First slide

Gravitaional potential energy

Question

The ratio of escape velocity at the surface of earth to the escape velocity at the surface of a planet whose radius and mean density are twice as that of earth is

Easy

A

1 : 2
B

1 : 2√2
C

1 : 4
D

1 : √2

Solution

As escape velocity from earth
$\large {V_e} = \sqrt {\frac{{2GM}}{R}} = \sqrt {\frac{{2G}}{R}\frac{4}{3}\pi {R^3}d}$
$\large = \sqrt {\frac{{8\pi G{R^2}d}}{3}} \,\,\, \Rightarrow {V_e}\alpha \sqrt {{R^2}d}$
$\large \frac {V_1}{V_2}=\sqrt {\left ( \frac {R_1}{R_2} \right )^2\frac {d_1}{d_2}}$
$\large =\sqrt {\left ( \frac {R}{2R} \right )^2\frac {d}{2d}}=\frac {1}{\sqrt 8}=\frac {1}{2\sqrt 2}$

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