Areocentric orbits

[Ford Prefect]

Well, technically speaking I'm particularly curious about the areostationary orbit; ie. an areosynchronous orbit directly above the Mars' equator, with a period equal to the Mars' rotational period and an orbital eccentricity of approximately zero. Information regarding the various geocentric orbits (LEO, HEO, whatever) is readily available and easy enough to find. However, equivalent information for Mars is much harder to locate. Can anyone help me out?

[Surlethe]

The fact you want an orbital eccentricity of zero makes things very easy. Here's how you figure out the equations. Centripetal acceleration is given by Mars' gravitational field: a = GM/R^2, where R is the radius you're looking for. Meanwhile, using a fact from circular motion, a = w^2R, where w = 2Pi/T is the angular frequency of the orbit and T is the period of one Martian day. So equate them, solve for R, and you've got your orbital characterization.

[Ford Prefect]

Those equations don't look too daunting, so I'll try them out. Are there any other equations which can be used to determine different orbits? If I can do them for Mars I can do them for anywhere, really.

[Ariphaos]

Venus and Mercury don't really have them, neither does our moon but you can make 'moonstalks' pointing directly to and away from Earth.

It's actually a bit arbitrary - making them a bit faster than the rotation around gas giants, for example, can be quite useful.

[Surlethe]

Venus and Mercury don't really have them, neither does our moon but you can make 'moonstalks' pointing directly to and away from Earth.

It's actually a bit arbitrary - making them a bit faster than the rotation around gas giants, for example, can be quite useful.

Could you be a little clearer? Are you talking about space elevators on the other planets?

[Kuroneko]

The only caveats you might worry about is having T as a sidereal day rather than a solar day (8.864244e4 s for Mars), and that the standard gravitational parameter μ = GM is often known more accurately than either G or M individually, though not really for Mars (4.2828e13 m³/s²). The physical altitude would be less the equatorial radius, of course.

The resulting solution R³ = μT²/(4π²) is actually a particular case of Kepler's third law, and is valid even for elliptic orbits, where R is re-interpreted to be the semimajor axis of the orbit. This would not be an stationary orbit in the strict sense, but over the equator it would come back to the same place every sidereal day.

As a sanity check, make sure R is significantly less than the Hill-Roche sphere of H = a(1-e) [ m/(3M) ]^{1/3}, where a is the semimajor axis of the planet's orbit, e is its orbital eccentricity, and m/M is the ratio of planetary to stellar masses (or standard gravitational parameters, μ' = 1.32712440018e20 m³/s² for Sol). Orbits inside this sphere would not be in danger of being captured by the star, but such perturbation would cause precession, deforming it away from being stationary unless R/H is very small.

Venus has μ = 3.24859e14 m³/s², a = 1.0821e11 m, e = 0.00677323, T = 2.0997e7 s, giving an equatorial stationary orbital radius of R = 1.5366e9 m, which is unfortunately outside the Hill-Roche sphere H = 1.0e9 m. For the Moon, μ = 4.9027779e12 m³/s², a = 3.84399e8 m, e = 0.0549, T = 2.3606208e6 s, giving R = 8.84528e7 m, which is actually inside H = 5.8e7 m, but close enough for an attempted stationary orbit to significantly precess.

[Ariphaos]

Could you be a little clearer? Are you talking about space elevators on the other planets?

Err sorry. Exactly the opposite. You're not required to make the ring 'Jovisynchronous' or however you'd morph the cytherean term. So if you spin it faster than the planet's rotation, you can make it arbitrarily close to the planet's outer atmosphere.

This especially makes sense on gas giants, since there's nothing to tie a space elevator to.