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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
detailed solution
Correct option is A
ve=2GMR where M=43πR3d∴ ve=8GπR3d3R=8GπR2d3=8πGd3R∝RNow (v)earth veplanet =(R)earth (R)planet 11ve=R2R or ve=22km/secTalk to our academic expert!
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