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The radius of planet is twice the radius of earth. Both have almost equal averge mass densities $V_P$ and $V_E$ are escape velocities of the planet and the earth respectively, then:
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Verified Answer
The correct answer is:
$V_P=2 \mathrm{~V}_{\mathrm{E}}$
According to the question
$$
R_P=2 R_E, \rho_E=\rho_P
$$
Escape velocity of earth
$$
\begin{aligned}
V_E & =\sqrt{\frac{2 G M_E}{R_E}}=\sqrt{\frac{2 G}{R_E}\left(\frac{4}{3} \pi R_E^3 \rho_E\right)} \\
& =R_E \sqrt{\frac{8}{3} \pi G \rho_E}
\end{aligned}
$$
Similarly
Escape velocity of planet
$$
\begin{aligned}
V_P & =\sqrt{\frac{2 G M_P}{R_P}}=\sqrt{\frac{2 G}{R_P}\left(\frac{4}{3} \pi R_p{ }^3 \rho_p\right)} \\
& =R_P \sqrt{\frac{8}{3} \pi G \rho_p}
\end{aligned}
$$
from eq. (i) and eq. (ii)
$$
\frac{V_E}{V_P}=\frac{R_E}{R_P} \sqrt{\frac{\rho_E}{\rho_P}}=\frac{R_E}{2 R_E} \sqrt{\frac{\rho_Q}{\rho_E}}=\frac{1}{2}
$$
(given)
$$
\Rightarrow V_p=2 V_E
$$
$$
R_P=2 R_E, \rho_E=\rho_P
$$
Escape velocity of earth
$$
\begin{aligned}
V_E & =\sqrt{\frac{2 G M_E}{R_E}}=\sqrt{\frac{2 G}{R_E}\left(\frac{4}{3} \pi R_E^3 \rho_E\right)} \\
& =R_E \sqrt{\frac{8}{3} \pi G \rho_E}
\end{aligned}
$$
Similarly
Escape velocity of planet
$$
\begin{aligned}
V_P & =\sqrt{\frac{2 G M_P}{R_P}}=\sqrt{\frac{2 G}{R_P}\left(\frac{4}{3} \pi R_p{ }^3 \rho_p\right)} \\
& =R_P \sqrt{\frac{8}{3} \pi G \rho_p}
\end{aligned}
$$
from eq. (i) and eq. (ii)
$$
\frac{V_E}{V_P}=\frac{R_E}{R_P} \sqrt{\frac{\rho_E}{\rho_P}}=\frac{R_E}{2 R_E} \sqrt{\frac{\rho_Q}{\rho_E}}=\frac{1}{2}
$$
(given)
$$
\Rightarrow V_p=2 V_E
$$
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