A body is projected vartically upwards from the bottom of a crater of moon of depth `( R)/(100)` where R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body formt eh surface of the moon.

Answer

h=99R

Explanation

Speed of particle at `A, v_(A)` = escape velocity on the surface of moon
`v_(A) = sqrt((2GM)/(R ))`
At highest point `v_(B) = 0`
Mechnical energy is conserved in the process.
`-(GMm)/(R-(R//100))+(1)/(2)mv_(A)^(2) = -(GMm)/(R+h)+0`
or `(1)/(2)mv_(A)^(2)=U_(B)-U_(A)=m(v_(B)-v_(A))` or `(v_(A)^(2))/(2) = V_(B)-V_(A)`
`therefore (GM)/(R )=-(GM)/(R+h) -[-(G)/(R^(3))(1.5R^(2)-0.5(R-(R )/(100))^(2))]`
Solving this equation, we get
`h = 99.5 R`