A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reache its maximum height, its acceleration due to the planet's gravity is `1//4^(th)` of its value at the surface of the planet. If the escape velocity from the planet is `v_(sec)=vsqrt(N)`, then the value of N is (ignore energy loss due to atmosphere)

Answer

2

Explanation

For a particle at a distance r from the centre of Earth, force is given by,
`F=m("GM"/r^2)`
Force becomes one-fourth, when r = 2R (R = radius of Earth)
Escape velocity `V_e=sqrt((2GM)/R)`
Using conservation of energy for the given particle: `1/2mV^2 - (GmM)/R =(GmM)/(2R)`
This gives , `V=sqrt((GM)/R) rArr V_e=Vsqrt2` Hence, n=2