Question

Two satellites \(A\) and \(B\) have ratio of masses as \(3:1\) in circular orbits of radii \(r\) and \(4r\). The ratio of total mechanical energy of \(A\) to \(B\) is \(\dfrac{m}{n}\). Find \((m+n)\).

Answer 1

Formula used:
\(E=-\dfrac{GMm}{r}\)
Given,
\(\dfrac{m_A}{m_B}=\dfrac{3}{1}\)

\(\dfrac{r_A}{r_B}=\dfrac{r}{4r}\)

Total energy of a satellite,

\(E=-\dfrac{GMm}{r}\)

\(\dfrac{E_A}{E_B}=\dfrac{m_A}{m_B}.\dfrac{r_B}{r_B}\)

\(=\dfrac{3}{1}\times\dfrac{4r}{r}=12:1\)

\(m+n=12+1=13\)

Final Answer: \(13\)