The escape velocity of a body from the surfac | vedclass

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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Since the mass $M$ of a planet can be expressed in terms of its density $

Vedclass · Accepted Answer

$22.4$

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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Since the mass $M$ of a planet can be expressed in terms of its density $ho$ and radius $R$ as $M = \frac{4}{3} \pi R^3 ho$,we substitute this into the formula:$v_e = \sqrt{\frac{2G(\frac{4}{3} \pi R^3 ho)}{R}} = \sqrt{\frac{8}{3} G \pi R^2 ho} = R \sqrt{\frac{8}{3} G \pi ho}$.From this expression,we see that $v_e \propto R$ when the density $ho$ is constant.Given that the planet has the same density as the earth but twice the radius $(R' = 2R_e)$,the new escape velocity $v_e'$ is:$v_e' = 2 	imes v_e = 2 	imes 11.2 \,km/s = 22.4 \,km/s$.

The escape velocity of a body from the surface of the earth is $11.2 \,km/s$. The escape velocity of a body from a planet having the same mean density as the earth but twice the radius of earth is: (in $\,km/s$)

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