15.4 Geostationary Satellites

A geostationary satellite is an artificial satellite that orbits Earth in a circular path directly above the equator such that it appears stationary relative to a fixed point on Earth's surface. This is achieved because the satellite's orbital period exactly matches Earth's rotational period.

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Conditions for a Geostationary Orbit

For a satellite to be geostationary, three conditions must be satisfied simultaneously:

1. Circular orbit — the orbit must be perfectly circular.
2. Period of 24 hours — the orbital period must equal one sidereal day ($T=24\text{ h}=86,400\text{ s}$).
3. Equatorial plane — the orbital plane must lie directly over Earth's equator (the orbit is called the Geostationary Earth Orbit, GEO).
4. West-to-East direction — the satellite must rotate in the same direction as Earth's rotation.

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Derivation of Geostationary Orbital Radius

For a satellite in a circular orbit, the gravitational force provides the centripetal force:

$F_g=F_c$

$\frac{G{M}_{E}m}{r^2}=\frac{m{v}^2}{r}=mr{\omega }^2$

where:

• $G=6.67\times 10^{-11}{\text{ N m}}^2{\text{ kg}}^{-2}$ is the gravitational constant
• $M_E=5.97\times 10^{24}\text{ kg}$ is the mass of Earth
• $m$ is the mass of the satellite
• $r$ is the orbital radius from Earth's centre
• $\omega =\frac{2\pi }{T}$ is the angular velocity

Substituting $\omega =\frac{2\pi }{T}$:

$\frac{G{M}_E}{r^2}=r{\left(\frac{2\pi }{T}\right)}^2$

$G{M}_E=\frac{4{\pi }^2{r}^3}{T^2}$

Solving for $r$:

$\boxed{r=\sqrt[3]{\frac{G{M}_E{T}^2}{4{\pi }^2}}}$

Key observation: The orbital radius $r$ depends only on $G$, $M_E$, and $T$ — it is independent of the satellite's mass.

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Numerical Values for a Geostationary Satellite

Substituting $T=86,400\text{ s}$, $G=6.67\times 10^{-11}{\text{ N m}}^2{\text{ kg}}^{-2}$, $M_E=5.97\times 10^{24}\text{ kg}$:

$r=\sqrt[3]{\frac{(6.67\times 10^{-11})(5.97\times 10^{24})(86,400{)}^2}{4{\pi }^2}}$

$r\approx 4.23\times 10^7\text{ m}=42,300\text{ km}$

Since Earth's radius $R_E\approx 6,400\text{ km}$, the altitude above Earth's surface is:

$h=r-R_E\approx 42,300-6,400\approx 36,000\text{ km}$

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Orbital Velocity of a Geostationary Satellite

The orbital velocity is given by:

$v=\frac{2\pi r}{T}=\frac{2\pi \times 4.23\times 10^7}{86,400}\approx 3.07{\text{ km s}}^{-1}$

Alternatively, using the orbital velocity formula:

$v_o=\sqrt{\frac{G{M}_E}{r}}$

This is much slower than a low-Earth orbit satellite (which travels at ~7.9 km/s), consistent with the fact that higher satellites move slower ($v\propto \frac{1}{\sqrt{r}}$).

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Summary Table

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Applications of Geostationary Satellites

• Communications — TV broadcasting, telephone relay, internet (e.g., INTELSAT)
• Weather monitoring — continuous imaging of the same region (e.g., METEOSAT)
• GPS and navigation — though GPS uses medium Earth orbit, geostationary satellites assist
• Military surveillance

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