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

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

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$15.55$

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(C) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.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 ratio of escape velocities is $\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} 	imes \frac{R_e}{R_p}}$.Substituting the values: $\frac{v_p}{v_e} = \sqrt{\frac{M_e/2}{M_e} 	imes \frac{R_e}{R_e/4}} = \sqrt{\frac{1}{2} 	imes 4} = \sqrt{2}$.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 the 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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