long the satellite
will stay in the space before it falls onto the planet's surface.
Solution:
GM
Air resistance F = – av2, where orbital velocity v =
r
GMa
r = the distance of the satellite from planet's centre  F = –
r
(GM ) a
3/2
GMa GM
The work done by the resistance force dW = Fdx = Fvdt = dt = dt ...(i)
r r r 3/2
dE d  GM m  GMm GMm
The loss of energy of the satellite = dE  =  −  = 2  dE =dr ...(ii)
dr dr  2r  2r 2r 2
GMm ( GM ) dt
3/2

Since, dE = – dW (work energy theorem) – dr =

2r 2 r 3/2

m R ( n −1 )=
m
( ) a mgR
R
dr
 t=–  = n −1
2a GM nR r a GM

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Illustration 35:
Two satellites S1 and S2 revolve round a planet in coplanar circular orbits in the same sense. Their periods of
revolution are 1 h and 8h respectively. The radius of the orbit of S1 is 104 km. When S2 is closest to S1, find:

(a) the speed of S2 relative to S1 and

(b) the angular speed of S2 as observed by an astronaut in S1.

Solution:
Let the mass of the planet be M, that of S1 be m1 and S2 be m2. Let the radius of the orbit of S1 be R1(= 104 km) and of
S2 be R2.

Let v1 and v2 be the linear speeds of S1 and S2 with respect to the planet. Figure shows the situation.

As the square of the time period is proportional to the cube of the radius.
3 2 2
 R2   T2   8h 
  =   =   = 64 v1
 R1   T1   1h 
R2 v2
R2
or, =4
R1
S1 S2
or, R2 = 4R1 = 4 × 104 m. R1

Now the time period of S1 is 1 h. So,

2R1
= 1h
v1

2Ri
or, v1 = = 2 104 km / h
1h
2R2
Similarly, v2 = =  104 km / h
8h

(a) At the closest separation, the are moving in the same direction. Hence the speed of S2 with respect to S1 is |v2
– v1| =  × 104 km/h.

(b) As seen from S1, the satellite S2 is a distance R2 – R1 = 3 × 104 km at the closest separation. Also, is moving at
 × 104 km/h in a direction perpendicular to the line joining them. Thus, the angular speed of S2 as observed
by S1 is

 104 km / h 
= = rad / h.
3  104 km / h 3

Illustration 36:
A space–ship is launched into a circular orbit close to the Earth's surface. What additional speed should now be
imparted to the spaceship so that orbit to overcome the gravitational pull of the Earth.
Solution:
Let K be the additional kinetic energy imparted to the spaceship to overcome the gravitation pull then by energy
conservation

GMm GMm
– + K = 0 + 0  K =
2R 2R

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GMm GMm GMm GMm

Total kinetic energy = + K = + =
2R 2R 2R R
1 GMm 2GM
Then, mv22 =  v2 =
2 R R

But, v1=.
GM
R
So Additional velocity = v2 – v1=
2GM
R
–
GM
R
= ( 2 –1 ) GM
R

Illustration 37:
For a particle projected in a transverse direction from a height h above Earth’s surface, find the minimum initial
velocity so that it just grazes the surface of Earth path of this particle would be an ellipse with center of Earth as the
farther focus, point of projection as the apogee and a diametrically opposite point on Earth’s surface as perigee.

Solution:
Suppose velocity of projection at point A is vA and at point B, the velocity of the particle is vB.
mv 2A GMe m mvB2 GMe m
then applying Newton’s 2nd law at point A and B, we get, = and =
rA ( R + n)2 rB R2

Where rA and rB are radius of curvature of the orbit at points A and B of the ellipse,
But rA = rB = r(say).
Now applying conservation of energy at points A and B
−GMe m 1 2 −GMe m 1 2
+ mv A = + mvB
R+h 2 R 2
1 1  1
 = ( mvB − mv A ) =  GMe m  2 −
1  1 1 
 GMe m  − 2 2
2 
 R ( R + h)  2  2  R ( R + h ) 

2R ( R + h) 2Rr rGMe R
or, r= =  VA2 = = 2GMe
2R + h R+r ( R + h) 2
(
r r + R)

where r = distance of point of projection from Earth’s centre = R + h.

Conditions for Different Trajectory

For a body being projected tangentially from above Earth’s surface, say at a distance r from Earth’s center, the
trajectory would depend on the velocity of projection v.
v
h /2
r
R

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Velocity Orbit

1. Velocity, v < GM  2R  Body returns to Earth

r r +R

GM
2. > v > GM  2R  Body acquires an elliptical orbit with Earth as the far-focus
r r r +R
w.r.t. the point of projection.
3. Velocity is equal to the critical velocity Circular orbit with radius r
GMe
of the orbit, v =
r
4. Velocity is between the critical and escape Body acquires an elliptical orbit with Earth as
velocity of the orbit the near focus w.r.t. the point of projection.
2GMe GMe
>v> i.e TE < 0
r r

2GMe
5. v = vesc = i.e TE = 0 Body just escapes Earth’s gravity, along a
r
parabolic path.
2GMe
6. v > vesc = i.e TE > 0 Body escape Earth’s gravity along a
r
hyperbolic path.

Double (Binary) Star System

If two stars of comparable mass revolve about their common centre of mass under the influence of mutual
gravitational force then this system is called binary star system.
m1 v1 m1

r1 r1

cm cm
r2 r2
m2 m2 v2

For centre of mass to remain at rest both the stars should rotate with same ''.
Let distance of centre of mass for m1 is r1 and m2 is r2. Then
m1r1 = m2r2 ....(i)
r1 + r2 = r ....(ii)
From (i) and (ii)
m2 r
r1 =
m1 + m2
m1 r
r2 =
m1 + m2

T1 = T2  1 = 2 ....(iii)

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Gm1 m2
= F = m1 12 r1 ...(1)
r2

Gm1 m2
= m222r2 ...(2)
r2

Gm1 (m1 + m2 )
 = 22
r2 m1 r

G(m1 + m2 )
= 12
r3

Angular momentum of system

L = m1v1r1 + m2v2r2

L = m1 r1 (r1 + r2)

L =  m1 m2 r  (r )
 m1 + m2 
v1

l = µr2 2d
COM d
m 3 3
2m
m1 m2 v2
where µ = is the reduced mass of system.
m1 + m2

A pair of stars rotating about their COM (centre of mass)

GM(2m)
F=
d2

Star-1 Star-2 Ratio

mv12 mv12 v1 2
1. F= F = =
2d / 3 d /3 v2 1

M2 m mv12 M2 m mv22
G = G =
d2 2d / 3 d2 d /3

Gm Gm
v1 = 2 v2 =
3d 3d

2 2d 2 d T1 1
2. T1 = T2 = =
v1 3 v2 3 T2 1

1 1 KE1 2
3. KE1 = mv12 KE 2 = mv22 =
2 2 KE2 1

2d d L1 2
4. L1 = mv1 L2 = 2mv2 =
3 3 L2 1

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Illustration 38:
In a double star, two stars (one of mass m and the other of 2m) distant d apart rotate about their common centre of
mass. Deduce an expression of the period of revolution. Show that the ratio of their angular momentum about the
centre of mass is the same as the ratio of their kinetic energies.
Solution:
The centre of mass C will be at distances d/3 and 2d/3 from the masses 2m and m respectively. Both the stars rotate
round C in their respective orbits with the same angular velocity . The gravitational force acting on each star due to
the other supplies the necessary centripetal force.
G ( 2m) m
The gravitational force on either star is . If we consider the rotation of the smaller star, the centripetal force
d2
 2d  2   2md2 
(m r 2) is m    and for bigger star   i.e. same
  3    3 
G (2m) m  2d 
 2
= m  2 or  =  3Gm 
3 
d  3   d 

Therefore, the period of revolution is given by


2
= 2  d 
3 d/3 2d/3
T= m
  3Gm  2m C

The ratio of the angular momentum is
2
( 2m)  d 
( I)big Ibig 3 =1
= =
( I)small I small  2d 
2
2
m 
 3 
1 2 
 I  Ibig 1
2 big
Since  is same for both. The ratio of their kinetic energies is = = , which is the same as the ratio of
1 2  I small 2
 I  
2 small
their angular momentum.

Gravitational Pressure
Illustration 39:
A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational
compression, as a function of the distance r from its centre. Evaluate p at the centre of the Earth, assuming it to be a
uniform sphere.
3
Ans. p= (1 – r2/R2) M2/R4. About 1.8 × 108 atmospheres.
8
Solution:
Consider an element of a cylinder extending from centre to surface.
r
dmg GMr
p= = −  3  dr
A R R

4G2  R − r 
2 2

=
3 2
2G2 2
=  R − r 2 
3 

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Illustration 40:
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r)
is the pressure at r(r < R), then the correct option(s) is(are) :-
P ( r = 3R / 4 ) 63 P ( r = 3R / 5) 16 P ( r = R / 2) 20
(A) P(r = 0) = 0 (B) = (C) = (D) =
P ( r = 2R / 3) 80 P ( r = 2R / 5) 21 P ( r = R / 3) 27

Ans. (BC)
Solution:
Gm
dF = [P – (P + dP)]A  dm = − ( dP ) A
r2 P+dP
Mr 3
P r G 3
4ar 2dr
R
  dP = −
0 R r 2 ( 4r 2 ) r
dr
GM  r2 
 P= 1 − 2 
2R  R 

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