Two satellites A and B,with a mass ratio of 3 | vedclass

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(D) The total mechanical energy $E$ of a satellite in a circular orbit is given by the formula $E = -\frac{GMm}{2r}$.Here,$G$ is the gravitational con

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$12:1$

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(D) The total mechanical energy $E$ of a satellite in a circular orbit is given by the formula $E = -\frac{GMm}{2r}$.Here,$G$ is the gravitational constant,$M$ is the mass of the planet,$m$ is the mass of the satellite,and $r$ is the radius of the orbit.For satellites $A$ and $B$,the ratio of their total mechanical energies is given by:$\frac{E_A}{E_B} = \frac{-\frac{GMm_A}{2r_A}}{-\frac{GMm_B}{2r_B}} = \frac{m_A}{m_B} 	imes \frac{r_B}{r_A}$.Given the mass ratio $\frac{m_A}{m_B} = \frac{3}{1}$ and the radius ratio $\frac{r_A}{r_B} = \frac{r}{4r} = \frac{1}{4}$,we have $\frac{r_B}{r_A} = \frac{4}{1}$.Substituting these values:$\frac{E_A}{E_B} = \frac{3}{1} 	imes \frac{4}{1} = \frac{12}{1}$.Therefore,the ratio is $12:1$.

Two satellites $A$ and $B$,with a mass ratio of $3:1$,are in circular orbits of radii $r$ and $4r$ respectively. The ratio of the total mechanical energy of $A$ to $B$ is:

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