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Question and Answer

Escape velocity of a body from the surface of earth is \(11\ldotp 2\;km/sec\). If the mass of earth becomes double of its present mass and radius becomes half of its present radius then escape velocity will become ( )
A. 5.6 km/s
B. 11.2 km/s (remains unchanged)
C. 22.4 km/s
D. 44.8 km/s

Answer

C
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Solution

Given: M'=2M and R'=R/2
Escape velocity, \(v=\sqrt{\dfrac{2GM}{R}}\)
\(v\text{'}=\sqrt{\dfrac{2G\times 2M}{R/2}}=\sqrt{\dfrac{4\times 2GM}{R}}=2v=2\times 11\ldotp 2=22\ldotp 4\;km/s\)
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