Given,

$r_p=$ radius of perihelion $=2 R$

$$r_a=\text { radius of apnelion }=6 R$$

Hence, we can write

$$ \begin{aligned} & r_a=a(1+e)=6 R \quad \text{... (i)}\\ & r_p=a(1-e)=2 R \quad \text{... (ii)} \end{aligned}$$

Solving Eqs. (i) and (ii), we get eccentricity, $e=\frac{1}{2}$

By conservation of angular momentum, angular momentum at perigee $=$ angular momentum at apogee

$$\begin{array}{ll} \therefore & m v_p r_p=m v_a r_a \\ \therefore & \frac{v_a}{v_p}=\frac{1}{3} \end{array}$$

where $m$ is mass of the satellite. Applying conservation of energy, energy at perigee $=$ energy at apogee

$$\frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a}$$

where $M$ is the mass of the earth.

$$\begin{aligned} &\begin{aligned} \therefore\quad v_p^2\left(1-\frac{1}{9}\right) & =-2 G M\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=2 G M\left(\frac{1}{r_p}-\frac{1}{r_a}\right) \quad\left(\text { By putting } v_a=\frac{v_p}{3}\right) \\ v_p & =\frac{\left[2 G M\left(\frac{1}{r_p}-\frac{1}{r_a}\right)\right]^{1 / 2}}{\left[1-\left(V_a / V_p\right)^2\right]^{1 / 2}}=\left[\frac{\frac{2 G M}{R}\left(\frac{1}{2}-\frac{1}{6}\right)}{\left(1-\frac{1}{9}\right)}\right]^{1 / 2} \\ & =\left(\frac{2 / 3}{8 / 9} \frac{G M}{R}\right)^{1 / 2}=\sqrt{\frac{3}{4}} \frac{G M}{R}=6.85 \mathrm{~km} / \mathrm{s} \\ v_p & =6.85 \mathrm{~km} / \mathrm{s}, v_a=2.28 \mathrm{~km} / \mathrm{s} \end{aligned} \end{aligned}$$

For circular orbit of radius $r$,

$$\begin{aligned} v_c & =\text { orbital velocity }=\sqrt{\frac{G M}{r}} \\ \text{For}\quad r & =6 R, v_c=\sqrt{\frac{G M}{6 R}}=3.23 \mathrm{~km} / \mathrm{s} \end{aligned}$$

Hence, to transfer to a circular orbit at apogee, we have to boost the velocity by $\Delta=(3.23-2.28)=0.95 \mathrm{~km} / \mathrm{s}$. This can be done by suitably firing rockets from the satellite.