MHT CET · Physics · Gravitation
The escape velocity of a body from any planet whose mass is six times the mass of earth and radius is twice the radius of earth will be
\(\left(\mathrm{V}_{e}=\right.\) escape velocity of a body from the earth's surface)
- A \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
- B \(2 \mathrm{~V}_{\mathrm{e}}\)
- C \(\frac{3}{2} \mathrm{~V}_{\mathrm{e}}\)
- D \(2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
(D)
\(\begin{aligned} \mathrm{V}_{\mathrm{e}} &=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ \mathrm{V}_{\mathrm{e}}^{\prime} &=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}}{2 \mathrm{R}}} \\ &=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}} \times 3} \\ &=\sqrt{3} \mathrm{~V}_{\mathrm{e}} \end{aligned}\)
\(\begin{aligned} \mathrm{V}_{\mathrm{e}} &=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ \mathrm{V}_{\mathrm{e}}^{\prime} &=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}}{2 \mathrm{R}}} \\ &=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}} \times 3} \\ &=\sqrt{3} \mathrm{~V}_{\mathrm{e}} \end{aligned}\)
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