Two satellites of same mass of a planet in circular orbits have periods of revolution 32 days and 256 days. If the radius of the orbit of the first is R, then the

Using kepler’s third law.

${\mathrm{R}}_2={\mathrm{R}}_1{\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{2/3}=\mathrm{R}{\left(\frac{256}{32}\right)}^{2/3}$

K.E. of satellite ${\mathrm{E}}_{\mathrm{K}}=\frac{1}{2}{\mathrm{mv}}_2^2=\frac{1}{2}\mathrm{m}\frac{\mathrm{GM}}{\mathrm{r}}=\frac{\mathrm{GMm}}{2\mathrm{r}}$

As ${\mathrm{E}}_{\mathrm{K}}\propto \frac{1}{\mathrm{r}}$

So, $\frac{E_{K_2}}{E_{K_1}}=\frac{r_1}{r_2}=\frac{R}{4R}=\frac{1}{4}\text{(or)}{E}_{K_2}<E_{K_1}$

Total mechanical energy, $E=PE + KE$

$=-\frac{\mathrm{GMm}}{\mathrm{r}}+\frac{\mathrm{GMm}}{2\mathrm{r}}=-\frac{\mathrm{GMm}}{2\mathrm{r}}$

Here $–ve$ sign shown that as $r$ increases $E$  increases. So total mechanical energy of the second is greater than of the first satellite.