Respuesta :
Given data:
* The mass of the Moon is 0.0123 times of mass of Earth as,
[tex]\begin{gathered} M^{\prime}=0.0123\times M_E \\ M^{\prime}=0.0123\times5.97\times10^{24} \\ M^{\prime}=0.0734\times10^{24}\text{ kg} \end{gathered}[/tex]where M_E is the mass of Earth,
* The radius of the Moon is 0.272 times the mass of Earth as,
[tex]\begin{gathered} R^{\prime}=0.272\times R_E \\ R^{\prime}=0.272\times6400\times10^3 \\ R^{\prime}=1740.8\times10^3\text{ m} \end{gathered}[/tex]where R_E is the radius of Earth,
Solution:
The escape velocity of the object from the surface of the Moon is,
[tex]v_e=\sqrt[]{\frac{2GM^{\prime}}{R^{\prime}}}[/tex]where G is the gravitational constant, M' is the mass of Moon, and R' is the radius of Moon,
Substituting the known values,
[tex]\begin{gathered} v_e=\sqrt[]{\frac{2\times6.67\times10^{-11}\times0.0734\times10^{24}}{1740.8\times10^3}} \\ v_e=0.0237\times10^5ms^{-1} \\ v_e=2.37\times10^3ms^{-1} \\ v_e=2.37kms^{-1} \end{gathered}[/tex]Thus, the escape velocity on the Moon is 2.37 Kilometers per second.
