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Satellite Motion

Level 1 - Physics topic page in Gravitation.

Principle

An orbit is sustained motion in which gravity continually turns the velocity toward the central body.

Notation

\(M\)
central mass
\(\mathrm{kg}\)
\(m\)
satellite mass
\(\mathrm{kg}\)
\(r\)
orbital radius from the source center
\(\mathrm{m}\)
\(v\)
orbital speed
\(\mathrm{m\,s^{-1}}\)
\(T\)
orbital period
\(\mathrm{s}\)
\(G\)
gravitational constant
\(\mathrm{N\,m^{2}\,kg^{-2}}\)
\(g(r)\)
gravitational field magnitude at radius \(r\)
\(\mathrm{N\,kg^{-1}}\)

Method

Derivation 1: Match gravity to the centripetal requirement

For a circular orbit, the satellite is always falling toward the center, so gravity must provide exactly the inward force needed for circular motion.

Gravity magnitude
\[F_g=G\frac{Mm}{r^2}\]
Circular-motion requirement
\[F_c=m\frac{v^2}{r}\]
Set the inward forces equal
\[G\frac{Mm}{r^2}=m\frac{v^2}{r}\]
Circular orbit speed
\[v=\sqrt{\frac{GM}{r}}\]

Derivation 2: Build the period

One orbit covers the circumference \(2\\pi r\). Dividing that distance by the orbital speed gives the time for one revolution.

One-orbit distance
\[2\pi r\]
Period definition
\[T=\frac{2\pi r}{v}\]
Substitute the orbit speed
\[T=2\pi\sqrt{\frac{r^3}{GM}}\]
Field form
\[g(r)=\frac{v^2}{r}=\frac{4\pi^2r}{T^2}\]
The same orbit can be described either by gravity or by its circular-motion kinematics.

The sketch is an instantaneous orbit picture. Velocity is tangent to the orbit while gravity points inward toward the central mass.

3.10 deg90 deg180 deg270 degvFg
The orbit persists because gravity keeps turning the velocity inward.

Because the inward force is perpendicular to the velocity in a circular orbit, gravity can keep changing direction without needing to reduce the speed.

Rules

These are the compact results from the method above.

Circular orbit balance
\[G\frac{Mm}{r^2}=m\frac{v^2}{r}\]
Orbital speed
\[v=\sqrt{\frac{GM}{r}}\]
Orbital period
\[T=2\pi\sqrt{\frac{r^3}{GM}}\]
Orbit field relation
\[g(r)=\frac{v^2}{r}=\frac{4\pi^2r}{T^2}\]

Examples

Question
Find the circular-orbit speed for a satellite at radius
\[8.4\times10^6\,\mathrm{m}\]
around a planet of mass
\[6.0\times10^{24}\,\mathrm{kg}\]
Answer
\[v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\times10^{-11})(6.0\times10^{24})}{8.4\times10^6}}=6.90\times10^3\,\mathrm{m\,s^{-1}}\]

Checks

  • Orbiting does not mean gravity is absent; it means gravity is providing the inward acceleration.
  • Higher circular orbits have lower speed but longer period.
  • The satellite mass cancels in the circular-orbit speed formula.
  • If gravity were stronger than the required centripetal force at that speed, the orbit would curve inward.

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Kepler Models