If it runs out of fuel then its free fall is just the current orbit in which it will stay essentially forever (since it is well above any atmospheric drag that might slow it down). But in the unlikely event that it has the extremely large amount of fuel needed to kill all of its orbital velocity, (and if it actually uses the fuel for that purpose), then it will fall straight down to the earth in a time that one can calculate – either with a bit of calculus, or by the clever trick of applying Kepler’s law for a very thin ellipse of major axis length equal to the orbital distance.

[Kepler’s third law says “The square of a planet’s orbital period is proportional to the cube of the length of the semi-major axis of its orbit” so the period is proportional to the 3/2 power of the major axis; and the full axis of a narrow ellipse from the satellite down to the earth is just the radius (ie*half*the major axis) of its previous circular orbit. So the period of that ellipse, if completed, would be the geostationary period of 24 hours divided by a factor of 2^{3/2}. But the satellite only gets (a bit less than) halfway round before crashing, so the actual time for the fall is a bit less than 24/2^{5/2}=3\sqrt{2} or about 4 hours.]

P.S. For the Moon, since the orbital period is 28 days, the time to fall would be more like 5 days, but I think the claim attributed to Galileo may arise from a misreading of his argument that the Earth is spinning on its axis (basically something like: If the Earth was NOT rotating and instead everything on the celestial sphere was going round once every 24 hours, then the Moon, Sun, and stars might all be at the geostationary distance but only those near the equator would actually be in balance and those at the pole would just fall to Earth in 4 hours)

P.P.S. It is not clear that Galileo actually had a good understanding of orbital mechanics and the inverse square law, so if he estimated a falling time of 4 hours that may have been little more than a lucky guess