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This comment says

Lower orbit also mean shorter ones; for observations (one of the ISS missions) it is an advantage. It also have shorter night times (ie smaller batteries)..

and the likely thinking is that for a point-source Sun at infinity the width of the Earth's shadow is constant, and so the lower the orbit, the the faster the velocity (shorter the period) and so the less time in the shadow.

but is that actually even true, or is it more complicated?

The context of the question Does the ISS still “need” to be at around 400 km? there is that the ISS' available orbits are between roughly 300 and 700 km (limited by drag, radiation respectively). Within that range, is the shorter eclipse duration really at the low end?

Use the simplifying assumptions of spherical earth in circular orbit without axis tilt or atmosphere. You're welcome to add them in later, but it gets complicated quickly.

"bonus points" for the the altitude of minimum duration (same assumptions, but ignore the 300-700 km).


The diagram might be helpful. $f$ is the fraction of the orbit in the Earth's shadow, $GM_E$ is the standard gravitational parameter of the Earth and is about 3.986E+05 km^3/s^2 and $T$ is the period of one orbit (from this answer).

enter image description here

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    $\begingroup$ It's also more complicated because of the nodal precession :) This answer gives some elements of response: space.stackexchange.com/questions/4686/… $\endgroup$ – Antzi Aug 27 '18 at 6:53
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    $\begingroup$ This is probably too basic an observation, but R^3/T^2 is constant for any object orbiting the Earth, so lower R means lower T. However, as you note, this doesn't necessarily mean less time in shadow. $\endgroup$ – barrycarter Aug 27 '18 at 17:17
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    $\begingroup$ @uhoh: A spherical Earth, in a vacuum... $\endgroup$ – Sean May 10 at 3:52
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    $\begingroup$ @uhoh: It was a play on the joke of "a spherical cow in a vacuum". $\endgroup$ – Sean May 10 at 3:59
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    $\begingroup$ @Sean ;-) :O etc. $\endgroup$ – uhoh May 10 at 4:00

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