Two satellites A and B having masses in the ratio 4: 3 are...

Given that $\dfrac{{m}_{1}}{{m}_{2}} = \dfrac{4}{3}, \dfrac{{r}_{1}}{{r}_{2}} = \dfrac{3}{4}$

Now the total energy of a satellite $TE = \dfrac{1}{2}m{{v}_{o}}^{2} + \left(\dfrac{- GMm}{r}\right)$

Since the orbital speed of a satellite ${v}_{o} = \sqrt{\dfrac{GM}{r}}$

$⇒ TE = - \dfrac{GMm}{2r} ∝ \dfrac{m}{r}$

$⇒ \dfrac{T{E}_{1}}{T{E}_{2}} = \dfrac{{m}_{1}}{{m}_{2}} \cdot \dfrac{{r}_{2}}{{r}_{1}} = \dfrac{4}{3} \times \dfrac{4}{3} = \dfrac{16}{9}$