Lessons · gravitation

Elliptical orbits: Kepler's laws fall out of the inverse-square law

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A real orbit is not a circle but an ellipse with the planet at one focus ( $\mu = GM = 4\times10^{14}$ m³/s², semi-major axis $a = 1.6\times10^{7}$ m). The satellite races through its closest point (perihelion) and crawls at its farthest (aphelion). There is no algebra formula for where it is at time $t$ — so the path is integrated numerically from $\ddot{\mathbf r}=-\mu\mathbf r/r^3$ , and SymPy checks it conserves energy and angular momentum. Slide the eccentricity from a circle to a long ellipse: the shape changes completely, but because every orbit shares the same semi-major axis, they all take the same time to go around — Kepler's third law.

Period

T = 2\pi\sqrt{a^3/\mu}

(same for every

e

)

20110 s

Perihelion distance

r_p = a(1-e)

9600000 m

Aphelion distance

r_a = a(1+e)

22400000 m

Speed at perihelion (fastest)

7638 m/s

Speed at aphelion (slowest)

3273 m/s

Solve the central-force ODE numerically

\ddot{\mathbf r} = -\frac{\mu}{r^3}\,\mathbf r,\qquad \mathbf s_{n+1}=\mathbf s_n+\tfrac{h}{6}(k_1+2k_2+2k_3+k_4)

The same inverse-square law, but now the radius varies, so there is no elementary time-parameterisation (Kepler's equation is transcendental). Step the position and velocity forward with RK4 from perihelion. The producer checks the result conserves energy and angular momentum and that the orbit closes — verification, for a numerical solution.

integrate $\ddot{\mathbf r}=-\mu\mathbf r/r^3$ by RK4; check energy $\tfrac12 v^2-\mu/r$ and angular momentum $x\dot y-y\dot x$ are conserved; check the orbit closes; same $a\Rightarrow$ same $T$

✓ The specific energy $\tfrac12 v^2-\mu/r=-\mu/2a$ is conserved along the orbit (max relative drift $4.7e-13$ ). [numeric]
✓ The angular momentum $x\dot y-y\dot x$ is conserved (max relative drift $1.1e-14$ ) — Kepler's second law: the line to the body sweeps equal areas in equal times. [numeric]
✓ After one period the path returns to its start (gap $<7.5e-12\,a$ ) — a closed ellipse (Kepler's first law). [numeric]
✓ Every orbit shares the semi-major axis $a$ , so all have the same period $T=2\pi\sqrt{a^3/\mu}$ — Kepler's third law depends on $a$ , not the shape. [numeric]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A planet moves at constant speed around its orbit, and a more eccentric orbit takes longer to complete.”

Neither is true. By Kepler's second law (angular momentum is conserved), the planet sweeps equal areas in equal times — so it moves fastest at perihelion and slowest at aphelion; at $e=0.5$ it is three times faster at its closest point than its farthest. And by Kepler's third law the period depends only on the semi-major axis $a$ , not the eccentricity: a circle and a long thin ellipse with the same $a$ take exactly the same time to go around. Slide the eccentricity and watch the shape stretch while the period stays put.

Modeling assumptions — author-asserted, disclosed not discharged

Two-body problem: a point (or spherical) planet far more massive than the satellite, whose mass cancels; no other bodies, no drag.
All orbits share the semi-major axis $a$ (so the same period), and stay above the planet's surface (perihelion $r_p = a(1-e) \ge 8\times10^6$ m even at $e=0.5$ ).

The trajectory, fully annotated

A static rendering (Matplotlib) at the default launch — the path y vs x, with the apex and range marked. The interactive version with launch-angle and speed sliders is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):