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 $11.2\,km/s$ . 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)

2.5 km/s

Answer 2

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 $V_{e s(e)}=11.2\, km / s$ The relation for escape velocity is $v=\sqrt{\frac{2 G M}{R}} \propto \sqrt{\frac{M}{R}}$ Hence $\frac{v_{e s}(m)}{v_{e s}(e)}=\sqrt{\frac{M_{m}}{R_{m}} \times \frac{R_{e}}{M_{e}}}$ $=\sqrt{\frac{M_{e}}{81} \times \frac{4}{R_{e}} \times \frac{R_{e}}{M_{e}}}=\frac{2}{9}$ $v_{e s(m)}=\frac{2}{9} \times 11.2=2.5\, km / s$