Electrostatics and Gravitation

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 m1 m2
F or F =G
r2 r2 r
where G = 6.67 × 10–11 Nm2 kg–2 is the universal gravitational constant.

Dimensional Formula of G :

Fr 2  MLT −2  L2   −1 3 −2 
G= = = M LT
m1m2 M2 
Vector Form of Newton's Law of Gravitation
m1 F21
Y
Let r12 = Displacement vector from 𝑚1 to 𝑚2 F21
r12
r12 m2
r21 = Displacement vector from 𝑚2 to 𝑚1 r1

r2
F21 = Gravitational force exerted on 𝑚2 by 𝑚1 O X

F12 = Gravitational force exerted on 𝑚1 by 𝑚2 Z

Gm1m2 Gm m
F12 = − 2
rˆ21 = − 13 2 r21
r21 r21

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Negative sign shows that :

(i) The direction of F12 is opposite to that r21

(ii) The gravitational force is attractive in nature

Gm1m2 Gm1m2
Similarly, F21 = − 2
rˆ12 or F21 = − 3
r12  F12 = −F21
r12 r12

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

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.
27
For Illustration gravitational force between the Sun and the Earth is of the order of 10 N although
distance between them is 1.5 × 107 km

Illustration 118:
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 = 2
= 2
= 5.3  10−10 N
r (0.5)

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.

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Electrostatics and Gravitation
Illustration 119:

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:
M1 m M2
GM1m
The force on m towards M1 is F1 =
r2 r
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 

Principle of superposition

The force exerted by a particle on other particle remains unaffected m1

F1
by the presence of other nearby particles in space. Total force acting m F2 m2
on a particle is the vector sum of all the forces acted upon by the
F3
individual masses when they are taken alone. m3

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

Illustration 120:

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

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Solution:
The resultant force on particle at A due to other two particles is
2 2 Gm2  Gm2 
FA = FAB + FAC + 2FAB FAC cos60 = 3 ...(i) F
 AB = FAC = 
a2  a2 
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,
mv2 mv2
Now = 3 ...(ii)
r a
mv 2 Gm2 3 Gm
From (i) and (ii) 3 =  v=
a a2 a
2r 2a a a3
Time period, T = = = 2
v 3 Gm 3Gm

a
Illustration 121: m m
Four point masses each of mass 'm' are placed on the corner of square
of side 'a'. Calculate magnitude of gravitational force experienced by a
each particle.
Solution: m m
F
m m
45°
45°
F1
F

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
Rod (M,)
Illustration 122: dM
Find gravitational force exerted by point mass ‘m’ on  a m
dM
uniform rod (mass ‘M’ and length ‘’). 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 =  2
=  ( x + a )2 =  − ( + a ) + a  = ( + a ) a
0 ( x + a) 0

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Electrostatics and Gravitation
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.

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
m r P
F GM
Hence E =  E= 2
m0 r r

GM
In vector form, E = − rˆ
r2
F  MLT −2   0 −2 
Dimensional formula of intensity of gravitational field = = = M LT
m M 

Illustration 123:
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 .
6  1024 kg 7.4  1022 kg
or, =
x2 ( 4.0  105 km − x )2
x 6  1024
or, = =9
4.0  105 km − x 7.4  1022
or, x = 3.6 × 105 km.

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Illustration 124:
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:
Consider any particle of mass dm. Gravitational field at point P due to dm
Gdm
dE = along PA
r2
Gdm A dm
Component along PO is dE cos  = cos
r2
R
Net gravitational field at point P is θ P
O x
Gdm G cos 
E =
r2 
2
cos  = dm
r
GMx
= towards the center of ring
( R2 + x 2 )3/2

Illustration 125:
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 =
( x 2 + r 2 )3/2 Disc (M, R)
M 2M dr
Here dM = 2
.2rdr = 2 rdr
R R
r
G.2Mxrdr P
 dE = dE
R ( x2 + r2 )
3/2
2
x
R
 2GMx  rdr
 E =  2 
0 R  ( x 2 + r 2 )3/2

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

Illustration 126:
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

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Electrostatics and Gravitation
Solution:
GdM Ring of radius = R sin 
dE = 2
.cos  r R
Rd 
M d
dM =  2R sin Rd ℓ
4R2
M  
dM = sin  d P
2 r
GM sin  cos  d
 dE =
2 2

Now  2 = R2 + r 2 – 2rR cos ...(i)

R 2 = 2 + r2 – 2r cos ...(ii)

2
+ r 2 − R2
 cos  =
2 r

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 )  2 
 E = 2 ,r R
4Rr  r + R  r
If point is inside the shell limit changes to [(R – r) to R + r]
E = 0 when r < R.

Illustration 127:
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
Gm 3 4G
then, E A = 2A = 2
=  ARA
RA RA 3

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GMB 4G
Similarly, E B = =  B RB
RB2 3

E A RB2 E A RA
(i) For mA = mB = (ii) For  A =  B =
E B R2A 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

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

GM
Thus, gravitational potential, V = − .
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 128: m
1
Find out potential at P and Q due to the two point mass system. Find out ℓ
ℓ/2
work done by external agent in bringing unit mass from P to Q. Also find
ℓ Q P
work done by gravitational force.
ℓ/2
Solution: ℓ
2 m
Gm
(i) VP1 = potential at P due to mass ‘m’ at ‘1’ = −

Gm
VP 2 = −

2Gm
 VP = VP1 + VP2 = −

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Electrostatics and Gravitation
Gm Gm
(ii) VQ1 = −  VQ2 = −
/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 =

Illustration 129:
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 can be considered to be made of large number of point

masses (m1, m2 ..........etc)
Ring R, M
Gm1 Gm2
VP = − − − ....... 2 2
R2 + x 2 R2 + x 2 R +x
R
G GM
=− ( m1 + m2 ..........) = , P
R2 + x 2 R2 + x 2
x
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 130:
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 131:
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 132:
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 133:
Find gravitational field at a point (x, y, z).
v = 2x2 + 3y2 + zx,
Solution:
𝐸⃗ = −𝛻𝑉
−v
Ex = = – 4x – z ; E y = –6y ; Ez = – x
x
 Field = 𝐸⃗ = – [(4x +z) 𝑖̂ + 6y𝑗̂ + x𝑘̂]

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
2
(r + x 2 )3/2 2
(r + 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
r +x2 2 r 2
r +x 2 r

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Electrostatics and Gravitation
(3) Uniform linear charge Uniform linear mass

𝐺𝜆
 𝑘𝜆 ሺcosβ + αሻ
r sinሺα + βሻ  𝑟
 𝑟 r 
𝑘𝜆 𝐺𝜆
ሺcosβ − cosαሻ ሺsinα − sin 𝛽ሻ
𝑟 𝑟

(4) Infinite Linear charge Infinite linear mass

 
r2 r2

r1 r1

A B A B

 

2K  2G
(a) E= g=
r r

r2 r 
(b) VB – VA = – 2K ln VB – VA = 2G ln  2 
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

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Case I. r>R
Q M
r P r P
R R

kQ GM
E= g=
r2 r2
kQ GM
V= VG = −
r r

Case II. r<R

Q M
R P R P
r r

E=0 g=0
kQ GM
V= VG = −
r 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
2R 3 ( )
3R2 − r 2 , r > R
2R 3 (
3R2 − r 2 , r > R)

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

charged solid sphere. solid sphere.

3 KQ2 3 GM 2
U= U=
5 R 5 R

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Electrostatics and Gravitation
Illustration 134:
A uniform solid sphere of density  and radius R has a spherical cavity of P
radius r inside it as shown. Find gravitation field at O C
(a) O (b) C (c) P (prove that field inside cavity is uniform)
4 
k  r 3  
3 4GOC 4
Ans. (a)  2
 CO (b) (c) GO1O2
(OC ) 3 3

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 g1 P
P
+ g2
O1 O2 O1 –  O2

Resultant field g = g1 + g2

4  4 
=  G  O1 P +  G  PO2
 3   3 
4 O1O2 = OC 
= GO1O2  
3
It is independent of position of point inside cavity

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

4 4 2 
G r 2  3 Gr  
= 3 2 CO =   CO
(CO ) (CO )
2

Illustration 135:
M 16M
(a) Find where is gravitational field intensity is zero.
(b) Find gravitational potential at P. D

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Solution:
GM G16M
(a) =
x 2 ( 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 

Illustration 136:
GFI at P = ?

R/2 P
R

Solution:

4
𝜌 𝜋𝑅3 = 𝑀
3
 R
4 𝑅 3
−𝜌 𝜋 ൬ ൰ = 𝑀′
3 2

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

Illustration 137:
x
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

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Electrostatics and Gravitation
Illustration 138:
x
A particle is released in a tunnel across Earth as shown in figure. Show that
it performs SHM and find T. r 
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. C
dr
Consider a body of mass m placed at a distance r from another body of mass M. The
B
GMm
gravitational force of attraction between them is given by, F = 2 A
r r
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) O Re
is given by,
r
GMm GMm
dW = F dr =
r2
dr   dW =  2
dr
 r

GMm
 Gravitational potential energy, U = −
r

Illustration 139:
Mass 𝑀1 is fixed and mass, m is released from rest at infinity, M1 
find its velocity, when separation between them is ′𝑟′. 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 140: 0
Particle is projected vertically with speed 0, particle reaches maximum
height of 3R, find 0 (where 𝑅 is radius of Earth)
R
Solution: M2
Height is always measured from surface. Earth
WG = KE f − KE i

0
Ui – Uf = 0 – KEi
GMe m  GMe m  1 2
− −−  = − mµ0
R  4R  2
GMe m GMe m 1
− + = − mµ02
R 4R 2
1 3GMe m
Mµ02 =
2 4R

3GMe
µ0 =
2R

Illustration 141:
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, 2
− =0  (10 a–r)2 = 16 r2
r (10a − r )2
M 16M
 10a – r = 4r  r = 2a P V
a 2a
Potential at point P,
r 10 a
−GM G.16M −GM 2GM −5GM
VP = − 2
= − =
r (10a − r ) 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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Electrostatics and Gravitation

v2 −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  rij
all pairs
j i

This expression can be written as

j =n m m
1 i =n
Us = − G  i j

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
M 1 dr
= −G  3  = − G ( 4 ) r 4dr
2
U sphere where  = R
r 4 3 3 r
  R
3 O
R R 2
2 r 
5
1 1 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 𝐸⃗ 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 mI𝑔 = mG𝐸⃗ .
Now the value of inertial and gravitational mass happen to be exactly same to a great degree of accuracy
for all bodies. Hence, 𝑔 = 𝐸⃗
m
The gravitational field intensity on the surface of Earth is therefore numerically
equal to the acceleration due to gravity ሺ𝑔ሻ, there. Thus, we get, mg

where, Me = Mass of Earth Re

Re = Radius of Earth 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  P
gh = = g1 +  g  1 −  when h << R. h
( Re + h)
2
 Re   Re 
Re
2 gh O
The decrease in the value of 'g' with height h = g – gh = .
Re

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

(b) Effect of Depth: m

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

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Electrostatics and Gravitation
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)

 d 
 gd = g  1 − 
 Re 

Important Points:

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

Thus weight (mg) of the body at the centre of the Earth is zero.

Re r

Earth
gr g  1/r2

g
g'

 g − gd  d
(ii) Percentage decrease in the value of 'g' with the depth =    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
Re2
the equator to the pole.

Equitorial
Polar Pole Radius
Radius

Equator

(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 .

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N

r = Re cos 

P(m) m2 r = m2Re cos 

C
 D
mg
W  E
O Q
Re
Equatorial plane

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 

 R 2 
At pole  = 90°  gpole = g, At equator  = 0  gequator = g 1 − e  .
 g 
Hence gpole > gequator
 R 2 
If the body is taken from pole to the equator, then g ' = g 1 − e  .
 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 142:
At some planet gapp at equator is equal to half of gapp of at poles.
Find  in terms of g and R.
Solution:
1 1 g g
ge = gp  g − R2 = g  2R =  =
2 2 2 2R

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Electrostatics and Gravitation
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
 r 
produce its own gravity) orbiting the Earth at a height h above its surface then
mGM mg
True weight = mgh = 2
= 2
; Apparent weight = m(gh – a)
(R + h)  h
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 143:
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 at r→
project it with a velocity V from the Earth's surface, so m0 v→
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 Me,R v
find positions (when the particle just reaches to r → ) m0
Ki + Ui = Kf + Uf
1 2  Gm   GMe 
mv + m0  − e  = 0 + m0  − ( 
2  R   r → ) 

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2GM0
 v=
R

2GMe
Escape speed from Earth is surface ve =
R

If we put the values of G, Me, R then 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.
Me,R
Illustration 144:

A very small groove is made in the Earth, and a particle of R/2

mass m0 is placed at R/2 distance from the centre. Find the m0
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 Me,R

1  GM  2 

2 R
m0v + m0  − 3  3R −     = 0 + 0
2
 e
R/2
at r → , v → 0
2  2R   2  
m0 v m0
11GMe
v= = ve at that position.
4R

Illustration 145:

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:

1 2 GMP m 1 2 GMP m
For a planet: mv − =0  mv −
2 Rp 2 Rp

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Electrostatics and Gravitation
1 2  GM 
On Earth → mv = m  2E h
2  RE 
GMpm GME .m M p ME h
 = 2
h  = 2
Rp RE Rp RE

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

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

m
→
r
Sun Area swept by
radius vector

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 JEE (Main + Advanced) : Physics
Suppose a planet is revolving around the Sun and at any instant its velocity is v, and angle between radius
vector (𝑟ሻ 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

A
m 
vdt
→
r m
O
P B
vdt sin

Sun

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 146:
x 2 y2
Suppose a planet is revolving around the Sun in an elliptical path given by + = 1 . Find time period of
a2 b2
revolution. Angular momentum of the planet about the Sun is L.

Sun

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

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Electrostatics and Gravitation
Kepler’s Law of Time Period
m0v 2 GM S m0
Suppose a planet is revolving around the Sun in circular orbit then =
r r2
GMs
v=
r
2r r v
Time period of revolution is T = = 2r Sun GM s m0
v GMs r2
m0
 42  3 r
2 2 3
T = r  T = r Ms
 GMs 
For all the planet of a Sun, T 2  r 3

Illustration 147:
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
r2=4×107m
T2 = ?

Sun

rt=107m
Earth
Tt = 365 days

Solution:
For both the planets
T2  r3
2 3 2 3
 TJupiter   rJupiter   TJupiter   4  107 
  =     = 7 
 TEarth   rEarth   365days   10 
Tjupiter = 8 × 365 days
Graph of T vs r : T
Graph of log T v/s log R :
Tr3/2
2 42  3  42 
T =  R  2log T = log   + 3log R
 GM s   GM s 

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

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 JEE (Main + Advanced) : Physics
log T

3
slope =
2
1  42 
C = log  
2  GM s 
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
v
attraction, it may collide to the planet.
Me
To avoid the collision, the satellite revolve around the planet, GMe m0
for circular motion of satellite. r m0
r
GMe m0 m0v 2
 = ....(i)
r2 r Planet

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 (i)
2

m0v 2 GMe m0
=
r r2
GMe m0
 m0v2 =
r
1 GMe m0
 KE = m0v 2 =
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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Electrostatics and Gravitation
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.

r 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 stationary. 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 m
 GMe 
m0
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.
• Its rotating direction should be same as that of Earth
(West to East).

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• Its orbit is called parking orbit and its orbital velocity is 3.1 km./sec.
 R 
• Maximum latitude at which message can be received by geostationary satellite is  = cos−1  e 
 Re + 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

• The area of Earth's surface escaped by geostationary satellite is
R
cos 2 = R
R+h 

4R2 − 2R2(1 − cos 2 ) = 2r 2 + 2r 2 cos(90 − )  R h

= 2R2(1 + sin )

Polar Satellite (Sun–Synchronous Satellite)

Polar orbit
It is that satellite which revolves in polar orbit around Earth. A polar Equatorial
orbit is that orbit whose angle of inclination with equatorial plane plane
of Earth is 90° and a satellite in polar orbit will pass over both the Equator
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 Equatorial
used for getting the cloud images, atmospheric data, ozone layer in orbit
the atmosphere and to detect the ozone hole over Antarctica.
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):

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

Work Done in Changing the Orbit of Satellite:

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

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Electrostatics and Gravitation
Illustration 148:
A satellite moves eastwards very near to 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  2R  4R2
Twest = and Teast =  T = Teast – Twest = 2R  2  =
v0 + R v0 − R 2 2 2 2 2
 v0 − R   v0 − R 
Illustration 149:
Find Wext to change orbital radius of satellite from r1 to r2.
Solution:
Wext + WG = KEf – KEi
Wext + Ui – Uf = KEf – KEi
Wext = (Kf + Uf) – (Ki + Ui) = TEf – TEi
−GM p MS  −GM p MS 
Wext = − 
2r2  2r1 
Illustration 150:
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)
Solution:
From law of energy conservation
Ui + K i = U f + K f
−GMm −GMm 1
+ Ki = + mV02
R 2R 2
2
GMm 1  Gm 
Ki = + m 
2R 2  2R 
3GMm 3
Ki = = mgR
4R 4

Illustration 151:
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

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(GM ) a dt
3/2
GMa GM
The work done by the resistance force dW = Fdx = Fvdt = dt = ...(i)
r r r 3/2
dE d  GM m  GMm GMm
The loss of energy of the satellite = dE ∴ = − =  dE = dr ...(ii)
dr dr  2r  2r 2 2r 2

(GM )
3/2
GMm
Since, dE = – dW (work energy theorem) – dr = adt
2r 2
r 3/2
R m R ( n −1 )=
 t=–
m
2a GM

nR
dr
r
=
a GM
( n −1 ) a mgR

Illustration 152:
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, they 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 at 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

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Electrostatics and Gravitation
Illustration 153:
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
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 154:
For a particle projected in a transverse direction from A m
a height h above Earth’s surface, find the minimum
h
initial velocity so that it just grazes the surface of Earth
path of this particle would be an ellipse with center of R Orbit of the body
Earth as the farther focus, point of projection as the Earth Just grazing earth’s surface
apogee and a diametrically opposite point on Earth’s B VB
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 + h)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 1  1 1 
 = ( mvB − mv A ) =  rGMe m  2 −
2 2
 GMe m  − 2 
( )
R R+h  2  R ( R + h)  
2

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.

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 JEE (Main + Advanced) : Physics
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

Velocity Orbit

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

GM GM  2R 
2. >v> Body acquires an elliptical orbit with Earth as the
r r  r + R 
far-focus 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.

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Electrostatics and Gravitation
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 m1
v1

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)
m2r
r1 =
m1 + m2
m1r
r2 =
m1 + m2
T1 = T2  1 = 2 ....(iii)
Gm1m2
= F = m112r1 ...(iv)
r2
Gm1m2
= m222r2 ...(v)
r2
Gm1 (m1 + m2 )
 = 22
r 2m1r
G(m1 + m2 )
= 12
r3
Angular momentum of system
L = m1v1r1 + m2v2r2
 mmr  v1
L =  1 2  (r )
 m1 + m2 
l = µr2 m
2𝑑 COM 𝑑 2m
mm 3 3
v2
where µ = 1 2 is the reduced mass of system.
m1 + m2
A pair of stars rotating about their COM (centre of mass)
GM(2m)
F=
d2

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 JEE (Main + Advanced) : Physics
Star-1 Star-2 Ratio
mv12 mv12 v1 2
1. F= F= =
2d /3 d /3 v2 1
M2m mv12 M2m 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 KE2 = mv22 =
2 2 KE2 1
2d d L1 2
4. L1 = mv1 L2 = 2mv2 =
3 3 L2 1

Illustration 155:
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
d2
  2d    2md2 
centripetal force (m r 2) is m   2  and for bigger star   i.e. same
  3    3 
G ( 2m) m  2d   3Gm 
 2
= m   2 or  =  3 
d  3   d 
Therefore, the period of revolution is given by
2  d3  
T= = 2   d/3 2d/3
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
1 2 Ismall 2
 I 
2 small
the ratio of their angular momentum.

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Electrostatics and Gravitation
Gravitational Pressure
Illustration 156:
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) GM2/R4. About 1.8 × 108 atmospheres.
8
Solution:
Consider an element of a cylinder extending from centre to surface.
r
dmg GMr
p= = −  3  4r 2dr
A R R

4G2 R − r  2G2 2 2

2 2

= = R − r 
3 2 3  

Illustration 157:
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:
dF = [P – (P + dP)]A
Gm
 dm = − ( dP ) A
r2

Mr 3
P r G 3
4ar 2dr r
R
  dP = −  r 2 ( 4r 2 ) dr
0 R

GM  r2 
 P=  1 − 
2R  R2 

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