I assume you are thinking a geostationary orbit. (geosynchronous orbit above the equator, something on this always remain over the same spot.)
[I am unfamiliar with mathjax, edit by someone, who is, is appreciated.]
Geostationary orbits satisfy this equation:
(2*pi/T)2 R = G M / R2 (where R is the radius of the orbit, T is the sidereal period, G is the gravitational constant, and M is the mass of the planet.)
You get the radius as:
R = 3th root( G M / (2*pi/T)2 ) This yields 42,267 kilometers in Earth's case, the sum of the famous 35,786 and Earth's radius.
The orbital velocity: v = (2*pi/T)*R, about 3074 m/s by Earth.
The lowest deltaV orbit from here, witch touches the planetary surface has periapsis at sea level, and apoapsis at 42,267km. (Sort of Hohmann transfer to surface orbit.)
Such orbit would have an orbital velocity at apoapsis of v1 = sqrt(G M ( 2/R - 2/(r+R) ) )
(where r is the radius of the planet, and R is the geostationary radius as above.) For Earth it is 1571 m/s. To deorbit, you have to lower your velocity by v-v1
So first of all you need 1500 m/s deltaV.
The time to periapsis will be t = pi*sqrt( ( (r+R)/2)3/ (G M) )
It would take 18874 s to touch down on Earth.
But on periapsis you would have an orbital velocity of v2 = sqrt(G M ( 2/r - 2/(r+R) ) )
The ground only moves with (2*pi/T)*r, so you have to decelerate the difference to land. (instead of hypervelocity crash.)
In Earths case you would need 9940 m/s additional deltaV for landing.
This is quite a lot. But if the planet has a suitable atmosphere, you could use aerobraking and parachutes to do the most of the deceleration.
Using these formulas, one can calculate deorbiting dV for any celestial body.
The ascent would ideally take the same time and dV. But since atmospheric and gravity drag work against you in liftoff case, a margin of some thousand m/s is needed. (Depending on planet size and atmosphere thickness.)
We need 11 hour for minimal dV landing and getting back on GEO.
A spacecraft with chemical engines to accomplish this on Earth would have a mass ratio of roughly 10. This is impossibly high for a compact, entry capable, non-staging spacecraft.
If we would have engines with better specific impulse and ALSO decent thrust-to -weight ratio, and capability to work efficiently in atmosphere,(Say 50,000m/s exhaust velocity and 20,000N thrust.) we could do it. But such a thing needs 0.5 GW thrust power, making it into a bulky fusion drive with large heat radiators.
My best idea (for futuristic but physically plausible solution) would be a solid core antimatter ramjet/rocket. It would use air as reaction mass during liftoff, (and also as coolant) and once orbit is achieved, make GEO transfer by low thrust - high efficiency plasma electromagnetic drive.
If the drive is overkill enough, you can make transfer with continuos thrust too, reducing needed time, but raising fuel consumption.
Of course, there is no way you could maintain 'syncroisedness' during ascent/descent, but if you have plenty of time to wait, or powerful continuos thrust drive, you can manage landing on the same spot you have orbited above before.
11 hour is the basic time, with no waiting (for landing on the same spot or getting back on the same GEO position) but minimal dV elliptical trajectory. If your engine is very efficient, it can be greatly
All distances in meter and all times in second. If I have miscalculted something (it is late) let me know.