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This answer to Is there a low Earth orbit with a 24-hour day night cycle? suggests that:

In fact, it's a rule of thumb that the higher you go the less time you spend in Earth's shadow: the shadow not only gets narrower, your orbit gets bigger.

Rules of thumb are never intended to be considered true in all cases; we use them in a pinch or when we plan to go back later and calculate rigorously (and often never do).

I'm wondering how often (if ever) this rule of thumb fails; if there's a class of circular Earth orbits where it's ever, or even frequently not true.

What's left out of the explanation is that when the satellite's orbit gets bigger, the satellite also moves more slowly.

So while the fraction of the orbit that's spent in eclipse is very likely to decrease with increasing semi-major axis, it's not clear that the absolute duration of a given eclipse will always decrease with increasing semi-major axis as well.

So I'd like to ask:

Question: Are there any Earth orbits where the duration of eclipse increases with increasing semi-major axis with all other parameters fixed?

uhoh
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Speculative back-of-a-napkin answer: the absolute duration of time spent in eclipse always increases.

For circular orbits about a body of mass $M$, orbital velocity is approximately a function of the semi-major axis ($a$):

$$v\approx\sqrt{\frac{GM}{a}}\ \text{or}\ v\approx\sqrt{\frac{GM}{r}}$$

Since we're already assuming the orbit's circular, let's just use the radius $r$ instead of the SMA from now on.

Assuming that the shadow is cast by a body of diameter $D_{earth}$ with distance $d$ from the light source of diameter $D_{sun}$, the width $w$ of the umbra at a radius $r$ away from the body is approximately:

$$w\approx\frac{r}{d}\left(D_{earth}-D_{sun}\right)+D_{earth}$$

For the actual earth & sun, this comes out to be something like $\frac{dw}{dr}\approx -0.009$, btw.

Of course, the time spent in the umbra is:

$$t=wv^{-1}$$

And plugging in really approximate forms of everything ($v\sim r^{-\frac{1}{2}}$, $w\sim r$) gives us:

$$t\sim r^\frac{3}{2}$$

Which means the time spent in the umbra roughly is an extralinear/subquadratic function of the radius of the orbit. In short, the higher you go, the more time you spend in shadow. Of course, this is really rough math, and probably doesn't hold at the extreme (I'd guess, within a few Earth radii of the ground). Who knows? It might not even kick in until you get past the Earth's Hill sphere, in which case it wouldn't hold at all. But for idealized situations (Sun's really far away/a point; Earth's really far away/a point) it makes sense that the umbra's not getting much smaller but your orbit is always getting slower.

Anton Hengst
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    Trivial violation of this answer: orbit further than the umbra distance of earth. 1.4million km. Well within the Earth's hill sphere, so still an Earth orbit. Maximum duration of total eclipse by the Earth is zero – CuteKItty_pleaseStopBArking Dec 21 '21 at 16:14
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    @CuteKItty_pleaseStopBArking yup, that comes from the "really approximate forms" part. The coefficient on $w$ is negative, since the umbra is shrinking. To actually see where the asymptotic decrease in orbital velocity is eventually outpaced by the linear decrease in umbra width would require... actual numbers! Heaven forbid: but you feel free to go ahead! – Anton Hengst Dec 21 '21 at 16:18
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    $w$ is not $\sim r$, it's approximately constant for reasonably small $r$, and it decreases with $r$. From what you wrote, $w\sim (\frac{D_{earth}}{0.009}+R_{earth}-r)$, so for $r$ within, let's say $10 R_{earth}$, $w$ changes only by about 5%. – Litho Dec 21 '21 at 16:54
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    Of course, it means that $t$ still increases with $r$ for small enough $r$, but not faster than $r$. – Litho Dec 21 '21 at 17:01
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    uhoh! I am getting old and forgetful; I'd asked a similar question three years ago. Would a lower LEO ISS orbit really have a shorter eclipse duration than a higher one? I kinda thought this was familiar territory but wasn't sure why. It looks like the duration turns around at about 1500 km and starts increasing again in the answer there, but there's no guarantee that's right. Do you think the two answers can be reconciled? – uhoh Dec 21 '21 at 23:13
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    @uhoh yeah, my math is highly approximated at best. Looks like below 1500km, the $+D_{earth}$ part I approximated away dominates; much further off the right of that graph the $-0.009\ \text{km}\cdot\text{km}^{-1}$ coefficient I approximated away will start to take over as your umbra just keeps getting smaller no matter how slow you're orbiting. EDIT: looks like that answer doesn't account for the narrowing umbra, actually – Anton Hengst Dec 21 '21 at 23:36
  • Plugging in a function for the radius of the umbra into the equations in that other answer gives me exactly what I'd expect: https://www.desmos.com/calculator/zdyc2edh1e – Anton Hengst Dec 21 '21 at 23:50