Physics/Option Modules/Astrophysics/Circular Orbits
Physics · Astrophysics · A2 Level

Circular Orbits

A moon circling a planet, a planet circling a star, a satellite circling the Earth — all of them are doing the same trick. Gravity pulls them inward, but they're moving fast enough sideways that they keep missing whatever they're falling toward, forever. An orbit is really just a very long, very controlled fall.

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Gravity as the centripetal force

Consider a body of mass $m$ — a planet, a moon, or a satellite — moving in a stable circular orbit of radius $r$ around a much more massive central body of mass $M$. For the orbit to stay circular, the orbiting body needs a net force directed constantly toward the centre: a centripetal force. Without such a force, an object in motion would simply travel in a straight line, drifting off into space rather than curving around anything.

In the case of orbital motion, that centripetal force is supplied entirely by gravity. There's no separate "orbital force" — the gravitational attraction between the orbiting body and the central mass is the centripetal force, pulling the orbiting body inward at exactly the rate needed to keep it moving on a circular path rather than flying off in a straight line or spiralling into the centre. This is worth sitting with for a moment: an orbiting satellite is not resisting gravity or "balancing" it against some outward force — it is continuously falling toward the central body, but moving sideways fast enough that the surface (or centre) it's falling toward continuously curves away beneath it.

  • Too slow for a given radius, and gravity overwhelms the motion — the orbiting body spirals inward.
  • Too fast for a given radius, and the required centripetal force is less than gravity supplies — the body moves outward onto a larger, or non-circular, path.
  • Exactly the right speed for that radius keeps the path circular indefinitely, assuming no other forces (such as atmospheric drag) act on the body.

Deriving the orbital speed

Because gravity provides the centripetal force, we can simply set Newton's law of gravitation equal to the centripetal force expression and solve for the speed needed to maintain a circular orbit at radius $r$.

DerivationOrbital speed for a stable circular orbit
1
Gravitational force provides the centripetal force:
$$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$$
Left side is Newton's law of gravitation; right side is the centripetal force needed to keep mass $m$ moving on a circle of radius $r$ at speed $v$.
2
Cancel one factor of $m$ from both sides:
$$\dfrac{GM}{r^2} = \dfrac{v^2}{r}$$
The orbiting body's own mass cancels completely — it appears on both sides of the original equation.
3
Multiply both sides by $r$ and take the square root:
$$v = \sqrt{\dfrac{GM}{r}}$$
This is the orbital speed required for a stable circular orbit of radius $r$ around a central mass $M$.
Result
The key insight

The orbital speed $v = \sqrt{GM/r}$ depends only on the central mass $M$ and the orbital radius $r$ — it does not depend at all on the mass of the orbiting body. A tiny satellite and a massive space station at the same orbital radius around the Earth must travel at exactly the same speed to stay in a stable circular orbit. This is the same underlying reason all objects fall with the same acceleration in a gravitational field, regardless of their own mass.

Orbital period, and a first look at Kepler's third law

Speed and period are linked for any object moving on a circular path: in one full orbit, the body covers a distance equal to the circumference $2\pi r$ in a time $T$, so $v = \dfrac{2\pi r}{T}$. Substituting this into the orbital speed relation lets us find an expression for the orbital period directly in terms of $r$, $M$, and fundamental constants.

DerivationOrbital period for a circular orbit
1
Start from the orbital speed relation and the definition of speed on a circular path:
$$v = \sqrt{\dfrac{GM}{r}} \qquad \text{and} \qquad v = \dfrac{2\pi r}{T}$$
The second expression is just distance (circumference) divided by time (period).
2
Set the two expressions for $v$ equal and square both sides:
$$\dfrac{4\pi^2 r^2}{T^2} = \dfrac{GM}{r}$$
Squaring removes the square root, making $T$ easier to isolate.
3
Rearrange to make $T^2$ the subject:
$$T^2 = \dfrac{4\pi^2 r^3}{GM}$$
Multiply both sides by $r T^2$ and divide by $GM$.
4
Take the square root to give the period itself:
$$T = 2\pi\sqrt{\dfrac{r^3}{GM}}$$
This is the orbital period for a circular orbit of radius $r$ around a central mass $M$.
Result

Notice that the period-squared expression rearranges to $T^2 = \dfrac{4\pi^2}{GM}\,r^3$. Since $4\pi^2/GM$ is a constant for orbits around any single central mass $M$, this means $T^2 \propto r^3$ — the square of the orbital period is proportional to the cube of the orbital radius. This relationship is significant enough to have its own name: it is Kepler's third law, and it gets a fuller treatment — including how it extends beyond perfectly circular orbits — later in this topic, in the Planetary Motion lesson.

Worked example

A satellite orbits the Earth in a stable circular orbit at a radius of $7.00\times10^{6}\,\text{m}$ from the Earth's centre. Given the Earth's mass $M = 5.97\times10^{24}\,\text{kg}$ and $G = 6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$, find the satellite's orbital speed and orbital period.

Worked exampleOrbital speed and period
1
Find the orbital speed using $v = \sqrt{GM/r}$:
$$v = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{7.00\times10^{6}}} \approx 7.55\times10^{3}\,\text{m s}^{-1}$$
Around 7.55 km/s — a typical low Earth orbit speed.
2
Find the orbital period using $T = 2\pi\sqrt{r^3/GM}$:
$$T = 2\pi\sqrt{\dfrac{(7.00\times10^{6})^3}{(6.67\times10^{-11})(5.97\times10^{24})}} \approx 5.82\times10^{3}\,\text{s}$$
About 97 minutes — consistent with the roughly 90-minute periods typical of low Earth orbit satellites.
Checks out

Key takeaways

  • A body in a stable circular orbit is kept on its curved path because gravity supplies exactly the centripetal force needed — there is no separate outward force to "balance."
  • Setting $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ and cancelling $m$ gives the orbital speed $v = \sqrt{GM/r}$.
  • Orbital speed depends only on the central mass $M$ and orbital radius $r$, never on the orbiting body's own mass.
  • Combining $v = \sqrt{GM/r}$ with $v = 2\pi r/T$ gives the orbital period $T = 2\pi\sqrt{r^3/GM}$.
  • This rearranges to $T^2 \propto r^3$ for circular orbits around a given central mass — this is Kepler's third law, covered fully in the next-but-one lesson.
  • Faster-than-required speeds push a body onto a larger or non-circular path; slower-than-required speeds let gravity pull it inward.
  • A satellite at low Earth orbit altitude typically has an orbital speed of several kilometres per second and a period of roughly 90 minutes.
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