The escape velocity from the earth's surface is 11 km,/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be
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a
22 km/sec
b
11 km/sec
c
5-5 km/sec
d
16.5 km/sec
answer is A.
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Detailed Solution
ve=2GMR where M=43πR3d∴ ve=8GπR3d3R=8GπR2d3=8πGd3R∝RNow (v)earth veplanet =(R)earth (R)planet 11ve=R2R or ve=22km/sec