KAHIBARO
Table of Contents
Synchronous orbit around Earth
A geostationary satellite is a special satellite that appears to remain fixed over one point on Earth's equator. To an observer on the ground, it seems not to move across the sky. This happens because the satellite orbits Earth with the same angular speed as Earth's rotation.
A satellite that matches Earth's rotation period is called geosynchronous. A geostationary satellite is a more specific case. It must be geosynchronous, it must move in a circular orbit, and its orbit must lie in the equatorial plane. Only then does it stay above the same point on Earth all the time.
A satellite is geostationary only if all three conditions are satisfied:
- Its orbital period equals Earth's rotation period.
- Its orbit is circular.
- Its orbit is in the equatorial plane.
If any one of these is not true, the satellite is not geostationary.
Why it stays above one point
Earth rotates from west to east. If a satellite also moves from west to east and completes one orbit in exactly the same time that Earth completes one rotation, then the satellite and the ground below stay aligned.
If the orbit were tilted, the satellite would appear to move north and south in the sky. If the orbit were elliptical, its speed would vary and it would drift east and west. So the orbit must be circular and directly above the equator.
Orbital period and altitude
To be geostationary, the orbital period must equal one sidereal day, which is about 23 hours 56 minutes, or about $86164 \, \text{s}$. This is the time Earth takes to rotate once relative to distant stars.
The gravitational force provides the centripetal force needed for the circular orbit. So we set
$$
\frac{G M_E m}{r^2} = \frac{m v^2}{r}
$$
where $M_E$ is Earth's mass, $m$ is the satellite's mass, and $r$ is the distance from Earth's center to the satellite.
Using $v = \frac{2\pi r}{T}$, we get
$$
\frac{G M_E}{r^2} = \frac{4\pi^2 r}{T^2}
$$
which gives
$$
r^3 = \frac{G M_E T^2}{4\pi^2}
$$
and therefore
$$
r = \left( \frac{G M_E T^2}{4\pi^2} \right)^{1/3}
$$
For Earth, this gives
$$
r \approx 4.22 \times 10^7 \, \text{m}
$$
from Earth's center.
Since Earth's radius is about
$$
R_E \approx 6.37 \times 10^6 \, \text{m}
$$
the altitude above Earth's surface is
$$
h = r - R_E \approx 3.58 \times 10^7 \, \text{m}
$$
or about
$$
h \approx 35{,}800 \, \text{km}
$$
This is the well known altitude of geostationary orbit.
For a geostationary satellite around Earth,
$$
T = 86164 \, \text{s}
$$
and
$$
r = \left( \frac{G M_E T^2}{4\pi^2} \right)^{1/3}
$$
Numerically,
$$
r \approx 4.22 \times 10^7 \, \text{m}, \qquad h \approx 3.58 \times 10^7 \, \text{m}
$$
Direction of motion
A geostationary satellite must orbit in the same direction as Earth's rotation. This is called a prograde orbit. If it orbited in the opposite direction, even with the same period, it would sweep rapidly across the sky instead of staying fixed.
Ground appearance
Because it appears fixed in the sky, a geostationary satellite is very useful for continuous communication with a large region of Earth. A ground antenna can point in one direction and remain there, instead of tracking the satellite across the sky.
From high latitudes, a geostationary satellite appears low above the horizon. Near the poles, geostationary satellites are very difficult or impossible to use because they lie near or below the horizon.
Comparison with other satellite orbits
Not all communication satellites are geostationary. Some satellites are geosynchronous but not geostationary, because their orbits are tilted or elliptical. These satellites return to the same general pattern each day, but they do not stay fixed at one point in the sky.
| Orbit type | Period | Orbit shape | Orbit plane | Appearance in sky |
|---|---|---|---|---|
| Geostationary | Equal to Earth's rotation period | Circular | Equatorial | Fixed |
| Geosynchronous | Equal to Earth's rotation period | May be elliptical | May be tilted | Repeats daily path |
| Low Earth orbit | Much shorter | Usually near circular | Various | Moves quickly across sky |
Main uses
Geostationary satellites are widely used for telecommunications, television broadcasting, weather monitoring, and some data relay systems. Their greatest advantage is continuous coverage of the same region. Their large altitude, however, means signals take a noticeable time to travel to and from the satellite.
Signal delay
Because geostationary satellites are far from Earth, communication signals need time to travel. A signal going from the ground to the satellite travels about $35{,}800 \, \text{km}$ upward, and often another similar distance back down.
Using the speed of light $c \approx 3.0 \times 10^8 \, \text{m/s}$, the time for one path of about $3.58 \times 10^7 \, \text{m}$ is roughly
$$
t \approx \frac{3.58 \times 10^7}{3.0 \times 10^8} \approx 0.12 \, \text{s}
$$
So a signal that goes up and back down takes about
$$
0.24 \, \text{s}
$$
not counting extra delays in equipment. This is why satellite phone or television links can have a small but noticeable delay.
A major practical feature of geostationary satellites is this tradeoff:
large altitude $\rightarrow$ wide coverage and fixed sky position,
but also
large altitude $\rightarrow$ noticeable signal delay.
Summary formula and idea
The key idea is simple. A geostationary satellite is placed at exactly the radius where gravity produces the circular motion needed for a period equal to Earth's rotation period. At that altitude, and only in a circular equatorial prograde orbit, the satellite appears motionless relative to the ground.
Essential result:
A geostationary satellite is a circular equatorial satellite with orbital period equal to Earth's rotation period. Its orbit radius is
$$
r = \left( \frac{G M_E T^2}{4\pi^2} \right)^{1/3}
$$
and for Earth its altitude is about
$$
h \approx 35{,}800 \, \text{km}
$$