Calculate the escape velocity on the surface of the Moon given the mass and radius of the Moon to be `7.34xx10^(22)` kg and `1.74xx10^(6)m` respectively.

Explanation

Given: Mass `(M)=7.34xx10^(22)kg`
Radius `(R)=1.74xx10^(6)m`
Gravitational acceleration on Moon `(g_(m))`
`=1/6g_(e)=9.8/6m//s^(2)=1.63m//s^(2)`
To find: `v_(esc)=?`
Formula : `v_(esc)=sqrt(2g_(m)R)`
Solution: `v_(esc)=sqrt(2xx1.63xx1.74xx10^(6))`
`=sqrt(5.67xx10^(6))`
`=2.38xx10^(3)m//s`
`v_(esc)=2.38km//s`
The escape velocity on Moon is `2.37km//s`