Question

A planet $A$ moves along an elliptical orbit around the Sun. At the moment when it was at the distance $r_0$ from the Sun its velocity was equal to $v_0$ and the angle between the radius vector $r_0$ and the velocity vector $v_0$ was equal to $\alpha$. Find the maximum and minimum distances that will separate this planet from the Sun during its orbital motion.

Answer 1

From the conservation of angular momentum about the Sun,
$m{v}_0{r}_0\sin \alpha =m{v}_1{r}_1=m{v}_2{r}_2$ or, $v_1{r}_1=v_2{r}_2=v_0{r}_0\sin \alpha$ (1)
From conservation of mechanical energy,
$\frac{1}{2}m{v}_0^2-\frac{\gamma m_{s}m}{r_0}=\frac{1}{2}m{v}_1^2-\frac{\gamma m_{s}m}{r_1}$
or, $\frac{v_0^2}{2}-\frac{\gamma m_s}{r_0}=\frac{v_0^2{r}_0^2{\sin }^2\alpha }{2r_1^2}-\frac{\gamma m_s}{r_1}$ (Using 1)
or, $(v_0^2-\frac{2\gamma m_s}{r_0})r_1^2+2\gamma m_s{r}_1-v_0^2{r}_0^2{\sin }^2\alpha =0$
So, $r_1=\frac{-2\gamma m_s\pm \sqrt{4{\gamma }^2{m}_s^2+4\left(v_0^2{r}_0^2{\sin }^2\alpha \right)(v_0^2-\frac{2\gamma m_s}{r_0})}}{2(v_0^2-\frac{2\gamma m_s}{r_0})}$
$=\frac{1\pm \sqrt{1-\frac{v_0^2{r}_0^2{\sin }^2\alpha }{\gamma m_s}(\frac{2}{r_0}-\frac{v_0^2}{\gamma m_s})}}{(\frac{2}{r_0}-\frac{v_0^2}{\gamma m_s})}=\frac{r_0[1\pm \sqrt{1-(2-\eta )\eta {\sin }^2\alpha }]}{(2-\eta )}$
where $\eta =\frac{v_0^2{r}_0}{\gamma m_s}$, ($m_s$ is the mass of the Sun).