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  • VITEEE - A particle of mass ‘m’ is moving in a circular orbit of radius ‘r’ under a central force field

\vec F = - \frac{k}{r^{2}} \hat r. The angular momentum of the particle about the centre is

proportional to

  1. r^{1/3}

  2. r^{1/2}

  3. r^{3/2}

  4. 1/r^{1/2}

Answers (1)

best_answer

\frac{mv^{2}}{r}=\frac{k}{r^{2}}

Angular momentum about centre would be given by L = mvr

\\\frac{mv^{2}}{r}=\frac{k}{r^{2}}\\ mv^{2}r^{2}=kr\\ m^{2}v^{2}r^{2}=mkr\\ mvr=\sqrt{kr}

L\ \alpha \ r^{1/2}

Posted by

Sayak

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