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?
- 2:1
- 1:2
- √2:1
- 4:1
The Correct Option is A
Solution and Explanation
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.
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