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For spherical weak-field gravity, a stationary clock at radius $r_g>R$ gains the displayed proper time relative to a surface clock over coordinate interval $t$. A geostationary orbit has $r_g=(GM/{\Omega }_{\oplus }^2{)}^{1/3}$. This is the potential contribution only. Combining it with orbital and surface rotational special-relativistic terms requires a consistent reference frame and must not be silently included in a gravitational-only comparison.

A geostationary orbit is a prograde equatorial circular Kepler orbit whose orbital period equals one sidereal day, so its satellite remains above a fixed location on Earth's equator. It is the terrestrial case of a planet-stationary orbit. Its radius follows from Newton's law of universal gravitation and circular acceleration: $r=(G{M}_E{T}^2/(4{\pi }^2){)}^{1/3}$, approximately $42164\mathrm{km}$ from Earth's centre. Its altitude above the surface is smaller by Earth's radius.

An elliptical transfer trajectory whose apogee is near the radius of a geostationary orbit. Apogee maneuvers can circularize the trajectory and reduce inclination.

An Earth orbit with period equal to one sidereal day. A geostationary orbit additionally requires circular, prograde, equatorial geometry.

The Geostationary Operational Environmental Satellite series, used for frequent weather observations from geostationary orbits.

A European meteorological satellite series including geostationary orbit missions for repeated regional weather observations.

A geostationary orbit has the same angular velocity as the Earth. Let $M_E=4\pi \rho R^3/3$ and $r_g=\eta R$. Balancing the required centripetal acceleration with the exterior gravitational field givesThe orbital plane must be equatorial and the motion prograde for the satellite to remain over one point. The uniform density assumption supplies $M_E$ in terms of the specified $\rho$ and $R$.

For the uniform unloaded space-elevator cable, let the constant linear mass density be $\lambda$ and the outer endpoint be $b={\eta }^{\prime }R$. In the co-rotating frame the outward effective force per unit length is $\lambda ({\omega }^{2}r-G{M}_E/r^2)$. The cable is just clear of the surface, with neither a ground load nor an endpoint counterweight, so both ends have zero applied tension. The integrated effective force must vanish:Using $G{M}_E={\omega }^2{\eta }^3{R}^3$, this becomesThe root ${\eta }^{\prime }=1$ is a zero-length cable. The physical nonzero cable hasIt extends well beyond the geostationary orbit so its outer centrifugal force balances the inner segment's weight. If a cable is anchored with a nonzero endpoint load or has a counterweight, that load changes the integrated balance; its length is not fixed by the zero-end-load formula.

An element of the rotating cable satisfiesWith $T(R)=0$, integration givesIt increases below $r_g$, decreases above $r_g$, and returns to zero at $b$. HenceThe tension is greatest at the geostationary orbit, where local effective gravity is zero. A zero local effective force does not mean zero tension: that section transmits the loads of the entire inner and outer cable portions.

Use the supplied planet mass $5.9\times 10^{24}\mathrm{kg}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, and adopt mean Earth radius $R_E=6371\mathrm{km}$. For a geostationary orbit, use one sidereal day, $T\simeq 86164\mathrm{s}$. The adopted radius and rotation period follow NASA's fact sheet.^[ref]The slightly smaller radius than the usual terrestrial value follows from using the question's smaller Earth mass. For coplanar prograde orbits, the preceding formulas giveThe escape transfer has $a\to \infty$, so its first tangential burn reaches the local escape speed $\sqrt{2\mu /r_0}$. No final circularization at infinity is needed:Thus the geostationary transfer requires about $0.687\mathrm{km}{\mathrm{s}}^{-1}$ more total impulse, or about 22 percent more, despite leaving the spacecraft gravitationally bound. Reaching the outer radius and circularizing there are separate requirements. Changing the starting orbit's inclination would add a plane-change cost absent from this calculation.

The wavelength at $12\mathrm{GHz}$ is $\lambda =c/f\simeq 0.0250\mathrm{m}$. For an ideal circular aperture of diameter $D=0.850\mathrm{m}$, the Rayleigh criterion places one source at the first minimum of the other's Airy disk:The customary full-ring packing estimate for the number of resolvable positions in a geostationary orbit isThis is an ideal angular-packing estimate. It identifies apparent angular spacing with spacing around the orbital ring and disregards Earth's occultation and station location. An individual ground station cannot see the whole ring; exact conversion from orbital longitude to sky direction requires its location and the orbital radius. Also, a practical tapered antenna illumination changes the ideal circular-aperture response.

Use a circular Kepler orbit to eliminate the terrestrial gravitational parameter: $T^2/r^3$ is constant for both satellites. The reference orbital radius is $r_1=(6400+150)\mathrm{km}=6550\mathrm{km}$, measured from Earth's centre. A geostationary orbit is circular, equatorial and prograde, with the sidereal rotation period, approximately $86164\mathrm{s}$. ThusIts altitude is about $3.61\times 10^4\mathrm{km}$ above the stated surface. The approximate low-orbit input sets the accuracy; replacing it by a modern measured gravitational parameter would answer a different numerical estimate. A $24$-hour approximation changes the radius by less than $0.2\%$. Matching the period alone would not make an inclined or eccentric orbit stationary above one point.

For two circular Kepler orbits about the same body, Kepler's third law gives $T^2/r^3=\text{constant}$. The reference radius is measured from Earth's centre:A geostationary orbit is prograde and equatorial, with one sidereal day as its orbital period. Using $T_g\simeq 1436\min$ and the supplied $T_1=109\min$,Using $24\mathrm{h}$ instead changes this estimate by less than one per cent. This is the orbital radius, not the altitude; the corresponding altitude is about $3.00\times 10^4\mathrm{km}$.

There is a numerical inconsistency in the source data. The quoted reference orbit implies $G{M}_E=4{\pi }^2{r}_1^3/T_1^2\simeq 2.56\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, appreciably below the measured terrestrial value, about $3.99\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. Using the latter instead gives the familiar geostationary orbit radius $4.22\times 10^4\mathrm{km}$. The boxed estimate consistently uses the period actually printed in the question.

The transfer is a Hohmann transfer orbit. Write $r_0$ for the original circular Kepler orbit radius and $a_t$ for the transfer ellipse's semi-major axis. The diagram measures both radii from Earth's centre, giving $r_1=1.2r_0$, and henceThe Hohmann transfer time is half the transfer orbital period, taking the satellite to the opposite end of the elliptic orbit. By Kepler's third law, relative to the original geostationary orbit's orbital period $T_0$,Using one sidereal day, $T_0=86164\mathrm{s}$, gives $t=4.9703\times 10^4\mathrm{s}=13.806\mathrm{h}$; the usual $24\mathrm{h}$ approximation gives $13.844\mathrm{h}$. The arrival impulse changes the velocity into that of the new circular Kepler orbit without adding coast time in the impulsive model.

There is a real discrepancy between the wording and the diagram: an increase in altitude above Earth's surface is different from the radius increase drawn. If altitude is interpreted literally, let $R_E$ be Earth's radius. Thenusing $R_E\simeq 6371\mathrm{km}$ and $r_0\simeq 42164\mathrm{km}$. Thus $13.8\mathrm{h}$ answers the explicitly labelled geometry; $13.5\mathrm{h}$ answers the literal altitude wording.

A geostationary orbit provides continuous viewing of approximately the same terrestrial region, favouring fast cloud evolution, storm tracking and the diurnal cycle. The large range makes fine spatial resolution demanding, high latitudes have poor viewing geometry, and a single platform has no global coverage. A polar orbit gives much smaller range and broad latitude coverage as Earth rotates beneath it, favouring detailed global mapping; any one location is sampled intermittently. A Sun-synchronous orbit can hold approximately fixed local solar time for comparing observations, but that sampling can miss much of the diurnal evolution. The science therefore sets a trade between spatial coverage, spatial resolution and temporal sampling.

The Moon's very slow, tidally locked rotation makes a lunar stationary orbit fundamentally problematic. Let its orbital angular velocity around Earth be $n$, approximately also its spin rate. For lunar mass $m$, terrestrial mass $M$, and separation $a$, the two-body estimates giveThus $r_{\mathrm{sync}}\simeq 3^{1/3}{r}_H$ lies outside the lunar Hill sphere. Earth's tidal perturbations prevent the desired simple stable, lunar-centred stationary orbit. This is stronger than merely saying that the required orbit is high; specially maintained three-body orbits are a different construction.

A geostationary orbit is a circular, prograde, equatorial orbit with period equal to one sidereal day, approximately $23\mathrm{h}56\min$. 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 givesOne 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.^[ref]

A synchronous circular Kepler orbit has an orbital period equal to its central body's sidereal rotation period $T$. For the ideal spherical planet of mass $M$, Kepler's third law givesThe satellite's own mass cancels in the test-particle approximation. The radius must exceed the planet's physical radius for an external orbit.

Matching periods alone does not keep the satellite above a fixed surface point. That stronger condition requires a prograde equatorial circular orbit, giving a planet-stationary orbit; the terrestrial case is a geostationary orbit. An inclined synchronous circular orbit moves north and south relative to the equator despite its matching period. This distinction is useful when choosing between repeated ground coverage and an apparently fixed satellite.

A radial co-rotating cable of constant linear mass density $\lambda$ extends from planetary radius $R$ to $b={\eta }^{\prime }R$, with zero applied endpoint tension. If the geostationary orbit has radius $\eta R$, integration of the effective force gives ${\eta }^{\prime }({\eta }^{\prime }+1)=2{\eta }^3$ for a nonzero cable. At position $r$ its tension is $\lambda [GM(1/R-1/r)-{\omega }^2(r^2-R^2)/2]$. Differentiation shows the maximum at the geostationary orbit. An anchor load or counterweight alters the integrated condition.