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Energy of an orbiting satellite
Appears in
Concepts tested here
- satellite-total-energy
- Total energy in circular orbit
All Questions
2016 Phase II 1 question
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A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of $g_0$, the value of acceleration due to gravity at the earth's surface, is:
$E = -GMm/2r$ with $GM = g_0R^2$.
Total energy $= -\frac{GM_em}{2(R + h)}$
$\because g_0 = \frac{GM_e}{R^2} \Rightarrow GM_e = g_0R^2$
$\therefore$ Energy $= -\frac{mg_0R^2}{2(R + h)}$
2010 AIPMT-MAINS 1 question
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The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2$ ($R_2 > R_1$) isTotal energy of a satellite in a circular orbit of radius R: $E = KE + PE = \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{2R}$
Energy in the first orbit: $E_1 = -\frac{GMm}{2R_1}$; energy in the second orbit: $E_2 = -\frac{GMm}{2R_2}$
Energy to be supplied: $E_1 + \Delta E = E_2$
$\Delta E = E_2 - E_1 = -\frac{GMm}{2R_2} + \frac{GMm}{2R_1}$
$\Delta E = \frac{1}{2}GmM\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$