The minimum and maximum distance of a satellite from the centre of the earth 2R and 4R respectively. Where R is the radius of the earth and M is the mass of earth. Its maximum speed is

From law of conservation of angular momentum

${\text{mv}}_{\text{1}}{\text{r}}_{\text{1}}={\text{mv}}_{\text{2}}{\text{r}}_{\text{2}}\Rightarrow {\text{m}}_{\text{1}}\text{4R}={\text{v}}_{\text{2}}\text{2R}$

$2{\text{v}}_{\text{1}}={\text{v}}_{\text{2}}\Rightarrow \left(1\right)$

From law of conservation of energy

${\left(\text{T}\text{.E}\right)}_1={\left(\text{T}\text{.E}\right)}_2$

$\frac{-\text{GMm}}{\text{4R}}+\frac{1}{2}{\text{mv}}_{\text{1}}^{\text{2}}=\frac{-\text{GMm}}{2\text{R}}+\frac{1}{2}{\text{mv}}_{\text{2}}^{\text{2}}$

$\text{By using}\left(1\right)$

$\frac{-\text{GM}}{\text{4R}}+\frac{{\text{v}}_{\text{2}}^{\text{2}}}{8}=\frac{-\text{GM}}{2\text{R}}+\frac{{\text{v}}_{\text{2}}^{\text{2}}}{2}$

${\text{v}}_{\text{2}}=\sqrt{\frac{\text{2GM}}{\text{3R}}}$

Second method:

In case (i) ${\text{U}}_1=\text{GMm}\left[\frac{1}{{\text{r}}_i}-\frac{1}{{\text{r}}_{\text{F}}}\right]$

$=\text{GMm}\left[\frac{1}{\text{r}}-\frac{1}{\text{2r}}\right]$

$=\frac{\text{GMm}}{\text{2r}}=\text{E}$

In case (ii) ${\text{U}}_2=\text{GMm}\left[\frac{1}{{\text{r}}_i}-\frac{1}{{\text{r}}_{\text{F}}}\right]$

$=\text{GMm}\left[\frac{1}{\text{r}}-\frac{1}{\text{3r}}\right]$

$=\frac{2}{3}\frac{\text{GMm}}{\text{r}}=\frac{1}{3}\text{E}$