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
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!
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.
