# Making A Geosynchronous Orbit Impossible

Here's the idea: A geosynchronous orbit is an orbit that makes sure that the orbiting satellite is always above the same spot on the planet, so its orbital period is equal to a local day there. I've been wondering how to make the orbital radius of a geosynchronous orbit less than the radius of the planet. How would that be possible?

• I think this is pretty straight-forward if the planet is simply spinning relatively fast, especially if it has a deep atmosphere and you're okay with your orbit being impossible because it would be in sufficiently thick atmosphere as to be practically unsustainable. If my back-of-napkin math is correct, all else being equal, you could have this if Earth's day was about 1.4 hours long. (Disclaimer: this might cause other problems...) Dec 11 '20 at 18:21
• At the other extreme, if your planet is tidally locked, the only way to achieve a "geosynchronous orbit" is to match the planet's orbit around its star, which would be difficult aside from sitting at one of the Lagrange points. However, that doesn't meet your criteria of "the orbital radius of a geosynchronous orbit [being] less than the radius of the planet". Dec 11 '20 at 18:25
• Making the geosynchronous orbit less than the radius of the planet means the surface of the planet would be moving at greater than orbital velocity. Apart from that, your question focuses on the radius of the orbit, but there are other ways of making geosynchronous orbit difficult or impossible. An inconveniently located moon could do the job, making such orbits highly unstable. Dec 11 '20 at 18:50
• Good site etiquette is to wait to accept an answer for 24 hours to give a chance for people in other time zones to answer the question. Good question, though. Dec 12 '20 at 0:57
• Note that geosynchronous orbit is impossible around ....................... our moon !! Literally the nearest example other than Earth! Coincidentally, there was a good question about this recently on, i believe, Astronomy site. Dec 12 '20 at 5:19

Easy: Put a moon right above geostationary orbit.

If you have a 24h day, and a moon that takes 25h to circle the planet, any satellite orbit at geostationary height would be totally unstable. The satellite would spend 12 days on the other side of the planet than the moon does, and then it will move very close to a really heavy body over the next 6 days, and either end up crashing into the moon or planet, or do a slingshot to escape the planet entirely.

Note that this setup would allow for satellites at the five Lagrange points of the planet/moon system, but these will have the same orbital period as the moon does, which is not geostationary.

• This is the best idea so far. Dec 12 '20 at 4:02
• Would a moon in Earth’s geostationary orbit be stable over astronomical time spans? Dec 12 '20 at 11:18
• @Michael Yes. In this case the rotation of the planet would be tidally locked to the moons orbit. Also, its Lagrange points would be geostationary. When the moon is higher than geosynchronous, it will slowly be pushed further out by absorbing angular momentum from the planet's rotation; and when the moon is lower than geosynchronous, it will slowly loose height by accelerating the planet's rotation. A tidally locked planet is actually the only stable configuration. Dec 12 '20 at 11:44
• You are a genius. I was thinking of doing stuff like changing the planet's density, rotational velocity, surface gravity, etc, but the most obvious solution was right in front of me. Dec 12 '20 at 17:14

You certainly can't make a planet spin too quickly for geostationary orbits to be possible; it turns out that the planet would be on the verge of instability.

I'll assume that you're interested in particular in a geostationary orbit, the particular case where an orbiting body stays above the same point on the planet's surface. For this to be the case, the planet's rotation period must equal the orbital period, so by Kepler's third law, the rotation period $$P$$, the mass of the planet $$M_p$$ and the orbital radius $$r$$ must be related by $$r^3=\frac{GM_p}{4\pi^2}P^2$$ For a geostationary orbit to be impossible, the orbital radius required for a geostationary orbit must be less than $$R_p$$, the radius of the planet. Therefore, the condition you desire is $$R_p>\left(\frac{GM_p}{4\pi^2}P^2\right)^{1/3}$$ or $$P<\left(\frac{4\pi^2R_p^3}{GM_p}\right)^{1/2}=\left(\frac{3\pi}{G\rho_p}\right)^{1/2}$$ with $$\rho_p$$ the density of the planet. For an Earth-like planet, the critical period is $$\sim$$1.40 hours.

On the other hand, at that point, the centifugal acceleration at the surface would be equal to the force of gravity from Earth itself, and it seems likely that the planet would be rotating nearly fast enough for the planet to rapidly become unstable. If you think about the problem you're describing, this makes sense. After all, you could also interpret the critical period as the period at which a piece of the planet at the surface would be moving fast enough to reach orbit.$$^{\dagger}$$

On the other hand, you could also consider slowing down the planet's rotation to the point where a geosynchronous orbit would be beyond the planet's Hill sphere. Outside the Hill sphere, the planet cannot retain satellites. Venus, for example, has a Hill sphere of roughly one million kilometers and a rotation period of 243 days. An object orbiting at the edge of Venus' Hill sphere would have a period of 127 days - meaning that a geosynchronous orbit should be impossible.

Now, in the question, you specified that you wanted a geosynchronous orbit to require a radius less than the planet's radius; we've seen that such a thing is impossible. However, you can make geosynchronous orbits impossible if you remove that requirement.

$$^{\dagger}$$ It also appears that the critical period is quite close to the dynamical timescale, certainly to within an order of magnitude - if it makes sense to talk about such a quantity for a solid planet. This, too, hopefully makes sense intuitively.

• Well, nice to see you came up with the same numbers as I did. I'm not entirely sure the planet would "tear itself apart", but its stability would be pretty shaky. Dec 11 '20 at 18:27
• Or, you could slow the planet's rotational speed so that any body attempting a geosynchronous orbit will orbit far away enough to be captured by the gravity of another member of its solar system. Or, make the planet's gravity so strong that any geosynchronous object will at least experience enough drag for it to lose speed until it plummets to the ground. Dec 11 '20 at 18:30
• @Matthew Thanks for the suggestion - I toned down the wording. Dec 11 '20 at 18:30
• OTOH, the OP specified "the orbital radius of a geosynchronous orbit [is] less than the radius of the planet" (emphasis added). In which case it will definitely try to at least reduce in radius, or something... That said, "rapidly become unstable" does seem better. So, short of time being highly warped in the vicinity (which is heavily into handwaving territory)... Dec 11 '20 at 18:36
• @TysonDennis I've added a section on extremely long periods for the sake of completeness, but they do violate your requirement that the orbital radius being less than the planetary radius, as Matthew pointed out. Dec 11 '20 at 18:42

For Venus, synchronous orbit is 1 533 883 km (How to calculate the synchronous orbit?) while its Hill sphere radius is only 1 000 000 km. That's a "slow spinner" preventing synchronous orbits.

For fast spinners, HDE 226868 put out calculations showing that for rocky planets such fast spinning is not likely possible. For gas giants, we can theorize a fluffier, faster spinning Saturn (real Saturn synchronous orbit radius is 112 043 km) so that atmospheric drag at that altitude can be too high to maintain a stable orbit.

Note that for a fast spinner, any atmosphere above the synchronous orbit would be lost to space, so this setup might not be stable for geological periods of time.

• Note that in such atmosphere the drag will not be really a problem. Dec 12 '20 at 23:07

There are two ways of preventing geosynchronous satellites.

The first is for the planet to spin too fast: assuming a planet identical to Earth, and ignoring distortion of the atmosphere due to the spinning, if a day was 87 minutes long an orbit equalling that would have an altitude of 133km; at that altitude, atmospheric drag will bring it down fairly fast. If the day was 86 minutes long, the altitude would be 82km; it is absolutely coming down. At 85 minutes long, the altitude is 32 km, and with a day 84 minutes long, it would need a negative altitude.

The second is for the planet to spin too slow. The slower the spin, the further out the satellite would need to be in order to be geosynchronous, and at some point Earth's gravity won't be sufficient to maintain it in a stable geosynchronous orbit.

• You cannot rotate a planet so fast that geostationary orbit is inside the atmosphere: The air above geostationary orbit would be moving too fast for its height and be thrown out into space. No matter how close you put geostationary orbit to the surface, you will be able to place a satellite there. There might not be enough air below geostationary orbit to sustain life, but satellites would happily hover a hundred meters over the surface. Dec 12 '20 at 0:26
• Which is why I specifically mentioned ignoring the effects on the atmosphere. Dec 12 '20 at 1:14
• @KeithMorrison It’s not just the effect on the atmosphere. Anything on the equator at ground level will be virtually weightless. There will be an enormous “tidal” bulge in the oceans. Dec 12 '20 at 18:35

Make a large enough and close enough satellite (a moon or a huge asteroid) that will rotate at a significant angle to the geostationary orbit. The influence of it's gravity should make the geostationary orbit unstable. It won't prevent stuff from staying there for short (or relatively short) periods of time though.