# What altitude would an aphrodiosynchronous satellite need?

A orbit has a period of 24 hours, and mostly stays over the same location on the surface of the Earth. According to this blogger, the equivalent term for Venus should be "aphrodiosynchronous".

Is such an orbit even possible? If so, what altitude (or radius) would such an orbit have?

• Scientists have stopped using planet-specific names for concepts that were originally developed for Earth and hence have the prefix geo. For example, scientists now use terms such as "geography of Mars", "geology of Mercury". Just use "Venus geosynchronous orbit". Emily Lakdawalla, the Solar System Specialist for The Planetary Society, agrees with me (at least with respect to Mars). – David Hammen Sep 22 '20 at 10:59
• A geosynchronous orbit for Earth has a period of one sidereal day (the time it takes for Earth to do a full 360° rotation around its axis), which is 23.933... hours, not 24 hours. This is due to the fact that as the Earth moves around the sun, it takes a little bit longer than the full rotation for the Sun to come back above the same meridian, which is what we usually call a day (the mean solar day). – jcaron Sep 22 '20 at 15:37
• cf, this Q&A: space.stackexchange.com/questions/40166/… -- I don't think planet-specific names like aphrodiosynchronous are really all that useful. – Polygnome Sep 22 '20 at 15:45
• @jcaron well, 23.933 does =24 to 2SF – Chris H Sep 23 '20 at 9:51
• @DavidHammen it's not quite true that scientists have generally stopped this sort of thing, at least the Juno team talks about perijoves all the time. apod.nasa.gov/apod/ap170603.html – leftaroundabout Sep 23 '20 at 11:27

Is such an orbit even possible?

TL;DR:
If the Sun wasn't around, yes, such an orbit is possible. But since the Sun is around, such an orbit is impossible.

About the name of the orbit

Quoting from Emily Lakdawalla, who has a bit more gravitas than some random file blogger,

What is a geostationary orbit like at Mars? I have to pause here for a brief discussion of semantics. The authors of this paper discuss "areostationary" for Mars orbits as opposed to "geostationary" for Earth, and Wikipedia uses the same convention, but I'm not a big fan of this sort of nomenclatural hair-splitting. You'd have to talk about "hermestationary" for Mercury, "cronostationary" for Saturn, "selenostationary" for the Moon, and so on. It gets tiresome. And while a very few people use "areology" to name the study of rocks on Mars and "selenology" to talk about rocks on the Moon, nearly everybody calls it all "geology" and a person who studies all that stuff a "planetary geologist." So I'm going to stick with calling it a "Martian geostationary orbit."

I'll call the subject of this question a Venus geostationary orbit.

The orbit's period and semi major axis length

A geosynchronous orbit has a period of 24 hours.

An Earth geosynchronous orbit has a period a bit less than 24 hours. In particular, geosynchronous satellites orbit the Earth once per sidereal day rather than once per mean solar day. The length of a Venus sidereal day is 243 days and 26 minutes, or 20996760 seconds.

The standard gravitational parameter for Venus is $$3.24858592\times10^{14}\, \mathrm{m}^3/\mathrm{s}^2$$. Via Kepler's third law, the semi-major axis length of a Venus geosynchronous orbit is $$a = \left(\mu\left(\frac T{2\pi}\right)^2\right)^{1/3}$$ or 1.53655 million kilometers.

Is such an orbit even possible?

Two widely used metrics for determining whether an orbit is possible are the Laplace sphere of influence and the Hill sphere. The Laplace sphere of influence approximates the distance at which the perturbing gravitational acceleration toward some other body (the Sun in this case) is equal to the gravitational acceleration toward the primary body (Venus in this case). The Hill sphere is a sphere whose radius is approximately the distance to the L1 and L2 Lagrange points. Arguing which of the two is correct is hair-splitting; they are both approximations.

The radius of the Laplace sphere of influence is given by $$r_L = R\left(\frac m M\right)^{2/5}$$ while the radius of the Hill sphere is given by $$r_H = R \left(\frac m{3M}\right)^{1/3}$$, where $$R$$ is the distance between the primary body and the third body, $$m$$ is the mass of the primary body, and $$M$$ is the mass of the third body. Plugging in the semi major axis length of the orbit of Venus about the Sun and the masses of Venus and the Sun yields radii of 0.616 million kilometers and 1.008 million kilometers for the radius of Venus's Laplace sphere of influence and Venus's Hill sphere.

Both the Laplace sphere of influence and Hill sphere are a bit generous with regard to stability. An orbit at greater than the Hill sphere radius is definitely not stable. Since a Venus geosynchronous orbit is more than 3/2 of the Hill sphere radius, such an orbit is not possible.

• +1 alone for just pointing out how silly it is to use terms like 'aphrodiosynchronous'! Of course, the rest of the answer is also excellent. – eps Sep 22 '20 at 16:28
• I must object by referring to a previous answer of yours! Laplace sphere of influence and Hill sphere answer different questions (not hair-splitting), as you yourself noted back then; it's the Hill sphere that bounds stable orbits. They scale differently and in general are not even close -- Hill sphere could be, e.g., 100x larger than Laplace sphere of influence for a small enough planet. – nanoman Sep 22 '20 at 20:53
• A most excellent hipologistic answer. – Schwern Sep 23 '20 at 18:00