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A planet is revolving ground the sun in an elliptical orbit. Its closest distance from the sun is $r_{min}$, the farthest distance from the sun is $r_{max}$. If the orbital angular velocity of the planet when it is the nearest to the sun is $\omega $, then the orbital angular velocity at the point when it is at the farthest distance from the sun is
$\sqrt {\frac{{{r_{min}}}}{{{r_{\max }}}}} \,\omega $
$\sqrt {\frac{{{r_{\max }}}}{{{r_{\min }}}}} \,\omega $
$\frac{{{r^2}_{\max }}}{{{r^2}_{\min }}}\,\omega $
$\frac{{{r^2}_{min}}}{{{r^2}_{\max }}}\,\omega $
Solution
According to law of conservation of angular momentum,
or $\mathrm{mr}_{1}^{2} \omega_{1}=\mathrm{mr}_{2}^{2} \omega_{2} \Rightarrow \mathrm{r}_{\min }^{2} \omega=\mathrm{r}_{\max .}^{2} \omega^{\prime}$
$\omega^{\prime}=\frac{\mathrm{r}_{\min }^{2}}{\mathrm{r}_{\max }^{2}} \omega$
