A satellite is moving in circular orbit of r with speed $v_0$  and time period $T_1$. A particle of small mass is projected from satellite in direction of it’s motion with relative speed $\left(\frac{2}{\sqrt{3}}-1\right){\mathrm{v}}_0$:

Particle velocity  $v=\frac{2}{\sqrt{3}}{\mathrm{v}}_0$

$=\frac{2}{\sqrt{3}}\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}=\sqrt{\frac{4\mathrm{GM}}{3\mathrm{r}}}$

Conservation of angular momentum,

$\mathrm{m}v\mathrm{r}=\mathrm{m}{v}^{\text{'}}{\mathrm{r}}^{ \text{'}}$

$v\mathrm{r}=v^{\mathrm{\prime }}{\mathrm{r}}^{\mathrm{\prime }}$ …(i)

And conservation of energy, 

$\frac{-\mathrm{GmM}}{\mathrm{r}}+\frac{1}{2}{\mathrm{mv}}^2=-\frac{\mathrm{GmM}}{\mathrm{r}}+\frac{1}{2}{\mathrm{mv}}^{\text{'}2}$  … (ii

Solving, the two equations, we get, $r^{\text{'}}=2r$

as r increases time period also increases.