Planetary orbits and Kepler’s laws

What is it about?

The planet is attracted only by the star’s gravitational force $F_G = G \frac{M m}{r^2}$ . The simulation does not know about ellipses: it computes acceleration, velocity and position in tiny time steps. Still, exactly Kepler’s orbits appear.

First law: planets move on ellipses with the Sun at one focus $F_1$ . For every point $r_1 + r_2 = 2a$ ( $a$ : semi-major axis, $c = \sqrt{a^2 - b^2}$ : focal distance, $e = \frac{c}{a}$ : eccentricity). Closest point: perihelion $r_P = a - c$ , farthest point: aphelion $r_A = a + c$ .
Second law: the line Sun–planet sweeps out equal areas in equal times, so $r_P v_P = r_A v_A$ (conservation of angular momentum).
Third law: for all planets of the same star

\frac{T^2}{a^3} = \frac{4\pi^2}{G M} \approx 2.97 \cdot 10^{-19}\,\tfrac{\text{s}^2}{\text{m}^3} \quad \text{(Sun)}

With the circular speed $v_K = \sqrt{\frac{G M}{r_0}}$ (perpendicular start) the orbit is a circle; from the escape speed $v_F = \sqrt{2} \cdot v_K$ on it is open (parabola, hyperbola). At 1 AU: $v_K \approx 29.8\,\tfrac{\text{km}}{\text{s}}$ , $v_F \approx 42.1\,\tfrac{\text{km}}{\text{s}}$ . The total energy $E = \frac{1}{2} m v^2 - G \frac{M m}{r}$ is negative for bound orbits.

Try it

Press Start and compare the speed at perihelion and aphelion. After one orbit the ellipse is divided into areas swept in equal times – compare the bars.
Drag the arrow tip across the rings $v_K$ and $v_F$ : circle · hyperbola
Watch $r_1 + r_2$ : string construction
Measure several orbits in the table: Kepler’s third law · two solar masses

Tasks

Task 1: Earth’s orbit

Find the circular speed at $r = 1.496 \cdot 10^{11}\,\text{m}$ from the Sun ( $M = 1.989 \cdot 10^{30}\,\text{kg}$ ) and the period. Load task

Show solution

$v_K = \sqrt{\frac{G M}{r}} \approx 29.8\,\tfrac{\text{km}}{\text{s}}$ , $T = \frac{2\pi r}{v_K} \approx 3.16 \cdot 10^{7}\,\text{s} \approx 365\,\text{d}$ .

Task 2: Perihelion and aphelion

$r_P = 1.00\,\text{AU}$ , $v_P = 36.0\,\tfrac{\text{km}}{\text{s}}$ , $r_A = 2.71\,\text{AU}$ . Find $v_A$ , $a$ and $e$ . Load task

Show solution

$v_A = v_P \frac{r_P}{r_A} \approx 13.3\,\tfrac{\text{km}}{\text{s}}$ , $a = \frac{r_P + r_A}{2} = 1.855\,\text{AU}$ , $e = \frac{r_A - r_P}{r_A + r_P} \approx 0.461$ (the simulation uses $r_A = 2.7099\,\text{AU}$ and shows $a \approx 1.85\,\text{AU}$ ).

Task 3: Mars

Mars has $a = 1.524\,\text{AU}$ . Find its period (Earth: $a = 1\,\text{AU}$ , $T = 1$ year). Load task

Show solution

$T = \sqrt{1.524^3}$ years $\approx 1.88$ years. Compare with the table

Ideas for the classroom

Use: gravitation and Kepler’s laws in upper secondary school (grades 11–13), from guessing a law to checking it with the table and the log–log graph.
Model: point masses, $m \ll M$ , only the star’s gravity; leapfrog (Verlet) integration with a fixed time step (energy conserved to better than $10^{-4}$ , angular momentum exactly). Star and planet are not to scale. Planet data: rounded mean orbital elements (JPL, E. M. Standish).
Common stumbling blocks: the Sun is at a focus, not at the centre; real planetary orbits are almost circles; the period does not depend on the planet’s mass; the escape speed does not depend on the direction.