Previous "method to the maddness" questions: 1, 2.

The image below is from Why does TDRS 1's inclination evolve so much differently than that of all the others starting in 1995? and I've used it a few other places as well:

Today I just noticed that in the early 2000's there was an existing triplet of TDRS satellites whose inclination was around 6 degrees and increasing and a new triplet deployed also at about 6 degrees, but also such that their inclination is decreasing!

Once those three (TDRS 8, 9, and 10) bottomed out and turned around and reached about 6 degrees continuing to increase, a second triplet was deployed at 6 degrees with decreasing inclination!


  1. When one deploys a satellite in geostationary orbit, how does one choose to deploy it in either an increasing or decreasing inclination state?
  2. Why are these TDRS triplets deployed in this manner?

enter image description here


If a satellite in GEO is let drifting and its position is not continuously corrected, its inclination will start to change. The main cause of the perturbation is the influence of Moon and Sun.

Orbits with a positive RAAN tend to have an increasing inclination, while those with a negative RAAN have a decreasing inclination. For reference, refer to figure 2 on page 4 of this paper, which can't be included here for copyright reasons. This plot shows the change of RAAN/inclination, averaged over a year to remove short-term fluctuations caused e.g. by the Moon.

The TDRS satellites can be let drifting (opposed to telecommunication satellites), because each satellite of a triplet covers about one third of the globe. If they all drift in roughly the same manner, their coverage will stay the same, no matter what their actual positions are.

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    $\begingroup$ I've seen this diagram discussed in other answers to other questions, and I never quite understand it. Right ascension is measured with respect to the celestial sphere rather than the position of the Sun or Moon, so is it possible to understand what is so "magic" about RAAN with respect to RA=0? $\endgroup$ – uhoh Feb 8 at 11:44
  • $\begingroup$ RAAN and inclination combined give the direction of the orbital plane. There is one stable direction at about RAAN=0 and i=8° where all perturbations even out. If your initial direction is different, you start "orbiting" this stable point in the RAAN/inclination plane, like you do with other equilibrium points. $\endgroup$ – asdfex Feb 8 at 14:11
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    $\begingroup$ But what puzzles me is that Right Ascension = 0 is a direction that is fixed with respect to the stars and the stars have nothing to do with the dynamics of the problem. The Sun and Moon do, but the Right Ascension of the Sun and Moon move 360 degrees over a year and month, respectively. I don't see what is so "magic" about RAAN with respect to RA=0. $\endgroup$ – uhoh Feb 8 at 16:30
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    $\begingroup$ okay I'm getting there. The reason that coordinates with respect to the celestial sphere are important is that Earth's spin axis is substantially tilted with respect to its orbital plane around the Sun and essentially fixed (short term at least), and Right Ascension is linked to it. RA=0 points towards Earth's vernal equinox or point of Aries. I'll go back and look at the equations again. $\endgroup$ – uhoh Feb 9 at 2:11
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    $\begingroup$ @asdfex interesting article! One bit I don't quite get is that I always saw the "conga line" of fig 2 as an approximate circle rather than a triangle. Does that arise only because we are looking at a diagram of Inc vs RAAN rather than the inclination vector of Iy vs Ix? $\endgroup$ – Puffin Feb 10 at 23:57

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