A geosynchronous satellite has an orbital period equal to the rotation period of the central body.
For Earth, this means the satellite repeats its position relative to Earth once per sidereal day.
A geostationary satellite is a special geosynchronous satellite. It must be:
- in a circular orbit,
- above the equator,
- moving in the same rotational direction as Earth,
- with an orbital period equal to Earth’s rotation period.
Then it appears fixed above the same point on Earth’s equator.
1. Circular-orbit period
For a circular orbit around a spherical body of mass \(M\), the orbital period is
\[
T=2\pi\sqrt{\frac{r^3}{GM}}.
\]
Here:
- \(T\) is the orbital period,
- \(r\) is the orbital radius measured from the center of the body,
- \(G\) is the gravitational constant,
- \(M\) is the central mass.
2. Synchronous-orbit condition
A synchronous orbit requires
\[
T_{\mathrm{orbit}}=T_{\mathrm{rot}}.
\]
So we set
\[
T_{\mathrm{rot}}=2\pi\sqrt{\frac{r^3}{GM}}.
\]
Squaring both sides gives
\[
T_{\mathrm{rot}}^2
=
4\pi^2\frac{r^3}{GM}.
\]
Rearranging gives
\[
r^3=\frac{GM T_{\mathrm{rot}}^2}{4\pi^2}.
\]
Therefore, the synchronous orbital radius is
\[
\boxed{
r_{\mathrm{sync}}
=
\sqrt[3]{\frac{GM T_{\mathrm{rot}}^2}{4\pi^2}}
}.
\]
3. Altitude above the surface
The calculator reports altitude above the surface. If \(R\) is the radius of the central body, then
\[
\boxed{
h_{\mathrm{sync}}=r_{\mathrm{sync}}-R
}.
\]
A positive altitude means the synchronous orbit is outside the surface.
If \(h_{\mathrm{sync}}<0\), the required circular synchronous orbit would be inside the body and is not physically possible as an external satellite orbit.
4. Orbital speed
The circular orbital speed at radius \(r\) is
\[
v=\sqrt{\frac{GM}{r}}.
\]
At the synchronous radius,
\[
\boxed{
v_{\mathrm{sync}}
=
\sqrt{\frac{GM}{r_{\mathrm{sync}}}}
}.
\]
The same speed can also be written using angular speed:
\[
\omega_{\mathrm{rot}}=\frac{2\pi}{T_{\mathrm{rot}}}.
\]
\[
v_{\mathrm{sync}}=\omega_{\mathrm{rot}}r_{\mathrm{sync}}.
\]
These two expressions agree only at the radius where gravity supplies exactly the centripetal acceleration needed for that angular speed.
5. Why only one altitude is geostationary
If a satellite tries to have Earth’s angular speed at an arbitrary radius, the required centripetal acceleration would be
\[
a_{\mathrm{required}}=\omega_{\mathrm{rot}}^2r.
\]
Gravity provides
\[
a_g=\frac{GM}{r^2}.
\]
A free circular orbit requires
\[
a_g=a_{\mathrm{required}}.
\]
Therefore,
\[
\frac{GM}{r^2}=\omega_{\mathrm{rot}}^2r.
\]
This equation has one positive solution for \(r\), so there is one true circular synchronous radius for a given mass and rotation period.
6. Earth geostationary example
For Earth, use approximately
\[
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg},
\qquad
R_{\oplus}=6.378137\times10^6\ \mathrm{m}.
\]
The sidereal rotation period is
\[
T_{\mathrm{rot}}\approx86164.1\ \mathrm{s}.
\]
The synchronous radius is
\[
r_{\mathrm{sync}}
=
\sqrt[3]{
\frac{(6.67430\times10^{-11})(5.9722\times10^{24})(86164.1)^2}
{4\pi^2}
}.
\]
\[
r_{\mathrm{sync}}\approx4.2164\times10^7\ \mathrm{m}.
\]
Convert this radius into altitude:
\[
h_{\mathrm{sync}}=r_{\mathrm{sync}}-R_{\oplus}.
\]
\[
h_{\mathrm{sync}}
\approx
4.2164\times10^7-6.378137\times10^6.
\]
\[
h_{\mathrm{sync}}\approx3.5786\times10^7\ \mathrm{m}.
\]
Therefore,
\[
\boxed{
h_{\mathrm{sync}}\approx35{,}786\ \mathrm{km}
}.
\]
The circular speed is
\[
v_{\mathrm{sync}}
=
\sqrt{\frac{GM_{\oplus}}{r_{\mathrm{sync}}}}
\approx
3.07\times10^3\ \mathrm{m/s}.
\]
\[
\boxed{
v_{\mathrm{sync}}\approx3.07\ \mathrm{km/s}
}.
\]
7. Geosynchronous versus geostationary
| Orbit type |
Requirement |
Ground appearance |
| Geosynchronous |
Orbital period equals Earth’s rotation period |
Repeats the same ground pattern each sidereal day |
| Geostationary |
Geosynchronous, circular, equatorial, prograde |
Appears fixed above one equatorial longitude |
| Inclined geosynchronous |
Same period but nonzero inclination |
Moves north and south in a daily figure-eight pattern |
8. Comparison with other Earth orbits
| Orbit |
Typical altitude |
Typical period |
Typical speed |
| ISS / low Earth orbit |
About 400 km |
About 92–93 min |
About 7.7 km/s |
| GPS orbit |
About 20,200 km |
About 12 h |
About 3.9 km/s |
| Geostationary orbit |
About 35,786 km |
About 23.93 h |
About 3.07 km/s |
Higher circular orbits have lower orbital speed, but the path is much larger, so the period is longer.
9. Summary formulas
| Goal |
Formula |
Meaning |
| Rotation angular speed |
\(\omega_{\mathrm{rot}}=2\pi/T_{\mathrm{rot}}\) |
Angular speed of the central body |
| Synchronous orbital radius |
\(r_{\mathrm{sync}}=\sqrt[3]{GM T_{\mathrm{rot}}^2/(4\pi^2)}\) |
Radius where orbital period matches rotation period |
| Synchronous altitude |
\(h_{\mathrm{sync}}=r_{\mathrm{sync}}-R\) |
Height above the surface |
| Synchronous speed |
\(v_{\mathrm{sync}}=\sqrt{GM/r_{\mathrm{sync}}}\) |
Circular orbital speed at the synchronous radius |
| Chosen-orbit period |
\(T=2\pi\sqrt{r^3/(GM)}\) |
Checks whether a chosen altitude is synchronous |
10. Assumptions and limits
- The orbit is circular.
- The central body is spherical or point-like.
- Atmospheric drag, thrust, and perturbations are ignored.
- For a truly geostationary satellite, the orbit must be equatorial and prograde.
- Real station-keeping requires corrections because Earth is not perfectly spherical and other bodies perturb the orbit.
Key idea: a geostationary satellite is not simply “high enough”; it must orbit at the unique radius where gravity gives exactly the period of Earth’s rotation.
