A bullet is fired vertically upwards with a velocity `upsilon` from the surface of a spherical planet when it reaches 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_("escape") = upsilon sqrt(N)`, then the value of `N` is : (ignore energy loss due to atmosphere).

Options

A. 2
B. 3
C. 4
D. 5

Answer

A

Explanation

(a) Given situation is shown in the figure.
Let acceleration due to gravity at the surface of the planet be g. At height h above planet's surface v=0.
According to question, acceleration due to gravity of the planet at height h above its surface becomes g/4.
`g_(h)=(g)/(4)=(g)/((1+(h)/(R ))^(2))`
`4=(1+(h)/( R))^(2)implies1+(h)/( R)=2`
`(h)/( R)=1 implies h=R`.

So, velocity of the bullet becomes zero at h=R.
Also `v_(esc)=vsqrt(N)impliessqrt((2GM)/( R))=vsqrt(N)`...(i)
Applying energy conservation principle,
Energy of bullet at surface of earth =Energy of bullet at highest point
`(-GMm)/(R )+(1)/(2)mv^(2)=(-GMm)/(2R)`
`(1)/(2) mv^(2)=(GMm)/(2R) :. v=sqrt((GM)/( R))`
Putting this value in eqn. (i) , we get
`sqrt((2GM)/(R ))=sqrt((NGM)/( R)) :. N=2`