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((2GM)/( R))`
B. `sqrt((GM)/( R))`
C. `sqrt((3GM)/( R))`
D. `sqrt((5GM)/(4R))`

Answer

B

Explanation

Applying the principle of conservation of energy,
`E_("surface of the earth")=E_("at a height h")` equal to R
`(1)/(2)mv^(2)-(GMm)/(R )=(1)/(2)m(0)^(2)-(GMm)/(R+R)`
[`because` The velocity of the highest point = 0]
`therefore (1)/(2)mv^(2)=(GMm)/(R )-(GMm)/(2R)=(GMm)/(2R)`
`v^(2)=(GM)/(R )` or `v=sqrt((GM)/(R ))`