# A bullet is fired vertically upwards with velocity υ from the surface of a spherical planet. When it reaches its maximum heights, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is υesc=υ√N, then the value of N is (ignore energy loss due to atmosphere)

#### We know that at the surface of the earth, value of g=GMR2

At height h above the Earth's surface, the value of acceleration due to gravity g′=GM(R+h)2

So it is given that g′=g4

When the bullet reaches maximum height, acceleration due to gravity is 14th of that at planet's surface.

That implies GM4R2=GM(R+h)2 →h=R

By conservation of mechanical energy, −GMmR+12mv2=−GMmh+R+0 since velocity is zero at max height. ⇒12mv2=GMm2R

v=√GMR=√2GM2R=1√2√2GMR=1√2vesc ⇒vesc=√2v

⇒N=2