KAHIBARO
Table of Contents
Meaning of Orbital Period
The orbital period is the time a planet, satellite, or other object takes to complete one full orbit around a central body. If the object returns to the same position in its path, one period has passed.
It is usually written as $T$ and measured in seconds in SI units. For planets, it is often also described in days or years.
Circular Orbit Formula
For a circular orbit, the orbital period can be found from the orbital speed and the circumference of the orbit. If the orbit has radius $r$, then the distance traveled in one revolution is
$$
2\pi r
$$
If the orbital speed is $v$, then
$$
T = \frac{2\pi r}{v}
$$
Using the circular orbital speed,
$$
v = \sqrt{\frac{GM}{r}}
$$
we substitute into the period formula:
$$
T = \frac{2\pi r}{\sqrt{GM/r}}
$$
which gives
$$
T = 2\pi \sqrt{\frac{r^3}{GM}}
$$
Here, $G$ is the gravitational constant and $M$ is the mass of the central body.
For a circular orbit around a body of mass $M$,
$$
T = 2\pi \sqrt{\frac{r^3}{GM}}
$$
This is one of the most important formulas for orbital motion.
Dependence on Orbital Radius
This formula shows that the orbital period increases as the orbital radius increases. Objects farther from the central body move more slowly and take longer to complete one orbit.
The dependence is not linear. Since
$$
T \propto r^{3/2}
$$
doubling the orbital radius does not merely double the period. Instead,
$$
T_{\text{new}} = 2^{3/2} T \approx 2.83 T
$$
So a much larger orbit gives a significantly longer period.
Relation to Kepler's Third Law
The orbital period formula for circular motion matches Kepler's third law. For objects orbiting the same central mass,
$$
T^2 \propto r^3
$$
More precisely,
$$
\frac{T^2}{r^3} = \frac{4\pi^2}{GM}
$$
This means that for all satellites orbiting the same planet, or all planets orbiting the same star, the ratio $T^2/r^3$ is constant.
For orbits around the same central body,
$$
\frac{T^2}{r^3} = \text{constant}
$$
This is the mathematical form of Kepler's third law for circular orbits.
Elliptical Orbits
Real planetary orbits are usually elliptical, not perfectly circular. For an elliptical orbit, the same basic period law still works, but the radius $r$ is replaced by the semi-major axis $a$:
$$
T = 2\pi \sqrt{\frac{a^3}{GM}}
$$
So the orbital period depends on the size of the orbit, not on where the object happens to be at a particular moment.
Comparing Different Orbits
If two objects orbit the same central mass, then dividing the period formulas gives a useful comparison rule:
$$
\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}
$$
or
$$
\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3
$$
This is helpful when one orbit is known and another must be found.
For elliptical orbits, use semi-major axes instead:
$$
\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{a_1}{a_2}\right)^3
$$
Example
Suppose a satellite orbits Earth at radius $r$. If another satellite orbits at radius $4r$, then
$$
\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{4r}{r}\right)^3 = 64
$$
so
$$
\frac{T_2}{T_1} = 8
$$
The outer satellite takes 8 times longer to complete one orbit.
Physical Interpretation
The farther an object is from the central body, the weaker gravity becomes. Because of this weaker gravitational pull, the object moves more slowly in orbit. At the same time, it must travel along a larger path. Both effects make the orbital period longer.
Common Quantities
| Quantity | Symbol | Meaning |
|---|---|---|
| Orbital period | $T$ | Time for one complete orbit |
| Orbital radius, circular orbit | $r$ | Distance from center of central body |
| Semi-major axis, elliptical orbit | $a$ | Size measure of ellipse |
| Central mass | $M$ | Mass of planet, star, or other attracting body |
| Gravitational constant | $G$ | Universal gravitation constant |
Simple Orbit Sketch
Key Result
The orbital period is determined by the size of the orbit and the mass of the central body. A larger orbit means a longer period, and a more massive central body means a shorter period for the same orbital size.
Orbital period increases with orbital size and decreases with central mass:
$$
T = 2\pi \sqrt{\frac{r^3}{GM}}
\quad \text{or} \quad
T = 2\pi \sqrt{\frac{a^3}{GM}}
$$
Use $r$ for circular orbits and $a$ for elliptical orbits.