Question

A particle of $3kg$ is moving under the action of a central force whose potential is given by $U\left(r\right)={10r}^3$ joule. For what energy will the orbit be a circle of radius $10m$?

Answer 1

$U=10r^{3}J$
$F=\frac{-dv}{dr}$
$=-30r^2$
force will be centipetal force
$F=\frac{m{v}^2}{r}=30r^2$
$K.E=\frac{1}{2}m{v}^2=\frac{1}{2}\times 30r^3$
$=15r^{3}J$
Total energy $=K.E+U.E$
$=15r^{3}J+10r^{3}J$
$=25r^{3}J$
$r=10n$
$T.E=25\times {10}^3=25000J$