Question

The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. $v_P$ and $v_E$ are escape velocities of the planet and the earth, respectively, then

• A. $v_p = 1.5 \,v_E$
• B. $v_p = 2 \,v_E$
• C. $v_E = 3\, v_p$
• D. $V_E = 1.5\, V_p$

Answer 1

Answer: (B)

Here, $R_{P}=2 R_{E}, \rho_{E}=\rho_{P}$
Escape velocity of the earth,
$V_{E}=\sqrt{\frac{2 G M_{E}}{R_{E}}}=\sqrt{\frac{2 G}{R_{E}}\left(\frac{4}{3} \pi R_{E}^{3} \rho_{E}\right)}$
$=R_{E} \sqrt{\frac{8}{3} \pi G \rho_{E}} \ldots$ (i)
Escape velocity of the planet
$V_{P}=\sqrt{\frac{2 G M_{p}}{R_{p}}}=\sqrt{\frac{2 G}{R_{p}}\left(\frac{4}{3} \pi R_{p}^{3} \rho_{p}\right)}$
$=R_{p} \sqrt{\frac{8}{3} \pi G_{p}} .$
Divide (i) by (ii), we get
$\frac{V_{E}}{V_{P}}=\frac{R_{E}}{R_{P}} \sqrt{\frac{\rho_{E}}{\rho_{P}}}$
$\frac{V_{E}}{V_{P}}=\frac{R_{E}}{2 R_{E}} \sqrt{\frac{\rho_{E}}{\rho_{E}}}=\frac{1}{2}$
or $V_{P}=2 V_{E}$