The escape velocity for the Earth is 11 km/s | vedclass

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(A) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Since $M = \rho V = \rho (\frac{4}{3}\pi R^3)$,we substitute this into th

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$22$

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(A) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Since $M = ho V = ho (\frac{4}{3}\pi R^3)$,we substitute this into the formula:$v_e = \sqrt{\frac{2G(\frac{4}{3}\pi ho R^3)}{R}} = \sqrt{\frac{8}{3}\pi ho G R^2} = R \sqrt{\frac{8}{3}\pi ho G}$.Given that the density $ho$ is the same,we have $v_e \propto R$.Let $v_{e1} = 11 \ km/s$ and $R_1 = R$. For the new planet,$R_2 = 2R$.Therefore,$\frac{v_{e2}}{v_{e1}} = \frac{R_2}{R_1} = \frac{2R}{R} = 2$.$v_{e2} = 2 	imes v_{e1} = 2 	imes 11 \ km/s = 22 \ km/s$.

The escape velocity for the Earth is $11 \ km/s$. The escape velocity for a planet having twice the radius and the same density as the Earth is .......... $km/s$.

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