Gravity and Orbits

Part 3: Newton’s Law of Gravitation

From Patterns to Mechanism

Now we arrive at Newton’s great insight: gravity is a universal force between all masses, following a precise law.

Each piece carries meaning. Gravity depends on the product of both masses (by the third law, each must feel the same force). The $r^2$ in the denominator is an inverse-square law: double the distance, and the force drops to one-quarter. And $G$ is a universal constant — the same everywhere in the universe. Solving $F_g = Gm_1m_2/r^2$ for $G$ confirms its units must be $\mathrm{cm^3\,g^{-1}\,s^{-2}}$ (equivalently $\mathrm{dyne\,cm^2\,g^{-2}}$). In the limits: as $r \to \infty$ the force approaches zero but never quite vanishes; if either mass is zero, there is no force.

Note on light and gravity: Newtonian gravity acts on mass. Since photons are massless, it doesn’t correctly predict how light bends near massive objects — Einstein’s General Relativity does. More on that in Part 6.

How Did Newton Know This?

Newton didn’t propose the inverse-square law out of thin air — he derived it by requiring that gravity reproduce Kepler’s observations. The reasoning, as a scaling argument:

Assume orbits are nearly circular ($r \approx a$), the force is a power-law central force $F \propto 1/r^n$ for unknown $n$, and the central mass dominates ($m \ll M$).

Step 1 — Kepler III as a constraint: $P^2 \propto a^3$.

Step 2 — Postulate a general force: $F = k\,Mm/r^n$.

Step 3 — Apply circular-motion physics: gravity provides the centripetal force, so $kMm/r^n = mv^2/r$. The planet’s mass $m$ cancels:

$\frac{kM}{r^n} = \frac{v^2}{r}$

Step 4 — Connect speed and period: for circular motion $v = 2\pi r/P$, so

$\frac{kM}{r^n} = \frac{4\pi^2 r}{P^2} \quad\Rightarrow\quad P^2 = \frac{4\pi^2}{kM}\, r^{n+1}$

Step 5 — Compare with Kepler III: $P^2 \propto r^3$ requires $n + 1 = 3$, so $n = 2$. The inverse-square law is the only power law consistent with Kepler’s Third Law.

Step 6 — Identify the constant: when $n = 2$, $k$ is Newton’s gravitational constant $G$, with units $\mathrm{cm^3\,g^{-1}\,s^{-2}}$, giving $F = GMm/r^2$. Newton didn’t just say “here’s a formula that works” — he showed the inverse-square law is required by the observed period-distance relation.

Newton’s Version of Kepler’s Third Law

Armed with the law of gravitation, Newton derived a complete version of Kepler III:

What’s different from Kepler’s empirical version? The constant of proportionality is now determined — it depends on $G$ and the total mass $(M+m)$ — and it applies to any two-body system, not just our solar system. A unit check confirms the right side has units of $\mathrm{s^2}$. When $M \gg m$ (Sun-planet systems), $P^2 \approx 4\pi^2 a^3 / (GM)$, recovering Kepler’s simpler form. And by measuring $P$ and $a$ we can solve for the total mass,

$M + m = \frac{4\pi^2 a^3}{G P^2},$

which is how we “weigh” stars in binaries, measure black-hole masses from orbiting stars, and estimate galaxy masses from rotation curves.

Two-Body Reality: The Center of Mass

Where does that $(M+m)$ factor come from? So far we’ve imagined the Sun sitting motionless. But Newton’s Third Law says if the Sun pulls on Earth, Earth pulls back equally — the Sun must move too. In reality, both bodies orbit their common barycenter.

For two bodies separated by distance $r$, the center of mass lies along the connecting line at distances

$r_1 = \frac{m_2}{m_1 + m_2}\, r \quad\text{and}\quad r_2 = \frac{m_1}{m_1 + m_2}\, r,$

with $r_1 + r_2 = r$. The more massive body orbits closer to the center of mass. When $M \gg m$, the barycenter lies very close to $M$ — which is why Kepler’s “Sun at one focus” works so well. For the Sun-Jupiter system the Sun wobbles by about 750,000 km (roughly one solar radius); that wobble is exactly how we detect exoplanets via the radial-velocity method. When both bodies are properly accounted for, the orbital dynamics depend on the total mass, which is where the $(M+m)$ in Kepler III comes from.