MHT CET · Physics · Gravitation
If the radius of a planet is 'R' and density ' \(\varrho\) ', then the escape velocity 'Ve' of any body from its surface will be proportional to
- A \(\mathrm{R}\)
- B \(\frac{\sqrt{\varrho}}{\mathrm{R}}\)
- C \(\mathrm{R} \sqrt{\rho}\)
- D \(\frac{\text{R}}{\sqrt\rho}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{R} \sqrt{\rho}\)
Step-by-step Solution
Detailed explanation
\(v_{e}=\sqrt{\frac{G M}{R}}=\sqrt{\frac{G \frac{4}{3} \pi R^{3} \rho}{R}}=k R \sqrt{\rho}\)
\(v_{e} \propto R \sqrt{\rho}\)
\(v_{e} \propto R \sqrt{\rho}\)
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