= Solution
A <geostationary orbit> is a circular, prograde, equatorial orbit with period equal to one <sidereal day>, approximately $23\ \mathrm h\ 56\ \mathrm{min}$. Its satellite therefore remains above one longitude on the equator. Merely having a one-day period gives a <geosynchronous orbit>; an inclined or eccentric version still moves across the ground observer's sky.
Balancing <gravitational acceleration> and <centripetal acceleration> gives
$$
r=\left(\frac{GM_ET^2}{4\pi^2}\right)^{1/3}\simeq42164\ \mathrm{km},\qquad h=r-R_E\simeq35786\ \mathrm{km}.
$$
One route is injection into an elliptical <geostationary transfer orbit>, followed by a burn near apogee to circularize and remove the remaining inclination; direct or low-thrust transfers are also possible. Subsequent <station keeping> controls longitude, inclination and eccentricity.
The fixed wide view permits frequent observations of weather, clouds and rapidly changing storms, as in <Meteosat> and <GOES>. The tradeoffs are large range, generally coarser spatial sampling for comparable optics, oblique views at high latitudes and poor polar coverage. It is valuable for following change through the day rather than obtaining global coverage with one spacecraft.<Catalog of Earth satellite orbits by NASA Earth Observatory>[“High Earth Orbit”: geosynchronous radius, the circular/equatorial conditions for geostationarity, and weather-monitoring applications.]{ref}
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