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JEE Main Physics Class 11 Gravitation Part 1 Questions
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© examsnet.com
Question : 57
Total: 100
A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is:
[10 Jan. 2019 I]
2
m
v
2
m
v
2
1
2
m
v
2
3
2
m
v
2
Validate
Solution:
At height r from center of earth, orbital velocity
v
=
√
G
M
r
By principle of energy conservation
K
E
of
c
m
′
+
(
−
G
M
m
r
)
=
0
+
0
(
∵
At infinity
,
P
E
=
K
E
=
0
)
or
K
E
of '
m
′
=
G
M
m
r
=
(
√
G
M
r
)
2
m
=
m
v
2
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