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Kepler Models of Orbits

Level 1 - Physics topic page in Gravitation.

Principle

Kepler's orbit model describes bound gravitational motion with ellipse geometry, constant areal sweep rate, and a period that scales with orbital size.

Notation

\(a\)
semi-major axis of an orbit
\(\mathrm{m}\)
\(r_p\)
periapsis distance
\(\mathrm{m}\)
\(r_a\)
apoapsis distance
\(\mathrm{m}\)
\(T\)
orbital period
\(\mathrm{s}\)
\(\Delta A\)
area swept in a short time
\(\mathrm{m^{2}}\)
\(L\)
orbital angular momentum magnitude
\(\mathrm{kg\,m^{2}\,s^{-1}}\)
\(M\)
central mass
\(\mathrm{kg}\)

Method

Derivation 1: Use the central-force property

Gravity acts along the line joining the orbiting body to the source, so the torque about the source is zero. Zero torque means angular momentum is conserved.

Central force
\[\vec F_g\parallel\vec r\]
Zero torque
\[\vec\tau=\vec r\times\vec F_g=0\]
Conserved angular momentum
\[\frac{d\vec L}{dt}=0\]

Derivation 2: Turn angular momentum into Kepler's second law

In a short time \(\\Delta t\), the swept region is approximately a triangle. Its area depends on the perpendicular part of the velocity.

Short-time swept area
\[\Delta A=\frac12 r(v_\perp\Delta t)\]
Angular momentum magnitude
\[L=mrv_\perp\]
Area rate
\[\frac{\Delta A}{\Delta t}=\frac{L}{2m}\]
Equal areas in equal times
\[\frac{dA}{dt}=\frac{L}{2m}=\text{constant}\]

Derivation 3: Build the period-size law

For a circular orbit, the inverse-square gravity model gives the full period formula. In the full Kepler problem, the same scaling survives when the orbit is elliptical, with radius replaced by the semi-major axis.

Circular orbit result
\[T^2=\frac{4\pi^2}{GM}r^3\]
Kepler third law
\[T^2=\frac{4\pi^2}{GM}a^3\]
Ellipse geometry
\[r_p+r_a=2a\]
The semi-major axis sets the overall orbital scale.

Solving the full inverse-square equations gives Kepler's first law: bound orbits are ellipses with the central mass at one focus, not at the geometric center.

Rules

These are the compact results from the method above.

Areal speed
\[\frac{dA}{dt}=\frac{L}{2m}=\text{constant}\]
Third law
\[T^2=\frac{4\pi^2}{GM}a^3\]
Semi-major axis
\[a=\frac{r_p+r_a}{2}\]

Examples

Question
Two bodies orbit the same star. One has semi-major axis \(a\), the other has semi-major axis
\[2a\]
What is the ratio of their periods?
Answer
\[T\propto a^{3/2}\]
\[\frac{T_{2a}}{T_a}=2^{3/2}=\sqrt{8}\approx2.83\]
The larger orbit takes about 2.8 times as long.

Checks

  • The central mass sits at a focus of the ellipse, not at the center of the ellipse.
  • Equal areas in equal times mean the orbiting body moves faster when it is closer to the source.
  • Kepler's third law compares orbits around the same central mass unless the constant is written explicitly.
  • The semi-major axis, not the instantaneous radius, sets the period for an elliptical orbit.

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