A satellite of mass m is in an elliptical orbit around the earth. The speed of the satellite at its nearest position is  $\sqrt{\frac{6{\mathrm{GM}}_{\mathrm{e}}}{5\mathrm{r}}}$ where r is the perigee (nearest point) distance from the centre of the earth. It is desired to transfer the satellite to the circular orbit of radius equal to its apogee (farthest point) distance from the centre of the earth. The change in orbital speed required for this purpose is

$\mathrm{E}=\frac{\mathrm{m}}{2}\times \frac{6{\mathrm{GM}}_{\mathrm{e}}}{5\mathrm{r}}-\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{2\mathrm{r}}=-\frac{2}{3}\frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{\mathrm{r}}$

which is the total energy of the earth-satellite system.

So, semi-major axis of the elliptical orbit is $\mathrm{a}=\frac{5\mathrm{r}}{4}$.
Speed of the satellite at the apogee position is ${\mathrm{v}}_{\mathrm{A}}=\frac{{\mathrm{v}}_{\mathrm{P}}\times \mathrm{r}}{2\mathrm{a}-\mathrm{r}}=\frac{2}{3}\sqrt{\frac{6{\mathrm{GM}}_{\mathrm{e}}}{5\mathrm{r}}}$ 

For orbit to change to a circle of radius 3r/2 = (2a - r), the rocket has to be fired when the satellite is at the apogee position.

New orbital speed is ${\mathrm{v}}_0=\sqrt{\frac{{\mathrm{GM}}_{\mathrm{e}}}{3\mathrm{r}/2}}=\sqrt{\frac{2{\mathrm{GM}}_{\mathrm{e}}}{3\mathrm{r}}}$

Required change in the orbital speed is

$\mathrm{\Delta v}={\mathrm{v}}_{\mathrm{A}}-{\mathrm{v}}_0=0.085\sqrt{\frac{{\mathrm{GM}}_{\mathrm{e}}}{\mathrm{r}}}.$