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  4. The escape velocity of a body on the surface of the earth is 11.2 km/s-1. If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

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Q. The escape velocity of a body on the surface of the earth is $11.2\, km/s^{-1}$. If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

AIPMTAIPMT 1997Gravitation

Solution:

Escape velocity of a body $(v_e) = 11.2 \,km/s$;
New mass of the earth $ {M'}_e = 2 M_e $ and new radius
of the earth $ {R'}_e = 0.5 R_e .$
Escape velocity $ (v_e) = \sqrt{\frac{ 2GM_e}{ R_e}} \propto \sqrt{\frac{ M_e}{ R_e}} $.
Therefore $ \frac{ v_e}{{v'}_e} = \sqrt{ \frac{ M_e}{ R_e} \times \frac{ 0.5 R_e}{ 2 M_e}} = \sqrt {\frac{1}{4} } = \frac{1}{2} $
or, $ {v'}_e = 2 v_e =22.4\, km / sec $.