Wikipedia on its page on efficiency of orbital inclination change writes:
However, maximum efficiency of inclination changes are achieved at apoapsis, (or apogee), where orbital velocity V, is the lowest.
Could somebody elaborate on this statement?
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2 Answers
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To change the inclination by an angle $\alpha$, you need to apply a velocity change of $\Delta v = v \times sin \alpha$, so the higher the velocity, the higher the velocity change that you need to apply, and thus the energy that you need.
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$\begingroup$ It seems as if more eccentric orbits have superior energy characteristics. Is there a catch? $\endgroup$ Mar 18, 2014 at 8:41
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I may be rusty in my understanding of orbital mechanics, but here's a try:
At apoapsis, an orbiting object will be travelling at the slowest velocity of its orbit. Hence, if you make a plane change at this point, assuming you are not changing any other orbital parameters, the velocity at your target orbit will also be the slowest velocity along that orbit. All you're really doing is changing direction, and so that's going to be easier when you're moving more slowly, as opposed to periapsis, at which point the the orbiting object will be at the fastest velocity of its orbit.
