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Manipal Entrance Test 2013 Physics Paper
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© examsnet.com
Question : 22
Total: 60
Escape velocity at surface of earth is 11.2 Km/s. Escape velocity from a planet whose mass is the same as that of earth and radius
1
4
that of earth, is
2.8 km/s
15.6 km/s
22.4 km/s
44.8 km/s
Validate
Solution:
At a certain velocity of projection the body will go out of the gravitational field of the earth and will never return to the earth, this velocity is known as escape velocity.
v
e
=
√
2
G
M
e
R
e
Given,
M
e
=
M
p
,
R
p
=
R
e
4
∴
v
ρ
v
e
=
√
M
e
M
e
×
R
e
R
e
∕
4
=
√
4
=
2
⇒
v
p
=
2
v
e
=
2
×
11.2
=
22.4
km
∕
s
© examsnet.com
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