GravitationKepler's Laws
Kepler's first law, the law of orbits, says: every planet moves in an ellipse, with the Sun at one focus of that ellipse. This broke with Copernicus, who allowed only circles. An ellipse is easy to draw. Fix two pins F₁ and F₂, tie the ends of a string to them, hold the string taut with a pencil T and move the pencil round. Throughout this chapter the Earth, potential energy and the g curve are blue, the moving body, its path and its velocity green, forces and total energy purple, and the Sun, swept areas and strings amber.
Drag the pencil round the pins.
The amber string never stretches, so for every point T on the green curve the two distances add up to the same length: TF₁ + TF₂ = constant. The two pins are the foci of the ellipse. The line through them meets the ellipse at P and A; its midpoint is the centre, and half of PA is the semi-major axis a, the dashed line. When the two foci merge, the ellipse becomes a circle and a becomes its radius.
For a planet, the Sun sits at one focus. The point of the orbit nearest the Sun, P, is the perihelion, and the farthest point, A, is the aphelion. The other focus is just an empty point in space.
Equal Areas in Equal Times
Kepler's second law, the law of areas: the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. It came from noticing that planets seem to move slower when they are far from the Sun than when they are near it. The orbit below is stretched far more than any real planet's, so that the effect is easy to see.
The amber areas are all the same size, however the planet's distance from the Sun changes. Near P the radius is short, so to sweep the same area in a month the planet must cover a long arc; near A the radius is long and a short arc is enough. Watch the green velocity arrow: it is longest at perihelion.
Why equal areas? Put the Sun at the origin, and let the planet of mass m have position r and momentum p = mv. In a short time Δt it moves by vΔt, and the thin triangle it sweeps has area
so the rate of sweeping is
where L = r × p is the planet's angular momentum about the Sun. Gravity is a central force: it always points along the line joining the planet to the Sun, so it has no torque about the Sun and L stays constant. A constant L means a constant ΔA/Δt. The law of areas is angular momentum conservation in disguise, and it holds for any central force.
Example 7.1: Halley's Comet
Example 7.1 At perihelion P, let a body be at distance rP from the Sun S with speed vP. (a) Relate rP and vP to rA and vA at aphelion A. (b) Does the body take equal times to go from B to C through A, and from C to B through P?
Solution (a) At P and at A the velocity is at right angles to the line from the Sun, as the green arrows show, so the angular momentum is just m r v at both points. It is conserved, so
Fig. 7.1(a) — an elliptical orbit with the Sun at S; the area SBAC and the area SCPB.
Halley's comet comes as close as rP = 8.8 × 10¹⁰ m to the Sun, moving at 54.6 km s⁻¹, and goes out to rA = 5.3 × 10¹² m. Its speed at aphelion is smaller by the factor rA/rP ≈
(b) Which journey takes longer?
Equal areas in equal times, so time is proportional to the area swept. The area SBAC, swept on the far side, is larger than the area SCPB on the near side, so the body takes longer to go from B to C through A. Halley's comet spends most of its 76-year trip far out near aphelion, moving slowly, and races round the Sun in a few months.
The Law of Periods
Kepler's third law, the law of periods: the square of the time a planet takes to go round the Sun is proportional to the cube of the semi-major axis of its orbit. Table 7.1 lists the semi-major axis a (in units of 10¹⁰ m) and the period T (in years) of the eight planets. The last column is the ratio Q = T²/a³ (in units of 10⁻³⁴ y² m⁻³).
Table 7.1 — planetary data that confirm the law of periods
| Planet | a | T | Q |
|---|---|---|---|
| Mercury | 5.79 | 0.24 | 2.95 |
| Venus | 10.8 | 0.615 | 3.00 |
| Earth | 15.0 | 1 | 2.96 |
| Mars | 22.8 | 1.88 | 2.98 |
| Jupiter | 77.8 | 11.9 | 3.01 |
| Saturn | 143 | 29.5 | 2.98 |
| Uranus | 287 | 84 | 2.98 |
| Neptune | 450 | 165 | 2.99 |
Kepler did not know which power of a to try, and he searched for about ten years. The graph plots T² against aⁿ for the eight planets, both scaled so that Neptune sits at (1, 1). Find the power n that puts every planet on the amber straight line.
Power: n =
With n = 3 the green points fall on one straight line through the origin: T² ∝ a³. That is why Q in Table 7.1 is almost the same for every planet. Check one: for Jupiter, Q = 11.9² ÷ 77.8³ =
The ratio comes out the same for planets whose years range from three months to 165 years. Newton later showed what that ratio is: Q = 4π²/(GMS), where MS is the mass of the Sun. We derive this for circular orbits in Section 7.9.
Speed Round: Kepler's Laws
Ellipse with the Sun at a focus; equal areas in equal times, so m v r is the same at perihelion and aphelion; T² ∝ a³, so with the Earth's orbit as the unit, T = a3/2 years.
Get five right to move on.