MHT CET · Physics · Gravitation
The escape velocity of a body from the surface of the earth is \(11.2 \mathrm{~km} / \mathrm{s}\). The escape velocity of a body from a planet having same mean density as the earth but twice the radius of earth is
- A \(5.5 \mathrm{~km} / \mathrm{s}\)
- B \(33.6 \mathrm{~km} / \mathrm{s}\)
- C \(22.4 \mathrm{~km} / \mathrm{s}\)
- D \(11.2 \mathrm{~km} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(22.4 \mathrm{~km} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
For escape velocity the formula is,
\(v_e=\sqrt{\frac{2 G M}{R}}=\frac{\sqrt{2 G\left(\frac{4}{3} \pi R^3\right) \rho}}{R}=\sqrt{\frac{8}{3} G \pi R^2 \rho}\)
Where, \(\rho\) is density of the planet and \(\mathrm{R}\) is its radius.
\(\begin{aligned} & \therefore v_e=R \sqrt{\frac{8}{3} G \pi \rho} \\ & \therefore v_e \propto R\end{aligned}\)
So, the escape velocity on the planet would be \(22.4 \mathrm{~km} / \mathrm{s}\).
\(v_e=\sqrt{\frac{2 G M}{R}}=\frac{\sqrt{2 G\left(\frac{4}{3} \pi R^3\right) \rho}}{R}=\sqrt{\frac{8}{3} G \pi R^2 \rho}\)
Where, \(\rho\) is density of the planet and \(\mathrm{R}\) is its radius.
\(\begin{aligned} & \therefore v_e=R \sqrt{\frac{8}{3} G \pi \rho} \\ & \therefore v_e \propto R\end{aligned}\)
So, the escape velocity on the planet would be \(22.4 \mathrm{~km} / \mathrm{s}\).
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