# Sun synchronous orbit and LTAN

I have to size the electric power system for a satellite on a circular, sun-synchronous orbit at 800 km altitude. In the problem I have been given the LTAN (Local Time of the Ascending Node) also, which is 11:00 am. How do I use the LTAN information? I think I should use it to compute the relative position between satellite and sun in order to check for eclipses during the orbit (which in turn will allow me to size the batteries), but I do not know how to use it.

EDIT: I forgot to say that I have to perform all the calculations by hand

• That information really seems like a red herring to me. You should size the power system for the pessimistic case of the longest possible eclipse regardless of when it happens instead of trying to hunt down individual eclipses. – SF. Feb 8 '16 at 1:56
• That makes sense, but how do I define the worst case? – Rhei Feb 8 '16 at 10:23
• I believe it will be when the satellite moves in the same direction as the Moon (so above one of the poles, not sure which one) and moving through the longest path through the umbra (entry and exit at end points of the diameter, not any shorter chord). But I'm not sure how penumbra affects your power supply vs requirements, and that would result in an optimization task, finding the curve with the lowest integral over a scalar field of irradiance in the Moon's shadow... It gets awfully complex. I can tell you how I see this approach but I'm not sure if I'm right. – SF. Feb 8 '16 at 11:12

Let me try. I'm not sure if I'm right so take everything with a grain of salt, but that's how I'd approach it.

Instead of trying to find the precise eclipses that will occur (based on LTAN) and trying to adapt to their duration, I'd choose a worst scenario eclipse: the satellite moving in the same direction as the Moon, as result staying in the shadow the longest, and also, with Moon at the perigee, resulting in largest umbra diameter and largest Moon linear velocity (again prolonging the eclipse). It's some simple trigonometry to obtain the diameter of umbra and penumbra at Earth's terminator line distance, with Moon at perigee.

The rest of the calculations happen in the orbital plane of the satellite, or the plane of terminator, or plane normal to the Sun-Moon line, at Earth's distance - which one is pretty insignificant as they are all very close to each other and differing by cosine of angle between them -> minimal error. Let's assume they are all the same plane so that you don't need to e.g. angle the umbra cross-section.

Calculate the curve of the trajectory of the satellite (vs time) in the reference frame of the (moving) center of umbra in that plane. Calculate the density of penumbra, say, in range [0,1] 1=full sun, 0=full shadow.

It will be awfully nonlinear - decreasing from outer edge to umbra edge with surface of circle partially obscured by another circle (relative displacement of the circles proportional to distance through penumbra, diameters of the circles proportional to angular size of the Sun and the (Super-)Moon as seen from Earth respectively). And a flat zero through Umbra.

Multiply by solar irradiation constant. That way you obtain the irradiation density field. Again, multiply by panel efficiency and you have your power input density field. Subtract your power output (constant) from the whole field function, then discard the positive part (more input than output while the batteries are full, power wasted). You have a field of "power debt"; how much power your satellite draws at each point of eclipse.

And then you apply your trajectory (f(t)->x,y) , offset (translated) by parameters $$x_0,y_0$$ relative to the field center. Apply integral (dt) along the curve over the scalar field, and optimize the resulting function for such $$x_0,y_0$$ that the integral value is lowest (most negative). It's important to integrate over time, not just location as your velocity isn't constant.

The value of the integral is your energy output during eclipse - the value of energy that must be kept in batteries to "survive" between break even points between input from the solar panels and power requirements.

EDIT:

As per your edit: the integral over such a nonlinear field would be a total overkill for calculation by hand.

• curvature of the trajectory across penumbra = straight line. The diameter in shadow relative to the diameter of the orbit is small. This reduces the whole optimization task to crossing the diameter of the shadow.
• ballpark diameter of "debt circle" within penumbra as ratio of power input to power requirement. Say, if your solar batteries produce 20% more power than necessary, reduce penumbra diameter by 20%. That way you drop the awful task of calculating "surface of circle partially obscured by another circle".
• Assume all, umbra and penumbra within the "debt circle" is flat zero - solar batteries don't contribute anything.
• assume constant velocity. That way your energy requirement will be just "time spent in 'debt circle'" times power requirements of the satellite.

Don't skip:

• penumbra and umbra size calculation
• satellite velocity in the umbra frame of reference if the Moon and the satellite move in the same direction.
• irradiation, panel efficiency, reducing size of penumbra to where panels don't suffice (although with much simplified function).

Explain what simplifications you've made. And also, that you'd want a bigger battery anyway, as a safety margin.

From the way the question is phrased, it would be referring to the eclipse of the spacecraft by the Earth (i.e. the sun is blocked by the Earth), not of the spacecraft by the Moon. Eclipse can refer to any body eclipsing any other body, not just the Moon.

LTAN is Local Time of the Ascending Node. The spacecraft crosses the equatorial plane of the Earth and, since it is in a sun-synchronous orbit, it will always cross that plane when it is the same time of day below. In this case, that means the spacecraft is always "looking down" at some point on Earth that currently has a time of 11 AM (or 11 PM on the other side of the Earth).

A 12 Noon LTAN would indicate that the plane of the spacecraft's orbit is exactly lined up with the direction of the sun. 11 AM is very close to this situation, so all this is telling you is that you should assume approximately the "worst case" maximum eclipse duration as the spacecraft passes behind the Earth each orbit (somewhere around 35 minutes for an 800 km orbit).

The 800 km orbit is much more important than the LTAN. At 800 km orbit, the satellite is going to have a orbital period of roughly 100 minutes. For a few of those successive 100 minute orbits, it's going to spend close to 1/2 of the orbit in eclipse. Without designing for any safety margin, the power system needs to be able to fully charge the batteries in roughly 50 minutes, and the batteries need to be able to hold enough charge to power the spacecraft for roughly 50 minutes.

From the table on the back cover of Wertz and Larson, an 800 km orbit would have an orbit period of 100.87 minutes and a maximum eclipse time of 35.1 minutes. The maximum eclipse occurs when the vector to the sun lies within the orbital plane of the spacecraft. This orbital sun angle is often referred to as orbital beta angle and usually represented by the greek letter beta. As another answer correctly stated, a Sun-Synchronous Orbit (SSO) with a Local Time of Ascending Node (LTAN) of 12 Noon would experience the maximum eclipse since the vector to the sun would lie within the orbital plane and an orbit with an LTAN of 11 AM would experience an eclipse time of close to that. The exact eclipse time could be calculated from the orbital sun angle using a formula available in many standard test books, e.g. Wertz and Larson, Space Mission Design and Analysis or Chobotov, Orbital Mechanics. These and many other references provide formulas for calculating orbit period and maximum eclipse time for an orbit at any altitude.