MHT CET · Physics · Gravitation
The escape velocity for a planet whose mass is six times the mass of earth and the radius is twice the radius of earth will be \(\left[\mathrm{V}_{\mathrm{e}}=\right.\) escape velocity from the earth \(]\)
- A \(\sqrt{2} \mathrm{~V}_{\mathrm{e}}\)
- B \(\frac{1}{2} \mathrm{~V}_{\mathrm{e}}\)
- C \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
- D \(2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
Escape velocity is defined using relation:
\(\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2=\frac{\mathrm{GMm}}{\mathrm{R}} \\ & \Rightarrow \mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\end{aligned}\)
If \(M\) is replaced \(6 M \& R\) is replaced by \(2 R\) then
\(\mathrm{V}_{\mathrm{e}}^{\prime}=\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
\(\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2=\frac{\mathrm{GMm}}{\mathrm{R}} \\ & \Rightarrow \mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\end{aligned}\)
If \(M\) is replaced \(6 M \& R\) is replaced by \(2 R\) then
\(\mathrm{V}_{\mathrm{e}}^{\prime}=\sqrt{3} \mathrm{~V}_{\mathrm{e}}\)
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