KVPY SX SB Exam 20141 Question Paper

Question : 23 of 120

Marks: +1, -0

A body moves in a circular orbit of radius R under the action of a central force. Potential due to the central force is given by

V(r) = kr

(k is a positive constant). Period of revolution of the body is proportional to

Solution:

F = -{dU }/{dr}=-d/{dr}[qV]\;\;\;

q → constant

F =-q[{dV}/{dr}]

F = -qk\ \ ← (\table v= kr;{dV}/{dr}=k)

mω^2R = - qk

m({2π}/T)^2R=-qk

{m(4π ^2)R }/T^2=-qk

⇒

Τ^2 ∝ R

⇒

Τ ∝ R^{1\/2}

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