Question

Determine the escape velocity of a rocket on the far side of a moon of a planet. The radius of the moon is $2.64\times {10}^6$m and its mass is $1.495\times {10}^{23}$. The mass of the planet is $1.9\times {10}^{27}$kg, and the distance between planet and the moon is $1.071\times {10}^9$m. Include the gravitational effect of planet and neglect the motion of the planet and the moon as they rotate about their CM.

• A. $1.560\times {10}^{2}m{s}^{-1}$
• B. $1.560\times {10}^{3}m{s}^{-1}$
• C. $2.8\times {10}^{4}m{s}^{-1}$
• D. $1.560\times {10}^{4}m{s}^{-1}$

Answer 1

The correct option is D

$1.560\times {10}^{4}m{s}^{-1}$

Total potential energy of the rocket is

$U=-G[\frac{M_{p}m}{(d+R_m)}+\frac{M_{m}m}{R_m}]$

If $v_e$ is the escape velocity,we can write

$\frac{1}{2}m{v}_e^2=-U$

$v_e^2=2G(\frac{M_p}{(d+R_m)}+\frac{M_m}{R_m})$

= $2\times 6.67\times {10}^{-11}(\frac{1.90\times {10}^{27}}{1.071\times {10}^9+2.64\times {10}^6}+\frac{1.495\times {10}^{23}}{2.64\times {10}^6})$

= $2.436\times {10}^8$

$v_e=1.560\times {10}^{4}m{s}^{-1}$