JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2026
- Medium
- MCQ
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is __________ m/s.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \( 100\sqrt{10} \)
Step-by-step Solution
Detailed explanation
\( V_{e}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2G\times\rho\times\frac{4\pi R^{3}}{3}}{R}}\Rightarrow V_{e}\propto\sqrt{\rho}\times R \)
\( \frac{(V_{e})_{B}}{(V_{e})_{A}}=\sqrt{\frac{\rho_{B}}{P_{A}}}\times\frac{R_{B}}{R_{A}}=\sqrt{\frac{0.1P_{A}}{P_{A}}}\times(\frac{0.1R_{A}}{R_{A}}) \)
\( \frac{(V_{e})_{B}}{(V_{e})_{A}}=\frac{1}{10}\times\frac{1}{\sqrt{10}} \)
\( (V_{e})_{B}=\frac{10\times1000}{10\sqrt{10}}=100\sqrt{10}m/sec \)
\( \frac{(V_{e})_{B}}{(V_{e})_{A}}=\sqrt{\frac{\rho_{B}}{P_{A}}}\times\frac{R_{B}}{R_{A}}=\sqrt{\frac{0.1P_{A}}{P_{A}}}\times(\frac{0.1R_{A}}{R_{A}}) \)
\( \frac{(V_{e})_{B}}{(V_{e})_{A}}=\frac{1}{10}\times\frac{1}{\sqrt{10}} \)
\( (V_{e})_{B}=\frac{10\times1000}{10\sqrt{10}}=100\sqrt{10}m/sec \)
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