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)

$v_p = 2 \,v_E$

Answer 2

Here, $R_P = 2 R_E$ , $\rho_E = \rho_P$
Escape velocity of the earth,
$V_E = \sqrt \frac {2GM_E}{R_E}= {\sqrt {\frac {2G}{R_E} \bigg(\frac {4}{3} \pi R_E^3\rho_E\bigg)}}$ $= R_E \sqrt {\frac {8}{3} \pi G\rho_E }\, \, \, ...(i)$
Escape velocity of the planet
$V_P = \sqrt \frac{2GM_P}{R_P} = {\sqrt {\frac {2G}{R_P} \bigg(\frac {4}{3} \pi R_P^3\rho_P\bigg)}}$ $=R_{P}\sqrt{\frac{8}{3}\pi G\rho_{P}}\ldots\left(ii\right)$
Divide (i) by (ii), we get
$\frac {V_E}{V_P} = \frac {R_E}{R_P} \sqrt \frac {\rho_E}{\rho_P}$
$\frac {V_E}{V_P} = \frac {R_E}{2R_E} \sqrt \frac {\rho_E}{\rho_E} = \frac {1}{2}$
or $V_P = 2V_E$