Question

If a planet has mass equal to 16 times the mass of earth, and radius equal to 4 times that of earth. The ratio of escape speed of a planet to that of earth is?

• A. 2:1
• B. 1:2
• C. √2:1
• D. 4:1

Answer 1

The Correct Option is A

Solution:
The escape velocity is given by the formula: $$ v_e = \sqrt{\frac{2GM}{R}} $$ where:

$v_e$ is the escape velocity

$G$ is the gravitational constant

$M$ is the mass of the celestial body

$R$ is the radius of the celestial body

Let:

$M_p$ be the mass of the planet

$R_p$ be the radius of the planet

$M_e$ be the mass of the Earth

$R_e$ be the radius of the Earth

$v_{ep}$ be the escape velocity of the planet

$v_{ee}$ be the escape velocity of the Earth

Given:

$M_p = 16 M_e$

$R_p = 4 R_e$

Escape velocity of Earth: $$ v_{ee} = \sqrt{\frac{2GM_e}{R_e}} $$ Escape velocity of the planet: $$ v_{ep} = \sqrt{\frac{2GM_p}{R_p}} $$ Ratio: $$ \frac{v_{ep}}{v_{ee}} = \frac{\sqrt{\frac{2GM_p}{R_p}}}{\sqrt{\frac{2GM_e}{R_e}}} = \sqrt{\frac{M_p R_e}{M_e R_p}} $$ Substitute the given values: $$ \frac{v_{ep}}{v_{ee}} = \sqrt{\frac{16M_e R_e}{M_e 4R_e}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2 $$ Therefore, the ratio of the escape velocity of the planet to the escape velocity of Earth is 2:1. 

The correct answer is (2) 2:1.