The escape velocities of two planets A and B | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The escape velocity is given by $V_e = \sqrt{\frac{2GM}{R}}$.Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G \rho \frac{4}

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(D) The escape velocity is given by $V_e = \sqrt{\frac{2GM}{R}}$.Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G \rho \frac{4}{3}\pi R^3}{R}} = \sqrt{\frac{8}{3}G\pi\rho} \cdot R$.Thus,$V_e \propto \rho^{1/2} R$.Given $\frac{V_{eA}}{V_{eB}} = \frac{1}{2}$ and $\frac{R_A}{R_B} = \frac{1}{3}$.From $\frac{V_{eA}}{V_{eB}} = \sqrt{\frac{\rho_A}{\rho_B}} \cdot \frac{R_A}{R_B}$,we get $\frac{1}{2} = \sqrt{\frac{\rho_A}{\rho_B}} \cdot \frac{1}{3}$,which implies $\sqrt{\frac{\rho_A}{\rho_B}} = \frac{3}{2}$,so $\frac{\rho_A}{\rho_B} = \frac{9}{4}$.The acceleration due to gravity is $g = \frac{GM}{R^2} = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}G\pi\rho R$.Therefore,$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B} = \left(\frac{\rho_A}{\rho_B}\right) \left(\frac{R_A}{R_B}\right) = \left(\frac{9}{4}\right) \left(\frac{1}{3}\right) = \frac{3}{4}$.

The escape velocities of two planets $A$ and $B$ are in the ratio $1:2$. If the ratio of their radii respectively is $1:3$,then the ratio of acceleration due to gravity of planet $A$ to the acceleration due to gravity of planet $B$ will be:

Similar Questions

An iron ball and a wooden ball of the same radius are released from the same height in vacuum. They take the same time to reach the ground. The reason for this is

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is [ $R$ = radius of earth,$g$ = acceleration due to gravity,$n=$ integer].

If the radius of the Earth decreases without any change in its mass,will the value of $g$ on its surface increase or decrease?

Vedclass Test Series

Exam Paper Generator

Online Exam Module