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Gravitation

Section: Physics

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Question : 12 of 22

Marks: +1, -0

A bullet is fired vertically upwards with velocity v from the surface of a spherical planet.When it reaches its maximum height,its acceleration due to planet's gravity is

\frac{1}{4}^{\text{th}}

of its value at the surface of the planet.I the escape velocity from the planet is

v_{esc}=v\sqrt{n}

.then the value of N is(ignore energy loss due to atmosphere)

[JEE Adv 2015 P1]

Your Answer:

Solution:

At height R fro the surface of planet acceleration due to planet's gravity is

\frac{1}{4}\text{th}

in comparison to the value at the surface

So.

-

GMm

\frac{1}{R} + \frac{1}{2} m v^2 = - \frac{GMm}{R+R} + \frac{1}{2} m v_{esc}^2 = 0

∴

V_{esc=} = v\sqrt{2}

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