Level - 3 DTS

 3Gm 
76.  
 a3 
 

Distance of the particle from center of mass of the system of

a
particles 
3
Net force of any particle  2 F cos 30

Gm 2
Where F 
a2

Gm 2 a 3Gm
 2  cos 30  m 2  ( Fcentrepetal  m 2r )  
2
a 3 a3

 GMm 
77.  
 (d  L )d 

Consider an element of length dx on the rod, mass of this element,

 Force on particle of mass m due to this element,
Gm dM GMm
dF   dx
2
x Lx 2
 Total force on the particle
GMm d L dx GMm
F 
L  d x 2

(d  L )d

r0   GM dm r GM dx
78.  2.5  10 4 s  Net force on rod 

r0 x 2

 r0 x2
1 1  GM 
 GM    
 r0 r0    r0 (r0   )

GM GM
a rod  and a Bead  where M is the mass of earth
r0 (r0  ) (r0  x 0 )2
In the frame of rod
 1 1  GM 
a rel  GM     ( x 0  r0 )
 (r  x )2 r0 (r0   )  r0 r0 (r0  )
 0 0

We assume that as bead moves out the rod, r0 does not change much.
2
gR 2   6400  103  10
 a rel   10    
r02 r0   
 4  10
8 
 4  108

Gravitation 199 Workbook -2| Solutions

 10  16  16  10 6  10

4  108

 64  10 12 m/s2
1
x0  a rel t 2
2
2  10 2  2 1
t    105  2.5  104 s
12 4
64  10

 GM 
79.  s  
 [( gR )  (GM )2 ]1/2
2 
 
GmM
f  cos 
R2 f
GmM
N sin   mg
R2
F  s N

GmM  GmM 
cos    s  mg  sin  
R2  R2 
 
GM cos  GM
s  
gR 2  GM sin  gR 2 sec   GM tan 

Max value of LHS will happen when gR 2 sec   GM tan  is minimum.

Differentiating w.r.t  and equating the derivative to zero.
gR 2 sec  tan   GM sec 2   0
GM
   sin 1
gR 2

Value of  for which gR 2 sec   GM tan  is maximum

(gR 2 )2  (GM )2
GM
 GM cos   gR 2 GM
s    ; s  ; s 
2
 gR  GM sin   max GM [(gR )  (GM )2 ]1/2
2
gR 2  GM 
2
gR
 M 
80.   Let’s first calculate the mass of the atmosphere between R  x  r
 2R 2 
r
4 x 2dx
m 
R

r x2  0 
 4 0
 dx    
R x  x 

Gravitation 200 Workbook - 2 | Solutions

  2 0 [r 2  R 2 ]
G (M  m )
Acceleration due to gravity at a distance r from the centre is g 
r2
Since, the whole mass distribution (planet + atmosphere is spherically symmetric)
GM G 20
 g  [r 2  R 2 ]
2 2
r r
GM G 20 R 2
   G .2 0
r2 r2
M GM
This expression is independent of r if GM  G .2 0 R 2  0   gconst  G .2 0 
2
2R R2
GM
Alter: Acceleration due to gravity on the surface g 
R2
g M 2 R
 
g M R
M 2R
But g  0  
M R
0
4 R 2 .R M
R  2 R  0 
M R 2 R 2

1
81. (a) (b) 7.6 cm
4

R0  radius of earth

10 R 0  radius of the planet [0  density of atmosphere; h 0  height of atmosphere]

4  3 
m0  
  R0  h 0   R 03  0
3  
 3 
4 h 
 R03 1  0   1 0
3  R  
 

4  3h 0 
 R03 1   1 0  4 R020h 0 ….. (a)
3  R0 

For planet m  4 R 2 .h  4 (10 R 0 )2 0h ….. (b)

Given m  10m 0

 100h  10h 0
h0
h 
10
(a) Ratio of atmospheric pressure

Gravitation 201 Workbook -2| Solutions

 P 0 gh g 1
  . ….. (c)
P0 0 g 0h 0 g0 10

4
GM G R 3 . 4 g R 1 5
 g  3  GR     10  
2
R R2 3 g0 R 0 0 4 2

P 5 1
Using (c)  
P0 2  10 4

(b) gh g 0  (76 cm )  P0

hg g  h  P

h g P h 5 1 38
       h   7.6 cm
76 g0 P0 76 2 4 5

3 5  GM  
82.    Let there are two stars 1 and 2 as shown below.
 2  a  
  

Let P is a point between C1 and C2 , where gravitational field strength is zero at P field strength due to
star 1 is equal and opposite to the fired strength due to star 2. Hence,
GM G (16 M ) r2
 or  4 also r1  r2  10a
r12 r22 r1

 4 
 r2    (10 a )  8a and r1  2a
 4 1
 
Now, the body of mass m is projected from the surface of larger star towards the smaller one. Between
C2 and P it is attracted towards 2 and between C1 and P it will be attracted towards 1. Therefore, the
body should be projected to just cross point P because beyond that the particle is attracted towards the
smaller star itself.
1
From conservation of mechanical energy mv 2min
2
= Potential energy of the body at P
– Potential energy at the surface of larger star.
1 2
 GMm 16GMm   GMm 16GMm 
 mv min        
2  r1 r2   10a  2a 2a 

Gravitation 202 Workbook - 2 | Solutions

  GMm 16GMm   GMm 8GMm 
       
 2a 8a   8a a 
1 2
 45  GMm 3 5  GM 
or mv min    v min   
2  8  a 2  a 
   
83.(99.5 R) Total energy at A = Total energy at B
( KE )A  ( PE )A  ( PE )B

  2 
1 2GM  GMm  2  99R   GMm
m  3 R     
2 R 3  R h
 2R 
  100  

On solving we get h = 99.5 R

 Gm GM 
84.  ve  4 v0  2 
 r0 r0 
 

For orbital motion, particles should at all times be under the influence of each others gravitational pull.
v 0 should be less than their escape velocity from each other.
Lets find the escape velocity for the system of two particles by applying law of conservation of energy in
frame of CM.
1 Gm 2 Gm GM
2 m (ve / 2)2  0 ; ve  4 v0  2
2 r0 r0 r0

 2GMR0 
85.  
 2GM  v 2 R 
 0 

As the cloud is homogenous, mass per unit volume is constant at all the instants
Assuming the cloud to be a solid sphere of uniform mass density, we can calculate the total energy of a
particle of mass m at the periphery when it has a velocity v and equate it to its energy when expansion
ceases.
Let us take any particle of mass m
1 GMm
T.E. of particle  mv 2 
2 R0
GMm
T.E. of particle  
R
GMm 1 GmM 2GMR 0
 mv 2  R
R 2 R0 2GM  v 2 R 0

Gravitation 203 Workbook -2| Solutions

86. (i) 6400 km (ii) 7920 m/s

(i) As the satellite is revolving in a circular orbit, the centripetal force is provided by the gravitational
force.
mv 2 GMm GM
   v2 
(R  h ) 2 R h
(R  h )

1 1 2GM 1  2GM  GM
But v  ve      2R + 2h = 4R  h = R = 6400 km.
2 2 R 4  R  (R  h )

(ii) If the satellite is stopped, its kinetic energy becomes zero. When it falls freely on the Earth, its
potential energy decreases and converts into kinetic energy.
 ( PE )i  ( PE ) f  ΔK

GMm  GMm  1 
     mv 2  0 
2R  R  2 
   
GM
 v   gR  9.8  6.4  10 6  7920 m / s
R

 1 
87. 
4 2 T

 

If the mass of the sun is M and radius of the planet’s orbit is r, then
GM
v0 
r
2r r
Thus, T   2 r
v0 GM

4 2r 3
 T2  …… (1)
GM
When stopped in the orbit, the planet will fall towards the sun. Let v be the velocity of the planet at a
distance x from the sun, then by conservation of mechanical energy
1  GMm   GMm 
mv 2      0   
2  x   r 
   
2
 dx  2GM r  x 
     
 dt  r  x 
   
dx 2GM r x
  
dt r x
t r 0 r x 
  dx

 dt  
0 2GM  r 

x 

On solving,
r  r  1
t    T.
2GM  2  4 2


Gravitation 204 Workbook - 2 | Solutions

 1 2
88. (i) E  mka 3 , (3km 2a 5 )1/2 (ii)
2 3ka

Potential energy of the particle,

dU
U  mv  mkr 3  F   3mkr 2
dr
(i) For circle of radius a

mv 2
  3mkr 2  mv 2  3mkr 2a
a

 3mka 3 ( r  a )
1 3
 Kinetic energy of the particle, K  mv 2  mka 3
2 2
Total energy of system
1
E  U  K  mka 3
2
and angular momentum
L  mva  m (3ka 3 )1/2 a
 (3km 2a 5 )1/2
2 a 2 a 2
(ii) T   
v (3ka 3 )1/2 3ka
1
89. (i) 1 (ii)
2
GM
(i) V0  …… (1)
r0

Speed after firing on the rocket is given by V 2  V02  V 2

1 GMm
For escape mV 2  0
2 r0

V02  V 2 GM V
  V  V0 , 1
2 r0 V0

(ii) Angular momentum does not change (with respect to the centre of the earth)
mV1(2r0 )  mV0r0

 2V1  V0 ….. (1)

Energy conservation
1 GMm 1 GMm
mV 2   mV12 
2 r0 2 2r0

2GM V02 GM
V2   
r0 4 r0

Gravitation 205 Workbook -2| Solutions

 V02 GM
V02  V 2    V 2  V 2  V 2 
0
4 r0  

3  GM 
V02  V 2  V02  V0  
4  r0 
 

V02 V 1
V 2  , 
4 V0 2

 7  2 4 7
90. (i)  R (ii)  R  R
 6   6 
   
2GM
(i) Escape velocity V0 
R
 Velocity of projection
1 2GM GM
V0   ….. (1)
2 R 2R
When R = radius of earth, M = Mass of earth
At the point where the body is farthest & nearest to
earth’s centre, its velocity is  r to the position with
respect to earth’s centre.
 Angular momentum when it is
farthest/nearest to earth’s centre is  mVr
Where r = distance from centre
V = velocity at fartest/nearest point
Angular momentum of body (about O) at the time of projection is
3
mV0 cos 30.R  mRV0
2
Since Angular momentum is conserved
3 3
mVr  mRV0  Vr  RV0
2 2
3 RV0
 V  ….. (2)
2 r
From conservation of energy
1 GMm 1 GMm 2GM 3 R2 2GM
mV02   mV 2  or V02   V02 
2 R 2 r R 4 r2 r

GM 2GM 3 R 2 GM 2GM
Using (i)   
2R R 4 r 2 2R r
1 2 3 R 2
   
2R R 8 r2 r
3 3R  16r
   12r 2  16 Rr  3R 2  0
2R 8r 2

Gravitation 206 Workbook - 2 | Solutions

 16 R  256 R 2  144 R 2 4 7 
or, r   R ….. (2)
24  6 
 
4 7   7 2
 h cannot be less then R hence.  h  R  R   R
 6   6 
   
4 7 
(ii) As rmin   R  R
 6 
 
Hence, the body will move on elliptical path but it will hit the surface of earth at some point.

 8 G  
91.  
 3 2 
 0 

The rod can stay above the same point only if it rotates with angular speed of earth from West to East
(just like geostationary satellites). The necessary centripetal force has to be provided by the gravitational
pull of the earth. Distance of centre of the rod from the centre of from the centre of the earth is
(n  1)R (n  1)
r  R  R
2 2
Mass of the rod m  (n  1)R
Where   mass per unit length.
Gravitational pull on the rod can be calculated by writing force on an element of length dx.
GM (dx )
dF  [M = mass of earth]
x2
Total force on the rod
nR
dx GM   n  1 
F  m 20r
F  GM 
x
R
2
 
R  n 
 

GM   n  1  2 n 1
 
   (n  1)R  0  R

R  n   2 
4 4 G  (n  1) 8 G 
But M  R 3   20  n2  n 
3 3 n 2 3 20

92. (i) 8 km/sec (ii) 0.25 m/s

(i) Speed of dust particles before collision with satellite  0

Gravitation 207 Workbook -2| Solutions

 Speed of dust particles after collision with satellite = V
Where V is the orbital velocity of the satellite. Satellite encounters (VS d) kg of dust per sec
 Force experienced = change in momentum per sec
F  (VS d )(V )  V 2Sd  (8  103 )2  0.5  1.6  10 11  5  10 4 N
[ orbital speed of near surface satellite  8 km/sec]
(ii) Total energy of the satellite
1 GMMe  R  radius of earth 
E  MV 2    
2 2R   path radius 
GMMe
E 
2R
GMMe 2R 2
 E   R  R  E
2R 2 GMM e
But change in energy in one revolution E  work done by F  2RF
4 3
2R 2 4 R 3 F 3  3  R  3F
 R  (2 R F ) ;    
GMM e GMM e GM  Me  GM 
 
 

3  5  10 4
Given   mean density of earth  5500 kg/m 3  R   0.4 km
6.67  10 11  10  5500
ve sign indicates reduction in radius.
1
Similarly, E   MV 2
2
E 2RF
E   MV V  V     0.25 m/s [Velocity increases]
MV MV
93. (i) No (ii) 1.11R

(i) No. The tunnel wall applies a normal force which produced a torque about the centre.
(ii) Let the speed of the projectile be V when it comes out of the tunnel.

Energy conservation can be applied between point C and A.

Gravitation 208 Workbook - 2 | Solutions

 GMm  3 2 1 2 
PE at a distance x ( x  R ) is given by   R  x 
R 3  4 2 
 K A  U A  KC  U C

1 GMm 1 2 GMm  3 2 R 2 
2
mV 2 
R

2
m  gR    R 
R 3  2

8 

GM
Simplifying after substituting g  , we get
R2
GM 1 GM
V2   V 
4R 2 R
When the projectile is at farthest distance its velocity u is perpendicular to its position vector
relative to the centre of the earth. Applying conservation of angular momentum between A and B
we get-
VR
mur  m (V cos )R  ur 
2
1
 ur  GMR ….. (i)
4
Energy conservation between A and B
1 GMm 1 GMm
mu 2   mV 2 
2 r 2 R
2
1  GMm  GMm 1 1 GM GMm
m    m 
2  4r  r 2 4 R R
 
R 1 7
   28r 2  (32R )r  R 2  0  r  1.11R
2 r 8R
32r

Gravitation 209 Workbook -2| Solutions