# Two satellites have their masses in the ratio $3:1$ The ratio of the orbit (circular ) are $1:4$ then the ratio of total mechanical energy of A and B

$(a)\;\frac{12}{1} \quad (b)\;\frac{10}{1} \quad (c)\; \frac{8}{1} \quad (d)\;\frac{5}{1}$

Energy $E=\large\frac{-GMm}{2r}$
$\large\frac{E_1}{E_2}=\frac{m_1}{r_1} \times \frac{r_2}{m_2}$
$\qquad=\bigg[\large\frac{m_1}{m_2}\bigg]\bigg[\large\frac{r_2}{r_1}\bigg]$
$\qquad=\large\frac{3}{1} \times \frac{4}{1}=\frac{12}{1}$
Hence a is the correct answer.

edited Feb 17, 2014 by meena.p