Gravitation Ques 3
3. The ratio of escape velocity at earth $\left(v_e\right)$ to the escape velocity at a planet $\left(v_p\right)$ whose radius and mean density are twice as that of earth is :
[2016]
(a) $1: 2$
(b) $1: 2 \sqrt{2} $
(c) $1: 4$
(d) $1: 2$
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Answer:
Correct Answer: 3.(b)
Solution: (b) As we know, escape velocity,
$ \begin{aligned} & v_e=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}} \cdot\left(\frac{4}{3} \pi \mathrm{R}^3 \rho\right)} \\ & v_e \propto \mathrm{R} \sqrt{\rho} \\ & \therefore \quad \frac{\mathrm{V} _{\mathrm{e}}}{\mathrm{V} _{\mathrm{p}}}=\frac{\mathrm{R} _{\mathrm{e}}}{\mathrm{R} _{\mathrm{p}}} \sqrt{\frac{\rho _{\mathrm{e}}}{\rho _{\mathrm{p}}}} \\ & \Rightarrow \quad \frac{\mathrm{V} _{\mathrm{e}}}{\mathrm{V} _{\mathrm{p}}}=\frac{\mathrm{R} _{\mathrm{e}}}{2 \mathrm{R} _{\mathrm{e}}} \sqrt{\frac{\rho _{\mathrm{e}}}{2 \rho _{\mathrm{e}}}} \\ & \therefore \quad \text { Ratio } \frac{\mathrm{V} _{\mathrm{e}}}{\mathrm{V} _{\mathrm{p}}}=1: 2 \sqrt{2} \end{aligned} $
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