Kepler's laws.

Step 1 of 6

Ellipse and focus

In the Unit on circular motion and gravity we worked out circular orbits. But a satellite moving a little too fast or a little too slow neither falls nor escapes: it moves on a stretched orbit. Johannes Kepler found, in 1609, from measurements of Mars, that this is what happens to every planet. Kepler’s first law: every planet moves on an ellipse, with the Sun at one of its two foci.

Two numbers describe the orbit. Size of the orbit is the semi-major axis $a$⁠, half the long diameter, in astronomical units: 1 AU is the Earth’s mean distance from the Sun, about $1.5 \times 10^{11}\ \mathrm{m}$⁠. Eccentricity $e$⁠ says how stretched the orbit is: $0$⁠ is a circle, and close to $1$⁠ is a long, thin ellipse. The foci are on the long axis, each $ae$⁠ from the centre.

In the Scene the eccentricity is $0.5$⁠, far more than any real planet’s, so that the shape stands out. The thin ring is the Earth’s orbit.

Step 2 of 6

Closest and farthest

The point of the orbit closest to the Sun is called perihelion, and the farthest aphelion. Both are on the long axis, and their distances from the Sun are:

$r_{\min} = a(1 - e) \qquad r_{\max} = a(1 + e)$$r_{\min} =$$a(1 - e) \qquad r_{\max} =$$a(1 + e)$⁠

The Earth’s orbit is almost a circle, $e = 0.017$⁠: in January our distance from the Sun is only about $3\%$⁠ less than in July. Eccentricity and Size of the orbit are now locked.

Step 3 of 6

Equal areas in equal times

Run the motion and watch the planet. Kepler’s second law says what happens to the speed: the line joining the Sun to the planet sweeps out equal areas in equal times. Close to the Sun the line is short, so to sweep the same area every week, the planet has to cover more path.

Today we know this is the conservation of angular momentum: gravity always pulls towards the Sun, so it does not turn the planet round it, and $r \cdot v$⁠ at perihelion equals $r \cdot v$⁠ at aphelion. Predict, and then Show equal areas will shade the area the line swept in the last month.

Step 4 of 6

Fast and slow

At perihelion and aphelion the motion is at right angles to the line to the Sun, so keeping $r \cdot v$⁠ gives the ratio of the speeds directly:

$\frac{v_{\max}}{v_{\min}} = \frac{r_{\max}}{r_{\min}} = \frac{1 + e}{1 - e}$$\frac{v_{\max}}{v_{\min}} =$$\frac{r_{\max}}{r_{\min}} =$$\frac{1 + e}{1 - e}$⁠

Now the orbit is locked on Mercury’s, the most stretched of the planets’ orbits. The wedges of Show equal areas show the same area, wide and short on one side and thin and long on the other.

Step 5 of 6

The third law

In the last Unit we derived Kepler’s third law for a circular orbit, $T^2 = \dfrac{4\pi^2}{GM}\, r^3$$T^2 =$$\dfrac{4\pi^2}{GM}\, r^3$⁠. It holds for an ellipse too, with the semi-major axis $a$⁠ in place of $r$⁠. Round the Sun, measuring $a$⁠ in astronomical units and $T$⁠ in years, the constant comes out as $1$⁠:

$T^2 = a^3$⁠

So the Earth, $a = 1$⁠, goes round in one year. Eccentricity is locked at $0.09$⁠, like Mars’s, and the Earth goes round in the Scene so you can compare.

Step 6 of 6

Shape and year

The third law does not contain the eccentricity. Size of the orbit is now locked at 1 AU, and the orbit is a circle. Predict what will happen to the year when we stretch it.