Understanding Kepler’s Laws
Kepler’s Laws are one of the most tested topics in orbital mechanics. This guide will help you understand all 3 powerful laws with formulas and exam-style examples.
Why do planets speed up near the Sun and slow down when they are far away?
How can we estimate a satellite’s orbital period from its distance?
The answers live in Kepler’s laws, elegantly explained today using Newtonian mechanics. This article walks you through the ideas, the math, and a couple of quick examples you can use right away.
Background & Geometry You’ll Need
Ellipses and orbital elements
Most bound orbits in the Solar System are well approximated by ellipses. An ellipse has:
- Semimajor axis \(a\) — half the long axis (sets the overall size of the orbit).
- Eccentricity \(e\) — a measure of “stretch,” with \(0 \le e < 1\) for ellipses. A circle has \(e=0\).
- Foci — two special points along the major axis. In a heliocentric orbit the Sun sits at one focus, not at the center.
The closest and farthest distances from the Sun are the perihelion and aphelion:
\(R_p = a(1-e)\) and \(R_a = a(1+e)\).
The semiminor axis is
\(b = a\sqrt{1-e^2}\).
Kepler’s Laws
- Law of Orbits: Planets move in elliptical orbits with the Sun at one focus.
- Law of Areas: A line from the Sun to a planet sweeps out equal areas in equal times.
- Law of Periods: The square of a planet’s orbital period is proportional to the cube of the semimajor axis: \(T^2 \propto a^3\).
Why the (Kepler’s) Second Laws Is Really Angular Momentum Conservation
The area-rate argument in Kepler’s Laws
Consider position vector \(\vec r\) from the Sun to a planet of mass \(m\).
In a short time \(dt\), the planet moves by \(d\vec r\) and the triangular area swept is approximately \(dA = \frac{1}{2} r^2 d\theta\). Dividing by time gives the area rate
\(\displaystyle \frac{dA}{dt} = \frac{1}{2} r^2 \dot\theta\).
Connect Kepler’s Laws to angular momentum
The planet’s angular momentum about the Sun is \(L = m r^2 \dot\theta\). Substitute into the area rate:
\(\displaystyle \frac{dA}{dt} = \frac{L}{2m}\).
If the only significant force is gravity (a central force pointing along \(\pm \vec r\)), the torque about the Sun is zero, so
\(L\) is constant.
Therefore \(\displaystyle\frac{dA}{dt}\) is constant, which is Kepler’s second law.
This also explains the speed change: near perihelion \(r\) is small, so \(\dot\theta\) must be larger to keep \(L\) fixed → the planet moves faster.
Deriving the Kepler’s Third Law
Start with a circular orbit
For a planet of mass \(m\) orbiting a mass \(M\) (e.g., the Sun), Newton’s law of gravitation provides the inward force:
\(\displaystyle \frac{GMm}{r^2}\),
which supplies the centripetal requirement \(m\omega^2 r\).
Set them equal:
\(\displaystyle \frac{GMm}{r^2} = m\omega^2 r \;\Rightarrow\; \omega^2 = \frac{GM}{r^3}\). Since \(\omega = \frac{2\pi}{T}\),
we obtain Kepler’s third law in constant form:
\(\displaystyle T^2 = \left(\frac{4\pi^2}{GM}\right) r^3\).
Generalize to ellipses
For elliptical orbits, replace \(r\) by the semimajor axis \(a\) (a result from averaging over the orbit in Newtonian mechanics). The same constant holds:
\(\displaystyle T^2 = \left(\frac{4\pi^2}{GM}\right) a^3.\)
The quantity in parentheses depends only on the central mass \(M\). Around the same central body, the ratio \(\displaystyle \frac{T^2}{a^3}\) is the same for all satellites.
Working With the Formulas
From period to semimajor axis (and back)
Given \(M\) and \(T\), solve for \(a\):
\(\displaystyle a = \left(\frac{GM}{4\pi^2}\right)^{\!1/3} T^{\,2/3}\).
Conversely, with \(a\) known you get \(T\) from the third law directly.
Perihelion, aphelion, and eccentricity
- \(R_p = a(1-e)\) (closest distance)
- \(R_a = a(1+e)\) (farthest distance)
- \(e = \displaystyle\frac{R_a – R_p}{R_a + R_p}\) (invert the above if distances are known)
Common Pitfalls
- Sun at a focus, not at center: The ellipse’s center is not the Sun’s location.
- Use meters and seconds: Keep SI units when plugging numbers into \(T^2 = \left(\displaystyle\frac{4\pi^2}{GM}\right)a^3\).
- Same \(\mathbf{M}\), same constant: Around Earth (for satellites), \(\displaystyle \frac{T^2}{a^3}\) is different than around the Sun because \(M\) is different.
Applications of Kepler’s Laws Beyond Planets
Artificial satellites and Kepler’s Laws
For Earth-orbiting satellites, use \(M = M_\oplus\). A low-Earth orbit with smaller \(a\) has a short period; a geostationary orbit satisfies \(T = 1\,\text{sidereal day}\), which fixes \(a \approx 42{,}164\,\text{km}\) from Earth’s center.
Rings, comets, and exoplanets
The same laws describe ring particles around Saturn, highly eccentric comets like Halley’s, and exoplanets around distant stars (with \(M\) equal to the host star’s mass).
Mini Checkpoints
- Two satellites orbit the same planet. Satellite A has \(a_A = 2R\) and Satellite B has \(a_B = R\). What is \(\displaystyle \frac{T_A}{T_B}\)?
- A planet’s perihelion is \(0.5\,\text{AU}\) and aphelion is \(1.5\,\text{AU}\). Find its \(a\) and \(e\).
- In a Sun-centred system, you measure \(T=8\,\text{yr}\). Estimate \(a\) in AU (assume other planets don’t matter).
Key Takeaways
- Orbits are ellipses with the Sun at one focus; \(a\) sets the size, \(e\) the shape.
- Equal areas in equal times ⇔ constant angular momentum: \(\displaystyle \frac{dA}{dt}=\frac{L}{2m}\).
- Third law in universal form: \(\displaystyle T^2=\left(\frac{4\pi^2}{GM}\right)a^3\) — one constant per central body.
Example
Given. Comet Halley orbits the Sun with period \(T = 76 \text{ years}\) and has a perihelion (closest approach) of \(R_p = 8.9 \times 10^{10}\ \text{m}\).
- (a) Find the aphelion distance \(R_a\) (farthest distance from the Sun).
- (b) Find the orbital eccentricity \(e\).
Key Ideas
- Kepler’s third law (Newtonian form): \(T^2 = \left(\displaystyle\frac{4\pi^2}{GM}\right)a^3\), where \(a\) is the semimajor axis and \(M\) is the Sun’s mass.
- Perihelion and aphelion for an ellipse: \(R_p = a(1-e)\) and \(R_a = a(1+e)\), so \(R_a + R_p = 2a\) and \(e = 1 – \displaystyle\frac{R_p}{a}\).
Constants
Solar mass \(M = 1.99 \times 10^{30}\ \text{kg}\), gravitational constant \(G = 6.67 \times 10^{-11}\ \text{N}\,\text{m}^2\text{/kg}^2\), and \(T = 76\ \text{yr} = 2.4 \times 10^9\ \text{s}\) (approx.).
(a) — Find \(R_a\)
Compute the semimajor axis.
\(a = \left(\displaystyle\frac{GM T^2}{4\pi^2}\right)^{1/3}\)
Substituting numbers:
\(a \approx \left(\displaystyle\frac{(6.67 \times 10^{-11})(1.99 \times 10^{30})(2.4 \times 10^{9})^2}{4\pi^2}\right)^{1/3} \approx 2.7 \times 10^{12}\ \text{m}.\)
Use \(R_a + R_p = 2a\):
\(R_a = 2a – R_p = 2(2.7 \times 10^{12}) – 8.9 \times 10^{10} \approx 5.3 \times 10^{12}\ \text{m}.\)
(b) — Find \(e\)
\(e = 1 – \displaystyle\frac{R_p}{a} = 1 – \displaystyle\frac{8.9 \times 10^{10}}{2.7 \times 10^{12}} \approx 0.97.\)
Interpretation
An eccentricity near unity indicates a very elongated ellipse: the comet spends most of its time far from the Sun and briefly swings through the inner Solar System near perihelion.
By mastering Kepler’s Laws, you’ll gain confidence in solving orbital mechanics problems for both AP exams and college physics.
For more details on Kepler’s Laws and related astronomy topics, see NASA Solar System Exploration, Wikipedia, and PhET.
For more articles about gravitational force, go to our previous article Gravitational Force of a Uniform Sphere: Ultimate Proof and Brilliant Explanation.
