Gravitation Question 140

Question: The escape velocity for a rocket from earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in km/sec[NCERT 1983; CPMT 1990; MP PMT 2000; UPSEAT 1999]

Options:

A) 11.2

B) 5.6

C) 22.4

D) 53.6

Show Answer

Answer:

Correct Answer: C

Solution:

$ [\frac{v_{p}}{v_{e}}=\sqrt{\frac{g_{p}}{g_{e}}\times \frac{R_{p}}{R_{e}}}] $ = $ [\sqrt{2\times 2}=2] $ Þ $ [v_{p}=2\times v_{e}=2\times 11.2=22.4\ km/s] $



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