# How fast is the onset of periodic eclipse for a spacecraft in LEO?

A spacecraft in low inclination LEO will undergo regular eclipsing of the Sun and this is a big issue for power management. How fast is the light-to-dark transition, say 90% to 10% sunlight intensity or something similar.

Maybe choose an orbit in the plane of the ecliptic for simple geometry, and an altitude of 500 km.

Is it milliseconds, seconds, tens of seconds? Is the transition time in LEO mostly determined by geometry or by meteorology?

When I saw eclipse duration exact to one millisecond in this answer I began to wonder just how to define when eclipse happens, and how long it takes.

## 2 Answers

About 8 seconds.

From the point of view of a satellite in LEO, orbiting in the same plane as the centres of the Earth and the Sun, the Sun goes around the Earth (and the satellite) once per orbit, so it moves 360 degrees in about 90 minutes. Entering eclipse is the same as sunset: the time from one edge of the Sun first touching the horizon to all of the Sun being below the horizon. That's the same as the time for the Sun to move its own diameter, about 0.5 deg. So entering eclipse takes about 1/720 of the orbital period.

• A LEO is from 200 to 2000 km height and from about 90 to 120 minutes. Eclipse duration is from 7.5 to 10 seconds then. – Uwe Mar 24 '18 at 19:44
• This is a concise and elegant explanation of how to do the calculation, but without naming a time, it doesn't exactly answer the question, so I've accepted the other answer. But this is very helpful as well, thanks! – uhoh Mar 25 '18 at 12:29
• I named the time in the first sentence! But I'm happy with concise and elegant - I was aiming for something more easily grasped to complement the more rigorous answer from @Prakhar, – djr Mar 25 '18 at 21:28
• @djr omg sorry! Indeed you did, I flew right past it. Thanks for calling this to my attention. – uhoh Mar 26 '18 at 11:28

Here are just on some rough pen&paper calculations...

Now eclipse entry could be thought as probably entring from penumbra to umbra.

So approximately d would be RE + H that is 6400 + 500 and theta is angular distance of sun, which is 0.53 degrees.

So the arc region of between penumbra and umbra comes about ~ 63.8 km

Which if a satellite at about 7km/s enters would cover in about 9~10 seconds.

Note : Calculations are approximate. Edit: Infact the calculation seems to be exact, the penumbra entry to umbra entry seems to be exactly (Re + h) * Theta, where Theta is angular distance of sun and Reis radius of earth and h is height of circular orbit. • I guess, the approximation becomes too off for LEO. – Prakhar Mar 24 '18 at 12:15
• I wonder if you can make some use of the diagram in this answer which I've also linked in the question here. – uhoh Mar 24 '18 at 12:21
• Okay if the Sun is 0.5 degrees wide, then we need the time between one side of the Sun lining up with the Earth's terminator and the other, wouldn't that just be the time it takes for the satellite to move 0.5 degrees around the Earth in its orbit? No diagrams needed? If the period of a 500 km orbit is about 5680 seconds, then it's 5680 / 720 = 7.9 seconds. Okay it checks out! – uhoh Mar 25 '18 at 9:07
• ^ does this argument holds true for highly elliptical orbit too ? Given the orbit parameters, what will be exact duration of the transition ? I feel this argument hides and highly simplifies the math – Prakhar Mar 26 '18 at 15:04
• I've asked for an approximate answer only, nothing exact. "Is it milliseconds, seconds, tens of seconds?" It's really hard to choose when there are two or more sufficient answers. Certainly if the orbit is highly elliptical, an actual equation using clearly represented mathematics would be better. it's absolutely not necessary for this question, but if you would like to add such an expression, that would be interesting! But I think it's way too much work. – uhoh Mar 26 '18 at 15:18