Question

Mass of moon is $1/81$ times that of earth and its radius is $1/4$ the radius of earth. If escape velocity on the surface of the earth is $11.2km/s$ . Then the value of escape velocity at surface of the moon will be

• A. 5 km/s
• B. 2.5 km/s
• C. 0.5 km/sec
• D. 0.14 km/ sec

Answer 1

Answer: (B)

Here : Mass of moon $M_m=\frac{M_e}{81}$
Radius of moon $R_m=\frac{R_e}{4}$
Escape velocity on the surface of earth $V_{es(e)}=11.2km/s$
The relation for escape velocity is
$v=\sqrt{\frac{2GM}{R}}\propto \sqrt{\frac{M}{R}}$
Hence $\frac{v_{es}(m)}{v_{es}(e)}=\sqrt{\frac{M_m}{R_m}\times \frac{R_e}{M_e}}$
$=\sqrt{\frac{M_e}{81}\times \frac{4}{R_e}\times \frac{R_e}{M_e}}=\frac{2}{9}$
$v_{es(m)}=\frac{2}{9}\times 11.2=2.5km/s$