Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceMedium2 minQB

PhysicsMediumsingle choice

A particle of mass m moves under the action of central force in a circular path, whose potential is given by

$V = - Kr^{3} r \to distance from center$

What is the time period of circular motion ?

Answer:

Solution:

Potential energy $U = - K r^{3}$

Force $F = - \frac{d U}{d r} = 3 K r^{2}$ (towards centre)

Centripetal force: $\frac{m v ^{2}}{r} = 3 K r^{2} \Rightarrow v = \frac{3 K r ^{3}}{m}$

Time period $T = \frac{2 π r}{v} = 2 π \frac{1}{3 rK}$

Answer: (C)

Stream:JEESubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: QB

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