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If the radius of a planet is 'R' and density ' $\varrho$ ', then the escape velocity 'Ve' of any
body from its surface will be proportional to
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body from its surface will be proportional to
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The correct answer is:
$\mathrm{R} \sqrt{\rho}$
$v_{e}=\sqrt{\frac{G M}{R}}=\sqrt{\frac{G \frac{4}{3} \pi R^{3} \rho}{R}}=k R \sqrt{\rho}$
$v_{e} \propto R \sqrt{\rho}$
$v_{e} \propto R \sqrt{\rho}$
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