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Question

Escape velocity at earth's surface is $11.2 k m / s$ . Escape velocity at the surface of a planet having mass $100$ times and radius $4$ times that of earth will be

A

$56 k m / s$

B

$112 k m / s$

C

$305 k m / s$

D

$11.2 k m / s$

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Solution

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Correct option is A)

$V_{e} = 2 g R$
$V_{e} = \frac{2 G M}{R} [∴ g = \frac{G M}{R ^{2}}]$

Now,
$11.1 K m / h r = \frac{2 G M}{R}$
$V_{e}^{1} = \frac{2 G ( 1 0 0 M )}{4 R}$
$\frac{1 1 . 1}{V _{e}^{1}} = \frac{1}{5}$

$V_{e}^{1} = 55.5$ Km/hr

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