How large would be the seasonal variation of eclipse duration for a spacecraft in LEO?

Question

The question How long will a 500 km altitude satellite spend in Earth - shade per orbit has several answers, and there is some discussion of constraints related to maximum eclipse, due to some issues related to Earth's axial tilt, shape and inclination of the satellite's orbit.

Assuming a circular orbit around a circular Earth and using only plane geometry, the equation in this answer when slightly rewritten look like:

$T_{eclipse} = 2T_0 \sin^{-1}(R_E / a_0) /2\pi$

Starting with a circular orbit with an altitude of 500 km, which means the radius = semi-major axis = 500 + 6378.13 km per this answer, we can estimate the amount of variation due to inclination. Plugging in Earth's polar and equatorial radius (6356.8 vs 6378.1 km) gives about 2130 vs 2145 seconds of eclipse for a polar vs equatorial orbit.

Question: Is there a way to calculate the duration of the eclipse based on the actual inclination of the orbit, that would somehow allow one to take into consideration "seasonal" and other variations regarding the Sun-Earth line and the line of nodes?

Interesting related question, nicely answered: How fast is the onset of periodic eclipse for a spacecraft in LEO?

Answer 1

A better formula is this one, which will calculate the fraction of an orbit in which a satellite is in eclipse.

$f_e=\cfrac{1}{\pi}\cos^{-1}(\cfrac{\sqrt{h^2+2R\cdot h}}{(R+h)\cos{\beta}})$

The key in this is the $\beta$. This term can vary by as much as 23 degrees more then the inclination. For a purely equatorial orbit of 500 km, the fraction of time in eclipse is 0.3778. For a 23 degree beta it is 0.3667. The orbital period is 94.5 minutes, thus the difference is 34.6 to 35.7, a difference of about 1.1 minutes. If, however, one assumes a larger $\beta$, as happens with a higher inclination, the difference can be much higher, and even reach 0 for a $\beta$ of around 60.