Article 1 — How Newton reached the 1/r² law

In 1687, in the Principia, Newton wrote down the form of the law of gravity:

\[ F = G\,\frac{M m}{r^2} \]

But how did he arrive at this? Not through a single flash of insight (the apple is a metaphor, not the reasoning). It was a clean mathematical argument: Kepler's laws + Newton's second law + the circular-orbit approximation = the 1/r² law. This article walks through that argument.

The starting point: Kepler's laws (~1609-1619)

Seventy years before Newton, Kepler had discovered three empirical laws about planetary orbits around the Sun:

1. Planetary orbits are ellipses, with the Sun at one focus.
2. The line joining a planet to the Sun sweeps equal areas in equal times. (Planets move faster when near the Sun.)
3. The square of the orbital period is proportional to the cube of the semi-major axis: \[ T^2 = k\, a^3 \] with the same k for every planet in the Solar System. That is a strong hint of a central force acting from the Sun.

The third law was Newton's key input. Let's see why.

Simplification: circular orbits

Most planetary orbits (except Mercury and Mars) are nearly circular — eccentricity is small. For Earth, e ≈ 0.017 — very nearly a circle. Take a circular orbit of radius r; the elliptical case reduces to the same result (with a in place of r).

In a circular orbit at constant speed v, the orbital circumference is 2πr and the period is T:

\[ v = \frac{2\pi r}{T} \]

The planet has centripetal acceleration toward the center (the Sun):

\[ a_c = \frac{v^2}{r} = \frac{(2\pi r/T)^2}{r} = \frac{4\pi^2 r}{T^2} \]

Combining Kepler's third law with centripetal acceleration

Kepler's third law: T² = k·r³. Substitute into the acceleration:

\[ a_c = \frac{4\pi^2 r}{k\, r^3} = \frac{4\pi^2}{k}\cdot \frac{1}{r^2} \]

The pillar result:

\[ \boxed{a_c \propto \frac{1}{r^2}} \]

A planet's acceleration toward the Sun falls off as the inverse square of the distance. That is the 1/r² law — and we have not yet even mentioned "force"; we derived it from the geometry of orbits and Kepler alone.

Bringing in "force": Newton's second law

By F = ma, if the planet has mass m and undergoes centripetal acceleration a_c, then the force from the Sun on the planet:

\[ F = m a_c \propto \frac{m}{r^2} \]

So the Sun's gravitational pull on a planet of mass m at distance r is proportional to m/r².

Why M (the Sun's mass) also appears

The above only put m (the planet's mass) into the force. Where does the M in GMm/r² come from?

Newton's elegant answer — his third law (action–reaction):

• The Sun pulls the planet with F = GMm/r².
• By Newton's third law, the planet pulls the Sun with the same magnitude of force.
• But if we replay the argument from the Sun's perspective (the Sun being pulled by the planet), the force must be proportional to M (the mass of the object being pulled — i.e., the Sun).

Both must be true simultaneously:

• F ∝ m/r² (Sun acting on planet)
• F ∝ M/r² (planet acting on Sun)

The only way: F ∝ Mm/r². Both masses appear, symmetrically. A proportionality constant is needed — Newton called it G:

\[ F = G\,\frac{Mm}{r^2} \]

But Newton did not have G

Notice that nowhere in the above did we compute the number G. Kepler gave the constant k; centripetal acceleration is 4π²r/T²; but the ratio only fixed the form of the law, not its magnitude.

Newton died 71 years before Cavendish. In Newton's era, the number G was unknown — and therefore the absolute mass of the Sun or Earth was unknown. Only ratios were accessible (M_Sun/M_Earth ≈ 330,000).

If you combine Kepler's law with a_c = 4π²r/T², you can compute the product GM_Sun (called the standard gravitational parameter of the Sun):

\[ G M_\odot = 4\pi^2 r^3 / T^2 \approx 1.33 \times 10^{20}\ \mathrm{m^3/s^2} \]

But you cannot pull G and M_Sun apart. To separate them, one must measure G independently — a task Cavendish completed a century later (Article 2).

"The Moon test": Newton's crowning check

At this point, Newton had a personal challenge: is this "central force" that holds planets around the Sun the same gravity that pulls an apple? If yes, then we have a universal law.

Newton's argument:

If the 1/r² law is correct, the gravitational acceleration on an object at distance r from Earth's center is proportional to 1/r². So if an apple at Earth's surface feels g, the Moon at distance r_moon should feel a much smaller acceleration:

\[ \frac{a_\text{moon}}{g} = \frac{R_E^2}{r_\text{moon}^2} \]

Numbers available to Newton:

• g ≈ 9.8 m/s² (from Galileo — see his method)
• R_E ≈ 6.37 × 10⁶ m (from Eratosthenes and later refinements)
• r_moon ≈ 3.84 × 10⁸ m (from astronomical parallax)
• T_moon ≈ 27.3 days ≈ 2.36 × 10⁶ s (from observation)

Prediction from the 1/r² law:

\[ \frac{r_\text{moon}}{R_E} \approx 60.3 \]

\[ a_\text{moon}^\text{predicted} = \frac{g}{60.3^2} \approx \frac{9.8}{3636} \approx 2.70 \times 10^{-3}\ \mathrm{m/s^2} \]

Direct measurement from the Moon's orbit (assumed circular):

\[ a_\text{moon}^\text{measured} = \frac{4\pi^2\, r_\text{moon}}{T_\text{moon}^2} = \frac{4\pi^2 \cdot 3.84 \times 10^8}{(2.36 \times 10^6)^2} \approx 2.72 \times 10^{-3}\ \mathrm{m/s^2} \]

They match! Predicted 2.70, measured 2.72 — less than 1% discrepancy. This is the moment: the same force that pulls the apple holds the Moon in orbit. Gravity is a universal law.

Newton himself said he first saw this in 1666, when he was 23 — but he waited 20 years to publish, both to complete the mathematics and to secure a better measurement of R_E.

What you should be able to do

After this article, you should be able to:

• Combine Kepler's third law (T² ∝ r³) with centripetal acceleration (4π²r/T²) to derive a ∝ 1/r²
• Explain why in F = G·Mm/r², both masses appear — not just one (from Newton's third law)
• Distinguish between "the form of the law" (Newton) and "the constant G" (Cavendish)
• State why M_E was unknown until 1798
• Reproduce the Moon test with numbers — a 1/r² prediction vs. direct measurement

Preview of Article 2

Now that we have the law but not the number G, the only way forward is to measure it directly in the laboratory: compute the gravitational force between two known masses. The problem: this force is tiny — for two 1-kg spheres 1 m apart, about 10⁻¹⁰ N. How do you measure a force this weak in the presence of friction, air currents, and vibrations from the ground?

Cavendish's answer: the torsion balance — a beautifully creative design we'll see in Article 2.

📚 Primary source: Isaac Newton, Philosophiæ Naturalis Principia Mathematica, 1687 — Book III, Propositions 1-8. 📖 Open reference: OpenStax University Physics Vol 1 — §13.1: Newton's Law of Universal Gravitation. 📖 Feynman Lectures Vol I — Ch 7: The Theory of Gravitation (§7-4 is a beautiful telling of the Moon test).

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