# Is a lunar Sun-synchronous orbit possible at the “frozen” inclination of 86°?

I have been considering if it would be possible to place a satellite in a Sun-synchronous orbit in one of the lunar frozen orbits at 86°. According to NASA scientists there are four inclinations for orbits wherein a satellite can stay in low lunar orbit indefinitely:

"There are actually a number of 'frozen orbits' where a spacecraft can stay in a low lunar orbit indefinitely. They occur at four inclinations: 27º, 50º, 76º, and 86º"—the last one being nearly over the lunar poles.

My question is whether precession would still be possible in a frozen orbit. An orbit over the lunar poles could potentially allow a continuous supply of solar energy. I believe this orbit might be possible with zero eccentricity at a bit above 200 km altitude. This is provided that the satellite can remain in the frozen orbit and would still precess due to the oblateness of the moon. However I am unsure if there might be some reason that this kind of orbital precession might not work as well since I am not very familiar with frozen orbits. I'm also unaware how long it might take for the Earth to perturb this orbit. Is a lunar Sun-synchronous orbit possible when using a frozen orbit?

I also did find an interesting way of reaching a lunar polar orbit on Hop's blog (who had directed me to this site): http://hopsblog-hop.blogspot.com/2013/08/lunar-ice-vs-neo-ice.html

Kind regards,

Peter

The Moon's $J_2$ of $2.0330530\times 10^{-4}$ gives only very low, not very inclined retrograde orbits. The inclinations start at 130° at a zero altitude orbit, and go up to 180° at about 200 km altitude. See this answer for more details.
• No. You need to keep periapsis above the surface. Increasing $e$ also requires you to increase $a$ to hold periapsis. Overall, the rotation rate of the node goes down as you increase eccentricity. – Mark Adler Sep 18 '14 at 14:10
• Ack. No, I was using the wrong $J_2$ value! I just checked and the number I found was from a table of normalized coefficients, instead of unnormalized. The value you found is the right one to use. The most recent estimate is $2.0330530\times 10^{-4}$. I will update the answer. – Mark Adler Sep 18 '14 at 20:24