The mass of a planet is half that of the eart | vedclass

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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_

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$15.55$

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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.Given: $M_p = \frac{M_e}{2}$ and $R_p = \frac{R_e}{4}$.The escape velocity on the planet $v_p$ is given by $v_p = \sqrt{\frac{2GM_p}{R_p}}$.Taking the ratio of escape velocities: $\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} 	imes \frac{R_e}{R_p}}$.Substituting the values: $\frac{v_p}{v_e} = \sqrt{\frac{1/2}{1} 	imes \frac{1}{1/4}} = \sqrt{\frac{1}{2} 	imes 4} = \sqrt{2}$.Therefore,$v_p = v_e 	imes \sqrt{2} = 11 	imes 1.414 = 15.55 \ km \ s^{-1}$.

The mass of a planet is half that of the earth and the radius of the planet is one-fourth that of earth. If we plan to send an artificial satellite from the planet,the escape velocity will be,(escape velocity on earth $v_e = 11 \ km \ s^{-1}$) (in $km \ s^{-1}$)

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