Let us find force of attraction between two bodies lying 1 m apart Let the mass of each body
is 50 kg
Answers
Answer:
Two bodies of masses m
1
and m
2
are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is
A
[2G
r
(m
1
−m
2
)
]
1/2
B
[
r
2G
(m
1
+m
2
)]
1/2
C
[
2G(m
1
m
2
)
r
]
1/2
D
[
r
2G
m
1
m
2
]
1/2
HARD
Answer
By applying law of conservation of momentum,
m
1
v
1
−m
2
v
2
=0⇒m
1
v
1
=m
2
v
2
........ (i)
Where v
1
and v
2
are the velocities of masses m
1
and m
2
at a distance r from each other.
By conservation of energy,
Change in P.E= change in K.E.
r
Gm
1
m
2
=
2
1
m
1
v
1
2
+
2
1
m
2
v
2
2
........ (ii)
Solving eqn. (i) and (ii) we get
v
1
=
r(m
1
+m
2
)
2Gm
2
2
and v
2
=
r(m
1
+m
2
)
2Gm
1
2
Relative velocity of approach, v
R
=∣v
1
∣+∣v
2
∣=
r
2G
(m
1
+m
2
)
Answer ByavatarToppr
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