A small object is launched form the surface of a massive planet of mass M and radius 3R in a non-radial direction. During its journey the object attains a minimum speed of $\frac{1}{2}\sqrt{\frac{GM}{5R}}$ and rises to maximum height of $5R$ above the surface. What time after the launch the object will have a velocity anti-parallel to launch velocity. [Neglect air resistance]

$$\begin{gathered} T.\in =\frac{-Gmm}{8R}+\frac{1}{2}m\frac{1}{4}\frac{Gm}{5R} \\ T.\in =\frac{-Gmm}{10R} \therefore 2a=10R \end{gathered}$$

$$\begin{gathered} a = 5R \\ T=2\pi \sqrt{\frac{{\left(5R\right)}^3}{Gm}}=\frac{2\bcancel{\pi }5\sqrt{5{\bcancel{R}}^3}}{Gm} \\ 5R = 3 R \\ e=\frac{3}{5} \\ b=5R\frac{4}{5}=4R \end{gathered}$$

Area = half area of ellipse + Area of ${\Delta }^x F_{1}AC$
$=\frac{\pi \left(5R\right)4R}{2}+$ Area of triangle $F_1{F}_{2}A$
 $$\begin{gathered} =10\pi R^2+\sqrt{5(s-a)(s-b)(s-c)}=10\pi R^2+\sqrt{8R\left(5R\right)\left(2R\right)R} \\ =(10\pi +\sqrt{80})R^2 \end{gathered}$$
Areal velocity has to remain constant
$$\begin{gathered} \therefore \frac{(10\pi +\sqrt{80})R^2}{T}=\frac{20\pi R^2}{2\pi \sqrt{\frac{125R^3}{Gm}}} \\ \therefore T=\frac{5(2+\sqrt{5}\pi )R^{\frac{3}{2}}}{\sqrt{Gm}} \end{gathered}$$