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Question

Physics Question on laws of motion

A particle moves in a circular orbit of radius rr under a central attractive force F=kr,kF=-\frac{k}{r},k is constant. The time period of its motion shall be proportional to

A

r1/2r^{1/2}

B

rr

C

r3/2r^{3/2}

D

r2/3r^{2/3}

Answer

rr

Explanation

Solution

Central attractive force
F=krmrω2=krF=-\frac{k}{r} m r \omega^{2}=-\frac{k}{r}
mr(2πT)2=kTmr4π2T2m r\left(\frac{2 \pi}{T}\right)^{2}=-\frac{k}{T} \frac{m r 4 \pi^{2}}{T^{2}}
=kTT2=mr24π2kT2r2=-\frac{k}{T} T^{2}=\frac{m r^{2} 4 \pi^{2}}{k} T^{2} \propto r^{2}
or TrT \propto r