1. SOLUTION OF GRAVITATION
EXERCISE-1 : PART - I
SECTION (A)
A 1. F = G 2
2
1
r
m
m
= 6.67 × 10–11
1
× 2
6
22
)
10
378
(
10
34
.
7
4
.
1



= 4.8 × 10–5
N
A 3. tan =
6
8
=
3
4
 = 53°
F = 2
r
GmM
= 2
)
1
.
0
(
01
.
0
260
.
0
G

a =
m
cos
F
2 
= 2G 2
)
1
.
0
(
260
.
0






5
3
= 31.2 G m/s2
SECTION (B)
B 1. Ex
= –
x
v


= –
x


(20x + 40y) = – 20
Ey
= –
x
v


= – y


(20x + 40y) = – 40
E

= Ex
î + Ey
ĵ = – 20 î – 40 ĵ Ans.
It is independent of co ordinates
Force = F

= mE

= 0.25 {– 20 î – 40 ĵ } = – 5 î – 10 ĵ
magnitude of F

= 2 2
5 10
 = 5
5 N
SECTION (C)
C 1. Potential energy at ground surface
potential energy =
R
GMm
–
potential energy at a height of R is R
potential energy =
R
2
GMm
–
When a body comes to ground
Loss in potential energy = Gain in kinetic energy

R
2
GMm
–
– 





R
GMm
–
=
2
1
mv2

R
2
GMm
=
2
1
mv2
 gR = v2 





 g
R
GM
2
  v = gR

2. SECTION (D)
D 1. sin  =
h
R
R

or  = colatitude = sin–1 





 h
R
R
D 3. (a) F = 2
)
R
2
(
GMm
= 2
2
R
4
GM
(b)
R
Mv 2
= 2
2
R
4
GM
v =
R
4
GM
T =
v
R
2
=
R
4
GM
R
2
= 4
GM
R3
(c) Angular speed
 =
T
2
=










GM
R
4
2
3
= 3
R
4
GM
(d) Energy required to separate = – { total energy }
= – { Kinetic energy + Potential energy }
= –







R
2
GM
–
Mv
2
1
Mv
2
1 2
2
2
= –






R
2
GM
–
Mv
2
2
= –






R
2
GM
–
R
4
GM
M
2
= –






R
4
GM
–
2
=
R
4
GM2
Ans.
(e) Let its velocity = ‘v’
Kinetic energy =
2
1
mv2
Potential at centre of mass = –
R
GM
–
R
GM
= –
R
GM
2
Potential energy at centre of mass = –
R
GMm
2
For particle to reach infinity
Kinetic energy + Potential energy = 0
2
1
mv2
×
R
GMm
2
= 0
v =
R
GM
4
Ans.

3. D 5. T2
 r3
For the Saturn , r is more , So T will be more as  =
T
2
so  will be less, for the Saturn .
and orbital speed Vo
=
r
GM
as r is more So VO
will be less for the Saturn.
SECTION (E)
E 1. Field outside shell g1
= g 





R
h
2
–
1
So] 9 = g 





R
h
2
–
1 
20
1
R
h

Field inside shell E = g 





R
h
–
1
E = 10 





20
1
–
1 E = 10 – 0.5 = 9.5 m/sec2
Solving we get x  9.5 m/s2
PART - II :
SECTION (A) :
A-1. Net force is towards centre
Fnet
= F2
+
2
F
2 1
= F2
+ 2 F1











2
2
2
2
1
2
2
2
2
2
r
2
Gm
)
r
2
(
Gm
F
r
4
Gm
)
r
2
(
Gm
F
 Fnet
= 2
2
r
4
Gm
+ 2
2
r
2
Gm
2
=
r
mv2
 v = )
2
2
1
(
r
4
Gm
 Ans.
A-3. In horizontal direction
Net force = 2
d
12
mm
3
G
cos30° – 2
2
d
4
m
G
cos60°
= 2
2
d
8
m
G
– 2
2
d
8
m
G
= 0
in vertical direction
Net force = 2
2
d
12
m
3
G
cos60° + 2
2
d
3
m
3
G
+ 2
2
d
4
m
G
cos30°
= 2
2
d
24
m
G
3
+ 2
2
d
3
m
G
3
+ 2
2
d
8
m
G
3
= 2
2
d
m
G
3





 

24
3
8
1
= 2
2
d
2
m
G
3
along SQ

4. SECTION (B)
B-1. dv = – Edr =
r
k
dr
Integrating both sides
 v
vi
v =  r
i
d
r
n
k 
v – vi
= k n
i
d
r
 v = vi
+ k n
i
d
r
Ans.
B-3.
1m
2m
4m
8m
v = v1
+ v2
+ v3
+ v4
+ ...... = –
1
Gm
–
2
Gm
– 4
Gm
–
8
Gm
–
16
Gm
–
32
Gm
– ........
= – Gm 








 .......
8
1
4
1
2
1
1  G.P. of infinite series = – m 





2
/
1
–
1
1
v = – Gm (2) = – 2Gm Ans.
B-5. If we take complete spherical shell than gravitational field intensity at P will be zero hence for the hemi
spherical shell shown the intensity at P will be along c.
SECTION (C)
C 1. Initial total energy = Initial kinetic energy + initial potential energy
=
2
1
m (0)2
+ 







0
R
GMm
– =
0
R
GMm
–
Total energy, when it reaches the surface of earth =
2
1
mv2
+ 





R
GMm
–
Applying energy conservation,
2
1
mv2
–
R
GMm
=
0
R
GMm
–
v =






0
R
1
–
R
1
GM
2 Ans.

5. C 2. (a)
120°
F
m
F
F
m
m
120°
120°
Due to geometry net force is zero.
(b) By geometry, x2
+
4
a2
= a2
and F1
= F2
F
F2
F1
x
a
x2
=
4
a
3 2
x =
2
a
3
Fnet
= F = 2
2
x
Gm
=
3
4
2
2
a
Gm
(c) Initial potential energy = –












a
Gm
a
Gm
a
Gm 2
2
2
= –
2
3
a
Gm2
Work done on system = Final potential energy – intial potential energy
= –
2
3
a
Gm2
–










a
Gm
3
–
2
=
2
3
a
Gm2
Ans.
(d) Initial kinetic energy = 0
Initial potential energy = –
a
Gm2
–
a
Gm2
= –
a
Gm
2 2
Total initial energy = –
a
Gm
2 2
Now, kinetic energy =
2
1
mv2
Potential energy = –
2
/
a
Gm
2 2
–
2
/
a
Gm2
= –
a
Gm
4 2
Total energy =
2
1
mv2
–
a
Gm
4 2

a
Gm
2 2
=
2
1
mv2

a
Gm
4
= v
v =
a
Gm
2 Ans.

6. SECTION : (D)
D-1. 2
r
GMm
=
r
mv2
v = 2
r
GM
M
r
V
m
T =
v
r
2
=
GM
r
2 2
3

=
3
2
3
r
3
4
G
r
2




T  
1
Ans.
D-3*. Escape velocity =
R
GM
2
= Ve
Orbital velocity =
R
GM
= V0
Escape velocity = 2 × orbital velocity  (A)
2
1
mVe
2
= 2 ×
2
1
mV0
2
 (C)
D-5*. PE = –G m1
m2
/r, ME = – G m1
m2
/ 2r
On decreasing the radius of orbit PE and ME decreases
SECTION (E)
E 1. we
= 50 × 10 = 500 N  wp
= 50 × 5 = 250 N Hence option A is correct
PART - III
1. (A) Till the particle reaches the centre of planet, force on both bodies are in direction of their respective
velocities, hence kinetic energies of both keep on increasing . After the particle crosses the centre of planet,
forces on both are retarding in nature. Hence as the particle passes through the centre of the planet, sum of
kinetic energiesof both the bodiesismaximum. Therefore statement-1 isTrue, Statement-2 isTrue;Statement-
2 is a correct explanation for Statement-1.
EXERCISE-2
PART - I
1. E

= î
3 – ĵ
4 N / kg
Then F

= mE

= (1 kg) ( î
3 – ĵ
4 ) N / kg = 3 î – 4 ĵ N
Work done = W = F

. S

S

= Displacement vector
displacement is along y =
4
3
x +
4
9
Any point on this line can be written as
k1
î + 






4
9
k
4
3
1 ĵ = k2
î + 






4
9
k
4
3
2 ĵ
S

= Displacement vector = (k2
– k1
) î +
4
3
(k2
– k1
) ĵ
Now, W = (3 î – 4 ĵ ) .






 ĵ
)
k
–
k
(
4
3
î
)
k
–
k
( 1
2
1
2
= 3 (k2
– k1
) – 3 (k2
– k1
) = 0 Ans.

7. 3. (a) Force will be due to the mass of the sphere upto the radius r
In case (i) 0 < r < b ; Mass M = 0, therefore F(r) = 0
In case (ii) b < r < a ; Mass M = 
3
4
(r3
– b3
), therefore F(r) = 








 2
3
r
b
r
Gpm
3
4
(iii) a < r <  ; Mass M = 
3
4
(a3
– b3
), therefore F(r) = 






 
 2
3
3
r
b
a
Gpm
3
4
(b) Uf
– Ui
= – 
2
1
r
r
c dr
.
F

(i) 0 < r < b ; u(r) = - 2 Gm(a2
— b2
)
(ii) b < r < a ; u(r) =
r
3
m
G
2 


(3ra2
- 2b3
- r3
)
(iii) a < r <  ; u (r) = )
b
a
(
r
3
m
G
4 3
3




5. (a) The gravitation field isuniform inside the cavity and isdirected along ´
OO . Hence the particle will strike atA.
(b) The gravitational field at any point P inside cavity.
|E

| =
3
4
G –
3
4
G =
3
4
Gy ´
OO =
3
2
GR
Total workdone = m |E

| . S

= m .
3
2 GR .
2
R
Applying work - energy theorem
Workdone by all force = Change in kinetic energy
m .
3
2 GR .
2
R =
2
1
mv2
v =
3
R
G
2 2


Ans.

8. 7. Applying angular momentum conservation :
mv0
 = mvd
v0
= vd .......... (i)
Intial energy =
2
1
mv0
2
+ 0
Final energy =
2
1
mv2
–
d
GMs
Applying energy conservation ,
2
1
mv0
2
=
2
1
mv2
–
d
m
GMs
v0
2
= v2
–
d
GM
2 s
.......... (ii)
From equation (i) and (ii) :
v0
2
= 2
2
2
0
d
v 
–
d
GM
2 s
d2
+ 2
0
s
v
GM
2
d – 2
= 0
Solving this quadratic
d = – 2
0
s
v
GM
+
2
2
2
0
s
v
GM










= 2
0
s
v
GM


















 1
–
GM
v
1
2
2
0

Ans.
PART - II
1. Gravitational field at ‘m’ due to hollowed - out lead sphere
= { Field due to solid spere } – { Field due to mass that was removed }
Field due to solid sphere = 2
d
GM
= E1
= 2
R
4
GM
Field due to removed mass = 2
x
'
GM
= E2
M’ =
3
R
3
4
M

×
3
2
R
3
4






 =
8
M
And x = d –
2
R
So, E2
= 2
2
R
–
d
8
GM






= 2
2
R
3
8
GM






= 2
R
18
GM
Enet
= E1
– E2
= 2
R
GM






18
1
–
4
1
= 2
R
36
GM
7
Fnet
= mEnet
= 2
R
36
GMm
7
Ans.

9. 3.
F = 
cos
)
R
2
(
GMm
2 =
2
3
)
R
2
(
GMm
2
 = 2
R
8
GMm
3
5. Ve
=
R
GM
2
V = kVe
=
R
GM
2
K
Initial total energy =
2
1
mv2
–
GMm
R
=
2
1
m.k2
R
GM
2
–
GMm
R
Final total energy =
2
1
m02
–
GMm
x
Applying energy conservation
2
1
mk2
.
R
GM
2
–
GMm
R
= 0 –
GMm
x
x
1
=
R
1
–
2
k
R
 x = 2
k
–
1
R
Ans.
7. inside a uniform spherical shell
Ein
= 0
Vin
= constant =
R
Gm
9. Fg
= 3
R
GMmr
pressing force = Fg
cos  = 3
R
cos
GMmr 
= 2
R
2
GMm
= constant
a =
m
sin
Fg 
= 3
R
sin
GMr 
a = 3
R
GMy
11.* S
=
5
.
1
2
, E
=
24
2
west to east
= 2 





24
1
–
5
.
1
1
Twest to east
=
east
to
west
2


= 1.6 hours
Similarly
east to west
= 2 






24
1
5
.
1
1
 Teast to west
=
17
24
hours

10. PART - III
1. P.E. = –
r
GMm
 K.E. =
2
1
mV2
Total energy =
r
GMm
 +
2
1
mV2
T.E. = 0 if 0
mV
2
1
r
GMm 2


  v =
r
GM
2
For v <
r
GM
2
, T.E. is – ve
for v >
r
GM
2
, T.E. is + ve
If V is
r
GM
i.e. equal to orbital velocity, path is circular.
.
If T.E. is negative, path is elliptical.
If T.E. is zero, path is parabolic.
If T.E. is positive, path is hyperbolic.
PART - IV
4. Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then escape
velocity from earth e
= )
Rg
2
(
Velocity of satellite s
=
2
e

= 2
/
)
g
R
2
( ......... (1)
Further ] s
= 





r
GM
= 







 h
R
g
R2
 2
s
 =
h
R
g
R2

h = R = 6400 km
5. T2
=
3
2
x
Gm
4
Hence time period of revolution T is
T =
Gm
x
2
3
 (Put x = 2R )
 T = g
R
8
2
6. Now total energy at height h = total energy at earth's surface (from principle of conservation of energy)
 0 – G M
h
R
m

=
2
1
m2
– GM
R
m
or
2
1
m2
=
R
m
GM
–
R
2
GMm
( h = R)
 v = gR

11. 7 to 9. Let theangular speed of revolution of both starsbeaboutthecommoncentre , that is, centreof massof system.
The centripetal force on star of mass m is
m2
3
d
2
= 2
d
)
m
2
(
Gm
. Solving we get T=
3
2
d
Gm
3
4
The ratio of angular momentum is simply the ratio of moment of inertia about center of mass of system.
2
3
d
m
2
3
d
2
m
L
L
2
2
M
m
M
m

















I
I
Similarly, The ratio of kinetic energy is simply the ratio of moment of inertia about center of mass of
system.
2
3
d
m
2
3
d
2
m
K
K
2
2
2
M
2
m
M
m

















I
2
1
I
2
1
EXERCISE- 3
PART - I
1. (A) Gravitational field isa conservative force field. In a conservative force field work done ispath independent.
 WI
= WII
= WIII
2. speed of particle at A VA
= escape velocity on the surface of moon =
R
GM
2
At highest point B, VB
= 0
From energy conservation.
2
A
mV
2
1
= VB
– VA
= m 





m
U
–
m
U A
B
or 






m
U
–
m
U
2
V A
B
2
A
, also 3
A
R
2
GM
–
m
U
 [3R2
– r2
]
 























2
2
3 100
R
–
R
5
.
0
–
R
5
.
1
R
GM
–
–
h
R
GM
–
R
GM
or
2
1 –1 3 1 99 1
–
R R h 2R 2 100 R
 
   
  
h = R
)
99
(
)
100
(
)
99
(
)
100
(
2
2
2
2


 h = R
199
19801
or h = 99.5 R  99R Ans
3. A
B
mA mB
com
rA rB
C
 2
B
A
B
A
)
r
r
(
m
Gm

= mA
rA 2
A
2
T
4
= mB
rB 2
B
2
T
4
 2
B
B
B
2
A
A
A
T
r
m
T
r
m

As C is com  mA
rA
= mB
rB
hence TA
=TB

12. 4. (A) It is similar equation as v = 2
2
x
a 
ω in SHM.
(B) Particle on positive x-axis move towards origin with speed decreasing as x decreasing.
(C) It is spring mass system performing SHM.
(D) Object moves away from Earth so its speed will decrease, since its speed is greater than escape velocity
so it will never return back.
5.  = 0
r < R
= 0 r > R
Case Ir < R
FC
=
r
mV2
mg
r
mV
R
r 2
 (g = acceleration due to gravity at surface of sphere)
V =
R
g
r for r < R
Case II r > R
2
r
GMm
=
r
mV2
V = R
r
g
r
GM
 So Ans. (C)
6. If only gravitational force acts on astronaut (that is in state of free fall), he shall feel weightless. Thus
statement-2 is correct explanation of statement-1.
7. Wext = U – UP
Wext = 0 –  





1
.
x
Gdm
–
–
Wext = G 2
2
2
r
R
16
rdr
2
R
7
M




 = 2
R
7
GM
2
  2
2
r
R
16
rdr
= 2
R
7
GM
2
 z
zdz
= 2
R
7
GM
2
[Z]
r
3R
4R
dr
P
x
Wext = 2
R
7
GM
2 R
4
R
3
2
2
r
R
16





 
Wext = 2
R
7
GM
2
 
R
5
–
R
2
4
Wext = 2
R
7
GM
2
 
5
–
2
4 .
8.
A B
c.m.
5d
6
d
6 11 Ms
2.2 Ms
c.m.
about
B
of
momentum
Angular
c.m.
about
momentum
angular
Total
=








































6
d
6
d
)
M
11
(
6
d
6
d
)
M
11
(
6
d
5
6
d
5
)
M
2
.
2
(
s
s
s
= 6.

13. 9. g = 2
R
GM
= 2
3
R
R
3
4
)
G
( 







; g R
g
'
g
= 















R
'
R
'
= 





3
2






R
'
R
=
11
6
Given
R
'
R
=
22
6
3
Ve =
R
GM
=
R
R
3
4
))
(
)
G
( 3
















 Ve  R  ; Ve = 3 km/hr.
10. Ve
= 0
v
2
KE =
2
e
mv
2
1
=  2
0
v
2
m
2
1
= mv0
2
11. Ves
=
R
GM
2
=
R
R
3
4
.
G
.
2 3


=
3
G
4 
R
Ves
 R
Sarface area of P = A = 4RP
2
Surface area of Q = 4A = 4 RQ
2
 RQ
= 2Rp
mass R is MR
= MP
+ MQ
3
R
R
3
4

 =
3
P
R
3
4

 +
3
Q
R
3
4

  RR
3
= RP
3
+ RQ
3
= 9RP
3
RR
= 91/3
RP
 RR
> RQ
> RP
Therefore VR
> VQ
> VP
P
R
V
V
= 91/3
and
Q
P
V
V
=
2
1
12. 0
0
mv
2
1
L
m
2
.
GM 2





L
GM
4
v 
M M
m
L L
PART - II
1.
2
1
T
T
=
2
/
3
2
1
r
r








or
5
T
= (4)3/2
or T = 40 hr.
2. Escape velocity is independent of direction of projection.
3.
X
R
mv2

= 2
2
2
R
R
)
X
R
(
GM


or V =
2
/
1
2
X
R
R
g









4. T  R3/2
Time period is independent of mass of satellite
5. PE =
R
GMm
–
R
2
GMm
=
2
mgR
6. g 





R
d
–
1 = g 





R
h
2
–
1 or d = 2h
7. W =
r
GMm
= 6.67 × 10–11
× 100 × 10 / 0.1 = 6.67 × 10–7
J

14. 8. Electric charge on the moon = electric charge on the earth
9. V =
R
GM
2
=
10
R
10
GM
2
e
e 
= 10
e
e
R
GM
2
= 110 k m/s
10. Acceleration due to gravity at leight h from earth surface.
g'
=
2
R
h
1
g






 
2
R
h
1
g
9
g








 h = 2R
11. 2
x
Gm
= 2
)
x
r
(
)
m
4
(
G

x
1
=
x
r
2

r – x = 2x
3x =
3
r
x =
3
r
3
/
r
2
)
m
4
(
G
3
/
r
Gm


r
Gm
6
r
Gm
3

 =
r
Gm
9
Ans.
12. 2
2
)
R
2
(
Gm
= m2
R
3
2
R
4
Gm
= 2
 = 3
R
4
Gm
v = R
v = 3
R
4
Gm
× R =
R
4
Gm
13. W = 0 –
R
GMm
R
GMm








= gR2
×
R
m
= mgR = 1000 × 10 × 6400 × 103
= 64 × 109
J = 6.4 × 1010
11.
R
6
GMm
1
2
1
R
3
GMm
R
3
GMm
R
3
GM
m
2
1
R
3
GMm
mv
2
1
E 2
0
f














R
GMm
Ei

 + K
Ei
= Ef
R
6
GMm
5
K 