Question

The mass of the Moon is 0.0123 time the mass of the Earth and its radius is about 0.272 times the radius of the Earth.Calculate the escape velocity on the Moon. Show any work that you do.

Answer 1

Given data:

* The mass of the Moon is 0.0123 times of mass of Earth as,

`\begin{gathered} M^(\prime)=0.0123* M_E \\ M^(\prime)=0.0123*5.97*10^(24) \\ M^(\prime)=0.0734*10^(24)\text{ kg} \end{gathered}`

where M_E is the mass of Earth,

* The radius of the Moon is 0.272 times the mass of Earth as,

`\begin{gathered} R^(\prime)=0.272* R_E \\ R^(\prime)=0.272*6400*10^3 \\ R^(\prime)=1740.8*10^3\text{ m} \end{gathered}`

where R_E is the radius of Earth,

Solution:

The escape velocity of the object from the surface of the Moon is,

`v_e=\sqrt[]{(2GM^(\prime))/(R^(\prime))}`

where G is the gravitational constant, M' is the mass of Moon, and R' is the radius of Moon,

Substituting the known values,

`\begin{gathered} v_e=\sqrt[]{(2*6.67*10^(-11)*0.0734*10^(24))/(1740.8*10^3)} \\ v_e=0.0237*10^5ms^(-1) \\ v_e=2.37*10^3ms^(-1) \\ v_e=2.37kms^(-1) \end{gathered}`

Thus, the escape velocity on the Moon is 2.37 Kilometers per second.