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11.3 — Kepler From Newton, and Newton From Kepler
Chapter 1.9 derived Newton's inverse-square law from Kepler's third law. This chapter closes the loop by going the other way — starting from Newton and deriving all three of Kepler's laws — and then re-derives Newton's law from Kepler once more, so the logical relationship between them is completely explicit.
Why both directions matter. Kepler's laws are a description of what the planets do, extracted from data with no theory behind them. Newton's law is a claim about why. Deriving each from the other shows they contain the same information, and it shows exactly what extra assumptions each direction needs.
Kepler's three laws
First law (1609). Each planet moves in an ellipse with the Sun at one focus.
Second law (1609). The line from the Sun to the planet sweeps equal areas in equal times.
Third law (1619). The square of the orbital period is proportional to the cube of the semi-major axis.
T^2 \propto a^3
How they were obtained. Kepler had Tycho Brahe's observations — twenty years of naked-eye measurements accurate to about 2 arcminutes, the best ever made without a telescope. He spent eight years on Mars alone.
The crucial moment came when he fitted a circular orbit to Mars and the residuals reached 8 arcminutes. Tycho's data was good to 2. Kepler could have called it experimental error and published. He wrote instead that those eight minutes "led the way to a complete reformation of astronomy".
Eight arcminutes is 1/225 of a degree. The entire heliocentric revolution turned on refusing to ignore it.
Deriving Kepler from Newton
Start with Newton's second law and his law of gravitation:
\vec{F} = m\vec{a}, \qquad \vec{F} = -\frac{GMm}{r^2}\hat{r}
Second law first, because it is the most general
The second law follows from angular momentum conservation alone, and does not need the inverse square at all.
Angular momentum (Chapter 1.8):
\vec{L} = \vec{r}\times m\vec{v}
\frac{d\vec{L}}{dt} = \vec{r}\times\vec{F}
Gravity is a central force — it points along \vec{r} — so \vec{r}\times\vec{F} = 0 and:
\vec{L} = \text{constant}
Now the area. In time dt the planet moves \vec{v}\,dt, and the triangle swept has area:
dA = \frac{1}{2}\left|\vec{r}\times\vec{v}\,dt\right| = \frac{L}{2m}dt
\boxed{\frac{dA}{dt} = \frac{L}{2m} = \text{constant}}
Kepler's second law, derived in four lines.
And notice what it did not use. No inverse square, no specific force law — only that the force points along the line to the centre. The second law would hold for any central force, including a spring, and this is worth knowing because it means the second law tests less than the other two.
The physical reading: the planet moves fastest at perihelion and slowest at aphelion, in exactly the proportion that keeps the swept area constant. Earth is 3.4 % closer to the Sun in January and moves 3.4 % faster, which is why northern winter is five days shorter than northern summer.
First law: the ellipse
This one needs the inverse square, and it is the harder derivation.
Set up in polar coordinates. The radial equation of motion, with the centrifugal term from the angular momentum:
m\left(\ddot{r}-r\dot{\theta}^2\right) = -\frac{GMm}{r^2}
Use L = mr^2\dot{\theta} to eliminate \dot{\theta}:
\dot{\theta} = \frac{L}{mr^2} \quad\Longrightarrow\quad \ddot{r}-\frac{L^2}{m^2r^3} = -\frac{GM}{r^2}
The substitution that makes it solvable: let u = 1/r and change the independent variable from t to \theta.
\frac{dr}{dt} = \frac{dr}{d\theta}\frac{d\theta}{dt} = \left(-\frac{1}{u^2}\frac{du}{d\theta}\right)\left(\frac{Lu^2}{m}\right) = -\frac{L}{m}\frac{du}{d\theta}
Differentiate again:
\ddot{r} = -\frac{L}{m}\frac{d^2u}{d\theta^2}\cdot\frac{d\theta}{dt} = -\frac{L^2u^2}{m^2}\frac{d^2u}{d\theta^2}
Substitute into the radial equation, with r = 1/u:
-\frac{L^2u^2}{m^2}\frac{d^2u}{d\theta^2}-\frac{L^2u^3}{m^2} = -GMu^2
Divide by -L^2u^2/m^2:
\boxed{\frac{d^2u}{d\theta^2}+u = \frac{GMm^2}{L^2}}
This is the harmonic oscillator equation with a constant driving term, and Chapter 2.1 solved exactly this. The solution is a constant plus a sinusoid:
u = \frac{GMm^2}{L^2}\left(1+e\cos\theta\right)
Invert to get r:
\boxed{r = \frac{L^2/GMm^2}{1+e\cos\theta}}
And that is the polar equation of a conic section with the focus at the origin:
r = \frac{p}{1+e\cos\theta}
For e < 1 it is an ellipse. Kepler's first law, and with it something Kepler could not have known: the orbit is an ellipse only because the exponent is exactly 2.
Any other exponent gives a precessing orbit rather than a closed one, which is precisely what Chapter 6.9 exploited — general relativity adds a small term to this equation and the ellipse fails to close by 43 arcseconds per century for Mercury.
The eccentricity comes from the energy:
e = \sqrt{1+\frac{2EL^2}{G^2M^2m^3}}
| Energy | e | Orbit |
|---|---|---|
| E < 0, minimum | 0 | Circle |
| E < 0 | 0<e<1 | Ellipse |
| E = 0 | 1 | Parabola |
| E > 0 | >1 | Hyperbola |
Bound orbits have negative total energy, which makes sense: you would need to add energy to escape to infinity where E = 0.
Third law
Take the second law's constant areal rate and integrate over one full orbit.
\frac{dA}{dt} = \frac{L}{2m} \quad\Longrightarrow\quad A = \frac{L}{2m}T
The area of an ellipse is \pi ab, with b = a\sqrt{1-e^2}:
\pi a^2\sqrt{1-e^2} = \frac{LT}{2m}
And from the orbit equation, the semi-latus rectum is:
p = a(1-e^2) = \frac{L^2}{GMm^2} \quad\Longrightarrow\quad L^2 = GMm^2a(1-e^2)
Square the area equation and substitute:
\pi^2a^4(1-e^2) = \frac{L^2T^2}{4m^2} = \frac{GMm^2a(1-e^2)T^2}{4m^2}
The (1-e^2) cancels on both sides, which is the key step:
\pi^2a^4 = \frac{GMaT^2}{4}
\boxed{T^2 = \frac{4\pi^2}{GM}a^3}
Kepler's third law, with the constant of proportionality now known.
And the cancellation of (1-e^2) means the eccentricity does not appear. A highly elliptical orbit and a circular one with the same semi-major axis have exactly the same period. Kepler found this empirically and had no way to know why.
Newton from Kepler
Now the reverse, which is how Newton actually reasoned in the 1660s and 1680s.
Assume a circular orbit for simplicity — Newton did the general case, and the circular version contains the argument.
Centripetal acceleration (Chapter 1.5):
a = \frac{v^2}{r} = \frac{4\pi^2r}{T^2}
Now impose Kepler's third law, T^2 = ka^3 = kr^3:
a = \frac{4\pi^2r}{kr^3} = \frac{4\pi^2}{k}\cdot\frac{1}{r^2}
\boxed{a \propto \frac{1}{r^2}}
The inverse-square law, from Kepler's third law and nothing else.
Then Newton's second law gives F = ma \propto m/r^2, and the third law — that forces come in equal and opposite pairs — requires the force to be symmetric in the two masses, so it must contain M as well:
F = \frac{GMm}{r^2}
This is exactly the derivation in Chapter 1.9, and it is worth being clear about what each direction establishes.
Kepler → Newton shows that the inverse-square law is required by the observations. Newton → Kepler shows that it is sufficient to explain all three.
Together they show the two are equivalent, and that is the strongest statement you can make about a physical law.
The Moon test
Newton's own decisive check, and it is the moment gravity became universal rather than celestial.
The claim: the same force that makes an apple fall keeps the Moon in orbit.
Prediction. If a \propto 1/r^2, and the Moon is at 60.3 Earth radii:
a_{\text{Moon}} = \frac{9.81}{(60.3)^2} = \frac{9.81}{3636} = 2.70\times10^{-3}\ \text{m/s}^2
Measurement. The Moon's orbital period is 27.32 days = 2.360\times10^{6} s, and its distance is 3.844\times10^{8} m:
a = \frac{4\pi^2r}{T^2} = \frac{4\pi^2(3.844\times10^{8})}{(2.360\times10^{6})^2} = \frac{1.518\times10^{10}}{5.570\times10^{12}} = 2.73\times10^{-3}\ \text{m/s}^2
Predicted 2.70\times10^{-3}, measured 2.73\times10^{-3} — agreement to 1 %.
Newton wrote that he "compared the force requisite to keep the Moon in her Orb with the force of gravity at the surface of the earth, and found them answer pretty nearly."
And he sat on it for twenty years. The delay is usually attributed to his having a poor value for the Earth's radius at first, though the historical detail is contested. What is certain is that Halley had to press him to publish, and the result was the Principia in 1687.
One thing the test needed that Newton had to prove separately: that a sphere attracts as though all its mass were at the centre. The shell theorem (Chapter 1.9) — without it, "60.3 Earth radii" would not be the right number to use, and Newton reportedly delayed publication partly until he could prove it.
Two bodies, done properly
Both objects orbit their common centre of mass, and the derivations above quietly assumed the Sun is fixed.
m_1\vec{r}_1 = m_2\vec{r}_2
The fix is exact and simple. Define the reduced mass:
\mu = \frac{m_1m_2}{m_1+m_2}
and the two-body problem becomes a one-body problem of mass \mu orbiting a fixed centre at distance r = r_1+r_2.
The corrected third law:
\boxed{T^2 = \frac{4\pi^2a^3}{G(m_1+m_2)}}
For the solar system the correction is tiny, since Jupiter is 1/1047 of the Sun's mass. For binary stars it is everything, and this equation is the primary method of measuring stellar masses.
Worked example: the Sun's wobble from Jupiter.
r_{\text{Sun}} = \frac{m_J}{m_J+m_\odot}\times a = \frac{1.898\times10^{27}}{1.989\times10^{30}}\times7.785\times10^{11} = 7.42\times10^{8}\ \text{m}
The Sun's radius is 6.96\times10^{8} m, so the barycentre lies just outside the solar surface. The Sun genuinely orbits a point in space above its own photosphere, once every 11.86 years, at about 12.5 m/s.
And that 12.5 m/s is measurable from another star, by Doppler shift (Chapter 2.5). It is how the first exoplanets were found, and modern spectrographs reach below 1 m/s — which is what is needed to detect an Earth, whose effect on the Sun is 0.09 m/s.
Where this shows up in your life
Every satellite is placed using T^2 = 4\pi^2a^3/GM, and Chapter 11.4 computes geostationary altitude from it.
Measuring masses. The only way to weigh a distant object is to watch something orbit it. The Sun's mass, every planet's mass, the black hole at the galaxy's centre, and the dark matter in a galaxy cluster are all measured this way.
Exoplanet detection, from the barycentre wobble.
And the Voyager missions, still transmitting after 47 years and now beyond the heliopause, are flying trajectories computed from these equations with corrections at the eighth decimal place.
What the next chapter fixes
The orbit equation has been derived and not yet used. Chapter 11.4 turns it into the working tools of orbital mechanics: the six orbital elements that specify any orbit completely, the vis-viva equation that gives speed at any point, the escape and circular velocities, and a from-scratch computation of geostationary altitude — the number that decides where every television satellite in the world sits.