Can a satellite orbit Earth so that it always has the Moon in line of sight?

Could a satellite, without too much fuel consumption for station keeping, orbit the Earth so that it always has a line of sight to the Moon? It would need a near polar orbit which precesses in tandem with the Moon's orbit around Earth. Wouldn't that be the perfect communication satellite orbit for constantly staying in touch with Lunar missions? Have any satellites used such an orbit and does that type of orbit have a three letter name?

EDIT: I'm looking for the possibility of a satellite orbiting Earth's poles so that it can always have contact with the Earth facing side of the Moon, and with satellites in lunar polar orbit which precess so that they remain in line of sight with this satellite (and the rotating Earth). A potential use for this would be that such a polar Earth satellite would have a relatively small altitude and be able to relay data to Earth more easily than having to use something like the Deep Space Network spread across the equator.

• Technically, a satellite orbiting the moon is also orbiting Earth. . . . – imallett Oct 21 '14 at 0:14
• Or in the same orbit as the moon :) – Baldrickk Oct 21 '14 at 8:33
• Or just, you know, land it on the moon. Then nothing can get between them. – moopet Oct 21 '14 at 15:55
• Yes. We can put it on the moon's surface. Boo. – Joshua Jan 18 '18 at 1:14

If I understand your edited question, then no. While the Earth's J2 (oblateness) produces enough torque to rotate a Sun-synchronous orbit once a year, it does not produce enough torque to rotate a "Moon-synchronous orbit" once a month. So there is no such orbit.

I am not clear on what the utility of such an orbit would be, even if it did exist. If you're going to point antennas at the Moon from the near vicinity of Earth, you can have many more antennas and much larger antennas on the surface of the Earth than what you would be able to put into Earth orbit for an equal amount of money. You would only need to divide your total aperture by three to account for the downside of not always being able to see the Moon.

• +1. I had the same thought on reading the problem and quickly came to the same conclusion: A moon-synchronous orbit is not possible around an Earth with its 6378 km equatorial radius. – David Hammen Oct 20 '14 at 23:13
• Oh, so its the oblateness which makes such torque possible! So the kind of effect I was looking for would be more pronounced for an irregular body like Hyperion, at least if adjusted for its lower mass? And as for usefulness I have nothing in mind. Maybe non-atmospheric line of sight, for whatever reason. – LocalFluff Oct 21 '14 at 10:52
• Correct. Earth is very nearly spherical, so orbits are pretty simple (until you get near the Moon), with some corrections for the non-spherical gravity field, mostly J2. For highly irregular bodies like Hyperion, "orbits" close to the body can be much more complicated, and be noticeably affected by higher-order terms of the gravity field. – Mark Adler Oct 21 '14 at 15:13
• @LocalFluff - To see what high order deviations from sphericality can do to an orbit, read this article, Bizarre Lunar Orbits. – David Hammen Oct 21 '14 at 17:04
• That's similar to L5. You can look at the other answers for discussions of the Lagrangian points. The edited question was asking specifically for low polar orbits of Earth. – Mark Adler Oct 21 '14 at 22:47

http://en.wikipedia.org/wiki/Lagrangian_point

A satellite orbiting at the L1 point or L2 point would be in constant contact with the moon. Unfortunately, neither of these are particularly useful for keeping in contact with a lunar mission. The L1 point is between the earth and the moon, and would offer no advantages over just putting the reciever on earth. The L2 point is on the far side of the moon, and would therefore not be able to communicate with earth.

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19680015886.pdf

This article seems to be exactly what you want. They conclude that a satellite near L2 would work.

• That'd be a different way to solve a similar basic problem. I've edited my question to try to clarify what I was really looking for to begin with. And I'm not at all interested in the far side of the Moon here. – LocalFluff Oct 20 '14 at 19:02
• L4 and L5 also maintain a constant position relative to the moon and can see it. The only Lagrange point that doesn't always see the moon is L3. – Loren Pechtel Oct 20 '14 at 19:51

It is possible in several ways. Let's get the dumb ones out the way first.

1. As I commented, you can orbit the moon itself. Technically correct, since anything orbiting the moon is also orbiting the Earth.

2. You can make your satellite big enough so that it always has a line-of-sight. A giant pole somewhat longer than the Earth's diameter can be arranged to orbit so as have a point with in line-of-sight.

3. Similarly, you can make your satellite encircle the Earth (a Niven ring or Dyson sphere)

1. You can orbit at a Lagrange Point. As this diagram of exactly your situation shows, any except $L_3$ are fine:

1. I haven't run the numbers on this, but it should be possible to make a synchronized polar orbit with the moon. I thought up a simple example:

Say the orbital inclination of the moon were $0^{\circ}$ and circular. Set up your satellite in the same orbit but at inclination $90^{\circ}$ and behind by a quarter orbit. It might be a little hard to visualize, but in this configuration the line of sight never passes through the Earth.

• While not directly specified in the question, it appears the OP had in mind a considerably lower altitude orbit than 4 or 5 would admit, as in his EDIT, he clarifies: "...such a polar Earth satellite would have a relatively small altitude...". By Kepler's law, any satellite fully synchronous with the moon must have a semi-major axis equal to the moon's, which would hardly qualify as low-altitude (excepting L1 & L2 which arise from the more complex n-body problem). – Caleb Hines Oct 21 '14 at 1:24

If the purpose is satellite communication between Earth and the Moon, here's my ranking of the options, from simple and cheap to exotic and expensive:

1. No satellites, a set of laser ground stations operates from the Earth surface to one or two sites on the Moon. The Moon's Earth-facing side always has line-of-sight with Earth, it's a lot cheaper to create a ground-based laser network, and they don't have to be put on the end of a huge firework either. Moon bounce [wikipedia]
2. Earth-Moon L4/L5 Lagrange point satellite. One satellite, always has visibility of Earth and Moon. Pros are that you only needed one thing on a huge firework, and these points are stable so it should stay there without needing fuel for course correction. The downsides are that you still need an Earth-based network of ground stations or satellites to link to as the Earth turns, and data will be completing 2 sides of a triangle, so the already bad latency (1300ms) will be made worse (~2300ms). E-M L-points [hyperphysics]
3. 1 Earth sat in a wide polar orbit. By synchronizing the orbit with that of the Moon, you could ensure that the Earth never blocks line of sight. However, this would require both a wide orbit to minimise the angular size of the Earth and a long period to permit the Moon to transit the angular size of the Earth during a fraction of its orbit. Helpfully long periods and large orbital radii go hand in hand. I think an odd fraction of the Moon orbital period (1/3, 1/5, 1/7) would permit an arrangement which prevented direct alignment of the Earth between them. I don't know if the Sun would drag this out of line, so occasional propulsion might be needed.
4. Earth-Moon L1. Again one satellite, so one giant rocket, but this time it will need propulsion as L1 is not a stable point (when it drifts away from L1, it will need to propel itself back). This should be doable with ion thrusters though, so only a moderately enlarged firework. Being directly between Earth and the Moon would have 2 Pros - easy to find, and the lowest latency.
5. 2+ Earth sats, orthogonal Polar orbits or same orbit 180 degrees apart. One or other satellite can always see the Moon, as when one is obscured by the Earth, the other is not. Same requirement for a ground-based network. Lather, rinse repeat for more satellites.
6. 1 Earth sat in polar orbit (as per the question), but with a matter-scoop fuelled ion thruster performing continuous course correction. Making the satellite and putting it up there is straightforward, but the propulsion would be entirely experimental. It would be worth putting up a sister satellite with nuclear fuel so that you could spend a bit longer refining the technology when the first one seizes up.
• +1 for #3. It's a very elegant solution and the orbit wouldn't need to be all that terribly high; a polar orbit of altitude close to GEO (but polar) would easily suffice, with 1/27 moon period, moving in phase such that it's never eclipsed by Earth. – SF. Oct 21 '14 at 14:33
• ...correction. LEO of ~35700km is insufficient. An orbit of about 45000km with period of 1/20 Moon period would be sufficient; 1/20 of its circumference at its maximum distance converts to about 1.85 degree angular diameter from the Moon; Earth angular size is 1.84 degree when seen from the Moon. – SF. Oct 21 '14 at 15:04

Fillan Grady already gave the examples of orbiting a Lagrangian point, but based on your response to that, and subsequent edit, it seems you want a more traditional Earth orbit, rather than a Lagrangian "cheat". But I will point out that these points are valid ways to orbit the earth, and there is a reason they are so useful.

Bottom-line: If you discount the L-points, it's not very likely, and certainly not in a low orbit. Every satellite orbit must follow a great circle, and therefore, must cross the equator at two points. This is true, even for polar orbits. Furthermore, these crossing points will remain fixed, relative to the celestial sphere (unless you invoke some sort of gravitational precession), while the Moon revolves around the Earth. That means that at some point, the Moon will almost certainly be opposite of the Earth during this crossing.

What you have to do is find a polar orbit high enough (and with the correct period and phase, relative to the Moon's orbit) that the Earth is small enough to not occult the moon during any of these crossings, ever. Consider a satellite in geosynchronous orbit (24 hour period). A back of the envelope calculation says that the earth will appear to have an angular diameter of about 20 degrees. The angular portion of the Moon's orbit which this hides, on the opposite side the Earth, will be even greater than this. However, assuming a 27 day circular orbit, the Moon only covers about 13 degrees of its orbit per day, so if the Moon is just about to go behind the Earth during one orbit, it will not have come out the other side by the time your satellite returns to the same point 24 hours later.

Now, a geosynchronous polar orbit would already take quite a bit of energy to achieve, and it isn't even close to providing what you would need. You'd have to go quite a bit higher, and it would almost certainly be cheaper to launch to L1, or to use network of multiple satellites on lower-altitude, lower-inclination orbits. But you've said that you want a low-altitude orbit -- that will make the Earth appear bigger, which will increase the time the Moon is behind it, while simultaneously increasing the frequency with which the satellite is in this shadow.

One more option--not in orbit at all. You can use solar sails to hover a satellite over the north or south pole or Earth. The patent on this has expired, you're free to do it.

Note that such a satellite must be far out, the lag going through the satellite will be considerably greater than the Earth-Moon lag. On the flip side, it will retain communication with a equatorial-orbit satellite when it goes "behind" the moon so long as it's not in too low an orbit.

http://en.wikipedia.org/wiki/Statite

Yes. You could place it at any of the Lagrange points that have LOS to the Moon. The first solution I would explore would be to place it in orbit about the Moon itself on a route that is generally perpendicular to the Earth.

There should be a number of acceptable orbital solutions to this problem, and dealing with larger bodies farther out will cost the least fuel for maintenance (but a larger amount for launch).

Besides of orbiting Lagrangian points, orbiting around Earth or Moon of 90 degrees with respect to Earth-moon plane inclination can be a solution. There is a simple image I made.

First one (orbiting around Earth) can be the cheapest solution of all because it is low-orbit.

• This is exactly what isn't an option. Your orbit would have to precess at a rate of about 13° per day to constantly keep the Moon in direct line of sight, otherwise there's periods of the Moon's orbital position where it's not perpendicular to the satellite's orbital plane. The Earth's J2 only allows for precession rate good enough for Sun-synchronous (about 1° per day) but not Moon-synchronous (about 13° per day) as your image suggests. See here. – TildalWave Oct 21 '14 at 10:57
• A probe makes one complete revolution around the Earth in LOE in about 90 minutes (like ISS) not in 1 day. That corresponds about 13/24/60/60*90=0.013° per revolution. So only 0.013° shift per revolution is needed for a probe to always see a face of Moon. That IS an option. Am I right? – isidat Oct 21 '14 at 12:05
• No, it corresponds to about 13°/(24/1.5) or 0.8125° per 90 minute orbit (you calculated for a 90 s orbit). But this is irrelevant, it's just describing same rate change at different intervals. My point is that the satellite's orbit won't follow the Moon's rotation around the Earth, and during a bit less than half of the satellite's orbit the Moon will be obstructed by the Earth and not in direct line of sight. And you can't get such precession rate in orbit around Earth at orbital altitudes that are not deep within it. Your 90 s orbit is 433 km from Earth's center or 5,938 km below surface. – TildalWave Oct 21 '14 at 12:10
• I do not understand why satellite's orbit does not keep the Moon's rotation as long as its orbit shifts 0.8125° per satellite's revolution. The satellite won't move along a fixed path if you understand by looking at the image. 0.8125° change per revolution will keep the satellite always see the Moon. – isidat Oct 21 '14 at 12:35
• It would, but you simply can't get the precession rate needed around the Earth so it follows Moon's orbit. See e.g. Orbital perturbation analysis - The effect of the Earth flattening. Earth's $J_2$ simply isn't big enough to place a satellite in such an orbit above the Earth to follow precession rate of about 13° per day and be Moon-synchronous. So you either have to be co-orbital (Moon's orbital period at its altitude or its libration points), or use propulsion to phase in orbit. – TildalWave Oct 21 '14 at 12:46

Yes, it's possible .. A polar orbit that has a longitudinal transition of the exact amount of speed the moon circles the Earth would allow such a thing.

• Your answer is the same as Isidat's, see the comments to that answer for why this won't work. – Hobbes Oct 24 '14 at 12:18