GRAVITATION

1.1 Newton's law of Gravitation

Every particle in the universe attracts all 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 F 12  F 21  F , then
F12 F21
m1 m2
1
F m1m2 and F 2 r
r
m1m2
so F
r2
Gm1m2
  F [G = Universal gravitational constant]
r2
Note : This formula is applicable only for spherically symmetric masses or point masses.

1.2 Vector form of Newton's law of Gravitation :

Y m1
  
Let r12 = Position vector of m1 w.r.t. m2 = r1  r2 F12
r21
r1 r12
  
r21 = Position vector of m2 w.r.t. m1 = r2  r1 F21 m2
 r2
F21 = Gravitational force exerted on m2 by m1
O X

F12 = Gravitational force exerted on m1 by m2

 Gm1m2 Gm1m2 
F12   2
ˆr12   3
r12
r21 r21

Negative sign shows that :

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

(ii) The gravitational force is attractive in nature

 Gm1 m2  Gm1 m2     
Similarly F21   2
ˆr21 or F21   3
r21  F12  F21 , r12   r21
r12 r12

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

1.3 Universal Gravitational Constant "G"

 Value of G = 6.67  10–11 N–m2/kg2 ;

 Gravitational force is always attractive.

 Gravitational forces are developed in the form of action and reaction pair. Hence they obey Newton's third
law of motion.
 It is independent of the nature of medium between two masses.
 Gravitational forces are central forces as they act along the line joining the centre of gravity 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 for round the trip.
171
 Gravitational force is weaker than the electromagnetic and nuclear forces.
Illustration 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

Gm1m2 6.67  1011  1  2

Force exerted by one particle on another is F  2
 2
 5.34  1010 N
r (0.5)

F 5.3  1010
Acceleration of heavier particle =   2.67  1010 m / s2 .
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.
Illustration 3.

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

M m M2
GM1 m r
Force on m towards M1 is F1 = d
r2

GM2 m
Force on m towards M2 is F2 =
(d  r)2

Since net force on m is zero  F1 = F2

GM1 m GM2 m dr M d

2
M2  M1 
      2  1   r = d 
d  r
2 2
r  r  M 1 r M1  M1  M2 

Illustration 4.

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
  1
GM F2
Force on m = 1 due to masses at corners 1 and 3 are F1 and F3 with F1 = F3 = F3
a2 Fr
a
 
GM
resultant of F1 and F3 is Fr = 2 and its direction is along the diagonal M a M
a2
2 3
i.e. toward corner 2
GM GM
Force on m due to mass M at 2 is F2   ; Fr and F2 act in the same direction.
( 2a) 2
2a2

Resultant of these two is the net force :

2GM GM GM  1
Fnet =  2   ; it is directed along the diagonal as shown in the figure.
a2 a2  2

Illustration
=
a
Two particles each of equal mass (m) move along a circle of radius (r) under the action
m r r m
of their mutual gravitational attraction. Find the speed of each particle.

Solution

mv 2 Gmm Gm 1 Gm
For the circular motion of each particle,  2
 v2   v .
r (2r) 4r 2 r
 Illustration 6.

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

Gm 2  Gm2 
FA  FAB  FAC  2FAB FAC cos 60   FAB  FAC  2 
2 2
3 2
...(i)
a  a 

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
then  3 ...(ii) A m
r a FAB FAC
FA
mv 2 Gm 2 3 Gm
From (i) and (ii) 3  2
 v O
a a a
m m
B a
C
2r 2a a a3
Time period T   2 .
v 3 Gm 3Gm

Illustration 7.

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

Solution

The weights of the spheres may be assumed to be concentrated at their centres.

4  4 
G  R 3     R 3 
So F  3   3   4 (G2 2 )R 4 
2
(2R) 9

  F  R4

Illustration 8.
A mass (M) is split into two parts (m) and (M–m), which are then separated by a certain distance. What ratio
m
will maximise the gravitational force between them ?
M
Solution
m(M  m) G
If r is the distance between m and (M – m), the gravitational force will be F  G  2 (mM  m2 )
r2 r
dF d2 F d G 2 
For F to be maximum = 0 and 2
< 0 as M and r are constants, i.e.  r 2 (mM  m )   0
dm dm dm  
G  G 
 (M  2m)  0 i.e. M – 2m = 0  r 2  0
r 2
 
m 1
or  , i.e., the force will be maximum when the two parts are identical.
M 2
2. GRAVITATIONAL FIELD AND IT'S INTENSITY

2.1 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.
. M


2.2 Gravitational Field Intensity I
 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
 F r
I M
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   M1L1 T 2 
Dimensions of intensity =   [M0 L1 T 2 ] .
m M1 

2.3 Gravitational Field Intensity Due to a Particle (Point - Mass) :

M r I
m = 1 unit
r

Gravitational field intensity = gravitational force exerted on unit mass

 GM GM   GM
 I  2  ˆr   3 (  r )  I  2  r̂ 
r r r

where 'M' is the mass of that particle due to which intensity is to be found.

2.4 Gravitational field intensity due to spherical mass distribution

(I) For solid sphere M

 GM   R r
  Case I : When r > R, i.e. outside the sphere then I out  r̂ O P
r2


M
 GM
  Case II : When r = R, i.e. at the surface then I surface  2  r̂  r=R
P
R O

  Case III : When r < R, i.e. inside the sphere then M

 GM '  R M'
I in  2 ( r) ...........(1) r
P
r

 GMr 3  GMr
I in  2 3  r̂    I in  3  r̂ 
r R R

175
 Important conclusions :-

GM 1
(1) I out    I out 
r2 r2

GM
(2) I sur 
R2

GMr
(3) I in      I in  r
R3

I
GM
(4) So, Imax  I sur  2 r<R r=R r>R
R O r
Iinr
1
Ioutr
GM(0) r
2

(5) I centre  0   I min  Icentre  0 –

GM
R3 R2

 Graph between ' I ' and 'r' for a solid sphere :

(II) Gravitational field intensity due to a Spherical Shell

R r>R
 GM
Case I : If r > R, the point is outside the shell then Iout  2  ˆr 
O P
r

 GM r=R
Case II : If r = R, the point is on the surface then Isurface  2  ˆr  O
P
R

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


 I v/s r graph for hollow sphere :


I
r<R r=R r>R
O r
1
Iin=0 Iout 
r2
GM
–
R2

177
3. ACCELERATION DUE TO GRAVITY

3.2 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 from its is centre, is
GMm
F
r2
Now according to Newton's second law F = mag
GM
Therefore a g  ...(i)
r2
GM
At the surface of Earth, acceleration due to gravity g  2
 9.8 m/s2
R
GMe
gh= 2
(Re+h)

3.3 Variation in Acceleration due to gravity h

 Due to Altitude (height) : GMe
g= 2
From diagram Re

2 Earth
gh R2e R 2e  h 
= =  1   Me
g (R e  h)2  h 
2
 R e  Re
R 2e 1  
 R e  O
2
 h   2h 
By Binomial expansion  1    1   [If h << Re, then higher power terms become negligible]
 R e   Re 

 2h 
 gh  g 1  
 Re 

 Due to depth : g
g
gd

r=Re–d
 d  gr g
1
g d  g 1   valid for any depth O r2
 Re 
O r
R
(Taking towards direction
centre of earth as positive)
Illustration 10.

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

Illustration 11.

Which of the following statements are true about acceleration due to gravity?

(A) 'g' decreases in moving away from the centre of earth if r > R

(B) 'g' decreases in moving away from the centre of earth if r < R

(C) 'g' is zero at the centre of earth

Solution
178
1
Variation of g with distance : If r > R then g   (A) is correct
r2

If r < R then g  r  (B) is incorrect & (C) is correct

GRAVITATIONAL POTENTIAL ENERGY

 The gravitational potential energy of a particle situated at a point in some gravitational field is defined as the
amount of work required to bring it from infinity to that point without changing its kinetic energy.
GMm Gm1m2 M
W=U=– or U = –
r r
m m1 m2
(Here negative sign shows the boundness of the two bodies)
r

 It is a scalar quantity.
r
 The gravitational potential energy of a particle of mass 'm' placed on the surface of earth of mass 'M' and
m
radius 'R' is given by :

GMm R
U M
R

4.2 Gravitational Potential Energy for 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.
m3

2
Gm1m2 Gm2 m3 Gm1m3 r2
U system     3 r3
r1 r2 r3

r1 m2
m1 1
4.3 To find the change in potential energy of body or work done to raise a particle of
mass 'm' to 'h' height above the surface of earth. m –GMm
U f=
R+h
W = U = Uf – Ui Fg
h
GMm  GMm 
 W     
Rh  R  m
Ui =
–GMm
R
R
M

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'.

v =0
(K.E. + U)surface = (K.E. + U)final

1 GMm GMm h v
  mv 2  0
2 R Rh m v =?
sur

M
R

To find the maximum height attained by a body when it is projected with velocity 'v' from the
surface of earth. vh=0

h=?
Do it by energy conservation v

M R
5. GRAVITATIONAL POTENTIAL

Gravitational potential is the amount of work done by external agent in bringing a body of unit mass
from infinity to that point without changing its kinetic energy.
Wext
V=
m
GM(1) GM
Gravitational force on unit mass at (P) will be = = 2
x2 x

Work done by this force when the unit mass is displaced through the distance dx is

GM
dWext = Fdx = . dx
x2

Total work done in bringing the body of unit mass from infinity to point (P) is
r
 GM 
r
GM GM
Wext =  x2 dx = –  x  = – r .
GM
 VP = –
r
 If r =  then V = 0. Hence gravitational potential is maximum at infinity (as it is a negative quantity
at point P)
GM e
 If r = Re (on the surface of Earth) VS = –
Re

 Relation between intensity and potential gradient

    dV
V=–  I .dr  dV= – I . dr I = –
dr
= –ve potential gradient.

 Gravitational Potential due to solid sphere and spherical shell :

V
Solid Sphere
r<R R r>R
O r
GM
Case I r > R (outside the sphere) ; Vout = –
r
GM
GM –
Case II r=R (on the surface) ; Vsurface =– R
R
3GM
–
2R
GM
Case III r < R (inside the sphere) ; Vin = – [3R2 – r2]
2R 3

It is clear that the potential V will be minimum at the centre (r = 0) but maximum in magnitude.

3 GM 3
Vcentre = – , Vcentre = Vsurface
2 R 2

Spherical shell V

GM r<R R r>R
Case I r > R (outside the sphere) ; Vout = – O r
r
GM
Case II r = R (on the surface) ; Vsurface = –
R GM
–
R
Case III r < R (inside the sphere) ;
Potential is same every where and is equal to its value at the surface
GM
Vin = –
R
6. ESCAPE VELOCITY & ESCAPE ENERGY
6.1 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.
1  GMm  2GM
From conservation of mechanical energy 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
v es   { 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.

6.2 Escape energy

Minimum energy given to a particle in the form of kinetic energy so that
ve
it can just escape the Earth's gravitational field. surface
GMm
Magnitude of escape energy = (–ve of PE on the Earth's surface) Earth
R
M
Escape energy = Kinetic Energy corresponding to the escape velocity R

GMm 1
  mv 2e O
R 2

 Gravitational potential energy or potential is a –ve quantity whose maximum value is zero at infinite
separation.

 Relation between Force & Gravitational potential energy is

dU
F
dr

Above relation is valid only for all conservative forces.

187
Illustration 17.

Three solid spheres of mass M and radius R are placed in contact as shown in figure. Find the potential
energy of the system ?

Solution

PE = PE12 + PE23 + PE31

GM 2 GM 2 GM 2 3GM 2
=     PE =  .
2R 2R 2R 2R

Illustration 18.

Four bodies each of mass m are placed at the different corners of a square of side a. Find the work done on
the system to take any one body to infinity.

Solution

Initial potential energy of the system m a

m
PEi = PE12 + PE23 + PE34 + PE41 +PE13 + PE24 1 2
a
4 3
m m
4GM 2 2GM 2
PEi =  
a a 2

After taking any one body (say the mass placed at corner 4) to infinity only three bodies remain

 Final potential energy the system is

2GM 2 GM 2
PEf = P12 + PE13 + PE23 =  
a a 2

 2GM 2 GM 2   4GM 2 2GM 2  2GM 2 GM 2

Wext. = PEf – PEi =        =  .
 a a 2  a a 2  a a 2

Illustration 19.

A body of mass m is placed on the surface of earth. Find the work required to lift this body by a height

Re
(i) h = (ii) h = Re (Me = mass of earth, Re = radius of earth)
1000

Solution :

Re
(i) h= , as h << Re , so
1000

 GMe   R e  GMe m
we can apply Wext = mgh ; Wext = (m)  2  
 R e   1000  1000R e

(ii) h = Re , in this case h is not very less than Re, so we cannot apply U = mgh

 GM e   GM e  GMe m
Wext = Uf – Ui= m(Vf – Vi) ; Wext = m    –    ; Wext = .
 R e  R e   R e   2R e

Illustration 20.

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 + 0 189
2 R 2

 v2 = n2ve2 –
2GM
R
= n2ve2 – ve2 = (n2 – 1) v2e v =  
n2  1 ve.
Illustration 21.

Me,R

A narrow tunnel is dug along the diameter of the earth, and a particle of
R/2
R
mass m0 is placed at distance from the centre. Find the escape speed of
2
the particle from that place.

Solution

Suppose we project the particle with speed ve, so that it just reaches infinity (r ).

Applying energy conservation principle

Me,R
Ki + Ui = Kf + Uf

1  GM   R  
2
R/2 at r,v0
m 0 v 2e  m 0   e
3R 2
     0 m0 ve m0
2  2R 3   2   

11GMe
.  ve 
4R

Illustration 22.

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
(A) 1.5v e (B) (C) ve (D) zero
2

Solution (A)

1 3GMm 3GM
From mechanical energy conservation, 0 + 0 = mv 2  v  1.5 v e .
2 2R R

Illustration 23.

A particle is projected vertically upwards from the surface of the earth (radius Re) with a speed equal to one
fourth of escape velocity. What is the maximum height attained by it ?

16 Re 4
(A) Re (B) (C) Re (D) None of these
15 15 15

Solution (B)

1 GMm GMm
From conservation of mechanical energy, mv2 = 
2 Re R

1 1 2GM
Where R = maximum distance from centre of the earth Also v = ve 
4 4 Re

1 1 2GM GMm GMm 16 R

 m     R = Re h = R–Re = e .
2 16 Re Re R 15 15

190
Illustration 24.
A mass of 6 × 1024 kg (= mass of earth) is to be compressed in a sphere in such a way that the escape
velocity from its surface is 3 × 108 m/s (equal to the velocity of light). What should be the radius of the
sphere?
(A) 9 mm (B) 8 mm (C) 7 mm (D) 6 mm
Solution (B)

 2GM   2GM  2  6.67  1011  6  1024

As, v e    , R =  2  , R   9  103 m  9 mm
 R  v
 e   3  10 
8
2

Illustration 25.
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?
(A) 4 J (B) 5 J (C) 6 J (D) 7 J
Solution (A)

V 4
Gravitational field I g    J / kg  m
x 10

Work done in moving a mass of 2 kg from the surface to a point 5 m above the surface,

 4 J 
W = mIgh = (2 kg)   (5 m) = 4 J
 10 kg  m 

Illustration 26.
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 GMm  1 1  1 GMm 1 (gR 2 )m 1

 0   + K.E.  K.E.      mgR.
3R 2R R  2 3  6 R 6 R 6

Illustration 27.

With what velocity must a body be thrown from the earth's surface so that it may reach a height 4Re above
the Earth's surface ? (Radius of the Earth Re = 6400 km, g=9.8 m/s2).

Solution

1 GMm0 GMm0
By using conservation of mechanical energy m0v2 – =0–
2 Re (Re  4Re )

1 GMm0 GMm0 1 4 GMm0 8 GM 8 gR2e

m0v2 = – +  m0v2 = v2 = =
2 5Re Re 2 5 Re 5 Re 5 Re

8
v2 = × 9.8 × 6400 × 103 = 108  v = 10 km/s.
5

191
7. 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
semi minor axis
b

A B
F1 F2
rmax=r2 Aphelion
Perihelion (Orbital point
(Orbital point
closest to sun) a farthest from sun)
semi major axis

(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.
 = angular speed of the planet. B
C A
Let L be the angular momentum of the planet about the Sun S d
 r
L = I = mr2 = mvr ...(ii)
where I (=mr2) is moment of inertia of the planet about the Sun S. S
D

dA L
 ...(iii)
From eq. (i) and (ii),
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
V2=Vmin

LA = LB r mvmaxrmin = mvminrmax A B
F1 F2
r vmax rmin = vmin rmax rmax= r2
v1=vmax
(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  R 3

192
8. 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.

8.1 Essential Conditions for Satellite Motion

FCP = Fgcos

O' S
cp
Fg S
O Fg sin O FCP = Fg

(i) (ii)
Unstable orbit Stable orbit
(Due to Fg sin , orbit will shift)

A satellite can revolve round the earth only in those circular orbits whose centres coincide with
the centre of earth.

8.2 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 GM GM
Fcp = Fg  =  v0 = = (r = Re + h)
r r 2
r (R e  h)
r m
For a satellite very close to the Earth's surface h << Re  r Re v0

M
GM
 v0 = = gR e  8 km/s
Re

Time Period of a Satellite

3 3
2r 2r 2 2r 2 42 3
T= = =  T2  r T  r (r = R+h)
2 3

v0 GM R g GM

Energy of a satellite

1 GMm L2
Kinetic energy K.E. = mv 20   (L = mrv0 = m GMr )
2 2r 2mr 2
GMm L2
Potential energy P.E. =   mv 20   2
r mr
mv 20 GMm L2
Total mechanical energy T.E. = P.E.  K.E.     .
2 2r 2mr 2

 Binding energy :

Total mechanical energy (potential + kinetic) 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)

1 GMm L2
B.E. = – T.E. B.E. = mv 20   .
2 2r 2mr 2

 P.E.
Hence B.E. = K.E. = – T.E. =
2

Work done in Changing the Orbit of a Satellite

 GMm
W = Change in mechanical energy of the system but E =
2r

GMm  1 1 
so W = E2 – E1 =  – 
2  r1 r2 
 Graphs :
GMm K.E.
KE  E
2r N
E r
GMm
TE   R
2r G
Y
GMm P.E.
PE  
r

Illustration 28.
Two satellites S1 and S2 are revolving round a planet in coplanar and concentric circular orbits of radii R1 and
R2 in the same sense respectively. Their respective periods of revolution are 1 h and 8 h. The radius of the
orbit of satellite S1 is equal to 104 km. Find the relative speed in km/h when they are closest.
Solution

T2 T12 T22 1 64
By Kepler's 3rd law, = constant    or  or R2 = 4 × 104 km
R3 R13 R23 (104 )3 R23

Distance travelled in one revolution, S1 = 2R1 = 2 × 104 and S2 = 2R2 = 2 × 4 × 104

S1 2  104 S 2  4 104
v1    2  104 km / h and v 2  2     104 km / h
t1 1 t2 8

 Relative velocity = v1 – v2 = 2 × 104 –  × 104 =  × 104 km/h

Illustration 29.
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
Additional velocity = Escape velocity – Orbital velocity = ves – v0

2GM GM
= –
R R

=  2 –1  GM
R
Illustration 30.

An astronaut, inside an earth's satellite experiences weightlessness because:

(A) he is falling freely (B) no external force is acting on him
(C) no reaction is exerted by the floor of the satellite (D) he is far away from the earth's surface

Solution (Ans. A, C)

As astronaut's acceleration = g; so he is falling freely. Also no reaction is exerted by the floor of the satellite.
Illustration 31.

If a satellite orbits as close to the earth's surface as possible

(A) its speed is maximum
(B) time period of its revolution is minimum
(C) the total energy of the 'earth plus satellite' system is minimum
(D) the total energy of the 'earth plus satellite' system is maximum
Solution (Ans. A, B, C)

GM
For (A) : orbital speed v 0  , rmin = R so v0 = maximum
r

For (B) : Time period of revolution T2  r3

GMm
For (C/D) : Total energy = 
2r
Illustration 32.
A planet is revolving round the sun in an elliptical orbit as shown in figure. Select correct alternative(s)
(A) Its total energy is negative at D. C
B

(B) Its angular momentum is constant

(C) Net torque on the planet about sun is zero D A
O
(D) Linear momentum of the planet is conserved
E
Solution F (Ans. A, B, C)
For (A) : For a bound system, the total energy is always negative.
For (B) : For central force field, angular momentum is always conserved.
For (C) : For central force field, torque = 0.
For (D) : In presence of external force, linear momentum is not conserved.
Illustration 33.
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of
escape velocity from the earth.
(a) Determine the height of the satellite above the earth's surface.
(b) If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, find the speed
with which it hits the surface of earth. Given M = mass of earth and R = Radius of earth
Solution
(a) Let height above the earth's surface = h then
GM 1 1 2GM GM
vorbital = = ve = =  R + h = 2R  h = R
Rh 2 2 R 2R
GMm
(b) If the satellite is stopped suddenly then it total energy E1 = –
2R
Let its speed be v when it hits the earth's surface then its total energy on earth surface
GMm 1
E2 = – + mv2
R 2

GMm GMm 1 GM
conservation law for mechanical energy yields E1 = E2  =– + mv2  v =
2R R 2 R