Question

A planet has radius equal to $2R_E$ and density is equal to $4{\rho }_E$. If the value of escape speed at surface of planet is $\alpha$ times $v_E$ ($v_E$ is escape speed on earth, $R_E$ is radius of earth and ${\rho }_E$ is the density of earth) then,

• A. $\alpha =2$
• B. $\alpha =9$
• C. $\alpha =6$
• D. $\alpha =4$

Answer 1

The correct option is D $\alpha =4$

$v_E=\sqrt{\frac{2G{M}_E}{R_E}}=\sqrt{\frac{2G{\rho }_E\times 4\pi R_E^3}{3R_E}}=\sqrt{C{\rho }_E{R}_E^2}$
$\frac{v}{v_E}=\sqrt{\frac{4{\rho }_E\times 4R_E^2}{{\rho }_E{R}_E^2}}=4$
$v=4v_E(\alpha =4)$