JEE Advanced 2015 · Paper 1 · Physics Q3

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PhysicsInteger+4Question 3 of 60

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 the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is $v_{esc}=v\sqrt N$ , then the value of $N$ is (ignore energy loss due to atmosphere)

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Answer and worked solution

Official answer: 2

Maximum height. $g\propto1/r^2$ , so $g/4$ means $r=2R$ .

Energy conservation.
$\tfrac12v^2=\frac{GM}R-\frac{GM}{2R}=\frac{GM}{2R}\implies v^2=\frac{GM}R.$

Escape speed. $v_{esc}^2=\dfrac{2GM}R=2v^2$ , so $N=2$ .

Answer: 2

Topics:Escape velocity Variation of g

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JEE Advanced 2015 · Paper 1, Physics question 3