MHT CET · Physics · Gravitation
The mass of a planet is six time that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is ' \(\mathrm{V}_{\mathrm{e}}\) ', then the escape velocity from the planet is
- A \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
- B \(\sqrt{2} \mathrm{~V}_{\mathrm{e}}\)
- C \(\mathrm{V}_{\mathrm{e}}\)
- D \(\sqrt{5} \mathrm{~V}_{\mathrm{e}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & V_e=\sqrt{\frac{2 G M}{R}} ; V_p=\sqrt{\frac{2 \mathrm{GM}_p}{R_p}} \\ & \therefore \frac{V_p}{V_e}=\sqrt{\frac{M_p}{M}} \cdot \frac{R}{R_p}=\sqrt{6 \times \frac{1}{2}}=\sqrt{3} \\ & \therefore V_p=\sqrt{3} V_e\end{aligned}\)
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