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Q.

The escape velocity of an object from a planet whose mass is 9 times that of the earth and radius is 4 times that of the earth in $k m s^{- 1}$ is

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a

16.8

b

33.6

c

25.2

d

67.2

answer is C.

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Detailed Solution

v_{e} = \sqrt{\frac{2GM}{R}}

$\frac{v_{P}}{v_{e}} = \sqrt{\frac{M_{p}}{M_{e}} \times \frac{R_{e}}{R_{e}}}$

$\frac{v_{p}}{v_{e}} = \sqrt{\frac{{9M}_{e}}{M_{e}} \times \frac{R_{e}}{{4R}_{e}}} = \sqrt{\frac{9}{4}}$

$v_{p} = \sqrt{\frac{G \times 9M}{4R}}$

$v_{p} = \sqrt{\frac{9}{4} \frac{GM}{R}} = \frac{3}{2} v_{e}$

$v_{p} = \frac{3}{2} \times v_{e}$

$v_{p} = \frac{3}{2} \times 11.2 = 16.8 km s^{- 1}$

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