Question:

Gravitational acceleration on the surface of a planet is $\frac{\sqrt6}{11}g$ where g, is the gravitational acceleration oil the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be $11kms^{-1},$ the escape speed on the surface of the planet in $kms^{-1}$ will be

Updated On: Feb 2, 2024
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The Correct Option is A

Solution and Explanation

let the gravitational acceleration on the surface be $g^{\prime}$ and density be $\rho^{\prime}$ $ \frac{ g ^{\prime}}{ g }=\frac{\sqrt{6}}{11} ; \frac{\rho^{\prime}}{\rho}=\frac{2}{3} $ Hence, $\frac{ R ^{\prime}}{ R }=\frac{3 \sqrt{6}}{22}$ $ \begin{array}{l} \frac{ v _{ esc }^{\prime}}{ v _{ esc }} \propto \sqrt{\frac{ R ^{\prime 2} \rho^{\prime}}{ R ^{2} \rho}}=\frac{3}{11} \\ v _{ esc }^{\prime}=3 km / s \end{array} $
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].