A body moves in a circular orbit of radius R | vedclass

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(A) The force on the particle is given by the negative gradient of the potential:$F = -\frac{dV}{dr} = -\frac{d}{dr}(kr) = -k$.Taking the magnitude of

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$R^{1/2}$

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(A) The force on the particle is given by the negative gradient of the potential:$F = -\frac{dV}{dr} = -\frac{d}{dr}(kr) = -k$.Taking the magnitude of the force,we have $F = k$.For circular motion,this force acts as the centripetal force:$F = mR\omega^2 = k$.Substituting $\omega = \frac{2\pi}{T}$,we get:$mR\left(\frac{2\pi}{T}\right)^2 = k$.$mR \cdot \frac{4\pi^2}{T^2} = k$.Rearranging for $T^2$:$T^2 = \frac{4\pi^2 m}{k} \cdot R$.Since $\frac{4\pi^2 m}{k}$ is a constant,we have $T^2 \propto R$,which implies $T \propto R^{1/2}$.

$A$ body moves in a circular orbit of radius $R$ under the action of a central force. The potential due to the central force is given by $V(r) = kr$ ($k$ is a positive constant). The period of revolution of the body is proportional to:

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