A circular orbit occurs when gravity supplies exactly the centripetal acceleration needed to keep an object moving around a central body.
For a satellite orbiting a spherical body of mass \(M\), the main formulas are
\[
\boxed{
v=\sqrt{\frac{GM}{r}}
}
\]
and
\[
\boxed{
T=2\pi\sqrt{\frac{r^3}{GM}}
}.
\]
Here \(r\) is the distance from the center of the central body to the satellite.
1. Radius from altitude
If the satellite is at altitude \(h\) above a body of radius \(R\), then the orbital radius is
\[
\boxed{
r=R+h
}.
\]
This is important because the formulas use the center-to-center distance, not just the altitude above the surface.
2. Deriving the circular orbital speed
Gravity provides the centripetal force:
\[
F_g=F_c.
\]
The gravitational force is
\[
F_g=\frac{GMm}{r^2}.
\]
The centripetal force required for circular motion is
\[
F_c=\frac{mv^2}{r}.
\]
Set them equal:
\[
\frac{GMm}{r^2}=\frac{mv^2}{r}.
\]
The satellite mass \(m\) cancels:
\[
\frac{GM}{r^2}=\frac{v^2}{r}.
\]
Multiply both sides by \(r\):
\[
v^2=\frac{GM}{r}.
\]
Therefore,
\[
\boxed{
v=\sqrt{\frac{GM}{r}}
}.
\]
3. Orbital period
The distance traveled in one circular orbit is the circumference:
\[
C=2\pi r.
\]
Period is distance divided by speed:
\[
T=\frac{2\pi r}{v}.
\]
Substitute \(v=\sqrt{GM/r}\):
\[
T=\frac{2\pi r}{\sqrt{GM/r}}.
\]
This simplifies to
\[
\boxed{
T=2\pi\sqrt{\frac{r^3}{GM}}
}.
\]
4. Connection with Kepler’s third law
Squaring the period formula gives
\[
T^2=4\pi^2\frac{r^3}{GM}.
\]
For circular orbits, \(r\) is the semi-major axis \(a\), so
\[
\boxed{
T^2=\frac{4\pi^2a^3}{GM}
}.
\]
This is Kepler’s third law in Newtonian form for a small satellite orbiting a central mass.
5. Two-body correction
For ordinary artificial satellites, the satellite mass is negligible compared with the planet:
\[
m\ll M.
\]
In that case, \(M+m\approx M\).
For binary stars or two comparable masses, use
\[
\boxed{
\mu=G(M+m)
}.
\]
Then
\[
\boxed{
v=\sqrt{\frac{\mu}{r}}
},
\qquad
\boxed{
T=2\pi\sqrt{\frac{r^3}{\mu}}
}.
\]
6. Centripetal acceleration and gravity
In circular orbit, the centripetal acceleration is
\[
a_c=\frac{v^2}{r}.
\]
Using \(v^2=GM/r\), this becomes
\[
\boxed{
a_c=\frac{GM}{r^2}
}.
\]
This is the same as the gravitational field strength at that radius.
Orbiting objects are continuously falling around the planet.
7. Solving for orbital radius from period
Sometimes the period is known, and the required radius must be found.
Start from
\[
T=2\pi\sqrt{\frac{r^3}{GM}}.
\]
Square both sides:
\[
T^2=4\pi^2\frac{r^3}{GM}.
\]
Rearranging gives
\[
r^3=\frac{GMT^2}{4\pi^2}.
\]
Therefore,
\[
\boxed{
r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}
}.
\]
The altitude is then
\[
\boxed{
h=r-R
}.
\]
8. Worked example: ISS at 400 km altitude
Take Earth’s radius and mass as
\[
R_{\oplus}=6.371\times10^6\ \mathrm{m},
\qquad
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg}.
\]
At \(400\ \mathrm{km}\) altitude,
\[
h=400\ \mathrm{km}=4.00\times10^5\ \mathrm{m}.
\]
The orbital radius is
\[
r=R+h
=
6.371\times10^6+4.00\times10^5
=
6.771\times10^6\ \mathrm{m}.
\]
The circular orbital speed is
\[
v=\sqrt{\frac{GM}{r}}.
\]
\[
v
=
\sqrt{
\frac{(6.67430\times10^{-11})(5.9722\times10^{24})}
{6.771\times10^6}
}.
\]
\[
v\approx7.67\times10^3\ \mathrm{m/s}.
\]
So
\[
\boxed{
v\approx7.67\ \mathrm{km/s}
}.
\]
The orbital period is
\[
T=\frac{2\pi r}{v}.
\]
\[
T
=
\frac{2\pi(6.771\times10^6)}
{7.67\times10^3}
\approx5.54\times10^3\ \mathrm{s}.
\]
Convert to minutes:
\[
T\approx\frac{5.54\times10^3}{60}\approx92.4\ \mathrm{min}.
\]
This agrees with the common ISS period of about 92 to 93 minutes.
9. Comparing LEO, GPS, and geostationary orbits
| Orbit type |
Typical altitude |
Typical speed |
Typical period |
| Low Earth orbit |
About 300–500 km |
About 7.6–7.8 km/s |
About 90–95 min |
| GPS orbit |
About 20,200 km |
About 3.9 km/s |
About 12 h |
| Geostationary orbit |
About 35,786 km |
About 3.1 km/s |
About 23.93 h |
Higher orbits have lower orbital speeds, but their periods are longer because the orbit is much larger.
10. Summary table
| Goal |
Formula |
Meaning |
| Orbital radius from altitude |
\(r=R+h\) |
Convert height above surface into center distance |
| Circular orbital speed |
\(v=\sqrt{GM/r}\) |
Speed needed to maintain circular orbit |
| Orbital period |
\(T=2\pi\sqrt{r^3/(GM)}\) |
Time for one full orbit |
| Centripetal acceleration |
\(a_c=v^2/r=GM/r^2\) |
Acceleration supplied by gravity |
| Radius from period |
\(r=\sqrt[3]{GMT^2/(4\pi^2)}\) |
Find orbital radius for a target period |
| Two-body correction |
\(\mu=G(M+m)\) |
Use when both masses are important |
11. Assumptions and limits
- The orbit is circular.
- The central body is treated as spherical or point-like.
- The satellite is outside the central body.
- Atmospheric drag is ignored.
- Thrust, orbital maneuvers, and perturbations are ignored.
- For real spacecraft, inclination, Earth’s oblateness, and drag can slightly change the actual orbit.
Key idea: a circular orbit happens when the satellite’s sideways speed is exactly large enough for gravity to bend the path into a circle.
