Two satellites $$A$$ and $$B$$ having masses in the ratio $$4:3$$ are revolving in circular orbits of radii $$3r$$ and $$4r$$ respectively around the earth. The ratio of total mechanical energy of $$A$$ to $$B$$ is

We need to find the ratio of total mechanical energy of satellites A and B. The total mechanical energy of a satellite of mass $$m$$ in a circular orbit of radius $$r$$ around Earth (mass $$M$$) is given by $$E = -\frac{GMm}{2r}$$.

For satellite A, which has mass $$4m_0$$ and orbits at radius $$3r$$, the energy becomes $$E_A = -\frac{GM(4m_0)}{2(3r)} = -\frac{4GMm_0}{6r} = -\frac{2GMm_0}{3r}$$. Similarly, for satellite B with mass $$3m_0$$ at radius $$4r$$, one finds $$E_B = -\frac{GM(3m_0)}{2(4r)} = -\frac{3GMm_0}{8r}$$.

Taking the ratio yields $$\frac{E_A}{E_B} = \frac{-\frac{2GMm_0}{3r}}{-\frac{3GMm_0}{8r}} = \frac{2}{3} \times \frac{8}{3} = \frac{16}{9}$$, so the correct answer is Option B: $$16:9$$.