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 $\alpha =4$

$v_E=\sqrt{\frac{2G{M}_E}{R_E}}=\sqrt{\frac{2G\times {\rho }_E\times 4\pi R_E^3}{3R_E}}$
$\Rightarrow v_E=\sqrt{\frac{8\pi G{\rho }_E{R}_E^2}{3}}$

If radius is $2R_E$ and density is $4{\rho }_E$ then,
$v_E^{{}^{\prime }}=\sqrt{\frac{8\pi G\left(4{\rho }_E\right)(2R_E{)}^2}{3}}$
$\Rightarrow v_E^{{}^{\prime }}=\sqrt{16\times \frac{8\pi G{\rho }_E{R}_E^2}{3}}$
$\Rightarrow v_E^{{}^{\prime }}=4v_E$

Therefore, $\alpha =4$.