Calculate the orbital and escape velocities for the Moon. Compare the results obtained with the corresponding velocities for the Earth.

Explanation

From Newton's second law
`(gammaMm)/(R^2)=(mv_0^2)/(R)` or `v_0=sqrt((gammaM)/(R))=1*67km//s` (1)
From conservation of mechanical energy
`1/2mv_e^2-(gammaMm)/(R)=0` or, `v_e=sqrt((2gammaM)/(R))=2*37km//s^2` (2)
In Eq. (1) and (2), M and R are the mass of the moon and its radius. In Eq. (1) if M and R represent the mass of the earth and its radius, then, using appendices, we can easily ge
`v_0=7*9km//s` and `v_c=11*2km//s`.