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Q.

Mass of planet is ${\frac{1}{10}}^{t h}$ of the mass of the earth. If the escape velocity from the surface of the planet is $\frac{1}{\sqrt{2}}$ times that from the earth, the radius of that planet in terms of earth’s radius R is

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a

$5 R$

b

$\frac{R}{5}$

c

$2 R$

d

$\frac{2}{R}$

answer is B.

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Detailed Solution

Escape velocity

$V_{e} = \sqrt{\frac{2 G M}{R}} \frac{V_{e}}{V_{p}} = \sqrt{2} = \sqrt{\frac{M_{E} R_{P}}{M_{P} R_{E}}} = \sqrt{10 \frac{R_{P}}{R_{E}}}$

R_E=R/5

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Mass of planet is 110th of the mass of the earth. If the escape velocity from the surface of the planet is 12 times that from the earth, the radius of that planet in terms of earth’s radius R is