Q160 / 200 · Manipal (MET) 2011 · Physics

Escape velocity from earth's surface is $11 \mathrm{~km} / \mathrm{s}$. If radius of a planet is double than that of earth and density is same as that of earth, then the escape velocity from this planet will be

A
$5.5 \mathrm{~km} / \mathrm{s}$
B
$11 \mathrm{~km} / \mathrm{s}$
C
$16.5 \mathrm{~km} / \mathrm{s}$
Correct answer
D
$22 \mathrm{~km} / \mathrm{s}$

Solution

Escape velocity $\begin{aligned} v_e & =\sqrt{\frac{2 G M}{R}} \\ & =R \sqrt{\frac{8}{3} G \rho}\end{aligned}$ $\therefore \quad v \propto R$ We can say, if radius of a planet is double than that of earth then escape velocity from this planet will be $22 \mathrm{~km} / \mathrm{s}$.

Hence, the correct answer is option: 'D'.

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