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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

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    $\begingroup$ Look up Areostationary orbit. $\endgroup$ – DylanSp Dec 16 '16 at 16:45
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    $\begingroup$ @DylanSp Thank you! It helped a lot. Its hard to look up when you do not know how it is called :) $\endgroup$ – Pavel Janicek Dec 16 '16 at 16:50
  • $\begingroup$ It took me a while to read it areo instead of aero $\endgroup$ – Antzi Dec 16 '16 at 17:36
  • $\begingroup$ Sure it's possible, nobody's done it because there's no use cases for it. Here we use geostationary orbits for communications satellites but in order to study a planet you want an orbit that sees all parts of the place, and lets you get close to features. $\endgroup$ – GdD Dec 17 '16 at 12:53
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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!

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  • $\begingroup$ And even though it is between Mars' both moons, they are way too small to make an areostationary orbit unstable, right? $\endgroup$ – LocalFluff Dec 16 '16 at 21:49
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    $\begingroup$ 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. $\endgroup$ – hughes Dec 16 '16 at 22:01
  • $\begingroup$ 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! $\endgroup$ – hughes Dec 16 '16 at 22:03
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    $\begingroup$ 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. $\endgroup$ – uhoh Dec 16 '16 at 23:36
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    $\begingroup$ 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. $\endgroup$ – Mark Adler Dec 19 '16 at 5:28
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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.

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  • $\begingroup$ Interesting article - I was looking for a link or discussion about the station keeping required to manage the gravitational interaction with the two moons, but instead found this article which is probably now not based on the most recent data. $\endgroup$ – uhoh Dec 16 '16 at 23:46
  • $\begingroup$ I've asked this follow-up question. $\endgroup$ – uhoh Dec 17 '16 at 0:23

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