A body is projected from the surface of the earth with thrice the escape velocity `(V_(e))` from the surface of the earth. What will be its velocity, when it will escape the gravitational pull?

Options

A. `V_(e)`
B. `3V_(e)`
C. `sqrt(2 ) V_(e)`
D. `2 sqrt(2) V_(e)`

Answer

D

Explanation

Ifv be the velocity in the inter planetary space and `U_(i) ,U_(f)` initial and final PE
Then, by the law of conservation of energy,
`U_(i) +K_(i) = U_(f) +K_(f)`
`(-GM_(e)m)/(R_(e)) +(1)/(2) m (3v_(e))^(2) =0 +(1)/(2) mv^(2)`
`(-GM_(e))/(R_(e)) +(9)/(2) v_(e)^(2) = (1)/(2) v^(2)`
`implies (-2GM_(e))/(R_(e)) + 9V_(e)^(2) =v^(2)`
`implies (-2 gR_(e)^(2))/(R_(e)) + 9v_(e)^(2) =v^(2) " " [ because GM_(e) = gR_(e)^(2)]`
`-2 g R_(e) + 9xx 2 = gR_(e) =v^(2) , " " [ because v_(e)^(2) = 2g R_(e)]`
`16 g R_(e) =v^(2) `
`16 gR_(e) =v^(2)`
`8.2 gR_(e) =V^(2)`
`implies v = sqrt(8). sqrt(2gR_(e)) = 2sqrt(2) v_(e)`