Question

Solve the problem.

If the mass of a planet is eight times the mass of the Earth and its radius is twice the radius of the Earth, what will be the escape velocity for that planet?

Answer 1

Escape velocity for Earth is

$$v_{esc}=\sqrt{\frac{2G{M}_e}{R_e}}=11.2km/s$$

Given:
Mass of the planet, M_p = 8 ${\displaystyle \times }$ Mass of the Earth (M_e) 
Radius of the planet, R_p = 2 ${\displaystyle \times }$Radius of the Earth (R_e)
Thus, the escape velocity of the planet is

$$v_{esc}^{\prime }=\sqrt{\frac{2G{M}_p}{R_p}}=\sqrt{\frac{8}{2}}\sqrt{\frac{2G{M}_e}{R_e}}=\sqrt{4}\times 11.2=22.4km/s$$

Answer 2

Given:

Mass of the planet, M_P = 8 M_E
Radius of the planet, R_P = 2 R_E
Escape velocity for the Earth, V_escE = 11.2 km/s
Escape velocity for the planet, V_escP =? 

${\displaystyle V_{\text{escP}}=\sqrt{\frac{2{\text{GM}}_{\text{p}}}{{\text{R}}_{\text{p}}}}}$

=${\displaystyle \sqrt{\frac{2\text{G}\left(8{\text{M}}_{\text{E}}\right)}{2{\text{R}}_{\text{E}}}}}$

${\displaystyle =\sqrt{\frac{8}{2}\times \frac{2{\text{GM}}_{\text{E}}}{{\text{R}}_{\text{E}}}}}$

${\displaystyle =\sqrt{\frac{8}{2}}\times \sqrt{\frac{2{\text{GM}}_{\text{E}}}{{\text{R}}_{\text{E}}}}}$

${\displaystyle =\sqrt{4}\times {\text{V}}_{\text{escE}}}$    

∵ ${\displaystyle {\text{V}}_{\text{escE}}=\sqrt{\frac{2{\text{GM}}_{\text{E}}}{{\text{R}}_{\text{E}}}}}$

${\displaystyle =2\times 11.2}$

${\displaystyle {\text{V}}_{\text{escP}}}$ = 22.4 Km/s

Escape velocity for the planet = 22.4 km/s