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

`g'=(g)/((1+(h)/(R))^(2))=(g)/(4)`
`1+(h)/(R)=2`
`h=R` …(i)
So velocity of particle becomes zero at h=R
give `v_(prop)=vsqrt(N)`
so `sqrt((2GM)/(R))=vsqrt(N)` …(ii)
Applying conservation of energy
`(-GMm)/(R)+(1)/(2)mv^(2)=-(GMm)/(2R)+0`
on solving
`v^(2)=(GM)/(R)`
so `v=sqrt((GM)/(R))` putting in equation (ii)
`sqrt((2GM)/(R))=sqrt((GM)/(R))sqrt(N)`
comparing `N=2`