Gravitation - 2 Practice Questions - DPP for NEET

 Page 1


(1) (a) A body projected up with the escape velocity v
e
will go
to infinity. Therefore, the velocity of the body falling
on the earth from infinity will be v
e
. Now, the escape
velocity on the earth is
v
e
 = 
e
gR = m) 10 (6400 (9.8m/s  2
3 2
´ ´ ´ )
     = 1.2 × 10 10
4
 m/s = 11.2 km/s.
The kinetic energy acquired by the body is
K = 
1
2
m v
e
2 
= 
1
2
 × 100 kg × (11.2 × 10
3
 m/s)
2
   = 6.27 × 10
9
 J.
(2) ( d) We know that  
2
GMm
r
= m w
2
r   or    
2
GM
r
 = w
2 
r ..
\ r
3
 = 
2
GM
w
where w is the angular velocity of the satellite
In the present case, w = 2w
0
,
where w
0
 is the angular velocity of the earth.
\  w = 2 × 7.3 × 10
–5
 rad/ sec.
      G = 6.673 × 10
–11
 n-m
2
/kg
2
and  M = 6.00 × 10
24
kg.
Substituting these values in equation (A), we get
  r
3 
= 
11 24
52
(6.673 10 )(6.00 10 )
(2 7.3 10 )
-
-
´´
´´
Solving we get   r = 2.66 × 10
7
m.
(3) (a)
From Kepler's Law , T
2
 µ r
3
\  
2
1
2
T
T
æö
ç÷
èø
 = 
3
1
2
r
r
æö
ç÷
èø
  Þ 
2
1
8
æö
ç÷
èø
 = 
4
2
10
r
æö
ç÷
èø
Þ  r
2
 
= 4x 10
4
 km
v = wr = 
2r
T
p
\  | v
2
 – v
1
| = 2p 
12
12
rr
TT
æö
-
ç÷
èø
  = p × 10
4
 km/hr
(4) (a) When S
2
 is closest to S
1
, the speed of S
2
 relative to S
1
is v
2
 – v
1
 = p × 10
4
 km/hr.  The angular speed of S
2
 as
observed from S
1
 (when closest  distance between them
is r
2
 – r
1
 = 3 × 10
4
 km)
w = 
21
21
vv
rr
-
-
 = – 
4
4
10
3 10
p´
´
 = – 
3
p
 rad/hr ,
| w | = 
3
p
rad/hr
(5) (c) Period of revolution of earth around sun
22
2 e
e
s
4R
T
GM
p
=
Period of revolutions of moon around earth
22
2 m
n
e
4R
T
GM
p
=
\  
2
e
m
T
T
æö
ç÷
èø
 = 
e
s
M
M
æö
ç÷
èø
 
3
e
m
R
R
æö
ç÷
èø
\  
s
e
M
M
 = 
2
m
e
T
T
æö
ç÷
èø
 
3
e
m
R
R
æö
ç÷
èø
 = 
3
2
(393)
13
= 3.56 × 10
5
(6) (a) According to law of conservation of angular momen-
tum,  mvr = constant
Þ  vr = constant
v
max
 .r
min 
=  v
min
. r
max
Þ  
B
A
V
V
 = 
max
min
v
v
 = 
max
min
r
r
 =  x
(7) (a) Angular momentum of satellite,  J = mvr.  But,
2
GMm
r
 = 
2
mv
r
  Þ v = 
GM
r
\   J = m 
GMr
(8) (a) The orbital velocity of space ship, v
0
 = 
GM
r
If space, ship is very near to earth's surface,
r = Radius of earth = R   \ v
0
 =  
GM
R
  = Rg =
6
6.4 10 9.8 ´´
   = 7.9195 × 10
3
 m/sec = 7.195 km/sec
The escape velocity of space-ship
v
e 
= 2Rg = 7.9195 2 = 11.2 km/sec
Additional velocity required = 11.2– 7.9195=3.2805 km/
sec.
­
­
R
1
S
1
S
2
v
2
v
1
R
2 ¬ ®
Page 2


(1) (a) A body projected up with the escape velocity v
e
will go
to infinity. Therefore, the velocity of the body falling
on the earth from infinity will be v
e
. Now, the escape
velocity on the earth is
v
e
 = 
e
gR = m) 10 (6400 (9.8m/s  2
3 2
´ ´ ´ )
     = 1.2 × 10 10
4
 m/s = 11.2 km/s.
The kinetic energy acquired by the body is
K = 
1
2
m v
e
2 
= 
1
2
 × 100 kg × (11.2 × 10
3
 m/s)
2
   = 6.27 × 10
9
 J.
(2) ( d) We know that  
2
GMm
r
= m w
2
r   or    
2
GM
r
 = w
2 
r ..
\ r
3
 = 
2
GM
w
where w is the angular velocity of the satellite
In the present case, w = 2w
0
,
where w
0
 is the angular velocity of the earth.
\  w = 2 × 7.3 × 10
–5
 rad/ sec.
      G = 6.673 × 10
–11
 n-m
2
/kg
2
and  M = 6.00 × 10
24
kg.
Substituting these values in equation (A), we get
  r
3 
= 
11 24
52
(6.673 10 )(6.00 10 )
(2 7.3 10 )
-
-
´´
´´
Solving we get   r = 2.66 × 10
7
m.
(3) (a)
From Kepler's Law , T
2
 µ r
3
\  
2
1
2
T
T
æö
ç÷
èø
 = 
3
1
2
r
r
æö
ç÷
èø
  Þ 
2
1
8
æö
ç÷
èø
 = 
4
2
10
r
æö
ç÷
èø
Þ  r
2
 
= 4x 10
4
 km
v = wr = 
2r
T
p
\  | v
2
 – v
1
| = 2p 
12
12
rr
TT
æö
-
ç÷
èø
  = p × 10
4
 km/hr
(4) (a) When S
2
 is closest to S
1
, the speed of S
2
 relative to S
1
is v
2
 – v
1
 = p × 10
4
 km/hr.  The angular speed of S
2
 as
observed from S
1
 (when closest  distance between them
is r
2
 – r
1
 = 3 × 10
4
 km)
w = 
21
21
vv
rr
-
-
 = – 
4
4
10
3 10
p´
´
 = – 
3
p
 rad/hr ,
| w | = 
3
p
rad/hr
(5) (c) Period of revolution of earth around sun
22
2 e
e
s
4R
T
GM
p
=
Period of revolutions of moon around earth
22
2 m
n
e
4R
T
GM
p
=
\  
2
e
m
T
T
æö
ç÷
èø
 = 
e
s
M
M
æö
ç÷
èø
 
3
e
m
R
R
æö
ç÷
èø
\  
s
e
M
M
 = 
2
m
e
T
T
æö
ç÷
èø
 
3
e
m
R
R
æö
ç÷
èø
 = 
3
2
(393)
13
= 3.56 × 10
5
(6) (a) According to law of conservation of angular momen-
tum,  mvr = constant
Þ  vr = constant
v
max
 .r
min 
=  v
min
. r
max
Þ  
B
A
V
V
 = 
max
min
v
v
 = 
max
min
r
r
 =  x
(7) (a) Angular momentum of satellite,  J = mvr.  But,
2
GMm
r
 = 
2
mv
r
  Þ v = 
GM
r
\   J = m 
GMr
(8) (a) The orbital velocity of space ship, v
0
 = 
GM
r
If space, ship is very near to earth's surface,
r = Radius of earth = R   \ v
0
 =  
GM
R
  = Rg =
6
6.4 10 9.8 ´´
   = 7.9195 × 10
3
 m/sec = 7.195 km/sec
The escape velocity of space-ship
v
e 
= 2Rg = 7.9195 2 = 11.2 km/sec
Additional velocity required = 11.2– 7.9195=3.2805 km/
sec.
­
­
R
1
S
1
S
2
v
2
v
1
R
2 ¬ ®
DPP/ P 19
57
(9) (b) The escape velocity  v
e
 = 2gR
Now,  (V
e
)
moon
 = 2gR
(V
e
)
earth
 = 2 6g 10R ´´ ,
So   
e earth
e moon
(V)
8
(V)
=
(10) (b) Escape velocity   = 
2GM
2gR
R
=
\ 
p
e
V
V
 = 
p
e
ep
g
R
gR
´
 = 10 1 10 ´=
V
p
 = 
10
 VV
e
(11) (a) We know that T
2 
µ a
3
Given that (12 T)
2  
µ a
1
3 
and T
2 
µ a
2
3
\   
3
1
3
2
a
a
 = 
2
2
(12T)
T
 = 144
or   
1
2
a
a
 = (144)
1/3
 = 5.242
Hence the jupiter's distance is 5.242 times that of the
earth from the sun.
(12) ( b) We know that T
2
 µ a
3
 Þ  T µ (a)
3/2
\ 
mars
earth
T
T
 = 
3
2
mars
earth
a
a
æö
ç÷
èø
 = (1.524)
3/2
 = 1.88
As the earth revolves round the sun in one year and
hence,  T
earth 
= 1 year.
\ T
mars 
= T
earth
 × 1.88 = 1 × 1.88  =  1.88 earth-year.
(13) (d)
mercury
mars
T
T
 = 
3/2
mercury
mars
a
a
æö
ç÷
èø
=  
3/2
0.387
1.524
æö
ç÷
èø
\  T
mars
  = T
mercury 
×  
3/2
1.524
0.387
æö
ç÷
èø
     = (0.241years) × (7.8) =  1.9 years.
(14) (a)
2
3
T
r
 = 
2
0
3
2r
v
r
æö p
ç÷
èø
 = 
2
3
(2 r) 1
r
GM
r
p
 = 
2
4
GM
p
[\ 
2
0
2
mv GMm
r
r
=
, 
2
0
GM
v
r
= ]
Slope  of T
2
 – r
3
 curve = tan q   = 
2
3
T
r
 = 
2
4
GM
p
(15) (c) Total energy of the particle at P
 E = E
kP
 + U  = 
2
e
1
mv
2
 – 
12
GM m GM m
d/2 d/2
-
   = 
2
e
1
mv
2
 – 
2Gm
d
(M
1 
+ M
2
)
At infinite distance from M
1 
and M
2
, the total energy
of the particle is zero.
\  
2
e
1
mv
2
 = 
2Gm
d
 (M
1 
+ M
2
),
\ v
e 
=
12
4G
(M M)
d
+
(16) (d) v   = 
GM
r
 = 
2
gR
r
 = 
2 12
6
9.8 6.4 10
8 10
´´
´
        = 7.08 km/sec.
(17) (b) From conservation of energy,
The energy at height h =  Total energy at earth's sur-
face
0 – 
GMm
Rh +
 = 
1
2
 mv
2
 – 
GMm
R
 ,
1
2
mv
2
 = 
GMm
R
 – 
GMm
Rh +
  = 
GMm
R
 – 
GMm
2R
Þ  v    = 
GM
R
 = 
2
Rg
R
 = Rg
             = 
3
6400 10 9.8 ´´ = 7.919 × 10
3
 m/s
= 7.919 km/sec
(18) (a) If a body is projected from the surface of earth with a
velocity v and reaches a height h, then using law of
conservation of energy,  
1
2
 mv
2
 = 
mgh
1 h /R +
.
 Given v = Kv
e
 = K 2gR and h = r –  R
Hence,
1
2
 mK
2
 2gR = 
mg(r R)
rR
1
R
-
-
+
  or r = 
2
R
1K -
(19) (a) Orbital speed,
v
0 
= 
e
g R = ) 10  (6.4  9.8
6
´ ´
     = 7.2 × 10
3
 m/s = 7.2 km/s.
Period of revolution,
T  = 2p R/g
     = 2 × 3.14 )/9.8 10 (6.4
6
´ = 5075 s = 84.6 minutes.
Page 3


(1) (a) A body projected up with the escape velocity v
e
will go
to infinity. Therefore, the velocity of the body falling
on the earth from infinity will be v
e
. Now, the escape
velocity on the earth is
v
e
 = 
e
gR = m) 10 (6400 (9.8m/s  2
3 2
´ ´ ´ )
     = 1.2 × 10 10
4
 m/s = 11.2 km/s.
The kinetic energy acquired by the body is
K = 
1
2
m v
e
2 
= 
1
2
 × 100 kg × (11.2 × 10
3
 m/s)
2
   = 6.27 × 10
9
 J.
(2) ( d) We know that  
2
GMm
r
= m w
2
r   or    
2
GM
r
 = w
2 
r ..
\ r
3
 = 
2
GM
w
where w is the angular velocity of the satellite
In the present case, w = 2w
0
,
where w
0
 is the angular velocity of the earth.
\  w = 2 × 7.3 × 10
–5
 rad/ sec.
      G = 6.673 × 10
–11
 n-m
2
/kg
2
and  M = 6.00 × 10
24
kg.
Substituting these values in equation (A), we get
  r
3 
= 
11 24
52
(6.673 10 )(6.00 10 )
(2 7.3 10 )
-
-
´´
´´
Solving we get   r = 2.66 × 10
7
m.
(3) (a)
From Kepler's Law , T
2
 µ r
3
\  
2
1
2
T
T
æö
ç÷
èø
 = 
3
1
2
r
r
æö
ç÷
èø
  Þ 
2
1
8
æö
ç÷
èø
 = 
4
2
10
r
æö
ç÷
èø
Þ  r
2
 
= 4x 10
4
 km
v = wr = 
2r
T
p
\  | v
2
 – v
1
| = 2p 
12
12
rr
TT
æö
-
ç÷
èø
  = p × 10
4
 km/hr
(4) (a) When S
2
 is closest to S
1
, the speed of S
2
 relative to S
1
is v
2
 – v
1
 = p × 10
4
 km/hr.  The angular speed of S
2
 as
observed from S
1
 (when closest  distance between them
is r
2
 – r
1
 = 3 × 10
4
 km)
w = 
21
21
vv
rr
-
-
 = – 
4
4
10
3 10
p´
´
 = – 
3
p
 rad/hr ,
| w | = 
3
p
rad/hr
(5) (c) Period of revolution of earth around sun
22
2 e
e
s
4R
T
GM
p
=
Period of revolutions of moon around earth
22
2 m
n
e
4R
T
GM
p
=
\  
2
e
m
T
T
æö
ç÷
èø
 = 
e
s
M
M
æö
ç÷
èø
 
3
e
m
R
R
æö
ç÷
èø
\  
s
e
M
M
 = 
2
m
e
T
T
æö
ç÷
èø
 
3
e
m
R
R
æö
ç÷
èø
 = 
3
2
(393)
13
= 3.56 × 10
5
(6) (a) According to law of conservation of angular momen-
tum,  mvr = constant
Þ  vr = constant
v
max
 .r
min 
=  v
min
. r
max
Þ  
B
A
V
V
 = 
max
min
v
v
 = 
max
min
r
r
 =  x
(7) (a) Angular momentum of satellite,  J = mvr.  But,
2
GMm
r
 = 
2
mv
r
  Þ v = 
GM
r
\   J = m 
GMr
(8) (a) The orbital velocity of space ship, v
0
 = 
GM
r
If space, ship is very near to earth's surface,
r = Radius of earth = R   \ v
0
 =  
GM
R
  = Rg =
6
6.4 10 9.8 ´´
   = 7.9195 × 10
3
 m/sec = 7.195 km/sec
The escape velocity of space-ship
v
e 
= 2Rg = 7.9195 2 = 11.2 km/sec
Additional velocity required = 11.2– 7.9195=3.2805 km/
sec.
­
­
R
1
S
1
S
2
v
2
v
1
R
2 ¬ ®
DPP/ P 19
57
(9) (b) The escape velocity  v
e
 = 2gR
Now,  (V
e
)
moon
 = 2gR
(V
e
)
earth
 = 2 6g 10R ´´ ,
So   
e earth
e moon
(V)
8
(V)
=
(10) (b) Escape velocity   = 
2GM
2gR
R
=
\ 
p
e
V
V
 = 
p
e
ep
g
R
gR
´
 = 10 1 10 ´=
V
p
 = 
10
 VV
e
(11) (a) We know that T
2 
µ a
3
Given that (12 T)
2  
µ a
1
3 
and T
2 
µ a
2
3
\   
3
1
3
2
a
a
 = 
2
2
(12T)
T
 = 144
or   
1
2
a
a
 = (144)
1/3
 = 5.242
Hence the jupiter's distance is 5.242 times that of the
earth from the sun.
(12) ( b) We know that T
2
 µ a
3
 Þ  T µ (a)
3/2
\ 
mars
earth
T
T
 = 
3
2
mars
earth
a
a
æö
ç÷
èø
 = (1.524)
3/2
 = 1.88
As the earth revolves round the sun in one year and
hence,  T
earth 
= 1 year.
\ T
mars 
= T
earth
 × 1.88 = 1 × 1.88  =  1.88 earth-year.
(13) (d)
mercury
mars
T
T
 = 
3/2
mercury
mars
a
a
æö
ç÷
èø
=  
3/2
0.387
1.524
æö
ç÷
èø
\  T
mars
  = T
mercury 
×  
3/2
1.524
0.387
æö
ç÷
èø
     = (0.241years) × (7.8) =  1.9 years.
(14) (a)
2
3
T
r
 = 
2
0
3
2r
v
r
æö p
ç÷
èø
 = 
2
3
(2 r) 1
r
GM
r
p
 = 
2
4
GM
p
[\ 
2
0
2
mv GMm
r
r
=
, 
2
0
GM
v
r
= ]
Slope  of T
2
 – r
3
 curve = tan q   = 
2
3
T
r
 = 
2
4
GM
p
(15) (c) Total energy of the particle at P
 E = E
kP
 + U  = 
2
e
1
mv
2
 – 
12
GM m GM m
d/2 d/2
-
   = 
2
e
1
mv
2
 – 
2Gm
d
(M
1 
+ M
2
)
At infinite distance from M
1 
and M
2
, the total energy
of the particle is zero.
\  
2
e
1
mv
2
 = 
2Gm
d
 (M
1 
+ M
2
),
\ v
e 
=
12
4G
(M M)
d
+
(16) (d) v   = 
GM
r
 = 
2
gR
r
 = 
2 12
6
9.8 6.4 10
8 10
´´
´
        = 7.08 km/sec.
(17) (b) From conservation of energy,
The energy at height h =  Total energy at earth's sur-
face
0 – 
GMm
Rh +
 = 
1
2
 mv
2
 – 
GMm
R
 ,
1
2
mv
2
 = 
GMm
R
 – 
GMm
Rh +
  = 
GMm
R
 – 
GMm
2R
Þ  v    = 
GM
R
 = 
2
Rg
R
 = Rg
             = 
3
6400 10 9.8 ´´ = 7.919 × 10
3
 m/s
= 7.919 km/sec
(18) (a) If a body is projected from the surface of earth with a
velocity v and reaches a height h, then using law of
conservation of energy,  
1
2
 mv
2
 = 
mgh
1 h /R +
.
 Given v = Kv
e
 = K 2gR and h = r –  R
Hence,
1
2
 mK
2
 2gR = 
mg(r R)
rR
1
R
-
-
+
  or r = 
2
R
1K -
(19) (a) Orbital speed,
v
0 
= 
e
g R = ) 10  (6.4  9.8
6
´ ´
     = 7.2 × 10
3
 m/s = 7.2 km/s.
Period of revolution,
T  = 2p R/g
     = 2 × 3.14 )/9.8 10 (6.4
6
´ = 5075 s = 84.6 minutes.
58
DPP/ P 19
(20) (d) If the period of revolution of a satellite about the earth
be T, then
T
2
 = 
23
e
e
4 (R h)
GM
p+
where h is the height of the satellite from earth's sur-
face.
\ M
e
 = 
23
e
2
4 (R h)
GT
p+
The satellite is revolving just above the earth, hence h
is negligible compared to R
e
.
\ M
e 
= 
23
e
2
4R
GT
p
But M
e
 = 
4
3
p R
e
3
 r where r  is the density of the earth.
Thus  
4
3
p R
e
3
 r = 
23
e
2
4R
GT
p
\   r T
2
 = 
3
G
p
.
which is universal constant. To determine its value,
r T
2
 = 
3
G
p
 = 
–1132
3 3.14
6.67  10 m / kg-s
´
´
(21) (a)
KQ
KP
E
E
 = 
2
Q
2
P
v
v
.
Linear velocity of earth,
V
e
= 
e
e
2R
T
p
 = 
6
6.28 6.4 10
24 3600
´´
´
 = 463 m/s
Orbital velocity ,   V
0
 =
e
Rg = 7.9 × 10
3
 m/s
According to question,
V
P
 =  V
0
 + V
e 
= 7900 – 463 = 7437 m/s
V
Q
 = V
0 
+ V
e
 =  7900 + 463 = 8363 m/s
\ 
KQ
KP
E
E
 = 
2
8363
7437
æö
ç÷
èø
(22) (b)
e
2GM
v i.e.
R
= escape velocity depends upon the
mass and radius of the planet.
(23) (a)
e
2GM8
v RG
R3
= = pr
If mean density is constant then 
e
vR µ
p
ee
e
pp
v
vR 1
v
vR22
= = Þ=
(24) (a)
0
GM
v
r
=
(25) (d)
23
TR µ
2 323 23
;;
e emm
T KRTkRTkR = ==
2
em
RR
R
+
=
3
2/3 2/3
2
1/3 1/3
1
2
em
TT
Tk
kk
éù
Þ = +´ êú
êú
ëû
3/2
2/3 2/3
2
em
TT
T
éù
+
Þ=êú
êú
ëû
(26) (a) –
22
ses
e
e
GMM GMM
E
RR
=-=
     
2
2
2
ee
em e
RE M
RR M
=´
+ æö
ç÷
èø
     
2
e
e
e em
R M
E
M RR
æö
=
ç÷
+ èø
(27) (c) Areal velocity of the artificial planet around the sun
will be more than that of earth.
(28) (a)
0e
e
g
vR
Rh
=
+
For satellite revolving very near to earth 
ee
R hR +=
As ( ) hR <<
5 31
0e
v R g 64 10 10 8 10 m / s 8 kms ;
-
= ´´=´=
Which is independent of height of a satellite.
(29) (d) Due to resistance force of atmosphere, the satellite
revolving around the earth losses kinetic energy.
Therefore in a particular orbit the gravitational
attraction of earth on satellite becomes greater than
that required for circular orbit there. Therefore satellite
moves down to a lower orbit. In the lower orbit as the
potential energy ( ) U GMm/r =- becomes more
negative, Hence kinetic energy 
( )
k
E GMm/2r =
increases, and hence speed of satellite increases.
(30) (a) Because gravitational force is always attractive in nature
and every body is bound by this gravitational force of
attraction of earth.