Gravitation

1. Introduction
The motion of celestial bodies such as the moon the earth, the planets, etc, has been a
subjected of great interest for a long time. Famous Indian astronomer and mathematician,
Aryabhatta, studied these motions in great detail, most likely in the 5th century A.D., and wrote
his conclusions in his book Aryabhatta. He established that the earth revolves about its own
axis. He also gave description of motion of other celestial bodies as seen from the earth.
After Aryabhatta, further developments in the study of gravitational forces are as under :
(i) The hypothesis about planetary motion given by Nicolaus Copernicus (1473-1543).
(ii) The careful experimental measurements of the positions of the planets and the sun by
Tycho Brahe (1546 - 1601)
(iii) Analysis of the data and the formulation of empirical laws by Johannes Kepler (1571 - 1630).
(iv) The development of a general theory by Isaac Newton (1642 - 1727)

2. Newton’s Law of Gravitation

It states that every particle in the universe attracts every other particle with a force which is
directly proportional to the product of their masses and is inversely proportional to the square
of the distance between them.
 
If | F12 | = | F21 | = F, then
1
F ∝ m1m2 and F ∝
r2
 
F12 F21
m1 m2

m1m2
so F ∝
r2
Gm1m2
∴ F=
r2
Newton further generalised the law by saying that not only the earth but all material bodies in
the universe attract each other according to equation with same value of G. The constant G is
–11 2 2
called universal constant of gravitation and its value is found to be 6.67 × 10 N–m /kg .
Equation is known as the universal law of gravitation.
Note : This formula is applicable only for spherically symmetric masses or point masses.

2.1 Vector form of Law of Gravitation

 
Let r 12 = Displacement vector from m1 to m2 F12 
 m1  F
Y  21
r21 = Displacement vector from m2 to m1 r12 r 21 m2
 
F2,1 = Gravitational force exerted on m2 by m1 r1 
 r2
F1,2 = Gravitational force exerted on m1 by m2 O X
 Gm1m2  Gm1m2 
F1,2 = r21 = – r21 Z
r21
2
r213
 Negative sign shows that

(i) The direction of F12 is opposite to that of r21
(ii) The gravitational force is attractive in nature
 Gm1m2 
Similarly F12 = – r21
r212
 Gm1m2   
or F21 = – r 12 ⇒ F 12 = – F21
r123
The gravitational force between two bodies are equal in magnitude and opposite in directions.

• Universal Gravitational Constant ‘G’

• Universal Gravitational Constant is a scalar quantity.
–11 2 2
• Value of G : SI : G = 6.67 × 10 N-m /kg ;
–8 2 2 –1 3 –2
CGS : G = 6.67 × 10 dyne-cm /g Dimensions : [M L T ]
• Its value is same throughout the universe; G does not depend on the nature and size of the
bodies; it does not depend even upon the nature of the medium between the bodies.
• Its value was first found out by the scientist “Henry Cavendish” with the help of “Torsion
Balance” experiment.

2.2 Properties of Gravitational Force

• Gravitational force is always attractive.
• Gravitational forces are developed in the from of action and reaction pair. Hence they obey
Newton’s third law of motion.
• It is independent of the nature of medium in between two masses.
• Gravitational forces are central forces as they act along the line joining the centres of the
two bodies.
• Gravitational forces are conservative forces so work done by gravitational force does not
depends on path.
• If any particle moves along a closed path under the action of gravitational force then the
work done by this force is always zero.
• Gravitational force is weaker than the electrostatic and nuclear forces.
• Force developed between any two masses is called gravitational force and force between
Earth and any body is called force of gravity.
• Gravitational force holds good over a wide range of distances. It is found true from
interplanetary distances to interatomic distances.
• It is a two body interaction i.e., gravitational force between the two particles is independent
of the presence or absence of other bodies or particles.

Example 1:
Two spherical balls of mass 10 kg each are placed 100 m apart. Find the gravitational force of
attraction between them.
Solution:
Gm1m2 6.67 × 10–11 × 10 × 10 –13
F= = = 6.67 × 10 N
r 2
(100) 2

Example 2:
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
the heavier particle.
Solution:
Force exerted by one particle on another is
Gm1m2 6.67 × 10–11 × 1 × 2 –10
F= = = 5.34 × 10 N
r 2
(0.5)2
F 5.3 × 10–10
Acceleration of heavier particle = = = 2.6 × 10–10 ms–2
m2 2

Note : This example shows that gravitational force is quite weak but this is the only force keep binds
our solar system and also the universe comprising of all galaxies and other interstellar system.

3. Law of Superposition of Forces

Gravitational force follows law of superposition which states
The total gravitational force on a particle due to a number of particles is the resultant of the
forces of attraction exerted on the given particle due to the individual particles i.e.
   
F = F1 + F2 + F3 ....

Key Points

• It may be worth noting that a uniform spherical shell of matter or a uniform solid sphere attracts
a particle outside their periphery as if their mass is concentrated at their centre.

Example 3:
Three masses, each equal to M are placed at the three corners of a square of side a. Calculate
the force of attraction on unit mass placed at the fourth corner.
Solution: 
  M F1 m = 1

Force on m due to masses at corners 1 and 3 are F1 and F3 1 F2

GMm  F3
with F1 = F3 = Fr
a2 a
  GMm
resultant of F1 and F3 is Fr = 2 , and its direction is along the
a2 a M
M
diagonal 2 3
i.e., towards corner 2
Force on m due to mass M at 2 is
 GMm GMm
Fr = = ;
( ) 2a
2
2a

Fr and F2 act in the same direction. Resultant of these two is the net force :
GMm  1
Fnet = Fr + F2 =  2+ 
a 2
 2
along the diagonal
GM  1
Since m = 1 kg So. Fnet = 2 
2+ 
a  2 
Example 4:
Three particles of mass 2kg each are placed as shown. Find out net gravitational force on ‘A’.

10m 10m

B 10m C
Solution:
60º
Net force on ‘A’ = 2 F cos =F 3 A
2
F F
Where F is the force exerted by B & C on A 30º 30º
G(2)(3)
Fnet = 3
(10)2
–11
Fnet = 0.46 × 10 N

Example 5:
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 intended
that each particle moves along a circle while maintaining their original separation ‘a’. Determine
the initial velocity that should be given to each particle and the time period of the circular
motion.
Solution:
The resultant force on particle at A due to other two particles is

FA = FAB
2
+ FAC
2
+ 2 FABFAC cos 60°

Gm2
= 3 ......(i)
a2
 Gm2 
 F=
AB
F=
AC 
 a2 

a
Radius of the circle r =
3
If each particle is given a tangential velocity v, so that the resultant force acts as the centripetal
force,
mv 2 mv 2 Am
then = 3 ....(ii)
r a FAB FAC
From (i) and (ii) FA
O
mv 2
Gm 3
2
Gm
3 = ⇒ v= m m
a a2 a B a C

2πr 2πa a a2
Time period T = = = 2π
v 3 Gm 3Gm
Example 6:
Two stationary particles of masses M1 and M2 are ‘d’ distance apart. A third particle lying on the
line joining the particles, experiences no resultant gravitational force. What is the distance of
this particle from M1?
Solution:
Let m be the mass of the third particle
M1 m M2
GM1m
Force on m towards M1 is F1 = r
r2 d
GM2m
Force on m towards M2 is F2 =
(d– r)2
Since net force on m is zero ∴ F1 = F2
2
GM1m GM2m  d–r  M2
⇒ = ⇒   =
r2 (d– r)2  r  M1

d M2  M1 
⇒ –1= ⇒r = d  
r M1  M1 + M2 
 

Example 7:
A mass (M) is split into two parts (m) and (M-m) which are then separated by a certain distance.
m
What ratio will maximise the gravitational force between them?
M
Solution:
If r is the distance between m and (M - m), the gravitational force will be
m(M– m) G
F=G = 2 (mM – m2)
r 2
r
dF d2F
for F to be maximum = 0 and < 0 as
dm dm2
M and r are constants,
d G 2 
i.e.,  (mM– m ) = 0
dm  r2 
 G 
i.e., M – 2m = 0  2 ≠ 0
 r 
m 1
or = , i.e., the force will be maximum when the two parts are identical
M 2

Example 8:
Two solid spheres of same size of a certain metal are placed in contact with each other. Prove
that the gravitational force acting between them is directly proportional to the fourth power of
their radius.
M M

R
Solution:
The weights of the spheres may be assumed to be concentrated at their centres.
4  4 
G  πR3ρ  ×  πR3ρ 
3 3  = 4 (Gπ2ρ2) R4
So F =   
(2R) 2
9
4
∴ F∝ R

Concept Builder-1

Q.1 Two bodies of mass 3 kg and 4 kg each placed 50 m apart. Find the gravitational force of
attraction between them in C.G.S. unit.

Q.2 Gravitational force on 2 kg and 5 kg bodies is 10 newton then find separation between bodies ?

Q.3 Four masses, each equal to M are placed at the four corners of a square of side a. Calculate
the force of attraction on another mass m1 kept at the centre of the square.

Q.4 Three identical particles each of mass (m) are placed at the three corners of an equilateral
triangle of side “a”. Find the gravitational force exerted on one body due to the other two.

Q.5 Three identical point masses, each of mass 1 kg lie in the x-y plane at points (0, 0) (0, 0.2m)
and (0.2m, 0) respectively. The gravitational force on the mass at the origin is :
(1) 1.67 × 10 (i + j) N
–10 
(i + j) N
–11
(2) 3.34 × 10

(i + j) N (i – j) N

–9 –10
(3) 1.67 × 10 (4) 3.34 × 10
Q.6 Two particles each of equal mass (m) move along a circle of radius (r) under the action of their
mutual gravitational attraction. Find the speed of each particle.

m r r
m

Q.7 Where a mass of 3 kg has to be placed near a combination of two masses 9 kg and 16 kg placed
14 m apart, such that it experiences no force.

Q.8 There identical sphere (density ρ0 & radius r0) are arranged as shown what is the net force
experienced by any one of the spheres

r0 r0

r0
4. Gravitational Field and it’s Intensity Gravitational Field
The gravitational field is the space around a mass or an assembly of masses within which it can
exert gravitational forces on other masses.

Gravitational field
Theoretically speaking, the gravitational field extends up to infinity. However, in actual practice
the gravitational field may become too weak to be measured beyond a particular distance.

• Gravitational Field Intensity

The gravitational field intensity at a point within a gravitational field is defined as the
gravitational force exerted on unit mass placed at that point.
m

F

 r
 F M
Ι=
m
Gravitational field intensity is a vector quantity whose direction is same as that of the
gravitational force. Its SI unit is ‘N/kg’.
F  M1L1T–2 
Dimensions of intensity =
  = [M0L1T–2 ]
m M 
1
 

5. Gravitational Field Intensity Due to a Particle (Point-Mass)


Ι m = 1unit
M
r

Gravitational field intensity = gravitational force exerted on unit mass

 GM GM
⇒ Ι = 2 (– r ) = 3 (– r )
r r
where ‘M’ is the mass of that particle due to which intensity is to be found.

Example 9:
Infinite particles each of mass ‘M’ are placed at positions x = 1 m, x = 2m, x = 4m ..... ∝. Find the
gravitational field intensity at the origin.
Y M M M M
∞ X
O
1m
2m
4m
8m

Solution:
    
Ιnet = Ι 1 + Ι2 + Ι 3 + Ι 4 + ......∞ terms
GM  GM  GM i
i + i + .......∞
(1)2 (2)2 (4)2
  1 1 
terms = GM i  1 + + + .......∞ 
 4 16 
 1
Here in the GP a = 1 and r = 
 4 

   
    
So, Ιnet = GM i 
1 
 = GM i  1 
1– 1   3  
  
 4   4  
 4
⇒ Ιnet = GM i
3

Example 10:
Draw the variation of Intensity along x-axis If two masses are arranged as shown in figure.

A B
m m

Solution:
Ι
Since two masses are identical. Intensity will be I → + ve x − direction
Mid point
zero at the mid point of AB. Closer to A intensity of AB
x
will be directed towards A & Closer to B intensity A B
Ι =0
will be directed towards A & closer to B intensity I → –ve x − direction
Ι
will be directed towards B. As we know I ∝ so
r2
variation will be as follows

6. Gravitational Field Intensity Due to a Spherical Shell

Case I : If r > R, the point is outside the shell then

 GM r >R

Ιout = 2 (–r) R
r O P

Case II : If r = R, the point is on the surface then

 GM r =R
Ι surface = 2 ( –r ) O P
R

Case III : If r < R, the point is inside the shell then I = 0


⇒ Ι v/s r graph for hollow sphere :

Ι

o r <R r =R r >R r
Ιin =0
Ι out ∝ 1/ r 2

–GM / R2
7. Gravitational Field Intensity Due to Spherical Mass Distribution
If the observation point is located on the surface or outside the surface then the spherical mass
can be taken as a particle which is situated at the centre of the sphere. ie. point mass.
For solid sphere
Let ‘M’ be the mass of sphere, ‘R’ the radius of sphere and ‘r’ the distance of the point under
consideration from the centre of sphere.
M
R

 GM
⇒ Case I : When r > R, i.e., outside the sphere then Ιout = 2 (– r )
r
M

R r
O P

 GM
⇒ Case II : When r = R, i.e., at the surface then Ι surface = 2 (– r )
R
M

r =R
P
O

 GM'
⇒ Case III : When r < R, i.e., inside the sphere then Ιin = 2 (– r ) ........(i)
r
4 3
πr × ρ
M' V '× ρ ' 3 r3 Mr3
We know, = = = 3 ⇒ M' =
M V×ρ 4 3 R R3
πR × ρ
3
M

R M'
r
P

Putting the expression for M’ in above equation (i) we get

GMr3 GMr
Iin = (– r ) ⇒ Iin = (– r )
rR
2 3
R3
Important conclusions
GM 1
(1) Iout = 2 ∴ Ιout ∝ 2
r r
GM
(2) Isur = 2
R
GMr
(3) Iin = ∴ Ιin ∝ r
R3
GM
(4) So, Imax = Isur = 2
R
 GM(0)
(5) Icentre = ∴ Imin = Icentre = 0
R3

Ι

o r <R r =R r >R r
Ιin ∝ r
Ι out ∝ 1/ r 2

–GM / R2


Graph between Ι and ‘r’ for a solid sphere

Example 11:
A solid sphere of uniform density and radius R exerts a gravitational force of attraction F1 on a
particle P, distant 2R from the centre of the sphere. A spherical cavity of radius R/2 is now
formed in the sphere as shown in figure. The sphere with cavity now applies a gravitational
force F2 on the same particle P. Find the ratio F2/F1.

O P
R R

Solution:
GMm
F1 = , F2 = force due to whole sphere – force due to the sphere forming the cavity
4R2
GMm GMm 7GMm F2 7
= – ⇒ ∴ =
4R2
18R2
36R2 F1 9

8. Gravitational Field Intensity Due to a Ring

8.1 Intensity Due to a Ring at Axial Point
Gdm
dI =
( R2 + x2 )
dm
dΙ
θ
x
R

Now only axial component will be added other components will be concluded.
Gdm x
So I= ∫ dl cos θ = ∫R 2
+x 2
.
(R + x2 )1/2
2

GMx
I=
(R + x2 )3/2
2

E
+ Emax
x = +R / 2
x =R/ 2 r
– Emax
 • When x <<< R
GMx
Ι=
R3
At the centre of ring x = 0 so E0 = 0
dΙ R
Ι will be maximum when at x = ± ,
dx 2
2GM
Ιmax =
3 3R2

Example 12:
Find the force exerted by the particle on the ring as shown in figure.
M

m
4R
3
R

Solution:
Intensity due to the ring at the location of point mass
 4R 
GM  
Ι=  3 
3/2
  4R  
2

R2 +   
  3  
36 GMm
Ι=
125 R2
Force on point mass
F=
Now according to newton’s third law,
|Force exerted by ring on particle| = |Force exerted by particle on ring|
36 GMm
So FRing =
125 R2

Concept Builder-2

Q.1 Two concentric shells of masses M1 and M2 are having radii r1 and r2. Which of the following is
the correct expression for the gravitational field at a distance r ?
M2
r2
M1
r1

G(M1 + M2 ) G(M1 + M2 )
(1) , for r < r1 (2) , for r < r2
r 2
r2
GM2 GM1
(3) , for r1 < r < r2 (4) , for r1 < r < r2
r 2
r2
Q.2 From a solid sphere of mass M & radius r, as sphere of diameter r is cut off as shown. What is
the intensity at a point x distance away from the centre.

P
x

9. Acceleration Due to Gravity

9.1 Gravity
In Newton’s law of gravitation, the force of attraction between any two bodies is gravitation
force. If one of the bodies is Earth then the gravitation is called ‘gravity’. Hence, gravity is the
force by which Earth attracts a body towards its centre. It is a special case of gravitation.

9.2 Acceleration Due to Gravity Near Earth’s Surface

Let us assume that Earth is a uniform sphere of mass M and radius R. The magnitude of the
gravitational force of Earth on a particle of mass m, located outside the Earth at a distance r
GMm
from its centre, is F = . Now according to Newton’s second law F = mag
r2
GM
Therefore ag = ......(i)
r2
GM
At the surface of Earth, acceleration due to gravity g == 9.8 m/s2
R2
However any ag value measured at a given location will differ from the ag value calculated
according to equation due to any three reasons :-
(i) Earth’s mass is not distributed uniformly.
(ii) Earth is not a perfect sphere and
(iii) Earth rotates.

Percentage Change in ‘g’

GM G 4
In terms of density g = = 2 × πR3 × ρ
R2
R 3
4
∴ g= πGRρ If ρ is constant then g ∝ R
3
1 ∆g ∆R
• If M is constant then g ∝ ; For % variation in ‘g’ upto 5% = –2
R2 g R
• If mass (M) and radius (R) correspond to a planet and if small changes ∆M and ∆R occur in
(M) and (R) respectively then
GM ∆g ∆M ∆R
by g = ∴ = –2
R2 g M R
∆g ∆M
If R is constant then = ;
g M
∆g ∆R
If M is constant then =–2
g R
Example 13:
What will be acceleration due to gravity at a planet having diameter twice the diameter of earth
& mass 3 times that of earth.
Solution:

GMe GMP
As we know ge = & gP =
R 2
e
R2e
2
gp Mp  Re  3
So =   =
ge Me  Rp 
 4

3 2
⇒ gp = g = 7.362 m/s
4 e

Example 14:
What will be the acceleration due to gravity on a planet having density 3 times that of earth &
radius half of earth.
Solution:
As we know

4 4
ge = πGρeRe, gp = πGρpRp,
3 3

gp ρp  Rp  3 3
⇒ =   = ⇒ gp = ge
ge ρe  Re  2 2

Example 15:
Draw a rough sketch of the variation in weight of a spacecraft which moves from earth to moon.
Me Fe m Fm Mm

Solution: r
d

GMe GMm
Net weight = Fe ~ Fm = – m
r 2
(d– r)2

mg

mg / 6
weight of
space craft
r
Earth dis tance moon
10. Variation in Acceleration Due to Gravity
10.1 Above the Surface of Earth Due to Altitude (Height)
From diagram
–2
gh R2e R2e  h 
= = =  1 + 
g (Re + h)2  h
2
 Re 
R 1 + 
2

 Re 
e

=gh GMe / (Re + h)2

h
g = GMe / R2e
Earth
Me
Re

O
–2
 h  2h 
By binomial expansion  1 +
  =  1 – 
 Re   Re 
[If h << Re, then higher power terms become negligible]
 2h 
∴ gh = g  1 – 
 Re 
Note :
(i) This formula is valid if h is upto 5% of earth’s radius. (320 km from earth’s surface)
GMe
(ii) If h is greater than 5% of the earth’s radius we use gn =
(Re + h)2

Example 16:
At what height above the earth surface value of gravity is 98% of gravity at earth surface :
Solution:
∆g 2h
As we know =–
g Re
2h
So –2% = – ⇒ h = 64Km
Re

Note : It can be understood (& memorised) that value of g decreases by 1% after moving every 32 Km.

Example 17:
th
At what height above the Earth’s surface will be the acceleration due to gravity 1/9 of its value
at the Earth’s surface ? (Radius of Earth is 6400 km)
Solution:
Acceleration due to gravity at height h is
g g  h 
g' = = ⇒  1 +  = 3
9  Re
2
 h 
1+ 
 Re 
⇒ h = 2Re = 12800 km.
10.2 Variation in ‘g’ inside Earth Surface Due to Depth
Assuming that the density of Earth remains same throughout the volume
4
At Earth’s surface : g = πGReρ ......(i) g
3
P
At a depth d inside the Earth : d gd
r = Re – d

GM' O
For point P only mass of the inner sphere is effective gd =
r2
Mass of sphere of radius r = M’

4 3 4 3 Me M
M' = pr r = pr × = M' = 3e r3
3 3 4 / 3 πR3
R

G Mer
3
GM r GMe R –d g
gd = × = 2e × = 2 × e
r 2
Re
3
Re Re Re Re

 d
gd = g  1 –  valid for any depth
 Re  r
O R
∆gd = g – gd = decrement in g with depth
(Taking direction towards
centre of earth as positive)
 d ∆g e d
= g – g 1 –  ⇒ =
 Re  g Re

Example 18:
At what depth below the Earth’s surface the acceleration due to gravity is decreased by 1%?
Solution:
∆g d d 1 d
= ⇒ = ∴ d = 64 km
g Re 100 6400

Example 19:
At which height above earth’s surface is the value of ‘g’ same as in a 100 km deep mine?
Solution:
At height ‘h’ above the earth surface (h < 5% Re)

∆g –2h
=
g Re

At depth x below the earth surface

∆g –x
=
g Re

For g to be same, ∆g will also be same

2h x x 100
So = ⇒ h= = km, So h = 50 km
Re Re 2 2
10.3 Variation in ‘g’ Due to Rotation
Net force on particle at P
2 mω2r cosr
mg’ = mg – mrω cosλ N
2
g’ = g – rω cosλ from ∆OMP r = Re cosλ where λ = Latitude r λ
M λ
2 2
Substituting for r we have g’ = g – Re ω cos λ Re
W λ E
• At the equator (λ = 0°)
O
2 2
geq = g – ω Rcos (0°) gmin or
• At the poles (λ = 90°)
2 2 2
gpole = g – ω Rcos (90°) = g – ω R(0) : gmax or gpole = g S

It means that acceleration due to gravity at the poles does

not depend upon the angular velocity or rotation of earth.

Condition of weightlessness on Earth’s surface:

If apparent weight of body is zero then angular speed of Earth can be calculated as mg’ = mg
2 2
– mRe ω cos λ

2 2 1 g
0 = mg – mRe ω cos λ ⇒ w =
cos λ Re
But at equator λ = 0°
g 1 –3
∴w= = rad/s = 0.00125 rad/s = 1.25 × 10 rad/s
Re 800
Note:
• If Earth were to rotate with 17 times of its present angular speed then bodies lying on
equator would fly off into the space. Time period of Earth’s rotation in this case would be
1.4 h.
• If Earth stops rotating about its own axis, then the apparent weight of bodies or effective
acceleration due to gravity will increase at all the places except poles.

Example 20:
Determine the speed with which Earth would have to rotate about its axis so that a person on
3
the equator weighs th of its present value. Write your answer in terms of g and R.
5
Solution:
Weight on the equator
3 3 2g
W' = W ⇒ mg = mg – mω2R ⇒ω=
5 5 5R

Example 21:
Which of the following statements are true about acceleration due to gravity?
(1) ‘g’ decreases in moving away from the centre of earth if r > R
(2) ‘g’ decreases in moving away from the centre of earth if r < R
(3) ‘g’ is zero at the centre of earth
(4) ‘g’ decreases if earth stops rotating on its axis
Solution:
1
Variation of g with distance : if r > R then g ∝
r2
∴ (1) is correct
If r < R then g ∝ r
∴ (2) is incorrect & (3) is correct
2 2
variation of g with ω : g’ = g – ω Rcos λ
If ω = 0 then g will not change at poles where cosλ = 0. While at other points g increases
∴ (4) is incorrect

10.4 Variation due to Shape of Earth:

From the diagram
polar axis N
GMe
Rp = Re (Re = Rp + 21 km) gp =
Rp2 Rp
W E
GMe O equatorial
and ge = Re axis
(Rp + 21)2
∴ ge < gp S
21 km
By putting the values (Not to scale)

Example 22:
At sea level the value of g is minimum at :
(1) the equator (2) 45° north latitude
(3) 45° south latitude (4) the pole
Solution:
At the equator, radius of earth is largest so g at equator will be minimum.

Concept Builder-3

Q.1 The radius of Earth is about 6400 km and that of Mars is 3200 km. The mass of Earth is 10
times that of Mars. An object weighs 200 N on the surface of Earth. Its weight on the surface
of Mars will be :-
(1) 80 N (2) 40 N (3) 20 N (4) 8 N

Q.2 What will be the weight of a person on a planet having mass M (=4 Me) & diameter D(= 2Re). If
he weighs 600 N on earth.

Q.3 What will be the acceleration due to gravity on a planet having density 2 times that of earth &
diameter half of radius of earth ?

Q.4 A stone dropped from a height ‘h’ reaches the Earth’s surface in 1 s. If the same stone is taken
to Moon and dropped freely from the same height then it will reach the surface of the Moon in
a time (The ‘g’ of Moon is 1/6 times that of Earth) :-
(1) 6 seconds (2) 9 seconds (3) 3 seconds (4) 6 seconds
Q.5 The maximum vertical distance through which an astronaut can jump on the earth is 0.5 m.
Estimate the corresponding distance on the moon.

Q.6 Find the percentage decrement in the weight of a body when taken to a height of 16 km above
the surface of earth. (radius of earth is 6400 km)

Q.7 At which height from the earth’s surface does the acceleration due to gravity decrease
by 1%

Q.8 If earth is assumed to be a sphere of uniform density then plot a graph between acceleration
due to gravity (g) and distance from the centre of earth.

Q.9 What is the value of acceleration due to gravity at a height equal to half the radius of earth,
2
from surface of earth? [take g = 10 m/s on earth’s surface]

Q.10 At which height from the earth’s surface does the acceleration due to gravity decrease by 75%
of its value at earth’s surface?

Q.11 At what depth below the surface does the acceleration due to gravity becomes 70% of its value
on the surface of earth ?
1
Q.12 At what depth from earth’s surface does the acceleration due to gravity becomes times that
4
of its value at surface ?

Q.13 Weight of a body decreases by 1% when it is raised to a height h above the Earth’s surface. If
the body is taken to a depth h in a mine, then its weight will :
(1) decrease by 0.5 % (2) decrease by 2% (3) increase by 0.5% (4) increase by 1%

Q.14 Calculate the angular velocity of earth so that an object at equator may feel weightlessness.

Q.15 Calculate the angular speed of earth such that an object at latitude 45° feels half of its weight.

Q.16 As we go from the equator to the poles, the value of g :

(1) remains the same
(2) decreases
(3) increases
(4) decreases up to latitude of 45° and then increases

11. Gravitational Potential Energy

11.1 Potential Energy Corresponding to a Conservative Field
The potential energy of a system corresponding to a conservative force was defined as
f  
Uf – Ui = ∫ F .dr .
i

The change in potential energy is equal to the negative of the work done by the internal forces.
11.2 Gravitational Potential Energy of a Two Mass System
Change in potential energy of the system of the two particles as the distance changes from
r1 to r2.
r
dr
r1
A B D E C
r2

Consider a small displacement when the distance between the particles changes from r to r +
dr. In the figure, this corresponds to the second particle going from D to E.
The force on the second particle is
Gm1m2 
F= along DA .
r2
The work done by the gravitational force in the displacement is
Gm1m2
dW = – dr
r2
The increase in potential energy of the two-particle system during this displacement is
Gm1m2
dU = –dW = dr
r2
The increase in potential energy as the distance between the particles change from r1 to r2 is
U(r2) – U(r1) = ∫ dU
r2 r2
Gm1m2 1
= ∫
r1 r 2
dr = Gm1m2 ∫r
r1
2
dr

r
 1 2
= Gm1m2 – 
 r r1
1 1
= Gm1m2  –  . ...(11.2)
 r1 r2 
We choose the potential energy of the two particle system to be zero when the distance
between them is infinity. This means that we choose U(∞) = 0. By (11.2) the potential energy U(r),
when the separation between the particles is r, is
U(r) = U(r) – U(∞)
 1 1 Gm1m2
= Gm1m2  –  = –
∞ r r
The gravitational potential energy of a two particle system is
Gm1m2
U(r) = – ...(11.3)
r
where m1 and m2 are the masses of the particles r is the separation between the particles and
the potential energy is chosen to be zero when the separation is infinite.

11.3 Gravitational Potential Energy of a Three Particle System

If there are more than two particles in a system then the net gravitational potential energy of
the whole system is the sum of gravitational potential energies of all the possible pairs in that
system.
For eg :
 m3
 Gm1m2   Gm2m3   Gm1m3 
Usystem =  –  +  –  +  – 
 r1   r2   r3  r3 2
r2
Gm1m1 Gm2m3 Gm1m3 3
Usystem = – – –
r1 r2 r3

m1 r1 m2
1
Example 23:
Three particles each of mass m are placed at the corners of an equilateral triangle of side d.
Calculate (a) the potential energy of the system, (b) work done on this system if the side of the
triangle is changed from d to 2d.
Solution:
(a) As in case of two-particle system potential energy is given by –(Gm1m2/r), so
UA = U12 + U23 + U31 C
m
Gmm 3Gm2
So, UA = –3 =– 60º
d d d
3Gm2
(b) When d is changed to 2d, UB = –
2d
Am B
3Gm2 m
So, work done = UB – UA =
2d

11.4 Change in Gravitational Potential Energy of Earth - Mass System

(a) We change in gravitational potential energy of the earth-particle system when the particle
was raised through a small height over earth’s surface. In this case the force Mg may be treated
as constant and the change in potential energy is
Uf – Ui = mgh
where the symbols have their usual meanings.

(b) W = ∆U = Uf –Ui
m GMm
Uf = –
GMm  GMm  Fg R+h
⇒W = – – – 
R+h  R  h
m GMm
1 1  Ui = –
⇒ W = GMm  –  R
 R R + h
R
R + h–R  M
⇒ W = GMm  
 R(R+ h) 
 
 
h
⇒ W = gR2m   2
[ GM = gR ]
 2 h
 R  1 +  
  R
mgh
W=
 h
1+ 
 R
 Special cases
h mgh
(i) If h << R, then ∈0 ∴ W∈ = mgh
R 1+0
mgh mgR
(ii) If h = R, then W = =
 R 2
1+ 
 R 

11.5 Application based on Earth-Mass System

(a) The velocity required to project a particle to a height ‘h’ from the surface of earth.
Applying ‘COME’ on the surface and at a height ‘h’.
(K.E. + U)surface = (K.E. + U)final
1 GMm GMm
⇒ mv2 – =0–
2 R R+h
1 GMm  GMm 
⇒= mv2 – =–  
2 R R+h
1 mgh
⇒ mv2 =
2 h
1+
R

2gh 2gh
⇒ v2 = ⇒v =
h h
1+ 1+
R R

Note : If a body is released from a height ‘h’ above the surface of earth, then its velocity on reaching
the earth’s surface is also given by :
2gh
v=
h
1+
R

(b) To find the maximum height attained by a body when it is projected with velocity ‘v’ from
the surface of earth.
2gh
Form v2 =
h
1+
R
v 2h
⇒ v2 + = 2gh
R
v 2h
⇒ v2 = 2gh –
R
 v2 
⇒ v2 = h  2g – 
 R

v2 v 2R v 2R
⇒h = = ⇒h=
v
2
2gR – v 2 2gR – v 2
2g –
R
Example 24:
With what velocity must a body be thrown from the earth’s surface so that it may reach a height
–2
4Re above the Earth’s surface? (Radius of the Earth Re = 6400 km, g = 9.8 ms ).
Solution:
By using conservation of mechanical energy
1 GMm0 GMm0
m0v2 – =0–
2 Re
(Re
+ 4Re )

1 GMm0 GMm0
m0v2 = – +
2 5Re Re

4 GMm0 8 GM 8 gRe
2
1
⇒ m0v2 = ⇒ v2 = =
2 5 Re 5 Re 5 Re
8 3 8 4
v2 = × 9.8 × 6400 × 10 = 10 ⇒ v = 10 km/s.
5

Example 25:
A body of mass m kg starts falling from a distance 2R above the earth’s surface. What is its
kinetic energy when it has fallen to a distance ‘R’ above the earth’s surface ? (Where R is the
radius of Earth)
Solution:
By conservation of mechanical energy,
GMm GMm
– +0=– + K.E.
3R 2R
GMm  1 1 1 GMm
⇒ K.E. =  –  =
R  2 3  6 R
1 (gR2 )m 1
= = mgR
6 R 6

12.1 Definition of Gravitational Potential:

Gravitational field around a material body can be described not only by gravitational intensity

Ι g , but also by a scalar function, the gravitational potential V. Gravitational potential is the
amount of work done by external agent in bringing a body of unit mass from infinity to that
Wext
point without changing its kinetic energy. V =
m
GM
Gravitational force on unit mass at (P) will be = 2
x
Work done by this force when the unit mass is displaced through the distance dx is
GM
dWext = Fdx = . dx
x2

x dx

M r
 Total work done in bringing the body of unit mass from infinity to point (P) is
r
r
GM  GM  GM
Wext = ∫∞ x2 dx = –  x  = – r
∞

GM
This work done is the measure of gravitational potential at point (P) ∴ VP = –
r
• If r = ∞ then V∞ = 0. Hence gravitational potential is maximum at infinity (as it is a negative
quantity at point P)
GMe
• If r = Re (on the surface of Earth) VS = –
Re

12.2 Relation Between Intensity and Potential Gradient:

   
V = – ∫ Ι . dr ⇒ dV = Ι .dr

dV
∴Ι = – = – ve potential gradient.
dr

12.3 Gravitational Potential Due to Regular Bodies:

Solid Sphere:
GM
Case I : r > R (outside the sphere); Vout = –
r
V
r <R R r >R
O r

–GM / R

–3GM / 2R

GM
Case II : r = R (on the surface); Vsurface = –
R
Case III : r < R (inside the sphere);
GM
Vin = – [3R2 – r2]
2R3
It is clear that the potential |V| will be maximum at the centre (r = 0)
3 GM 3
| Vcentre | = , Vcentre = Vsurface
2 R 2

Spherical Shell:
GM
Case I: r > R (outside the sphere) Vout = –
r
V
r <R R r >R
O r

–GM / R
 GM
Case II: r = R (on the surface); Vsurface = –
R
Case III: r < R (inside the sphere);
GM
Potential is same every where and is equal to its value at the surface Vin = –
R

Example 26:
In a certain region of space gravitational field is given by I = – (K/r) (Where r is the distance
from a fixed point and K is constant). Taking the reference point to be at r = r0 with V = V0. Find
the potential at a distance r.
Solution:
K
∫ dV =– ∫ E.dr , ∫ dV = ∫ r dr
V = K log r + c at r = r0; V = V0
⇒ V0 = K log r0 + c ⇒ c = V0 – K log r0
By substituting the value c in equation
r
V = K log   + V0
 r0 

Example 27:
Two bodies of respective masses m and M are placed d distance apart. The gravitational
potential (V) at the position where the gravitational field due to them is zero is :-
G G GM G
(1) V = – (m + M) (2) V = – (3) V = – (4) V = – ( m + M)2
d d d d
Solution:
Equilibrium position of the neutral point from mass ‘m’ is :
 m 
=  d
 m + M
 
–Gm1 –Gm2
V1 = ; V2 =
r1 r1

–Gm
V1 = ( M + m );
md

–Gm
V2 = ( M + m)
md

–G
V1 = M ( M + m );
d
–G
V2 = M ( M + m)
d
–G
V= ( M + m)
d
 Concept Builder-4

Q.1 Three particles each of mass m are placed at the corners of an equilateral triangle of side ‘a’.
Energy is given to their system so that they are now at the corners of an equilateral triangle of
side 20a. Energy given to the system in this process is :
Gm2 19Gm2 3Gm2 57Gm2
(1) (2) (3) (4)
20a 20a 20a 20a

Q.2 What is the potential energy of a body of mass m relative to the surface of earth at a (a) height
h = R above its surface ? (b) depth d = R below its surface ?

Q.3 Gravitational potential difference between a point on the surface of a planet and point 10 m
above is 4 J/kg. Considering the gravitational field to be uniform, how much work is done in
moving a mass of 2 kg from the surface to a point 5 m above the surface ?
(1) 4 J (2) 5 J (3) 6 J (4) 7 J

Q.4 The gravitational acceleration on the surface of earth is g. Find the increase in potential energy
in lifting an object of mass m to a height equal to the radius of earth.

Q.5 A particle is fired vertically upwards from the surface of earth and reaches a height 6400 km.
2
Find the initial velocity of the particle if R = 6400km and g at the surface of earth is 10 m/s .

Q.6 The intensity of gravitational field at a point situated at a distance 8000 km from the centre of
Earth is 6.0 N/kg. The gravitational potential at that point in N-m/kg will be :-
7 5 2
(1) 6 (2) 4.8 × 10 (3) 8 × 10 (4) 4.8 × 10

K
Q.7 The gravitational field due to a certain mass distribution is E = in the x-direction (K is a
x3
constant). Taking the gravitational potential to be zero at infinity, its value corresponding to
distance x is :-
K K K K
(1) (2) (3) (4)
x 2x x2 2x2

2 3
Q.8 Two masses of 10 kg and 10 kg are separated by 1 m distance. Find the gravitational potential
at the mid point of the line joining them.

13. Escape Energy & Escape Velocity

Escape Velocity (Ve)
It is the minimum velocity required for an object located at the planet’s surface so that it just
escapes the planet’s gravitational field.
Consider a projectile of mass m, leaving the surface of a planet (or some other astronomical
body or system), of radius R and mass M with escape speed ve.
when the projectile just escapes to infinity it has neither kinetic energy nor potential energy.
 From conservation of mechanical energy
1  GMm  2GM
mv 2e +  –  = 0 + 0 ⇒ ve =
2  R  R
The escape velocity of a body from a location which is at height ‘h’ above the surface of planet,
we can use:
2GM 2GM
ves = = ( r = R + h)
r R+h
Where, r = distance from the centre of the planet.
h = Height above the surface of the planet.

Escape speed depends on

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

Escape speed does not depend on

(i) Mass (m) of the body which is projected
(ii) Angle of projection.
If a body is thrown from the Earth’s surface with escape speed, it goes out of earth’s
gravitational field and never returns back to the earth’s surface.

• Escape energy
Minimum energy given to a particle in the form of kinetic energy so that it ve
can just escape the Earth’s gravitational field. surface
Earth

GMm Me
Magnitude of escape energy = (– ve of PE on the Earth’s surface) Re
R
Escape energy = Kinetic Energy corresponding to the escape velocity O

GMm 1
⇒ = mv 2e
R 2
Note : In the above discussion it can be any planet for that matter

Key Points

2GM 1
• ve = If M = constant then ve ∝
R R
• ve = 2gR If g = constant ve ∝ R

8πGp
• ve = R If ρ = constant then ve ∝ R
3
• Escape velocity does not depend on the mass of the body being projected, angle of projection
or direction of projection.
0 0
ve ∝ m and ve ∝ θ
• Escape velocity at : Earth’s surface ve = 11.2 km/s, Moon surface ve = 2.31 km/s.
• Atmosphere on Moon is missing because root mean square velocity of gas particles is greater
than escape velocity. i.e., vrms > ve
Example 28:
24
A mass of 6 × 10 kg (= mass of earth) is to be compressed in a sphere in such a way that the
8
escape velocity from its surface is 3 × 10 m/s (equal to the velocity of light). What should be
the radius of the sphere?
(1) 9 mm (2) 8 mm (3) 7 mm (4) 6 mm
Solution:

 2GM   2GM 
As, ve =   , R =  2  ,
 R   ve 
2 × 6.67 × 10–11 × 6 × 1024
∴R= = 9 × 10–3 m = 9mm
(3 × 108 )2

Example 29:
The masses and radii of the earth and moon are M1, R1 and M2, R2 respectively. Their centres are
at distance d apart. What is the minimum speed with which a particle of mass m should be
projected from a point midway between the two centres so as to escape to infinity ?
Solution:
Potential energy of m when it is midway between M1 and M2,
 GM1 –GM2 
U = m(V1 + V2) = m – + 
 d/2 d/2 
–2Gm
= [M1 + M2]
d
And as potential energy at infinity is zero, so work required to shift m from the given position
to infinity,
W = 0 – U = 2Gm (M1 + M2)/d
As this work is provided by initial kinetic energy,

1 2 2Gm(M1 + M2 ) G(M1 + M2 )
mυ = or u = 2
2 d d

Example 30:
A narrow tunnel is dug along the diameter of 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.
Me ,R

R/2

m0
Solution:
Suppose we project the particle with speed ve, so that it Me ,R just
reaches infinity (r → ∞).
Applying energy conservation principle
R/2 at r → ∞, v → 0
Ki + Ui = Kf + Uf
 GM m0 m0
 R  
 2 ve
1
m0 v 2e + m0 – 3e 3R –    = 0
2

2  2R
   2  

11GMe
⇒ ve =
4R

Example 31:
If velocity given to an object from the surface of the Earth is n times the escape velocity then
what will be its residual velocity at infinity?
Solution:
1 GMm 1
Let the residual velocity be v, then from energy conservation m(nve)2 – = mv2 + 0
2 R 2
2 2 2 2GM
⇒ v = n ve –
R
2 2 2 2 2
= n ve – ve = (n – 1)ve

⇒v = ( n2 – 1 ve )
Concept Builder-5

Q.1 A planet has mass 100 times that of earth, what has to be its Diameter so that no physical body
can escape from it.

Q.2 A body of mass m is situated at a distance 4Re above the Earth’s surface, where Re is the radius
of Earth. What minimum energy should be given to the body so that it may escape ?
mgRe mgRe
(1) mgRe (2) 2mgRe (3) (4)
5 16

Q.3 The escape velocity for a planet is ve. A particle starts from rest at a large distance from the
planet, reaches the planet only under gravitational attraction and passes through a smooth
tunnel through its centre. Its speed at the centre of the planet will be :-
ve
(1) 1.5v e (2) (3) ve (4) zero
2

Q.4 A rocket is fired with a speed v = 2 gR near the earth’s surface and directed upwards. (a) Show
that it will escape from the earth. (b) Show that in interstellar space its speed is v = 2gR .

Q.5 A projectile is fired vertically upward from the surface of earth with a velocity Kve where ve is
the escape velocity and K < 1. Neglecting air resistance, show that the maximum height to which
2
it will rise measured from the centre of earth is R/(1 – K ) where R is the radius of the earth.
14. Kepler’s Laws of Planetary Motion
Kepler found important regularities in the motion of the planets. These regularities are known
as ‘Kepler’s three laws of planetary motion.
(a) First Law (Law of Orbits)
All planets move around the Sun in elliptical orbits having the Sun at one focus of the orbit.

rmin = r1 P

b
A B
F1 F2
Aphelion
Perihelion rmax = r2

a
semi major axis

When a particle moves with respect to two fixed points in such a way that the sum of the
distances from these two points is always constant then the path of the particle is an ellipse.
And the two fixed points are called focal points.
According to Figure :-
PF1 + PF2 = AF1 + AF2 = BF1 + BF2 = constant
But in ellipse AF1 = BF2 (minimum distance from both focal is same)
PF1 + PF2 = BF2 + AF2 = BF1 + AF1 = 2a = length of major axis
r1 + r2 = rmin + rmax = 2a
r1 + r2 rmin + rmax
∴ a= = (Mean distance)
2 2

(b) Second Law (Law of Areas)

A line joining any planet to the Sun sweeps out equal areas in equal intervals of time, i.e., the
areal speed of the planet remains constant.
According to the second law, if a planet moves from A to B in a given time interval, and from C
to D in the same time interval then the areas ASB and CSD will be equal.
dA = area of the curved triangle
1 1 1
SAB = (AB × SA) = (rdθ × r) = r22θ
2 2 2
B
C
dθ A
θ r

S
D

Thus, the instantaneous areal speed of the planet is

dA 1 2 dθ 1 1
= r = r2 ω = rv .....(i)
dt 2 dt 2 2
where ω is the angular speed of the planet
Let L be the angular momentum of the planet about the Sun S and m is the mass of the planet.
2
Then, L = Iω = mr ω = mvr .......(ii)
2
where I (= mr ) is the instantaneous moment of inertia of the planet about the Sun S.
dA L
From equation (i) and (ii), = .......(iii)
dt 2m
Now, the areal speed dA/dt of the planet is constant according to Kepler’s second law. Therefore
according to eq. (iii), the angular momentum L of the planet is also constant, that is, the angular
momentum of the planet is conserved. Thus, Kepler’s second law is equivalent to conservation
of angular momentum.
Applying conservation of angular momentum between points A and B
rmin = r1
v 2 = v min

A B
F1 F2
rmax = r2
v1 = v max

LA = LB ⇒ mvmaxrmin = mvminrmax
⇒ vmaxrmin = vminrmax

A planet moves around the sun in an elliptical orbit of semi major axis a and eccentricity e
rmin = r1 P

b
A B
F1 F2
rmax = r2

a
semi major axis

For an ellipse : its general equation is given as

x2 y2
+ =1
a2 b2
If a > b then a is semi major axis b is semi minor axis and e is eccentricity where
2
2 2 2 b
b = a (1 – e ) ⇒e = 1–  
a
Applying the conservation of angular momentum (COAM) at the perihelion and aphelion
mvprp = mvara

va
ae ae
F1
A B
rp ra F2
Perihelion Aphelion
vp

a a
semi major axis

rmax = a(1 + e); rmin = a(1 – e)

vp vmax 1 + e
⇒= = .....(i)
va vmin 1– e
 By conservation of mechanical energy
1 GMm 1 GMm GMm
mvp2 – = mv 2a – =– .....(ii)
2 rp 2 ra 2a
By solving equation (i) and (ii)
GM  1 – e  GM  1 + e 
⇒ va =   ; vp =  
a  1+ e a  1– e 

(c) Third Law (Law of Periods)

The square of the period of revolution of any planet around the Sun is directly proportional to
the cube of the semi-major axis of its elliptical orbit.
T2 ∝ a 3
Note : For a circular orbit semi major axis = Radius of the orbit
T2 ∝ R3

Example 32:
8
Calculate the mass of the sun if the mean radius of the earth’s orbit is 1.5 × 10 km and G =
–11 2 2
6.67 × 10 N × m /kg .
Solution:
In case of orbital motion as υ = (GM/ r) so

2πr r 4π2r3
T= = 2πr , i.e., M =
υ GM GT2
4 × π2 × (1.5 × 1011 )3
∴ M=
6.67 × 10–11 × (3.15 × 107 )2
7
[as T = 1 year = 3.15 × 10 s]
30
i.e., M 2 × 10 kg

Example 33:
A planet is revolving round the sun in an elliptical orbit as shown in figure. Select correct
alternative(s)
B
C

D A
O

E F
(1) Its total energy is negative at D
(2) Its angular momentum is constant
(3) Net torque on the planet about sun is zero
(4) Linear momentum of the planet is conserved
Solution:
(1, 2, 3 )
For (1) : For a bound system, the total energy is always negative.
For (2) : For central force field, angular momentum is always conserved.
For (3) : For central force field, torque = 0
For (4) : In presence of external force, linear momentum is not conserved.
 Concept Builder-6
11
Q.1 The mean radius of the earth’s orbit around the sun is 1.5 × 10 metres. The mean radius of the
10
orbit of mercury around the sun is 6 × 10 metres. Calculate the year of the mercury.

Q.2 If earth describes an orbit round the sun of double its present radius, what will be the year on

earth?

Q.3 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear

speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy (f) total

energy throughout its orbit ? Neglect any mass loss of the comet when it comes very close to
the Sun.

Q.4 A planet is revolving around the Sun in an elliptical orbit. Its closest distance from the Sun is
rmin. The farthest distance from the Sun is rmax. If the orbital angular velocity of the planet when,

it is nearest to the Sun is ω, then the orbital angular velocity at the point when it is at the

farthest distance from the Sun is :-

2 2
 r   r  r  r 
(1)  min ω (2)  max  ω (3)  max  ω (4)  min  ω
 rmax   rmin   rmin   rmax 
   

Q.5 Let the speed of the planet at the perihelion P in Fig. be vp and the Sun-planet distance SP be

rp. Relate (rp, vp) to the corresponding quantities at the aphelion (rA, vA). Will the planet take

equal times to traverse BAC and CPB ?

B

2b
P A
S S'

C
2a

th
Q.6 If the gravitational force were to vary inversely as m power of the distance, then the time
period of a planet in circular orbit of radius r around the Sun will be proportional to :-
–3m/2 3m/2 m+1/2 (m + 1)/2
(1) r (2) r (3) r (4) r
15. Satellite Motion
A light body revolving round a heavier planet due to gravitational attraction, is called a satellite.
Moon is a natural satellite of Earth.
Satellite

Natural Satellite Artificial Satellite

(eg. Moon revolving
around earth)

Hellosynchronous Geo Stationary

or Satellite
Sun Synchronous
or
Polar
or
Geo - Satellite

15.1 Essential Conditions for Satellite Motion

• The centre of satellite’s orbit should coincide with the centre of Earth.
• Plane of the orbit of satellite should pass through the centre of Earth.
FCP Fg cos θ
=

O' θ S

Fg O'
O F=
g sin θ S
(i) (ii) O FCP = Fg

Unstable orbit
(Due to Fg sinθ, orbit will shift) Stable orbit

• It follows that a satellite can revolve round the earth only in those circular orbits whose
centres coincide with the centre of earth. Circles drawn on globe with centres coincident
with earth are known as great circles. Therefore, a satellite revolves around the earth along
circles concentric with great circles.

15.2 Orbital Velocity & Time Period of a Satellite

• Orbital Velocity (v0)
A satellite of mass m moving in an orbit of radius r with speed v0. The required centripetal
force is provided by gravitation.
mv 20 GMm
Fcp = Fg =
r r2
GM GM
⇒ v0 = = (r = Re+ h)
r (Re + h)

r m
v0

M
 For a satellite very close to the Earth’s surface h << Re ∴ r ∈ Re

GM
∴ v0 = = gRe = 8 km/s
Re
• If a body is taken to some height (small) from Earth and given a horizontal velocity of
magnitude 8 km/s then it becomes a satellite of Earth.
• v0 depends upon mass of planet, Radius of the circular orbit of satellite and density of
planet
• If orbital velocity of satellite becomes (or increased by 41.4%) or K.E. is doubled then it
escapes from the gravitational field of Earth.

• Time Period of a Satellite

2πr 2πr3/2 2πr3/2 4π 3 2 3
T= = = ⇒ T2 = r ⇒ T ∝ r (r = R + h)
v0 GM R g GM

• For a Satellite Close to Earth’s Surface

GMe
v0 = = 8 km/s
Re

Re
T0 = 2π = 84 minutes = 1 hour 24 minute = 1.4 h = 5063 s
g
In terms of density
2π(Re )1/2 3π
T0 = =
(G× 4 / 3πRe × ρ) 1/2
Gρ
Time period of a near by satellite only depends on the density of the planet.

For Moon hm = 380,000 km and Tm = 27 days

2p(Re + hm ) 2π(386400 × 103 )
v0 = = = 1.04 km
Tm 27 × 24 × 60 × 60

Example 34:
A small satellite revolves round a planet in an orbit just above planet’s surface. Taking the mean
density of planet as ρ, calculate the time period of the satellite.
Solution:
Let the radius of the planet be R then time period
2πR 2πR 2πR3/2
T= = =
v GM GM
R
4 3
mass of planet M = πR ρ
3
2πR3/2 3π
∴T= =
4 3 Gρ
G. πR . ρ
3
Example 35:
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half
the magnitude of escape velocity from the earth. (a) Determine the height of the satellite above
the earth’s surface. (b) If the satellite is stopped suddenly in the orbit and allowed to fall onto the
–2
earth, find the speed with which it hits the surface of the earth. (g = 9.8 ms and RE = 6400 km)
Solution:
(a) We know that for satellite motion
GM g
υ0 = =R
r (R+ h)
 GM 
as g = 2 and r= R + h
 R 
1 1
In this problem υ0 = υe = 2gR
2 2
[as υe = 2gR ]
R2 g 1
So, = gR
R+h 2
i.e., 2R = h + R or h = R = 6400 km
(b) By conservation of ME
 GMm  1 2  GMm 
0 + –  = mυ +  – 
 r  2  R 
2 1 1 
or υ = 2GM  – 
R 2R 
[as r = R + h = R + R = 2R]
GM
or υ =
R
= gR = 10 × 6.4 × 106 = 8 km/s

Example 36:
A satellite moves in a circular orbit around the earth. the radius of this orbit is one half that of
the moon’s orbit. Find the time in which the satellite completes one revolution.
Solution:
3/2
T∝r
3/2 3/2
Ts r   1
=  s  =  
Tm  rm  2
⇒ Ts = 9.7 days

15.3 Energy of Satellite

1 GMm L2
Kinetic energy K.E. = mv 20 = =
2 2r 2mr2
GMm L2
Potential energy P.E. = – = – mv 20 = –
r mr2
Total mechanical energy T.E. = P.E. + K.E.
mv 20 GMm L2
=– =– =–
2 2r 2mr2
• 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 bound or the different parts of the system are bonded to each other.

Binding energy of a satellite (system)

B.E. = – T.E.
1 GMm L2
B.E. = mv 20 = =
2 2r 2mr2
–P.E .
Hence B.E. = K.E. = – T.E. =
2
Escape energy and ionisation energy are the practical examples of binding energy.
Here, at point A :
 | (P.E.) | > K.E.  KE + PE = –ve
K.E.
At A, B and C system is bounded.
At point D :
Energy r→
 |PE| = K.E. A B C D
TE = KE + PE = 0
So, the system is unbounded. P.E.

15.4 Work Done in Changing the Orbit

W = Change in mechanical energy of the system
–GMm
but E =
2r
GMm 1 1
so W = E2 – E1 =  – 
2  r1 r2 
Graphs
GMm
KE = – K.E.
2r
GMm Energy r
TE = –
2r T.E.
GMm
PE = –
r P.E.

Example 37:
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 it overcomes the gravitational pull of the
Earth.
Solution:
Let ∆K be the additional kinetic energy imparted to the spaceship to overcome the gravitation
pull then
GMm
∆K = – (total energy of spaceship) =
2R
GMm GMm GMm GMm
Total kinetic energy = + ∆K = + =
2R 2R 2R R
 1 GMm 2GM
then mv 22 = ⇒ v2 =
2 R R
GM
But v1 = . So additional velocity required
R
2GM GM GM
= v2 – v1 = – = ( 2 – 1)
R R R
Alternate solution
Additional velocity = Escape velocity – orbital velocity
= ves – v0

2GM GM GM
= – ⇒ ( 2 – 1)
R R R

Example 38:
If a satellite orbits as close to the earth’s surface as possible
(1) its speed is maximum
(2) time period of its revolution is minimum
(3) the total energy of the ‘earth plus satellite’ system is minimum
(4) the total energy of the ‘earth plus satellite’ system is maximum
Solution:
(1, 2, 3)
GM
For (1) : orbital speed v0 = , rmin = R so v0 = maximum
r
2 3
For (2) : Time period of revolution T ∝ r
GMm
For (3/4) : Total energy = –
2r

15.5 GEO-Stationary & Polar Satellite

GEO-Stationary Satellite
• It rotates in an equatorial plane.
• Its height from the Earth’s surface is 36000 km. (≈ 6Re)
Polar orbit

Equatorial
plane
Equator

Equatorial
orbit

• Its angular velocity and time period should be same as that of Earth.
• Its rotating sense should be same as that of Earth (West to East)
• Geo Stationary/ Telecommunication/ Parking / Synchronous/ Satellite are always projected
from equator (for example Singapore).
• Its orbit is called parking orbit and its orbital velocity is 3.1 km/s.
 Polar Satellite (Sun - Synchronous Satellite)
It is the satellite which revolves in the polar orbit around Earth. A polar orbit is one whose angle
of inclination with the equatorial plane of Earth is 90° and a satellite in polar orbit will pass
over both the north and south geographical poles once per revolution. Polar satellites are Sun-
synchronous satellites.
Polar satellites are employed to obtain the cloud images, atmospheric data, information
regarding ozone layer in the atmosphere and to detect the ozone hole over Antarctica etc.

15.6 Weightlessness
When the weight of a body becomes zero, the body is said to be in a state of weightlessness.
In a satellite around the earth, every part and parcel of the satellite has an acceleration towards
the centre of the earth which is exactly the value of earth’s acceleration due to gravity at that
position. Thus in the satellite everything inside it is in a state of free fall. If a body is in a satellite
(which does not produce its own gravity) orbiting the Earth at a height h above its surface then
mGM mg
True weight = mgh = =
(R+ h)2
 h
2

 1 + 
 R
and Apparent weight = m(gh – a)
v 20 GM GM
But a = = = = gh
r r 2
(R+ h)2
⇒ Apparent weight = m(gh – gh) = 0.

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

Key Points

• The time period of the longest pendulum on the surface of earth is given by
R
T = 2π = 84.6 minutes
g
Note : The time period of a satellite orbiting close to the earth’s surface is also 84.6 minutes.
• The angular velocity and time period of revolution of a G.S.S. is same as that of earth. It means
that a G.S.S. completes its revolution around the earth once in 24 hours.
• Height of a G.S.S. (Geo-stationary satellite) from the surface of earth is about 36000 km.
Therefore, its distance from the centre of earth is about R + H = 36000 + 6400 = 42400 km.
• It is used as a communication satellite. It is also known as parking satellite, telecommunication
satellite or synchronous satellite.
• One G.S.S. can cover nearly one third surface area of earth. Therefore a minimum of three G.S.S.
are required to cover the whole earth.
• Orbital velocity depends upon the mass of the central body and orbital radius (distance of
satellite from the centre of the central body). If the distance of satellite increases, then the
orbital velocity (v0) decreases.
• Orbital velocity does not depend on the mass of satellite.
• If a body is taken to a small height and given a horizontal velocity of 8 km/s, it will start revolving
around the earth in a circular orbit which means that it will also become a satellite close to the
earth’s surface.
 • If a body is released from a revolving satellite, then it will continue the move in the same orbit
with the same orbital velocity which means that it will also become a satellite close to the
earth.
• When the total energy of a satellite is negative, it will be moving in either a circular or an
elliptical orbit.
• When the total energy of a satellite is zero, it will escape away from its orbit and its path
becomes parabolic.
th
• If the gravitational force is inversely proportional to the n power of distance r, then the orbital
1–n n+ 1
velocity of a satellite v 0 ∝ r 2
and time period T ∝ r 2
.
• The total energy of any planet revolving around the sun is negative ( it is bounded)
• First Satellite of Earth is Sputnik I. First Geo-satellite of India is Aryabhatt I.
First Geo-stationary satellite of India is Apple I, Example of other Satellites of India are :
Bhaskar-I, Rohini-I, Bhaskar-II (Geo Satellite); Insat-I(A) Insat-I (B) (Geo Stationary Satellite)

Example 39:
An astronaut, inside an earth’s satellite experiences weightlessness because :
(1) he is falling freely
(2) no external force is acting on him
(3) no reaction is exerted by the floor of the satellite
(4) he is far away from the earth’s surface
Solution:
Ans. (1, 3)
As astronaut’s acceleration = g; so he is falling freely. Also no reaction is exerted by the floor
of the satellite.

Example 40:
Is it possible to place an artificial satellite in an orbit such that it is always visible over Kota?
Write down the reason.
Solution:
No, Kota is not in the equatorial plane.

Concept Builder-7

Q.1 The time period of revolution of moon around the earth is 28 days and radius of its orbit is 4 ×
5 –11 2 2
10 km. If G = 6.67 × 10 Nm /kg then find the mass of the earth.

Q.2 Two satellites are orbiting around the earth in circular orbits of the same radius. One of them
is 100 times greater in mass than the other. Compare their periods of revolution :
(1) 100 : 1 (2) 1 : 100 (3) 1 : 10 (4) 1 : 1

Q.3 Two satellites A and B, having ratio of masses 3 : 1 are in circular orbits of radius r and 4r.
Calculate the ratio of total mechanical energies of A to B.

Q.4 An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due
to small but continuous dissipation against atmospheric resistance. Then explain why its speed
increases progressively as it comes closer and closer to the earth.
Q.4 Write the answer of the following questions in one word :-
(a) What is the orbital speed of Geo-stationary satellite ?
(b) For a satellite moving in an orbit around the earth what is the ratio of kinetic energy to
potential energy?

Q.6 An object weighs 10 N at the north pole of the Earth. In a geostationary satellite distant 7R from
the centre of the Earth (of radius R), the true weight and the apparent weight are respectively:
(1) 0, 0 (2) 0.2 N, 0 (3) 0.2 N, 9.8 N (4) 0.2 N, 0.2 N

• Bound and Unbound Trajectories

Imagine a very tall building on the surface of the earth from where a projectile is fired with
a velocity v parallel to the surface of the earth. The trajectory of the projectile depends on
its velocity.

v < v0 Hyperbola
Earth v > ve

Parabola ve
v0
v0 < v < ve

Circle

Ellipse

Velocity Trajectory
Projectile may not orbit the arth in an elliptical
v < v0
path, or it may falls back on the earth's surface.
v = v0 Projectile orbits the earth in a circular path.
v0 < v < ve Projectile orbits in an elliptical path.
Projectile does not orbit. It escapes the
v = ve
gravitational field of earth in a parabolic path.
Projectile does not orbit. It escapes the
v > ve
gravitational field of earth in a hyperbolic path.

• Geostationary Satellites
The plane of the orbit lies in equatorial plane of earth.
Height from the earth surface is 36000 km. This orbit is called parking orbit.
Orbital speed is nearly 3 km/s.
Time period is equal to that of earth rotation i.e., 24 hours.
Rotating direction should be same as that of earth (west to east)
• Polar Satellite
Used for remote sensing, their orbit contains axis of rotation of earth. They can cover entire
earth surface for viewing.
• Binary Star System
Two stars of mass M1 and M2 form a stable system when they move in circular orbit about
their centre of mass under their mutual gravitational attraction.
 GM1M2
• F= ,
r2
V2
• M1r1 = M2r2 r
M1 M2
GM1M2 M1V12 M2 V22 r1
cm
r2
• = = V1
r2 r1 r2
M2r M1r
• r1 = , r2 =
M1 + M2 M1 + M2

G
• V1 = M2
(M1 + M2 )r

G
• V 2 = M1
(M1 + M2 )r
when M1 = M2

GM
• V1 = V2 =
2r
r
• r1 = r2 =
2
 ANSWER KEY FOR CONCEPT BUILDERS

CONCEPT BUILDER-1 CONCEPT BUILDER-4

1. F = 3.2 × 10 dyne. –8
1. (4)
2. r = 0.816 × 10 m –5
1 1
2. (a) mgR, (b) – mgR
3. Zero 2 2
1
Gm2 3. (1) 4. mgR
4. Fnet = 3F = 3 2
a2
5. 8 km/s 6. (2)
5. (3)
7. (4)
1 Gm 8. 1.47 × 10 J/kg
6. v=
–7

2 r
7. 6 m from 9 kg mass.
CONCEPT BUILDER-5
4π 2
1. 1.78 m 2. (3)
8. Gρ20r04
3 3
3. (1) 4. 2gR
2
5. R/(1 – K )
CONCEPT BUILDER-2

7 GM CONCEPT BUILDER-6
1. (4) 2.
8 x2
3/2
2
1. T =   year 2. 2 2 years
5
m

CONCEPT BUILDER-3
3. All quantities vary over an orbit except
1. (1) 2. 2400 N
angular momentum and total energy.
1 4. (4)
3. g 4. (1)
2
e

5. The planet will take a longer time to

5. 3m 6. 0.5% traverse BAC than CPB.
7. 32 km 6. (4)
g

CONCEPT BUILDER-7
1. 6.47 × 10 kg 24
2. (4)
8.
O R
r 12
3.
1
9. 4.44 m/s 2
10. 6400 km
4. Kinetic energy increases, but potential
3 energy decreases, and the sum decreases
11. 1920 km 12. R
4 due to dissipation against friction.
13. (1)
1
5. (a)  3.1 km/s, (b) –
14. 1.25 × 10 Rad/s
–3
2
15. 1.25 × 10 Rad/s
–3
16. (3) 6. (2)
 Exercise - I

Newton's Law of Gravitation Law of Superposition of Forces

1. Newton's law of gravitation :
6. Four particles of masses m, 2m, 3m and
(1) is not applicable out side the solar
4m are kept in sequence at the corners of
system
a square of side a. The magnitude of
(2) is used to govern the motion of
satellites only gravitational force acting on a particle of
(3) control the rotational motion of mass m placed at the centre of the square
satellites and planets will be :
(4) control the rotational motion of 24m2G 6m2G
(1) (2)
electrons in atoms a2 a2
4 2Gm2
2. Mass particles of 1 kg each are placed (3) (4) zero
a2
along x-axis at x = 1, 2, 4, 8, .........∞. Then
7. Three equal masses of 1 kg each are
gravitational force on a mass of 3 kg
placed at the vertices of an equilateral
placed at origin is (G = universal
triangle PQR and a mass of 2 kg is placed
gravitational constant) :
at the centroid O of the triangle. The force
4G in newton acting on the mass of 2 kg is :-
(1) 4G (2)
3
(1) 2 (2) 2
(3) 2G (4) ∞
(3) 1 (4) zero

3. Gravitational force between two masses

8. A spherical hole of radius R/2 is made in a
at a distance 'd' apart is 6N. If these solid sphere of radius R. The mass of the
masses are taken to moon and kept at sphere before hollowing was M. The
same separation, then the force between gravitational field at the centre of the hole
them will become :- due to the remaining mass is:
1
(1) 1N (2) N
6
R
(3) 36 N (4) 6 N

4. The value of universal gravitational GM

(1) zero (2)
constant G depends upon : 8R2
(1) Nature of material of two bodies
GM GM
(2) Heat constant of two bodies (3) (4)
2R2 R2
(3) Acceleration of two bodies
(4) None of these
9. Three uniform spheres of mass M and
5. The tidal waves in the seas are primarily
radius R each are kept in such a way that
due to : each touches the other two. The
(1) The gravitational effect of the sun on magnitude of the gravitational force on
the earth any of the sphere due to the other two is:
(2) The gravitational effect of the moon
3 GM2 3 GM2
on the earth (1) (2)
4 R2 2 R2
(3) The rotation of the earth
(4) The atmospheric effect of the earth it 3GM2 3 GM2
(3) (4)
self R2 2 R2
Gravitational Field Intensity Due to Particle, 13. A spherical shell is cut into two pieces
Spherical Shell, Spherical Mass Distribution, along a chord as shown in figure. For
points P and Q :
Ring
Q

10. If the distance between the centres of P

earth and moon is D and mass of earth is
81 times that of moon. At what distance
from the centre of earth gravitational field (1) ΙP > ΙQ (2) ΙP < ΙQ
will be zero : (3) ΙP = ΙQ = 0 (4) ΙP = ΙQ ≠ 0
D 2D
(1) (2)
2 3 14. Figure shows two concentric spherical
4D 9D shells of mass M1 and M2 and of radii R1
(3) (4)
5 10 and R2 respectively. Gravitational potential
at a point at a distance x from centre such
11. Following curve shows the variation of that R1 < x < R2 will be :
intensity of gravitational field (I) with M2
M1
distance from the centre of solid sphere
R1
(r) :-
O
R2
Ι Ι
R r
(1) O (2) O
R r
–GM2 M M 
(1) (2) –G  1 + 2 
R2  R1 x 

Ι Ι M M  M M 
R r (3) –G  1 + 2  (4) –G  1 + 2 
 x R1   x R2 
(3) O (4) O
R r

15. In Q.14, gravitational intensity at a

Distance x from the centre such that
12. The force between a hollow sphere and a
x > R2 is :
point mass at P inside it as shown in
figure: M + M  M M 
(1) G  1 2 2  (2) G  21 + 22 
R 
 x   1 R2 
P
M M  M M 
(3) G  21 + 22  (4) G  1 + 22 
C R   x R 
 1 x   1 

(1) is attractive and constant Acceleration Due to Gravity

(2) is attractive and depends on the
position of the point with respect to 16. The value of 'g' on earth surface depends :
centre C (1) only on earth's structure
(3) is zero (2) only on earth's rotational motion
(4) is repulsive and constant (3) on above both
(4) on none of these and is same
17. Diameter and mass of a planet is double 22. One can easily "weight the earth" by
that of earth. Then time period of a calculating the mass of earth using the
pendulum at surface of planet is how formula
much times of time period at earth (in usual notation)
surface : G 2 g 2
(1) R (2) R
1 g E G E
(1) times (2) 2 times
2 g G 3
(3) Equal (4) None of these (3) R (4) R
G E g E
th
18. Gravitation on moon is 1/6 of that on
23. Mars has a diameter of approximately 0.5
earth. When a balloon filled with hydrogen
of that of earth, and mass 0.1 of that of
is released on moon then this :
earth. The surface gravitational field
(1) Will rise with an acceleration less than
strength on mars as compared to that on
g earth is greater by a factor of :
 
6 (1) 0.1 (2) 0.2
g (3) 2.0 (4) 0.4
(2) Will rise with acceleration  
6
(3) Will fall down with an acceleration Variation in Acceleration Due to Gravity
 5g 
less than   24. Acceleration due to gravity at the centre
 6 
of the earth is :
g
(4) Will fall down with acceleration   g
6 (1) g (2)
2
(3) zero (4) infinite
19. The acceleration due to gravity g and
mean density of earth ρ are related by 25. The value of 'g' reduces to half of its value
which of the following relations? [G = at surface of earth at a height 'h', then :-
gravitational constant and R = Radius of (1) h = R
earth]: (2) h = 2R
4πgR2 4πgR2
(1) ρ = (2) ρ = (3) h = ( 2 + 1)R
3G 3G
(4) h = ( 2 – 1)R
3g 3g
(3) ρ = (4) ρ =
4πgR 4πGR3 –2
26. At any planet 'g' is 1.96 m sec . If it is safe
to jump from a height of 2m on earth, then
20. The mass of the moon is 1% of mass of the what should be corresponding safe height
earth. The ratio of gravitational pull of for jumping on that planet :
earth on moon to that of moon on earth (1) 5 m (2) 2 m
will be : (3) 10 m (4) 20 m
(1) 1 : 1 (2) 1 : 10
(3) 1 : 100 (4) 2 : 1 27. If the earth stops rotating suddenly the
value of g at a place other than poles
21. Imagine a new planet having the same would :-
density as that of earth but it is 3 times (1) Decrease
bigger than the earth in size. If the (2) Remain constant
(3) Increase
acceleration due to gravity on the surface
(4) Increase or decrease depending on the
of earth is g and that on the surface of the
position of earth in the orbit round the
new planet is g' then :
sun
(1) g' = 3g (2) g' = g/9
(3) g' = 9g (4) g' = 27 g
28. When you move from equator to pole, the 35. The imaginary angular velocity of the earth
value of acceleration due to gravity (g) :- for which the effective acceleration due to
(1) increases gravity at the equator shall be zero is
(2) decreases equal to
(3) remains the same (1) 1.25 × 10
–3
rad/s (2) 2.50 × 10
–3
rad/s
(4) first increases then decreases –3 –3
(3) 3.75 × 10 rad/s (4) 5.0 × 10 rad/s
2
29. When the radius of earth is reduced by 1% [Take g = 10 m/s for the acceleration due
without changing the mass, then the to gravity if the earth were at rest and
acceleration due to gravity will radius of earth equal to 6400 km]
(1) increase by 2% (2) decrease by 1.5%
(3) increase by 1% (4) decrease by 1% 36. More amount of sugar is obtained in 1 kg
weight:
30. Weight of a body of mass m decreases by (1) At North pole
1% when it is raised to height h above the (2) At equator
earth's surface. If the body is taken to a
(3) Between pole and equator
depth h in a mine, then its weight will
(4) At South pole
(1) decrease by 0.5% (2) decrease by 2%
(3) increase by 0.5% (4) increase by 1%
Gravitational Potential Energy
31. Acceleration due to gravity at earth's
–2 37. A particle falls on earth :
surface is 'g' ms . Find the effective value
(i) from infinity, (ii) from a height 10 times
of acceleration due to gravity at a height
the radius of earth. The ratio of the
of 32 km from sea level: (Re = 6400 Km)
–2 –2
velocities gained on reaching at the earth's
(1) 0.5 g ms (2) 0.99 g ms surface is :
–2 –2
(3) 1.01 g ms (4) 0.90 g ms (1) 11 : 10 (2) 10 : 11
(3) 10 : 11 (4) 11 : 10
32. The change in the value of 'g' at a height
'h' above the surface of the earth is same
38. Two different masses are dropped from
as at a depth 'd'. If 'd' and 'h' are much
same height. When these just strike the
smaller than the radius of earth, then ground the following is same :
which one of the following is correct? (1) kinetic energy
(1) d = h (2) d = 2h (2) potential energy
3h h (3) linear momentum
(3) d = (4) d =
2 2 (4) acceleration

33. If the rotational speed of earth is 39. If Me is the mass of earth and Mm is the
increased then weight of a body at the mass of moon (Me = 81 Mm). The potential
equator energy of an object of mass m situated at
(1) increases a distance R from the centre of earth and
(2) decreases r from the centre of moon will be :-
(3) becomes double
R  1
(4) does not changes (1) –GmMm  + r  2
 81  R
34. A body weights W newton at the surface  81 1 
(2) –GmMe  + 
of the earth. Its weight at a height equal  r R
to half the radius of the earth will be :
 81 1 
W W (3) –GmMm  + 
(1) (2)  R r
2 2
 81 1 
4W W (4) GmMm  – 
(3) (4)  R r
9 4
40. The gravitational potential energy is Gravitational Potential/Work Done
maximum at:
(1) infinity 45. Which of the following curve expresses
(2) the earth's surface the variation of gravitational potential
(3) the centre of the earth with distance for a hollow sphere of radius
(4) Twice the radius of the earth R:
R
41. A body attains a height equal to the radius
of the earth when projected from earth's (1) V
surface. The velocity of the body with
r
which it was projected is :
R
r
GM
(1) (2)
R
V
2GM
(2)
R R
r

5 GM (3)
(3)
4 R V

3GM R
(4) r
R (4)
V
42. The gravitational potential energy of a
body at a distance r from the center of the
earth is U. The force at that point is : 46. Gravitational potential difference between
U surface of a planet and a point situated at
(1) 2
r a height of 20 m above its surface is 2
U joule/kg. If gravitational field is uniform,
(2)
r
then the work done in taking a 5 kg body
(3) Ur
2 upto height 4 meter above surface will be:
(4) Ur
(1) 2 J (2) 20 J
43. A projectile of mass m is thrown vertically (3) 40 J (4) 10 J
up with an initial velocity v from the
surface of earth (mass of earth = M). If it 47. Two small and heavy spheres, each of
comes to rest at a height h, the change in mass M, are placed a distance r apart on
its potential energy is
a horizontal surface. The gravitational
(1) GMmh/R(R + h)
2 2 potential at the mid-point on the line
(2) GMmh /R(R + h)
(3) GMmhR /(R + h) joining the centre of the spheres is :-
(4) GMm/hR(R + h) (1) zero
GM
44. A particle falls from infinity to the earth. Its (2) –
r
velocity on reaching the earth surface is :
2GM
(1) 2Rg (3) –
(2) Rg r

(3) Rg 4GM
(4) –
r
(4) 2Rg
 Escape Velocity & Escape Energy 54. An electron moves in a circular orbit at a
distance from a proton with kinetic energy
48. Body is projected vertically upward from
E. To escape to infinity, the energy which
the surface of the earth with a velocity
must be supplied to the electron is:
equal to half the escape velocity. If R is
(1) E (2) 2 E
radius of the earth, the maximum height
attained by the body is :- (3) 0.5 E (4) 2E
R R
(1) (2) 55. A particle is fired upward with a speed of
6 3
20 km/h. The speed with which it will
2
(3) R (4) R move in interstellar space is:
3
(1) 8.8 km/h (2) 16.5 km/h
(3) 11.2 km/h (4) 10 km/h
49. Escape velocity for a projectile at earth's
surface is Ve. A body is projected from
earth's surface with velocity 2Ve. The
Kepler's Law
velocity of the body when it is at infinite 56. Kepler's second law is a consequence of :
distance from the centre of the earth is :- (1) conservation of kinetic energy
(1) Ve (2) 2Ve (2) conservation of linear momentum
(3) 2 Ve (4) 3 Ve (3) conservation of angular momentum
(4) conservation of speed
50. Potential energy of a 3 kg body at the
surface of planet is –54 J, then escape 57. In adjoining figure earth goes around the
velocity will be : sun in elliptical orbit on which point the
(1) 18 m/s (2) 162 m/s orbital speed is maximum :
(3) 36 m/s (4) 6 m/s B

A C
51. Escape velocity of a 1 kg body on a planet
is 100 m/s. Potential energy of body at that
D
planet is :
(1) –5000 J (2) – 1000J (1) On A (2) On B
(3) –2400J (4) –10000J (3) On C (4) On D

2 3
52. A particle on earth’s surface is given a 58. If a graph is plotted between T and r for
velocity equal to its escape velocity. Its a planet then its slope will be :-
total mechanical energy will be:
4π2 GM
(1) negative (2) positive (1) (2)
GM 4π2
(3) zero (4) infinite
(3) 4π GM (4) Zero
53. A hole is drilled from the surface of earth
to its centre. A particle is dropped from 59. A planet of mass m is moving in an
rest at the surface of earth. The speed of elliptical orbit about the sun (mass of sun
the particle when it reaches the centre of = M). The maximum and minimum
the earth in terms of its escape velocity distances of the planet from the sun are
on the surface of earth ve is: r1 and r2 respectively. The period of
ve revolution of the planet will be
(1) (2) ve
2 proportional to :
ve (1) r13/2 (2) r23/2
(3) ve (4)
2 (3) (r1 – r2)3/2 (4) (r1 + r2)3/2
60. The earth revolves around the sun in one 66. A body is dropped by a satellite in its
year. If distance between them becomes geo-stationary orbit :
double, the new time period of revolution (1) it will burn on entering in to the
will be :-
atmosphere
(1) years (2) 2 2 years
(2) it will remain in the same place with
(3) 4 years (4) 8 years
respect to the earth
61. The mean distance of mars from sun is 1.5 (3) it will reach the earth in 24 hours
times that of earth from sun. What is (4) it will perform uncertain motion
approximately the number of years
required by mars to make one revolution 67. Two ordinary satellites are revolving round
about sun? the earth in same elliptical orbit, then
(1) 2.35 years (2) 1.85 years
which of the following quantities is
(3) 3.65 years (4) 2.75 years
conserved :-
62. The maximum and minimum distances of (1) velocity
12
a comet from the sun are 8 × 10 m and (2) angular velocity
12
1.6 × 10 m respectively. If its velocity (3) Angular momentum
when it is nearest to the sun is 60 m/sec (4) none of above
then what will be its velocity in m/s when
it is farthest?
68. One projectile after deviating from its path
(1) 12 (2) 60
starts moving round the earth in a circular
(3) 112 (4) 6
path of radius equal to nine times the
Satellite Motion radius of earth R. Its time period will be :-

63. Binding energy of moon and earth is :- R R

(1) 2π (2) 27 × 2π
GMeMm GMeMm g g
(1) (2)
rem 2rem R R
(3) π (4) 0.8 × 3π
GMeMm GMeMm g g
(3) – (4) –
rem 2rem
69. A satellite launching station should be :
64. Two artificial satellites A and B are at a (1) near the equatorial region
distance rA and rB above the earth's (2) near the polar region
surface. If the radius of earth is R, then (3) on the polar axis
the ratio of their speed will be:
(4) all locations are equally good
1
2
r +R 2 r +R
(1)  B  (2)  B  70. If two bodies of mass M and m are revolving
 rA + R   rA + R 
1
around the centre of mass of the system in
2
r  r  2 circular orbit of radii R and r respectively
(3)  B  (4)  B 
 rA   rA  due to mutual interaction. Which of the
following formula is applicable :-
65. The average radii of orbits of mercury and GMm GMm
7 (1) = mω2r (2) = mω2r
earth around the sun are 6 × 10 km and (R+ r)2 R2
8
1.5 × 10 km respectively. The ratio of their
orbital speeds will be :- GMm GMm
(3) – = mω2R (4) = mω2r
r2 R2 + r2
(1) 5 : 2 (2) 2 : 5
(3) 2.5 : 1 (4) 1 : 25
71. The relay satellite transmits the television 77. If a satellite is revolving very close to the
programme continuously from one part of surface of earth, then its orbital velocity
the world to another because its : does not depend upon :
(1) Period is greater than the period of (1) Mass of satellite (2) Mass of earth
rotation of the earth about its axis (3) Radius of earth (4) Orbital radius
(2) Period is less than the period of
rotation of the earth about its axis 78. The minimum projection velocity of a body
from the earth's surface so that it
(3) Period is equal to the period of
becomes the satellite of the earth
rotation of the earth about its axis 6
(Re = 6.4 × 10 m).
(4) Mass is less than the mass of earth
3 –1 3 –1
(1) 11 × 10 ms (2) 8 × 10 ms
3 –1 3 –1
72. If the satellite is stopped suddenly in its (3) 6.4 × 10 ms (4) 4 × 10 ms
orbit which is at a distance equal to radius 79. Geostationary satellite :-
of earth from earth's surface and allowed (1) is situated at a great height above the
to fall freely into the earth. The speed with
surface of earth
which it hits the surface of earth will be :
(2) moves in equatorial plane
(1) 7.919 m/s (2) 7.919 km/s
(3) 11.2 m/s (4) 11.2 km/s (3) have time period of 24 hours
(4) have time period of 24 hours and
73. The gravitational force between two bodies moves in equatorial plane
1 1
is directly proportional to (not 2 ) where
R R 80. A satellite of mass m goes round the earth
'R' is the distance between the bodies. Then along a circular path of radius r. Let mE be
the orbital speed for this force in circular
the mass of the earth and RE its radius then
orbit is proportional to :
(1) 1/R
2
(2) R
o the linear speed of the satellite depends on.
(3) R (4) 1/R (1) m, mE and r (2) m, RE and r
(3) mE only (4) mE and r
74. What will be velocity of a satellite revolving
around the earth at a height h above surface
of earth if radius of earth is R : 81. Near the earth's surface time period of a
satellite is 1.4 hrs. Find its time period if it is at
g g
(1) R2 (2) R the distance '4R' from the centre of earth :-
R +H (R+ h)2
 1 
g R+h (1) 32 hrs. (2)   hrs.
(3) R (4) R 8 2 
R+h g
(3) 8 2 hrs. (4) 16 hrs.
75. Two artificial satellites of masses m1 and m2
are moving with speeds v1 and v2 in orbits of 82. A communication satellite of earth which
radii r1 and r2 respectively. If r1 > r2 then which takes 24 hrs. to complete one circular
of the following statements is true : orbit eventually has to be replaced by
(1) v1 = v2 (2) v1 > v2 another satellite of double mass. If the
(3) v1 < v2 (4) v1/r1 = v2/r2 new satellite also has an orbital time
period of 24 hrs, then what is the ratio of
76. Orbital radius of a satellite S of earth is
the radius of the new orbit to the original
four times that of a communication
satellite C. Period of revolution of S is : orbit ?
(1) 4 days (2) 8 days (1) 1 : 1 (2) 2 : 1
(3) 16 days (4) 32 days (3) 2:1 (4) 1 : 2
83. The orbital velocity of an artificial satellite 88. A satellite is moving in a circular orbit
in a circular orbit just above the earth's around earth with a speed v. If its mass is
surface is v0. The orbital velocity of
satellite orbiting at an altitude of half of m, then its total energy will be :
the radius is :- 3 2
(1) mv2 (2) mv
3 2 4
(1) v0 (2) v0
2 3
1 1
2 3 (3) mv2 (4) – mv2
(3) v0 (4) v0 2 2
3 2

84. An earth's satellite is moving in a circular 89. Two satellites of same mass m are revolving
orbit with a uniform speed v. If the round of earth (mass M) in the same orbit
gravitational force of the earth suddenly
of radius r. Rotational directions of the two
disappears, the satellite will :
(1) vanish into outer space are opposite therefore, they can collide.
(2) continue to move with velocity v in Total mechanical energy of the system
original orbit
(3) fall down with increasing velocity (both satellites and earth's) is (m << M) :-
(4) fly off tangentially from the orbit with GMm 2GMm
same velocity (1) – (2) –
r r
85. An artificial satellite moving in a circular GMm
(3) – (4) Zero
orbit around the earth has a total (kinetic 2r
+ potential) energy E0. Its potential energy
is :
(1) –E0 (2) E0 90. A planet is moving in an elliptical orbit. If
(3) –2E0 (4) 2E0 T, U, E and L are its kinetic energy,
potential energy, total energy and
Energy of Satellite magnitude of angular momentum
86. Potential energy and kinetic energy of a respectively, then which of the following
two particle system under imaginary force
field are shown by curves KE and PE, statement is true :-
respectively in figure. This system is (1) T is conserved
bound at :
(2) U is always positive
KE Dis tance (3) E is always negative
Energy
A B C D (4) L is conserved but the direction of
PE
vector will continuously change
(1) only point A
(2) only point D
(3) only point A,B and C 91. Two identical satellites are at the heights
(4) All points A, B, C and D R and 7R from the earth's surface. Then
which of the following statement is
87. A satellite of earth of mass 'm' is taken
from orbital radius 2R to 3R, then incorrect: (R = Radius of the earth)
minimum work done is :- (1) Ratio of total energy of both is 5
GMm GMm (2) Ratio of kinetic energy of both is 4
(1) (2)
6R 12R
(3) Ratio of potential energy of both is 4
GMm GMm
(3) (4) (4) Ratio of total energy of both is 4
24R 3R
92. For a satellite moving in an orbit around Energy
the earth, the ratio of kinetic energy to 1
potential energy is 0
r
(1) 2 (2) 1/2 2
3
1
(3) (4) 2
2 (1) 1 Potential, 2 Kinetic, 3 Total
(2) 1 Total, 2 Kinetic, 3 Potential
93. A satellite of mass m revolves in a circular (3) 1 Kinetic, 2 Total, 3 Potential
orbit of radius R round a planet of mass (4) 1 Potential, 2 Total, 3 Kinetic0
M. Its total energy E is :-
GMm GMm 95. A space shuttle is launched in a circular
(1) – (2) +
2R 3R orbit near the earth's surface. The
GMm GMm additional velocity be given to the space-
(3) – (4) +
R R shuttle to get free from the influence of
gravitational force, will be :
94. A satellite is orbiting earth at a distance r. (1) 1.52 km/s
Variations of its kinetic energy, potential (2) 2.75 km/s
energy and total energy, is shown in the (3) 3.28 km/s
figure. Of the three curves shown in figure,
(4) 5.18 km/s
identify the type of mechanical energy
they represent.

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ans. 3 1 4 4 2 3 4 3 1 4 1 3 4 4 1 3 2 4 3 1 1 2 4 3 4
Que. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Ans. 3 3 1 1 1 2 2 2 3 1 2 1 4 3 1 1 2 1 4 3 1 4 2 4 4
Que. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. 1 3 4 1 2 3 1 1 4 2 2 1 2 1 1 2 3 2 1 1 3 2 2 3 3
Que. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95
Ans. 2 1 2 4 4 3 1 3 4 4 3 2 4 1 3 1 2 1 3 3
 Exercise - II
1. Masses and radii of earth and moon are 5. A planet revolves around the sun in an
M1, M2 and R1, R2 respectively. The distance elliptical orbit. If vp and va are the
between their centre is 'd'. The minimum velocities of the planet at the perigee and
velocity given to mass 'M' from the mid apogee respectively, then the eccentricity
point of the line joining their centre so of the elliptical orbit is given by :
that it will escape :
vp
4G(M1 + M2 ) 4G M1M2 (1)
(1) (2) va
d d (M1 + M2 )
v a – vp
(2)
2G  M1 + M2  2G v a + vp
(3)   (4) (M1 + M2 )
d  M1M2  d
vp + v a
(3)
vp – v a
2. A projectile is fired vertically upward from
the surface of earth with a velocity KVe, vp – v a
where Ve is the escape velocity and K < 1. (4)
vp + v a
Neglecting air resistance, the maximum
height to which it will rise measured from
6. Two solid spherical planets of equal radii
the centre of the earth is : (Where
R = radius of earth) R having masses 4M and 9M their centre
are separated by a distance 6R. A
R R
(1) (2) projectile of mass m is sent from the
1 – K2 K2
planet of mass 4M towards the heavier
1 – K2 K2 planet. What is the distance r of the point
(3) (4)
R R from the lighter planet where the
gravitational force on the projectile is
3. If R is the average radius of earth, ω is its zero?
angular velocity about its axis and g is the (1) 1.4 R (2) 1.8 R
gravitational acceleration on the surface (3) 1.5 R (4) 2.4 R
of earth then the cube of the radius of
orbit of a geostationary satellite will be
7. Read the following statements :
equal to :
S1 : An object shall weigh more at pole
R2 g R2 ω2 than at equator when weighed by
(1) (2)
ω g using a physical balance.
R2 ω2 R2 g S2 : It shall weigh the same at pole and
(3) (4)
g ω2 equator when weighed by using a
physical balance.
4. A geostationary satellite is orbiting the S3 : It shall weigh the same at pole and
earth at a height of 6R from the earth's equator when weighed by using a
surface (R is the earth's radius). What is spring balance.
the period of rotation of another satellite S4 : It shall weigh more at the pole than at
at a height of 2.5 R from the earth equator when weighed using a spring
surface? balance.
(1) 6 2 hours Which of the above statements is/are
(2) 10 hours correct?
(1) S1 and S2 (2) S1 and S4
5 5
(3) hours (3) S2 and S3 (4) S2 and S4
3
(4) None of the above
8. Assume that a tunnel is dug through earth Weight (N)
from North pole to south pole and that the 600
earth is a non-rotating, uniform sphere of A
400
density ρ. The gravitational force on a 240
B
C
particle of mass m dropped into the 200
tunnel when it reaches a distance r from D Time
the centre of earth is :
 3   4π  (1) A (2) B
(1)  mGρ  r (2)  mGρ  r
 4π   3  (3) C (4) D

 4π   4π 2 
(3)  mGρ  r2 (4)  m Gρ  r 11. If the length of the day is T, the height of
 3   3 
that TV satellite above the earth's surface
which always appears stationary from
9. Three identical bodies (each mass M) are earth, will be :
placed at vertices of an equilateral 1/3
 4π2GM 
triangle of arm L, keeping the triangle as (1) h =  
 T
2
such by which angular speed the bodies 
1/2
should be rotated in their gravitational  4π2GM 
(2) h =  
fields so that the triangle moves along  T
2

circumference of circular orbit : 1/3
 GMT2 
3GM GM (3) h =  2 
–R
(1) (2)  4π 
L3 L3
1/3
 GMT2 
GM GM (4) h =  +R
(3) (4) 3 2 
3L3 L3  4π 

10. Suppose the acceleration due to gravity at 12. Imagine a light planet revolving around a
2 very massive star in a circular orbit of
the earth's surface is 10 m/s and at the
2 radius R with a period of revolution T. If
surface of mars it is 4.0 m/s . A 60 kg
the gravitational force of attraction
passenger goes from the earth to the mars
between the planet and the star is
in a spaceship moving with a constant -5/2 2
proportional to R , then T is
velocity. Neglect all other objects in the
proportional to :
sky. Which part of figure best represent
3 7/2
the weight (Net gravitational force) of the (1) R (2) R
3/2 9/2
passenger as a function of time : (3) R (4) R

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12
Ans. 1 1 4 1 4 4 4 2 1 3 3 2
 Exercise – III (Previous Year Question)

1. Two satellites of earth, S1 and S2 are 4. The radii of circular orbits of two satellites
moving in the same orbit. The mass of S1 A and B of the earth, are 4R and R,
respectively. If the speed of satellite A is
is four times the mass of S2. Which one 3V, then the speed of satellite B will be :
of the following statements is true? [AIPMT -2010]
[AIPMT -2007] (1) 3V/2 (2) 3V/4
(1) The kinetic energies of the two (3) 6V (4) 12V

satellites are equal 5. A particle of mass M is situated at the

(2) The time period of S1 is four times centre of a spherical shell of same mass
that of S2 and radius a. The gravitational potential at
a
(3) The potential energies of earth and a point situated at distance from the
2
satellite in the two cases are equal
centre, will be: [AIPMT-2010]
(4) S1 and S2 are moving with the same
4GM 3GM
(1) – (2) –
speed a a
2GM GM
(3) – (4) –
2. The additional kinetic energy to be a a
provided to a satellite of mass m revolving
6. The dependence of acceleration due to
around a planet of mass M, to transfer it
gravity 'g' on the distance 'r' from the
from a circular orbit of radius R1 to
centre of the earth, assumed to be a
another of radius R2(R2 > R1) is: sphere of radius R of uniform density, is
[AIPMT -2007] as shown in figure below: [AIPMT -2010]
g
 1 1   1 1  g
(1) GmM  –  (2) 2GmM  – 
 R1 R2   R1 R2 
(a) (b)
 1 r
1  1 1  1  R
r
(3) GmM  –  (4) GmM  2 – 2  R
R 
2  R1 R2   1 R2  g g

3. The figure shows elliptical orbit of a planet (c) (d)

m about the sun S. The shaded area SCD is r r
R R
twice the shaded area SAB. If t1 is the time The correct figure is :-
for the planet to move from C to D and t2 (1) (a) (2) (b)
(3) (c) (4) (d)
is the time to move from A to B then :
[AIPMT -2009] 7. A planet moving along an elliptical orbit is
m v closest to the sun at a distance r1 and
B C farthest away at a distance of r2. If v1 and
S v2 are the linear velocities at these points
A D
respectively, then the ratio is
[AIPMT -2011]
(1) t1 = t2 (2) t1 < t2 (1) (r1/r2)
2
(2) r2/r1
(3) t1 = 4t2 (4) t1 = 2t2 (3) (r2/r1)
2
(4) r1/r2
8. A spherical planet has a mass Mp and 12. If ve is escape velocity and v0 is orbital
diameter Dp. A particle of mass m falling velocity of a satellite for orbit close to the
freely near the surface of this planet will earth's surface, then these are related by :
experience an acceleration due to gravity [AIPMT (Main) 2012]
equal to : [AIPMT -2012]
(1) ve = 2v 0
(1) GMp / Dp2 (2) 4GMp m/ Dp2
(2) ve = 2 v0
(3) 4GMp / Dp2 (4) GMp m/ Dp2
(3) v0 = 2 v0
9. A geostationary satellite is orbiting the (4) v0 = ve
earth at a height of 5R above that surface
of the earth, R being the radius of the 13. A black hole is an object whose
earth. the time period of another satellite gravitational field is so strong that even
in hours at a height of 2R from the surface light cannot escape from it.To what
of the earth is : [AIPMT -2012]
approximate radius would earth
(1) 6 2 (2) 6 / 2 24
(mass = 5.98 × 10 kg) have to be
(3) 5 (4) 10
compressed to be a black hole ?
[AIPMT 2014]
10. The height at which the weight of a body
–9
th
becomes 1/16 of its weight on the surface (1) 10 m
–6
of earth (radius R), is : [AIPMT -2012] (2) 10 m
(1) 3R (2) 4R (3) 10 m
–2

(3) 5R (4) 15R (4) 100 m

11. Which one of the following plots represents

14. Dependence of intensity of gravitational
the variation of gravitational field on a
particle with distance r due to a thin field (E) of earth with distance (r) from
spherical shell of radius R? (r is measured centre of earth is correctly represented by :
from the centre of the spherical shell): [AIPMT 2014]
[AIPMT (Main) 2012] E
E
R
O
R r
O (1)
r
(1)

E
E

O
R r
(2)
(2)
O r
R
E E

R
O
(3) (3) r
O r
R

E
E

O
R r
(4) (4)
O r
R
15. Kepler's third law states that square of 19. The acceleration due to gravity at a height
period of revolution (T) of a planet around 1 km above the earth is the same as at a
the sun, is proportional to third power of depth d below the surface of earth. Then :
average distance r between sun and planet. [NEET 2017]
2 3
i.e. T = Kr here K is constant.
1
If the masses of sun and planet are M and (1) d = km
2
m respectively then as per Newton's law
(2) d = 1 km
of gravitation force of attraction between
3
GMm (3) d =
them is F = , here G is gravitational 2
r2
(4) d = 2 km
constant. The relation between G and K is
described as : [AIPMT 2015]
(1) GMK = 4π
2
(2) K = G 20. If the mass of the Sun were ten times
smaller and the universal gravitational
1 2
(3) K = (4) GK = 4π constant were ten time larger in
G
magnitude, which of the following is not
16. A satellite S is moving in an elliptical orbit correct ? [NEET 2018]
around the earth. The mass of the satellite (1) Raindrops will fall faster
is very small compared to the mass of the (2) Walking on the ground would become
earth. Then, [Re-AIPMT 2015] more difficult
(1) the acceleration of S is always (3) Time period of a simple pendulum on
directed towards the centre of the the Earth would decrease
earth. (4) 'g' on the Earth will not change
(2) the angular momentum of S about the
centre of the earth changes in
21. The kinetic energies of a planet in an
direction, but its magnitude remains
elliptical orbit about the Sun, at positions
constant.
A, B and C are KA, KB and KC respectively.
(3) the total mechanical energy of S varies
periodically with time. AC is the major axis and SB is
(4) the linear momentum of S remains perpendicular to AC at the position of the
constant in magnitude. Sun S as shown in the figure. Then
[NEET 2018]
17. A remote – sensing satellite of earth revolves
6
in a circular orbit at a height of 0.25 × 10 m
above the surface of earth. If earth's radius
6 –2
is 6.38 × 10 m and g = 9.8 ms , then the
orbital speed of the satellite is :
[Re-AIPMT 2015]
(1) KA < KB < KC (2) KA > KB > KC
–1 –1
(1) 6.67 km s (2) 7.76 km s
–1 –1
(3) KB < KA < KC (4) KB > KA > KC
(3) 8.56 km s (4) 9.13 km s

22. A body weighs 200 N on the surface of the

18. The ratio of escape velocity at earth (ve) to
earth. How much will it weigh half way
the escape velocity at a planet (vp) whose
down to the centre of the earth ?
radius and mean density are twice as that
[NEET 2019]
of earth is : [NEET 2016]
(1) 150 N (2) 200 N
(1) 1 : 2 (2) 1 : 2 2
(3) 250 N (4) 100 N
(3) 1 : 4 (4) 1 : 2
23. The work done to raise a mass m from the 27. What is the depth at which the value of
surface of the earth to a height h, which acceleration due to gravity becomes 1/n
is equal to the radius of the earth, is : times the value that at the surface of
[NEET 2019] earth? (radius of earth = R)
(1) mgR (2) 2 mgR [NEET_Covid_2020]
2
1 3 (1) R/n (2) R(n –1)/n
(3) mgR (4) mgR
2 2 (3) Rn/(n –1) (4) R/n

24. The time period of a geostationary 28. A particle is released from height S from
satellite is 24 h, at a height 6RE (RE is the surface of the Earth. At a certain
radius of earth) from surface of earth. The height its kinetic energy is three times its
time period of another satellite whose potential energy. The height from the
height is 2.5 RE from surface will be, surface of earth and the speed of the
[NEET 2019 (Odisha)] particle at that instant are respectively :
[NEET - 2021]
(1) 6 2 h (2) 12 2 h
S 3gS S 3gS
24 12 (1) , (2) ,
(3) h (4) h 4 2 4 2
2.5 2.5
S 3gS S 3gS
(3) , (4) ,
2 2 4 2
25. Assuming that the gravitational potential
energy of an object at infinity is zero, the
29. The escape velocity from the Earth's
change in potential energy (final – initial)
surface is v. The escape velocity from the
of an object of mass m, when taken to a
surface of another planet having a radius,
height h from the surface of earth (of
four times that of Earth and same mass
radius R), is given by,
density is: [NEET - 2021]
[NEET 2019 (Odisha)]
(1) v (2) 2v
GMm GMmh
(1) − (2) (3) 3v (4) 4v
R+h R (R + h)

GMm 30. A body of mass 60g experiences a

(3) mgh (4) gravitational force of 3.0 N, when placed
R+h
at a particular point. The magnitude of the
26. A body weighs 72 N on the surface of the gravitational field intensity of that point is:
earth. What is the gravitational force on it, [NEET - 2021]
at a height equal to half the radius of the (1) 0.05 N/kg (2) 50 N/kg
earth ? [NEET - 2020] (3) 20 N/kg (4) 180 N/kg
(1) 30 N (2) 24 N
(3) 48 N (4) 32 N

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ans. 4 3 4 3 2 4 2 3 1 1 4 2 3 1 1 1 2 2 4 4 2 4 3 1 2
Que. 26 27 28 29 30
Ans. 4 2 4 4 2