A body projected from the surface of the earth attains a height equal to the radius of the earth. The velocity with which the body was projected is

Options

A. `sqrt((GM_(e))/(R))`
B. `sqrt((2GM_(e))/(R))`
C. `sqrt((5)/(4)(GM_(e))/(R))`
D. `sqrt((3GM_(e))/(R))`

Answer

A

Explanation

By applying conservation of energy
`KE_(1)+PE_(1)=KE_(f)+PE_(f)`
`(1)/(2)mv^(2)=-(GM_(e)m)/(R)=0-(GM_(e)m)/(2R)`
`(1)/(2)mv^(2)=(GM_(e)m)?(R)[-(1)/(2)+1]`
`(1)/(2)mv^(2)=(GM_(e)m)/(2R)impliesu=sqrt((GM_(e))/(R))`