Why do orbits in the Iridium constellation have an inclination of 86.4°?

Question

Wikipedia's Iridium_satellite constellation; In-orbit spares says that Iridium satellites have an inclination of 86.4°:

Significant orbital inclination changes are normally very fuel-intensive, but orbital perturbation analysis aids the process. The Earth's equatorial bulge causes the orbital right ascension of the ascending node (RAAN) to precess at a rate that depends mainly on the period and inclination. The Iridium satellites have an inclination of 86.4°, which places every satellite in a prograde (inclination < 90°) orbit. This causes their equator crossings to steadily precess westward.

A spare Iridium satellite in the lower storage orbit has a shorter period so its RAAN moves westward more quickly than the satellites in the standard orbit. Iridium simply waits until the desired RAAN (i.e., the desired orbital plane) is reached and then raises the spare satellite to the standard altitude, fixing its orbital plane with respect to the constellation. Although this saves substantial amounts of fuel, it can be a time-consuming process.

Question: This replacement scheme would work at other high inclinations, and polar coverage could still (probably) be 100% anywhere between 75° and 105° (they're spaced every 30° at the equator). I can't figure out anything profoundly special about 86.4°. Not that there needs to be. Was this a complex yet uninteresting optimization of several things, or does this particular inclination angle reflect something specific and significant?

For more about the constellation see the excellent answers to What (actually) makes Iridium “the world's only truly global mobile satellite communications company”?

Answer 1

The probability of collisions over the poles was indeed the reason not to use 90° exactly.

The reason to choose precisely this value might be that at this particular inclination, the Iridium satellites cross the Earth's equator perpendicularly (as seen in an Earth-fixed frame). Back of the envelope:

$$\cos(86.4°) \frac{R_E}{h + R_E} v_{sat} \sim 460 \ \text{m/s}$$

where 460 m/s is velocity at equator, $v$ is the satellites velocity at altitude $h$ and $R_E$ is the Earth's radius. At 90° they would instead cross the equator at a small angle.

This property might have some operational interest for Iridium, as the Earth can be divided in nice longitude bands. To be confirmed...

(For information: 86.4° requires less launch delta-v than 93.6°.)

Answer 2

A colleague of mine from years ago, Jan King, told me that he was a consultant for Iridium in their early planning stages and that their original plan was to place the spacecraft in a precisely 90 deg inclined orbit. This has some special properties in terms of orbit perturbations because it means that most of the biggest gravity irregularities like earth's oblateness at the equator (J2) even out over the coarse of an orbit and thus don't build up over time to disturb the orbit. This can save fuel for orbit maintenance. Jan pointed out that making the inclination exactly 90 deg has the unfortunate effect of making collisions among the constellation not only possible, but likely over the mission lifetime. After this was confirmed by simulation, Iridium altered the orbit inclination to be as close to 90 deg as they could while maintaining a low probability of collision over the poles.

I don't know if this is the sole reason for the resulting inclination, or if it was a compromise between this and other potential factors, such as launch vehicle lift capacity within available launch azimuth ranges. Since this was merely a conversation with a friend, I'm afraid I don't have a particular reference I can point to for further information.