The minimum energy required to launch a satel | vedclass

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(A) The total energy of a satellite in a circular orbit at a distance $r = R + h$ from the center of the planet is given by $E_0 = -\frac{GMm}{2r}$.Gi

Vedclass · Accepted Answer

$\frac{5 GMm}{6 R}$

Vedclass · Answer

(A) The total energy of a satellite in a circular orbit at a distance $r = R + h$ from the center of the planet is given by $E_0 = -\frac{GMm}{2r}$.Given the altitude $h = 2R$,the orbital radius is $r = R + 2R = 3R$.Thus,the orbital energy is $E_0 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.The potential energy of the satellite on the surface of the planet is $E_i = -\frac{GMm}{R}$.The minimum energy required to launch the satellite is the difference between the final orbital energy and the initial surface energy:$\Delta E = E_0 - E_i = -\frac{GMm}{6R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{6R} = \frac{5GMm}{6R}$.

The minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ into a circular orbit at an altitude of $2R$ is

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