Kepler's Three Laws from Gravity

Derivation

Kepler's Three Laws from Gravity

D-030 Home PU-101 Threads force · fields · symmetry Depends on Effective Potential and the Orbit Equation, Angular Momentum, Torque, and Central Forces

Statement

For a body of mass \(m\) bound to a spherically symmetric mass \(M\) by the Newtonian inverse-square force \(\vec{F} = -\dfrac{GMm}{r^2}\,\hat{r}\), the central-orbit shape equation integrates exactly to the conic \(r(\theta)=\dfrac{p}{1+e\cos\theta}\). For total energy \(E<0\) (equivalently \(0\le e<1\)) this is an ellipse with the centre of force at one focus (First Law). Conservation of angular momentum makes the areal velocity constant, \(\dfrac{dA}{dt}=\dfrac{L}{2\mu}\) (Second Law). Integrating that constant rate over one complete ellipse gives \(T^2=\dfrac{4\pi^2 a^3}{G(M+m)}\) (Third Law).

Why it matters

The three empirical laws Kepler extracted from Tycho Brahe's naked-eye data are not three independent facts: they are three faces of one dynamical statement, that gravity falls off as the inverse square of distance. Recovering all three from \(F\propto 1/r^2\) was Newton's decisive vindication of universal gravitation, and it is the archetype of deriving global orbital structure from a purely local force law.

The derivation also isolates which law encodes which physics. The Second Law is pure angular-momentum conservation and holds for any central force; the First and Third are special to the inverse square. This dissection is the tool by which anomalies — Mercury's perihelion advance, galactic rotation curves — are read as evidence that the true potential deviates from \(1/r\).

Assumptions

Central inverse-square force, \(\vec F=-\dfrac{GMm}{r^2}\hat r\).For any other power law the orbit is generally an open, precessing rosette; only \(1/r^2\) and the linear (Hooke) force give closed orbits (Bertrand's theorem), so both the fixed ellipse and the exact \(T^2\propto a^3\) fail. Isolated two-body system.A third mass adds a non-central perturbation: the pair's total angular momentum is still conserved but each individual orbit precesses and is no longer a fixed ellipse. Point masses, or spherically symmetric mass distributions.By the shell theorem a spherical body acts externally as a point mass; if \(M\) is oblate (a nonzero \(J_2\)) or the body orbits inside the mass, the potential gains multipole terms and the perihelion drifts. Bound orbit: total mechanical energy \(E<0\), i.e. \(0\le e<1\).For \(E\ge 0\) the same conic solution is a parabola (\(e=1\)) or hyperbola (\(e>1\)): the path is unbounded and no period \(T\) exists, so the Third Law is vacuous. Non-relativistic speeds and weak field, \(v\ll c\), \(GM/(rc^2)\ll 1\).General relativity adds an effective \(1/r^3\) term to the potential; the ellipse precesses at \(\Delta\varpi = \dfrac{6\pi GM}{c^2 a(1-e^2)}\) per orbit — the famous \(43''\) per century for Mercury.

Derivation

\[ \frac{d^2 u}{d\theta^2}+u = -\frac{\mu}{L^2 u^2}\,F\!\left(\tfrac1u\right), \qquad L=\mu r^2\dot\theta=\text{const},\qquad u\equiv\frac1r \]

Both quoted from prior results: the Binet shape equation in \(u\equiv 1/r\) (binet-orbit-equation-effective-potential), and constancy of \(L\) because a central force exerts zero torque, \(\vec\tau=\vec r\times\vec F=\vec 0\) (angular-momentum-and-torque). Here \(\mu\) is the reduced mass. A

\[ F\!\left(\tfrac1u\right)=-k\,u^2 \;\Rightarrow\; \frac{d^2u}{d\theta^2}+u = -\frac{\mu}{L^2u^2}\bigl(-k\,u^2\bigr)=\frac{\mu k}{L^2} \]

Insert the inverse-square law with \(k\equiv GMm>0\), so \(F=-k/r^2=-k\,u^2\). The two factors of \(u^2\) cancel — the special feature of the inverse square — leaving a constant on the right. A

\[ u(\theta)=\frac{\mu k}{L^2}+A\cos(\theta-\theta_0) \]

The equation is linear with the harmonic operator \(\tfrac{d^2}{d\theta^2}+1\) and constant forcing: particular solution \(\mu k/L^2\) plus homogeneous \(A\cos(\theta-\theta_0)\). Choose the polar axis so \(\theta_0=0\) (measure \(\theta\) from perihelion) and take \(A\ge 0\). B

\[ r=\frac1u=\frac{p}{1+e\cos\theta}, \qquad p\equiv\frac{L^2}{\mu k}, \qquad e\equiv A\,p \]

Invert and factor out \(\mu k/L^2\). This is the standard polar equation of a conic with a focus at the origin (the centre of force). Classifying by \(e\): \(0\le e<1\Rightarrow\) ellipse — Kepler's First Law. The semi-latus rectum satisfies \(p=a(1-e^2)\). C

\[ \frac{dA}{dt}=\frac12 r^2\dot\theta=\frac{L}{2\mu}=\text{const} \]

The area swept by the radius vector in \(dt\) is the triangle \(dA=\tfrac12 r\cdot r\,d\theta\). Substituting the conserved \(L=\mu r^2\dot\theta\) makes the rate constant — Kepler's Second Law. Note this used only centrality, not the inverse square. B

\[ \int_0^T dA = \pi a b = \frac{L}{2\mu}\,T \;\Rightarrow\; T=\frac{2\pi\mu\,ab}{L} \]

Over one period the radius vector sweeps the whole ellipse, area \(\pi a b\). Because \(dA/dt\) is constant, total area equals rate times \(T\); solve for the period. B

\[ b=a\sqrt{1-e^2},\quad L=\sqrt{\mu k\,a(1-e^2)} \;\Rightarrow\; T=\frac{2\pi\mu\,a^2\sqrt{1-e^2}}{\sqrt{\mu k\,a(1-e^2)}}=2\pi\sqrt{\frac{\mu}{k}}\,a^{3/2} \]

Use \(b=a\sqrt{1-e^2}\) and invert \(p=L^2/\mu k=a(1-e^2)\) to get \(L\). Substituting, the factors of \(\sqrt{1-e^2}\) cancel exactly, so the period is independent of eccentricity — it depends only on \(a\). C

\[ T^2=\frac{4\pi^2\mu}{k}\,a^3 = \frac{4\pi^2 a^3}{G(M+m)} \]

Square, then insert \(\mu=\dfrac{Mm}{M+m}\) and \(k=GMm\), giving \(\dfrac{\mu}{k}=\dfrac{1}{G(M+m)}\) — Kepler's Third Law. The orbit shape \(e\) and orientation have dropped out entirely. C

Result

\[ r(\theta)=\frac{a(1-e^2)}{1+e\cos\theta} \quad\Big|\quad \frac{dA}{dt}=\frac{L}{2\mu} \quad\Big|\quad T^2=\frac{4\pi^2 a^3}{G(M+m)} \]

Reading. The single inverse-square shape equation produces all three laws at once. First: bound orbits are ellipses with the attracting mass at a focus (not the centre). Second: the line from focus to body sweeps area at a fixed rate, so the body moves fastest at perihelion. Third: the square of the period is set only by the semi-major axis and the total mass — not by the eccentricity, orientation, or (when \(M\gg m\)) the orbiting body's own mass. That mass-independence is why a feather and a moon at the same \(a\) share one period.

Units check. \([a^3/(GM)] = \dfrac{\mathrm{m^3}}{(\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg})}=\dfrac{\mathrm{m^3}}{\mathrm{m^3\,s^{-2}}}=\mathrm{s^2}=[T^2]\). The semi-latus rectum \([p]=\left[\dfrac{L^2}{\mu k}\right]=\dfrac{(\mathrm{kg\,m^2\,s^{-1}})^2}{(\mathrm{kg})(\mathrm{kg\,m^3\,s^{-2}})}=\mathrm{m}\), and \(e\) is dimensionless.

Limiting cases

\(e\to 0\): \(r=p=a\), a circle traversed at constant speed; the Second Law reduces to uniform angular motion.
\(M\gg m\): \(\mu\to m\) and \(M+m\to M\), so \(T^2=\dfrac{4\pi^2 a^3}{GM}\) — Kepler's original mass-independent form for planets round the Sun.
\(M=m\) (equal binary): \(T^2=\dfrac{4\pi^2 a^3}{2GM}\); the \(M+m\) correction is a factor of \(2\) and cannot be ignored.
\(e\to 1^-\): a needle-thin ellipse; the period is still fixed by \(a\) alone, while the perihelion-to-aphelion speed ratio diverges like \(\sqrt{(1+e)/(1-e)}\).

Breaks when

Strong field / relativistic speeds. The Schwarzschild potential adds an effective \(-\dfrac{GML^2}{\mu^2 c^2 r^3}\) term; the orbit no longer closes and the ellipse precesses (Mercury: \(43''\) per century beyond Newtonian).
Unbound energy, \(E\ge 0\). The conic solution is a parabola or hyperbola; the body escapes, there is no \(a>0\) in the closed sense and no period, so the Third Law is undefined.
Third bodies or non-spherical potentials. Planetary perturbations, an oblate primary (\(J_2\)), or an extended mass distribution all break the exact \(1/r^2\); orbits precess into rosettes and \(T\) acquires shape-dependent corrections.
Comparable-mass or many-body systems where "\(a\)" is ambiguous. In a hierarchical or resonant system the two-body ellipse is only an osculating approximation, valid over one orbit.

Failure modes

Radius-for-axis substitution. Plugging an instantaneous \(r\) (or the perihelion/aphelion distance) into \(T^2\propto a^3\). Only the semi-major axis \(a=(r_{\min}+r_{\max})/2\) belongs there.
Using \(M\) instead of \(M+m\). Fine for a planet round the Sun, wrong by up to a factor of \(2\) for binary stars or a planet-moon pair of comparable mass.
Focus vs centre. Placing the Sun at the geometric centre of the ellipse rather than at a focus; the two coincide only for \(e=0\).
Believing the period depends on \(e\) or launch direction. Two orbits with the same \(a\) but different eccentricities share one period exactly.
Averaging speeds instead of using \(L\). The Second Law fixes \(v_{\text{peri}}r_{\text{peri}}=v_{\text{apo}}r_{\text{apo}}\); a naive arithmetic mean of apsidal speeds is not the mean orbital speed.
Forgetting the reduced mass in \(L=\mu r^2\dot\theta\), then getting the two-body Third-Law constant wrong.

Discussion

The derivation makes explicit that the closed ellipse is a coincidence of the inverse square. Bertrand's theorem states that among all central forces only \(F\propto 1/r^2\) and \(F\propto r\) give orbits that close after one revolution; every other power law produces a precessing rosette. The exact cancellation of the \(u^2\) factors in Step 2, which turned the shape equation into simple harmonic motion in \(\theta\), is the algebraic fingerprint of that special status.

Each law carries different physical content, and separating them is diagnostically powerful. The Second Law is nothing but angular-momentum conservation and survives any central perturbation, so a body that violates equal areas must be feeling a non-central force. The First Law (closed ellipse, fixed orientation) is the most fragile: even a tiny extra radial term destroys closure, which is exactly why perihelion precession became a precision test-bed — first for undiscovered mass (Le Verrier's hunt for Vulcan) and finally for general relativity.

There is a deeper reason the Kepler ellipse is fixed in space: the inverse-square problem has a "hidden" conserved vector, the Laplace–Runge–Lenz vector \(\vec A=\vec p\times\vec L-\mu k\,\hat r\), which points along the major axis and has constant magnitude \(\mu k e\). Its conservation is equivalent to the orbit not precessing, and it reflects a larger dynamical symmetry (\(SO(4)\) for bound states) than the geometric \(SO(3)\) rotational symmetry alone. Any perturbation that lifts this accidental degeneracy makes \(\vec A\) rotate slowly — precession — which is why the LRL vector is the natural language for the perturbation theory of nearly-Keplerian orbits.

Common misconceptions. The Sun sits at a focus, never the centre, so the planet–Sun distance genuinely varies (Earth's perihelion is in January, unrelated to the seasons, which are driven by axial tilt). "Faster when closer" is a consequence of the Second Law, not a cause of the seasons. And the Third-Law constant \(4\pi^2/[G(M+m)]\) is a property of the central mass, which is precisely why measuring a satellite's \(a\) and \(T\) weighs the primary.

Worked examples

Example 1 — Earth's orbital period from its semi-major axis.

\[ T = 2\pi\sqrt{\frac{a^3}{GM_\odot}} \qquad (M_\odot\gg m_\oplus) \]

Third Law in the \(M\gg m\) limit, solved for \(T\). A

\[ a=1.496\times10^{11}\,\mathrm{m},\quad G=6.674\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}},\quad M_\odot=1.989\times10^{30}\,\mathrm{kg} \]

Insert numbers with units. A

\[ a^3=3.348\times10^{33}\,\mathrm{m^3},\quad GM_\odot=1.328\times10^{20}\,\mathrm{m^3\,s^{-2}},\quad \frac{a^3}{GM_\odot}=2.522\times10^{13}\,\mathrm{s^2} \]

Arithmetic; note the ratio already carries units \(\mathrm{s^2}\). A

\[ T=2\pi\sqrt{2.522\times10^{13}\,\mathrm{s^2}}=3.156\times10^{7}\,\mathrm{s}=365.3\ \text{days} \]

Reading. One year, recovered to four figures. The tiny discrepancy from \(365.256\) days is the neglected \(m_\oplus/M_\odot\approx 3\times10^{-6}\) correction plus rounding. Units check. \(\sqrt{\mathrm{s^2}}=\mathrm{s}\).

Example 2 — Mass of Jupiter from Io's orbit.

\[ M_J=\frac{4\pi^2 a^3}{G\,T^2} \qquad (M_J\gg m_{\text{Io}}) \]

Third Law solved for the central mass — the "celestial scale." A

\[ a=4.217\times10^{8}\,\mathrm{m},\qquad T=1.529\times10^{5}\,\mathrm{s}\ (1.769\ \text{days}) \]

Io's measured semi-major axis and sidereal period. A

\[ a^3=7.499\times10^{25}\,\mathrm{m^3},\quad T^2=2.338\times10^{10}\,\mathrm{s^2},\quad 4\pi^2=39.48 \]

Arithmetic. A

\[ M_J=\frac{(39.48)(7.499\times10^{25})}{(6.674\times10^{-11})(2.338\times10^{10})}=1.90\times10^{27}\,\mathrm{kg} \]

Reading. Within \(0.1\%\) of the accepted \(1.898\times10^{27}\,\mathrm{kg}\). A moon's orbit weighs its planet without ever touching it — the core technique behind measuring stellar and black-hole masses. Units check. \(\dfrac{\mathrm{m^3}}{(\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{s^2})}=\mathrm{kg}\).

Problems

Show directly, for a circular orbit of radius \(r\), that Newton's second law gives \(T^2=\dfrac{4\pi^2 r^3}{GM}\), i.e. the Third Law with \(a=r\).
Solution
For a circle gravity supplies the centripetal force: \(\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\), so \(v^2=\dfrac{GM}{r}\). The period is \(T=\dfrac{2\pi r}{v}\), hence \(T^2=\dfrac{4\pi^2 r^2}{v^2}=\dfrac{4\pi^2 r^2\cdot r}{GM}=\dfrac{4\pi^2 r^3}{GM}\). The \(m\) cancels, giving the mass-independence explicitly.
Jupiter's orbital period is \(11.86\,\mathrm{yr}\). Using AU–year units for the Sun (\(T^2=a^3\) with \(T\) in years, \(a\) in AU), find its semi-major axis.
Solution
\(a=T^{2/3}=11.86^{2/3}\). Take \(11.86^{1/3}=2.281\), then square: \(2.281^2=5.20\). So \(a=5.20\,\mathrm{AU}\), matching Jupiter's tabulated value. The AU–year form is just the Third Law with the constant \(4\pi^2/GM_\odot\) absorbed into the units.
Halley's Comet has a period of \(75.3\,\mathrm{yr}\) and a perihelion distance of \(0.586\,\mathrm{AU}\). Find (a) its semi-major axis, (b) its aphelion distance, and (c) its eccentricity.
Solution
(a) \(a=75.3^{2/3}\,\mathrm{AU}\); \(75.3^{1/3}=4.223\), squared \(=17.83\), so \(a=17.83\,\mathrm{AU}\). (b) From \(a=(r_{\text{peri}}+r_{\text{apo}})/2\), \(r_{\text{apo}}=2a-r_{\text{peri}}=35.66-0.586=35.08\,\mathrm{AU}\) (out past Neptune). (c) \(e=\dfrac{r_{\text{apo}}-r_{\text{peri}}}{r_{\text{apo}}+r_{\text{peri}}}=\dfrac{34.49}{35.66}=0.967\), a highly elongated ellipse.
A binary star system has an orbital period \(T=50\,\mathrm{yr}\) and a relative-orbit semi-major axis \(a=20\,\mathrm{AU}\). Find the total mass \(M+m\) in solar masses.
Solution
In Solar units \(\dfrac{a^3}{T^2}=M+m\) with \(a\) in AU, \(T\) in yr, mass in \(M_\odot\) (this follows because for the Sun–Earth case \(a=T=M=1\)). Then \(M+m=\dfrac{a^3}{T^2}=\dfrac{20^3}{50^2}=\dfrac{8000}{2500}=3.2\,M_\odot\). Only the sum follows; splitting it requires the individual orbit sizes about the centre of mass.
For an orbit of eccentricity \(e\), use the Second Law to find the ratio of the body's speed at perihelion to its speed at aphelion, and evaluate it for \(e=0.6\).
Solution
At both apsides the velocity is perpendicular to \(\vec r\), so \(L=\mu r v\) gives \(v_{\text{peri}}r_{\text{peri}}=v_{\text{apo}}r_{\text{apo}}\). Hence \(\dfrac{v_{\text{peri}}}{v_{\text{apo}}}=\dfrac{r_{\text{apo}}}{r_{\text{peri}}}=\dfrac{a(1+e)}{a(1-e)}=\dfrac{1+e}{1-e}\). For \(e=0.6\) this is \(\dfrac{1.6}{0.4}=4.0\): the body sweeps through perihelion four times faster than aphelion, the Second Law made quantitative.