A satellite is revolving round the earth in a circular orbit of radius ‘a’ with velocity v₀. A particle is projected from the satellite in forward direction with relative velocity $\mathrm{v}=(\sqrt{5/4}-1){\mathrm{v}}_0$ Calculate, during subsequent motion of the particle, the maximum distance from earth’s centre

${\mathrm{V}}_0=\sqrt{\frac{\mathrm{GM}}{\mathrm{a}}}$

Velocity of particle relative to satellite is ${\mathrm{V}}_{\mathrm{PS}}={\mathrm{V}}_0\left[\sqrt{\frac{5}{4}}-1\right]$

${\mathrm{V}}_{\mathrm{P}}-{\mathrm{V}}_{\mathrm{S}}={\mathrm{V}}_0\left[\sqrt{\frac{5}{4}}-1\right];{\mathrm{V}}_{\mathrm{P}}={\mathrm{V}}_0\sqrt{\frac{5}{4}}-{\mathrm{V}}_0+{\mathrm{V}}_0$

$\Rightarrow V_P=V_0\sqrt{\frac{5}{4}}\Rightarrow V_P>V_0\mathrm{\&}{V}_P<V_e\text{. }$

The particle revolves in elliptical orbit. At the positions minimum and maximum distances velocity vectors are perpendicular to the instant radius vectors. From Law of conservation of angular momentum

${\mathrm{mV}}_1\mathrm{r}={\mathrm{mV}}_{\mathrm{P}}\cdot \mathrm{a}\Rightarrow {\mathrm{V}}_1\mathrm{r}={\mathrm{V}}_{\mathrm{P}}\cdot \mathrm{a}$

$\Rightarrow {\mathrm{V}}_1\cdot \mathrm{r}=\sqrt{\frac{5}{4}}{\mathrm{V}}_0\cdot \mathrm{a}......\left(1\right)$

Apply law of conservation of energy

$\frac{1}{2}{\mathrm{mV}}_{\mathrm{I}}^2-\frac{\mathrm{GMm}}{\mathrm{r}}=\frac{1}{2}{\mathrm{mV}}_{\mathrm{p}}^2-\frac{\mathrm{GMm}}{\mathrm{a}}$

By using (1)

$\Rightarrow \frac{1}{2}\mathrm{m}\cdot \frac{5}{4}\frac{{\mathrm{V}}_0^2{\mathrm{a}}^2}{{\mathrm{r}}^2}-\frac{\mathrm{GMm}}{\mathrm{r}}=\frac{1}{2}\mathrm{m}\frac{5}{4}{\mathrm{V}}_0^2-\frac{\mathrm{GMm}}{\mathrm{a}}$

$\Rightarrow \frac{5\mathrm{a}}{8{\mathrm{r}}^2}-\frac{1}{\mathrm{r}}=\frac{-3}{8\mathrm{a}}\Rightarrow 3{\mathrm{r}}^2-8\mathrm{ar}+5{\mathrm{a}}^2=0$

$\Rightarrow r=a\left(\text{ or }\right)\frac{5a}{3}$

$\therefore$ The particle revolves in elliptical orbit with minimum distance a & maximum distance 5a/3 from the earth’s centre.