KAHIBARO
Table of Contents
The idea of a circular orbit
A circular orbit is a special kind of motion in which one object moves around another along a circle, while gravity continuously pulls it inward. In this chapter, we focus on the simplest case, a small body of mass $m$ moving in a circular path of radius $r$ around a much larger body of mass $M$.
The key idea is that gravity provides exactly the centripetal force needed to keep the orbiting body moving in a circle. The body is constantly falling toward the central object, but because it also has sideways motion, it keeps missing it.
Gravity as the centripetal force
For a circular orbit, the inward force needed is the centripetal force,
$$
F_c = \frac{mv^2}{r}
$$
The gravitational force between the two bodies is
$$
F_g = \frac{G M m}{r^2}
$$
For a circular orbit, these must be equal:
$$
\frac{mv^2}{r} = \frac{G M m}{r^2}
$$
The mass $m$ of the orbiting body cancels, giving
$$
v^2 = \frac{GM}{r}
$$
So the orbital speed is
$$
v = \sqrt{\frac{GM}{r}}
$$
This result is very important. It shows that for a circular orbit, the speed is determined only by the mass of the central body and the orbital radius.
For a circular orbit around a mass $M$ at radius $r$,
$$
\frac{mv^2}{r} = \frac{GMm}{r^2}
$$
so
$$
v = \sqrt{\frac{GM}{r}}
$$
The orbiting body's mass does not affect the orbital speed.
What the formula means
The formula
$$
v = \sqrt{\frac{GM}{r}}
$$
tells us several useful things. If the orbit is closer to the central body, meaning smaller $r$, the required orbital speed is larger. If the central body is more massive, meaning larger $M$, the required orbital speed is also larger.
This agrees with experience in astronomy. Satellites close to Earth move faster than satellites far away. Planets closer to the Sun move faster than planets farther away.
Orbital angular speed
If an object moves in a circular orbit of radius $r$ with speed $v$, its angular speed is
$$
\omega = \frac{v}{r}
$$
Using the circular orbit speed,
$$
\omega = \frac{1}{r}\sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{r^3}}
$$
This tells us how quickly the orbit sweeps out angle around the central body.
For a circular gravitational orbit,
$$
\omega = \sqrt{\frac{GM}{r^3}}
$$
Orbital period
The orbital period $T$ is the time needed to complete one full revolution. For a circle, the distance traveled in one orbit is the circumference:
$$
2\pi r
$$
Since speed is distance divided by time,
$$
v = \frac{2\pi r}{T}
$$
So
$$
T = \frac{2\pi r}{v}
$$
Substituting the circular orbital speed,
$$
T = \frac{2\pi r}{\sqrt{GM/r}}
$$
which becomes
$$
T = 2\pi \sqrt{\frac{r^3}{GM}}
$$
This is one of the most useful formulas for circular orbits.
Squaring both sides gives
$$
T^2 = \frac{4\pi^2}{GM} r^3
$$
This matches the form of Kepler's third law for circular motion.
For a circular orbit, the orbital period is
$$
T = 2\pi \sqrt{\frac{r^3}{GM}}
$$
and equivalently,
$$
T^2 = \frac{4\pi^2}{GM} r^3
$$
Visualizing a circular orbit
In a circular orbit, the velocity is tangent to the circle, while the gravitational force points toward the center.
Comparison of important quantities
The main quantities for circular orbits are closely related.
| Quantity | Formula | Meaning |
|---|---|---|
| Gravitational force | $F_g = \dfrac{GMm}{r^2}$ | Inward pull of gravity |
| Centripetal force needed | $F_c = \dfrac{mv^2}{r}$ | Force required for circular motion |
| Orbital speed | $v = \sqrt{\dfrac{GM}{r}}$ | Speed for a circular orbit |
| Angular speed | $\omega = \sqrt{\dfrac{GM}{r^3}}$ | Rate of angular motion |
| Orbital period | $T = 2\pi\sqrt{\dfrac{r^3}{GM}}$ | Time for one revolution |
Example with a satellite around Earth
Suppose a satellite orbits Earth in a circular orbit of radius $r$. If Earth has mass $M_E$, then the orbital speed is
$$
v = \sqrt{\frac{G M_E}{r}}
$$
If the satellite moves to a higher circular orbit, then $r$ increases. Because $r$ is in the denominator, the speed decreases. So higher circular orbits are slower, not faster.
This can seem surprising at first, but it follows directly from gravity becoming weaker at larger distances.
Common misunderstanding
It is easy to think that a larger speed always means a larger orbit. For circular gravitational orbits, that is not true. A lower orbit requires a higher speed because the object must curve more sharply, and that requires a larger centripetal acceleration.
Since
$$
a_c = \frac{v^2}{r}
$$
and gravity supplies that acceleration,
$$
\frac{v^2}{r} = \frac{GM}{r^2}
$$
we again get
$$
v^2 = \frac{GM}{r}
$$
So as $r$ decreases, $v$ must increase.
In circular gravitational motion, a smaller orbital radius means a larger orbital speed:
$$
v = \sqrt{\frac{GM}{r}}
$$
Therefore, lower circular orbits are faster.
Circular orbit near a planet's surface
If an object could orbit just above the surface of a planet of radius $R$, then the circular orbital speed would be
$$
v = \sqrt{\frac{GM}{R}}
$$
This is closely related to the idea of satellites orbiting Earth. The same rule applies whether the orbit is near the surface or far away, as long as the orbit is circular and the planet can be treated as roughly spherical.
Summary
A circular orbit happens when gravity exactly provides the centripetal force needed for circular motion. By setting gravitational force equal to centripetal force, we obtain the central formulas for orbital speed, angular speed, and period.
Core results for circular orbits around a mass $M$ at radius $r$:
$$
v = \sqrt{\frac{GM}{r}}
$$
$$
\omega = \sqrt{\frac{GM}{r^3}}
$$
$$
T = 2\pi\sqrt{\frac{r^3}{GM}}
$$
These formulas apply to ideal circular orbits under gravity.