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If the acceleration due to gravity $g$ doubles and the radius of earth becomes half that of the present value, then the value of escape velocity is (Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ and radius of earth, $R=6400 \mathrm{~km}$ )
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The correct answer is:
$12 \mathrm{~km} / \mathrm{s}$
Given, radius of earth, $R_e=6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m}$ and acceleration of gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$
$\therefore$ Escape velocity of the surface of earth,
$$
v_e=\sqrt{2 g R_e}=\sqrt{2 \times 10 \times 6.4 \times 10^6}=11.3 \times 10^3 \mathrm{~m} / \mathrm{s}
$$
Again, when $g^{\prime}=2 g$
$$
R^{\prime}=\frac{R}{2}
$$
Now, escape velocity is given by
$$
\begin{aligned}
\therefore \quad v_e^{\prime} & =\sqrt{2 g^{\prime} R^{\prime}}=\sqrt{2 \times 2 g \times \frac{R}{2}}=\sqrt{2 g R}=v_e \\
& =11.3 \times 10^3 \mathrm{~m} / \mathrm{s}=11.3 \mathrm{~km} / \mathrm{s} \approx 12 \mathrm{~km} / \mathrm{s}
\end{aligned}
$$
$\therefore$ Escape velocity of the surface of earth,
$$
v_e=\sqrt{2 g R_e}=\sqrt{2 \times 10 \times 6.4 \times 10^6}=11.3 \times 10^3 \mathrm{~m} / \mathrm{s}
$$
Again, when $g^{\prime}=2 g$
$$
R^{\prime}=\frac{R}{2}
$$
Now, escape velocity is given by
$$
\begin{aligned}
\therefore \quad v_e^{\prime} & =\sqrt{2 g^{\prime} R^{\prime}}=\sqrt{2 \times 2 g \times \frac{R}{2}}=\sqrt{2 g R}=v_e \\
& =11.3 \times 10^3 \mathrm{~m} / \mathrm{s}=11.3 \mathrm{~km} / \mathrm{s} \approx 12 \mathrm{~km} / \mathrm{s}
\end{aligned}
$$
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