Q154 / 180 · COMEDK 2013 · Physics
The radius of a planet is twice the radius of earth. Both have almost equal average
mass-densities. $v_{p}$ and $v_{e}$ are escape velocities of the planet and the earth, respectively, then
- A$v_{p}=1.5 v_{e}$
- Correct answerB$v_{p}=2 v_{e}$
- C$v_{e}=3 v_{p}$
- D$v_{e}=1.5 v_{p}$
Solution
The escape velocity of earth,
$$
v_{e}=\sqrt{\frac{2 G M_{e}}{R_{e}}}=R_{e} \sqrt{\frac{8}{3} \pi G \rho_{e}}
$$
Similarly for a planet,
$$
v_{p}=R{ }_{p} \sqrt{\frac{8}{3} \pi G \rho_{p}}
$$
$$
\therefore \quad \frac{v_{e}}{v_{p}}=\frac{R_{e}}{R_{p}} \sqrt{\frac{\rho_{e}}{\rho_{n}}}
$$
Here,
$$
R_{p}=2 R_{e}, \rho_{e}=\rho_{p} \Rightarrow \frac{v_{e}}{v_{p}}=\frac{1}{2} \text { or } v_{p}=2 v_{e}
$$
Hence, the correct answer is option: 'B'.