Given, $r_{p}=$ radius of perihelion $=2 \mathrm{R}$

$r_{a}=$ radius of aphelion $=6 \mathrm{R}$

Hence, we can write,

$r_{a}=a(1+e)=6 \mathrm{R}$

$r_{p}=a(1-e)=2 \mathrm{R}$

Solving equ. $(i)$ and $(ii)$, we get

eccentricity, $e=\frac{1}{2}$

Angular momentum remains unchanged.

$\therefore m v_{p} r_{p}=m v_{a} r_{a}$

$\therefore \frac{v_{a}}{v_{p}}=\frac{r_{p}}{r_{a}}=\frac{2 \mathrm{R}}{6 \mathrm{R}}=\frac{1}{3}$

Energy is same at perigee and apogee,

$\frac{1}{2} m v_{p}^{2}-\frac{\mathrm{GM} m}{r_{p}}=\frac{1}{2} m v_{a}^{2}-\frac{\mathrm{GM} m}{r_{a}}$

$\therefore v_{p}^{2}\left(1-\frac{1}{9}\right)=-2 \mathrm{GM}\left(\frac{1}{r_{a}}-\frac{1}{r_{p}}\right)=2 \mathrm{GM}\left(\frac{1}{r_{p}}-\frac{1}{r_{a}}\right)\left(\right.$ By putting $\left.v_{a}=\frac{v_{p}}{3}\right)$

${p}=\frac{\left[2 \mathrm{GM}\left(\frac{1}{r_{p}}-\frac{1}{r_{a}}\right)\right]^{\frac{1}{2}}}{\left[1-\left(\frac{v_{a}}{v_{p}}\right)^{2}\right]^{\frac{1}{2}}}=\left[\frac{\frac{2 \mathrm{GM}}{\mathrm{R}}\left(\frac{1}{2}-\frac{1}{6}\right)}{\left(1-\frac{1}{9}\right)}\right]^{\frac{1}{2}}$ $=\left(\frac{\frac{2}{3} \mathrm{GM}}{\frac{8}{9} \mathrm{R}}\right)^{\frac{1}{2}}=\sqrt{\frac{3}{4} \frac{\mathrm{GM}}{\mathrm{R}}}=6.85 \mathrm{~km} / \mathrm{s}$ 

$v_{p}=6.85 \mathrm{~km} / \mathrm{s}, v_{a}=2.28 \mathrm{~km} / \mathrm{s}$