Two satellites A and B,having a ratio of mass | vedclass

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(B) The total mechanical energy $E$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by the formula: $E = -\frac{GM

Vedclass · Accepted Answer

$12 : 1$

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(B) The total mechanical energy $E$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ is given by the formula: $E = -\frac{GMm}{2r}$.Given the masses $m_A = 3m$ and $m_B = m$,and the orbital radii $r_A = r$ and $r_B = 4r$.The ratio of the total mechanical energy is:$\frac{E_A}{E_B} = \frac{-\frac{GMm_A}{2r_A}}{-\frac{GMm_B}{2r_B}} = \frac{m_A}{m_B} 	imes \frac{r_B}{r_A}$.Substituting the given values:$\frac{E_A}{E_B} = \frac{3}{1} 	imes \frac{4r}{r} = \frac{12}{1}$.Thus,the ratio of the total mechanical energy of $A$ and $B$ is $12 : 1$.

Two satellites $A$ and $B$,having a ratio of masses $3 : 1$,are in circular orbits of radii $r$ and $4r$ respectively. Calculate the ratio of the total mechanical energy of $A$ and $B$.

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