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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2023
  • Medium
  • MCQ

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 of gravity of planet \(B\) will be:

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Answer & Solution

Correct Answer

(D) \(\frac{3}{4}\)

Step-by-step Solution

Detailed explanation

\(V _{ e }=\sqrt{\frac{2 GM }{ R }}=\sqrt{\frac{2 G \rho \frac{4}{3} \pi R ^{ S }}{ R }}= C \sqrt{\rho} \cdot R\)

\(\frac{ V _{ e _1}}{ V _{ e _2}}=\frac{ R _1}{ R _2} \sqrt{\frac{\rho_1}{\rho_2}}=\frac{1}{2}\)

\(\frac{ R _1^2}{ R _2^2} \times \frac{\rho_1}{\rho_2}=\frac{1}{4}\)

\(\frac{ R _1}{ R _2}=\frac{1}{3}\)

\(g =\frac{ GM }{ R ^2}=\frac{ G \frac{4}{3} \pi R ^3 \times \rho}{ R ^2} C \cdot \rho R\)

\(\frac{ g _1}{ g _2}=\frac{\rho_1 R _1}{\rho_2 R _2}=\frac{1}{4} \frac{ R _2^2}{ R _1^2} \times \frac{ R _1}{ R _2}\)

\(=\frac{1}{4} \times \frac{ R _2}{ R _1}=\frac{3}{4}\)

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