10

Hypothetical question: Are we able launch an orbiter from Earth to Mars in a way that such orbiter ends on "geostationary" position, hovering always one specific place over Mars?

My thought process behind this is that if we are able to do it, such orbiter could ease future communication with ground level missions

Pavel Janicek
  • 1,571
  • 1
  • 10
  • 19

2 Answers2

14

In general, (.*)stationary orbits are simply the altitude where a circular orbit has a period equal to the rotational period of the central body.

In order to find this altitude, we can transform Kepler's third law to solve for R:

Kepler's third law: $\frac{T^2}{R^3} = \frac{4\pi^2}{G M_{central}}$

Solved for $R$: $R = \sqrt[3]{\frac{T^2 G M_{central}}{4\pi^2}}$

Using $M_{earth}$ and $T_{earth}$:

$R = \sqrt[3]{\frac{(86164)^2 * (6.67e-11) * (5.97e24)}{4\pi^2}} = \sqrt[3]{\frac{2.97e24}{39.5}} = 4.22e7m = 42,200km$,

where 86164 seconds is the approximate length of Earth's sidereal day (~23h, 56m, 4s).

Subtracting the radius of Earth which is ~6370km, this is an altitude of ~35,800km. This can be confirmed by google.

Using $M_{mars}$ and $T_{mars}$ (which is 40 minutes longer than an Earth day):

$R = \sqrt[3]{\frac{(88643)^2 * (6.67e-11) * (6.39e23)}{4\pi^2}} = \sqrt[3]{\frac{3.36e23}{39.5}} = 2.04e7m = 20,400km$,

where 88643 seconds is the approximate length of a Martian sidereal day (~24h, 37m, 23s).

Subtracting the radius of Mars which is ~3390km, this gives an altitude of about 17,000km.

You can easily look up and substitute the numbers of any celestial body for this!

uhoh
  • 148,791
  • 53
  • 476
  • 1,473
hughes
  • 256
  • 1
  • 3
  • And even though it is between Mars' both moons, they are way too small to make an areostationary orbit unstable, right? – LocalFluff Dec 16 '16 at 21:49
  • 2
    I don't think the moons would exert more than a few micronewtons of force at their closest. To remain stable for months or years you'd probably need some kind of stationkeeping anyway - other forces like solar gravity, solar wind and radiation pressure would also contribute to drift. – hughes Dec 16 '16 at 22:01
  • Interestingly, since one moon is above this altitude and the other is below, their apparent motion from the surface of Mars would be in opposite directions! – hughes Dec 16 '16 at 22:03
  • 1
    Nice answer! Remember to use the sidereal day, about 23h 56m 4.09s (for Earth) and for Mars. Just a small difference, but it's a good concept to remember. – uhoh Dec 16 '16 at 23:36
  • Wikipedia says that Phobos causes some orbital resonance effects that would increase station-keeping costs, but doesn't list a source. – DylanSp Dec 16 '16 at 23:50
  • Good point about the sidereal day - it would only make about a 0.2% difference, but understanding why the sidereal day is used is a fun thought experiment! – hughes Dec 17 '16 at 00:25
  • It would be really great if you could go ahead and make the correction within your equation. People learn a lot from answers, so it's better to fix something if it's not correct. I know the correction is small, but the concept of the sidereal day vs solar day is really important here, and so if you are going to show the length of the day, it's not good to show the wrong one to future readers. – uhoh Dec 17 '16 at 01:08
  • 4
    Yes, you can get the numbers for any celestial body, but it wouldn't work for Venus or Mercury, since those orbits would not be stable, being outside of the Hill radius with respect to the Sun. – Mark Adler Dec 19 '16 at 05:28
7

Such an orbit is called an areostationary orbit. It's possible; the orbit would be about 17,000 km above Mars's surface. It's never been done, though.

DylanSp
  • 1,958
  • 1
  • 14
  • 25