Let's say that we want constant communication between the Lunar north pole and Earth. The 5.145° orbital inclination of the Moon means that the (mountain at) the Lunar north pole will have line-of-sight with Earth for direct communication during only about half of each monthly Lunar revolution.

In order to maintain uninterrupted communication with Earth, at least 3 communication satellites in polar orbits would be required. Could this instead be accomplished by using only 1 satellite in Earth orbit?

Could the communication satellite be put in an orbit which is very similar to that of the Moon, except for having another orbital inclination (say, twice that of the Moon). Same nodes, same semi-major axis, same eccentricity or almost. Couldn't it then remain straight "above" the Lunar north pole during the half of every orbit when the Earth is out of sight from the pole? Could it be achieved with minimal station keeping, maybe using a solar sail?


3 Answers 3


Could an relayer's orbit remain north of the moon? No. Such an orbit would have a center north of earth.

enter image description here

Orbits about the earth have a center within the earth. An inclined relayer orbital plane must intersect earth's plane along a line passing through the center.

A circular relayer orbit would spend half the time north of the moon's orbital plane and half the time south.

When the relayer is north of the orbital plane does it enjoy line of sight with a northern peak? Not necessarily.

Pictured is a northern lunar communications tower. As the satellite moves towards the moon's equatorial plane, it can fall into a moon shadow.

enter image description here

From the tower's point of view, something within the moon shadow would be below the horizon and out of line of sight.

For looking at eclipses, line of sight and similar problems I like to think of Dandelin Spheres. Borrowing from Dandelin Sphere models, the moon shadow falls within a tangent cone to the spherical moon.

Most orbits having the same periods as the moon would be quickly destabilized by the moon's influence. Exceptions are the Lagrange regions. EML4 and 5 are fairly stable.

If a EML4 or 5 probe where crossing through the moon's equatorial plane, how tall would a polar tower need to be enjoying line of sight?

enter image description here

EML4 distance from moon's center is 1 lunar distance, 384,400 km. Moon's radius is 1738 km

sin(angle)=opposite leg/hypotenuse. So the angle pictured is asin(1738/384400) or about .2º.

An EML4 relayer sitting on the moon's equatorial plane will be able to see surface points to within about .2º of the moon's north pole.

Of course a mountain would be visible even if closer to the pole. Let's imagine a mountain right on the pole. How tall would it have to be to be visible to the relayer?

enter image description here

1738.009 km - 1738 km is about 9 meters. So if your mountain is more than 9 meters high, it's top would be visible to the EML4 relayer sitting on the moon's equatorial plane.

Earth's trojan 2010 TK7 is inclined about 21 degrees to earth's orbital plane. So I would think it's possible for a lunar trojan inclined to the lunar orbit.

The angle between the moon's orbital plane and equatorial plane (a.k.a. obliquity) is 6.58º. See Orbit of the Moon. I am imagining a lunar trojan orbit inclined 6.58º with regard to the moon's orbital plane so that the relayer's plane is parallel to the moon's orbital plane.

enter image description here

I enlarged the moon to try to make the moon's equatorial plane more visible. When the moon passes through the line where the two planes intersect, they will be one and the same plane. At this point the relayer would be able to see a 9 meter tower on the north pole as well as a 9 meter tower on the south pole.

All other times the relayer would be either north or south of the moon's equatorial plane. So polar towers would be visible a little more than half the time.

It seems to me there's two possible trojan orbits that see the towers more than half the time: An EML4 orbit parallel to the moon's equatorial plane and an EML5 orbit also parallel to the moon's equatorial plane. But these two relayers would only be 120º apart. At first glance it seems to me they'd need to be 180º apart to provide a given pole with constant line of sight.

So at this time I can't imagine a pair trojan relayers that could provide constant line of sight to the poles.

Halo orbits about L1 and L2 are also possibilities. But these orbits are a lot closer than L4 and L5 and thus more vulnerable to falling into the moon's shadow as they approach the moon's equatorial plane. Thus less than half of their orbits would enjoy line of sight.

It seems to me 3 is the minimum number in any case.

  • $\begingroup$ EML4/5 certainly seems to be a good idea! But what about the type of concept which uses a highly inclined orbit with the same period as that of the Moon? Be it a polar orbit around the Moon, around Earth or around a Lagrange point. The point of such an orbit would be that a single satellite would cover each pole during its "winter" half-month every month. And much more of them than the summits of the crater ridges. $\endgroup$
    – LocalFluff
    Commented Feb 9, 2015 at 10:22
  • $\begingroup$ Given a semi-major axis of 384,400 km, L4 and L5 are the only stable locations I can think of. Inclined trojan orbits are possible. But no matter how inclined the orbits are, they must pass trough the moon's equatorial plane twice each orbit. I've tried to add to my answer to make this clear. $\endgroup$
    – HopDavid
    Commented Feb 9, 2015 at 18:02
  • $\begingroup$ @HopDavid: Don't two satellites in polar orbit around the moon 180 degrees apart cover the poles all the time? $\endgroup$ Commented Feb 10, 2015 at 16:22
  • $\begingroup$ @ThePopMachine No. As I mentioned when satellites near the moon's equatorial plane, they enter a radio shadow -- from the pole's point of view, they fall below the horizon. This is even more of a problem when the orbit is closer to the moon. $\endgroup$
    – HopDavid
    Commented Feb 10, 2015 at 17:17
  • $\begingroup$ @HopDavid: So a far out orbit. $\endgroup$ Commented Feb 10, 2015 at 17:24

It is not possible to do this with just one communication satellite orbiting around the Moon, due to the reasons described by @HopDavid. However, it can be done with just one comsat in Earth orbit.

Imagine a satellite in polar orbit of Earth, with a high apogee and a low perigee. If the apogee is over the north pole, the comsat will be over the equatorial plane in more than half of the time, with free line of sight to the Moon's pole. The short periods when the satellite is under the equatorial plane is covered by direct line of sight between the Moon and Earth. Such a high eccentricity orbit is somewhat analogue to a Molnya orbit, an orbit that is often used for Russian communication satellites to cover the high latitude regions of Russia most of the time of their orbit.


The orbit has to be high tough, with one revolution a month, but the stability for this kind of orbit is good. See for instance the planned orbit for TESS:

(at the end of the video)


Yes, if you'll accept a sufficiently loose definition of "orbit".

Put the satellite above Earth's north pole, deploy a solar sail to keep it there. Beware that it is going to be more than a million miles out, your lag time will be 10x what it is for a direct path transmission.

Robert L. Forward's patent on this (although he was intending it for reaching the polar regions of Earth, I would think his patent applies for reaching the moon) has run out, you're free to do it.

  • $\begingroup$ Wow, I didn't realize Robert L. Forward was so active in space industry as well as writing SF! Can you help find a patent number for this - it would be interesting to read! $\endgroup$
    – uhoh
    Commented Jan 29, 2016 at 3:30
  • 1
    $\begingroup$ @uhoh google.com/patents/US5183225 $\endgroup$ Commented Jan 29, 2016 at 3:33
  • $\begingroup$ This is fascinating! A statite (static + satellite) can hover near, rather than orbit the earth. The prior art is written well - concise but comprehensive. The Wikipedia article mentions alternate terms "non-Keplerian orbit" and "artificial Lagrange point". Thank you for bringing this up! $\endgroup$
    – uhoh
    Commented Jan 29, 2016 at 3:47

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