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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

uhoh
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    It's also more complicated because of the nodal precession :) This answer gives some elements of response: https://space.stackexchange.com/questions/4686/how-often-does-the-iss-orbit-align-with-the-day-night-terminator – Antzi Aug 27 '18 at 06:53
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    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. –  Aug 27 '18 at 17:17
  • @Antzi which is why I've said to use a spherical Earth and other simplifying assumptions. – uhoh Aug 27 '18 at 18:45
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    @uhoh: A spherical Earth, in a vacuum... – Vikki May 10 '19 at 03:52
  • @Sean Thanks but I'm not sure what you mean, can you complete the sentence? Are you saying yes, lower would be shorter if "A spherical Earth, in a vacuum..." or no it wouldn't, or are you proposing I adjust the question? – uhoh May 10 '19 at 03:55
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    @uhoh: It was a play on the joke of "a spherical cow in a vacuum". – Vikki May 10 '19 at 03:59
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    @Sean ;-) :O etc. – uhoh May 10 '19 at 04:00
  • Why do you think it's not true? The sun is 150M km away, you're talking a 400km difference. I haven't done the math but my gut says it varies by no more than 3 parts per million--a handful of milliseconds. – Loren Pechtel Nov 21 '19 at 00:32
  • @LorenPechtel in the context of this and the linked question "eclipse" refers to an artificial satellite in low Earth orbit (LEO) entering Earth's shadow for roughly 30 to 40+ minutes every roughly 90 minutes (depending on inclination, some Sun-synchronous satellites avoid eclipse entirely). Satellites use their batteries for power every time this happens. For MEO (e.g. GPS) and GEO (e.g. communications satellites) this happens rarely to never, but for LEO its a major problem. – uhoh Nov 21 '19 at 00:58
  • @LorenPechtel As you go lower in altitude, the eclipse fraction increases, and would be 50% at zero altitude, but the orbit period also decreases, so the eclipse duration may or may not decrease. – uhoh Nov 21 '19 at 00:59
  • @uhoh The shadow fraction increasing doesn't mean the shadow time increases. The dominant factor is the lower you go the faster you go. – Loren Pechtel Nov 21 '19 at 01:09
  • @LorenPechtel they both matter and they move in opposite directions. If you can show which one is dominant with math, than you can write the answer to the question. – uhoh Nov 21 '19 at 01:12
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    @uhoh The thing is the shadow fraction is irrelevant here. Lots of extraneous stuff but it supports my position: https://www.sciencedirect.com/science/article/pii/S2090997715000206 – Loren Pechtel Nov 21 '19 at 01:26
  • @LorenPechtel I've posted an answer which depends upon the eclipse fraction and the period. I don't think there's any way to solve the problem without the fraction. It may be hidden in the math in the paper but it's got to be there – uhoh Nov 21 '19 at 05:44

1 Answers1

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Would a lower LEO ISS orbit really have a shorter eclipse duration than a higher one?

No it definitely would not!

Using the equations in the question:

$$f = \frac{1}{2} - \frac{1}{\pi}\arccos \left(\frac{R_E}{R_E+h} \right)$$

$$T = 2 \pi \sqrt{ \frac{(R+h)^3}{GM_E}}$$

where $f$ is the fraction of the orbit in eclipse, T is the period of an Earth orbit at altitude $h$, $GM_E$ is Earth's standard gravitational parameter and $R_E$ is Earth's equatorial radius, then the period of eclipse is a local maximum at the lowest possible orbit and only decreases as altitude increases until almost 1400 km.

note: the following assumes a circular orbit, either equatorial or if inclined, then passing through the antisolar point. Of course a few times a year the ISS experiences no eclipses at all, and neither would some Sun-synchronous orbits designed for continuous solar power or solar-illuminated Earth observation.

It's 42.24 minutes at zero altitude, 37.28 at 200 km, and 36.11 at 400 km. There is a very broad minimum at about 1370 km altitude with an eclipse duration of only 34.82 minutes. Above that the eclipse duration increases again.

enter image description here

def T_and_f(h):
    T  = twopi * np.sqrt((R+h)**3/GMe)
    f  = 0.5 - np.arccos(R/(R+h))/pi
    return T, f

import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
GMe               = 3.986E+14 # m^3/s^2
R                 = 6378137.  # meters

h = np.arange(0, 1001, 10) * 1000. # altitude in meters

T, f = T_and_f(h)

plt.figure()
plt.subplot(3, 1, 1)
plt.plot(h/1000., f, '-k')
plt.title('Eclipse fraction', fontsize=16)
plt.subplot(3, 1, 2)
plt.plot(h/1000., T/2/60., '--k')
plt.plot(h/1000., T*f/60., '-k')
plt.title('Eclipse duration (-- half-period) 0-1000km', fontsize=16)
plt.ylabel('minutes', fontsize=16)

plt.subplot(3, 1, 3)

h = np.arange(0, 10001, 10) * 1000. # altitude in meters
T, f = T_and_f(h)

plt.plot(h/1000., T/2/60., '--k')
plt.plot(h/1000., T*f/60., '-k')
plt.xlabel('altitude (km)', fontsize=16)
plt.ylabel('minutes', fontsize=16)
plt.title('Eclipse duration (-- half-period) 0-10000km', fontsize=16)
plt.ylim(None, 60.)
plt.show()
Antzi
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uhoh
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  • Awesome. I intuitively predicted the rebounce but I would not have thought about the initial dip. I wonder if we could show that in an animation to make it easier to understand intuitively – Antzi Nov 21 '19 at 06:52
  • @Antzi that should be straightforward; let's wait and see which one of "we" has the time to do it first! btw for me 0-1000km vs 0-10000km are hard for me to read and distinguish without a comma or the space that belongs between a number and its units. I forget that there are some people who don't examine a plots axes labels right away. – uhoh Nov 21 '19 at 07:14
  • Also, I prefer inline image links because doing that way prevents the SE post editor from eating my urls – uhoh Nov 21 '19 at 07:19