Kepler's laws and the ellipse

Builds on: Sections S4 and S3
Reading time: 40 min
Exercises: 6
Olympiad problems: 1

Earth is closest to the Sun in early January, at 147.1 million km, and furthest in early July, at 152.1 million km. In the northern hemisphere this surprises many people, since their winter comes when the Sun is nearest, but a change of about 3 per cent in distance is far too small to cause the seasons, which come from the tilt of Earth's axis. The change does show up in the calendar. The Sun takes about 186 days to go from the March equinox to the September equinox, and only about 179 days to come back, because Earth moves faster along the part of its orbit that is near the Sun. Johannes Kepler found the rules behind this in the early 1600s from Tycho Brahe's measurements of Mars, and Isaac Newton later showed that they follow from his law of gravitation ((S.17)). In this section we describe the shape of an orbit and Kepler's three laws.

3.1.1The ellipse

Before you readMake a guess first

Earth's orbit is drawn to scale on a page, 20 cm across. How would it look?

An ellipse is a stretched circle. It is the set of points whose distances from two fixed points, called its foci, add up to the same length. Tying a loop of string round two pins and running a pencil round inside the loop draws one. Half the longest diameter of the ellipse is its semi-major axis $a$ , and half the shortest is its semi-minor axis $b$ . How stretched the ellipse is shows in its eccentricity $e$ , which is the distance from the centre to each focus divided by $a$ . A circle has $e = 0$ , and as $e$ approaches 1 the ellipse becomes long and thin.

In an orbit the Sun sits at one focus and the other focus is empty. The closest point of the orbit to the Sun is the perihelion, at a distance $a - a e$ , and the furthest is the aphelion, at $a + a e$ . Round Earth the same points are called the perigee and the apogee, and round any body in general the pericentre and the apocentre.

ResultThe geometry of an ellipse(3.1)

\begin{aligned} r_{p} & = a (1 - e) \\ r_{a} & = a (1 + e) \\ b^{2} & = a^{2} (1 - e^{2}) = r_{p} r_{a} \end{aligned}

$a, b$	the semi-major and semi-minor axes
$e$	the eccentricity, with $0 \leq e < 1$
$r_{p}, r_{a}$	the distances from the focus to the pericentre and to the apocentre

Turned round, the first two lines give

a = (r_{p} + r_{a}) / 2

and

e = (r_{a} - r_{p}) / (r_{a} + r_{p})

, so the closest and furthest distances fix the whole ellipse. The area of the ellipse is

π a b

For Earth, with $a = 1.496 \times 10^{8}$ km and $e = 0.0167$ , the perihelion distance is $a (1 - e) = 147.1$ million km and the aphelion distance is $a (1 + e) = 152.1$ million km, the figures we began with. Since $b / a = \sqrt{1 - e^{2}} = 0.99986$ , the orbit is a circle to the eye, as the guess above found, and only the position of the Sun gives the ellipse away.

Quick checkEarth's furthest and closest distances

Earth's orbit has eccentricity 0.0167. What is the ratio of its aphelion distance to its perihelion distance?

3.1.2Kepler's three laws

Kepler's first law says that each planet moves in an ellipse with the Sun at one focus. His second law says that the line from the Sun to the planet sweeps out equal areas in equal times. His third law says that the square of the period is proportional to the cube of the semi-major axis, which Section S3 found by plotting the planets on logarithmic axes.

The second law is easiest to see in the figure. A planet near perihelion is close to the Sun, so it must move a long way along its orbit to sweep a given area, while near aphelion a short move sweeps the same area. The planet therefore moves fastest at perihelion and slowest at aphelion.

Figure 3.1An orbit and Kepler's second law

Eccentricity

e

Time since perihelion

Draw

Distance SX
Speed of X
Perihelion and aphelion

The figure shows an elliptical orbit with the Sun S at one focus, the empty focus F, the centre C, the perihelion P and the aphelion A. The planet X moves at its true speed, worked out from Kepler's equation (Section 3.4), and the shaded wedge is the area that the line SX swept in the last twelfth of a period. Turning on the equal-time wedges cuts the orbit into twelve pieces that the planet sweeps in equal times, and all twelve have the same area, long and thin near A and short and wide near P. The distances in the table are in units of the semi-major axis

a

. The figure can be set to the orbit of , , or .

The physical reason for the second law is that gravity pulls the planet straight towards the Sun, so it gives no twist, or torque, about the Sun. The planet's ((S.22)) therefore stays constant, and the rate at which the line from the Sun sweeps out area is half the angular momentum divided by the planet's mass.

ResultKepler's second and third laws(3.2)

\begin{aligned} \frac{d A}{d t} & = \frac{L}{2 m} = \frac{1}{2} r v_{⟂} \\ T^{2} & = \frac{4 π^{2}}{G (M + m)} a^{3} \end{aligned}

$d A / d t$	the rate at which the line from the central body to the orbiting body sweeps out area
$L, m$	the angular momentum and the mass of the orbiting body
$r, v_{⟂}$	its distance from the central body and the part of its velocity at right angles to the line between them
$T, a$	the period and the semi-major axis of the orbit
$M$	the mass of the central body

The third law is the law for circular orbits from (S.19) with the radius replaced by the semi-major axis, so all orbits with the same

a

round the same body have the same period, whatever their eccentricity. For the planets, with

T

in years and

a

in astronomical units, it becomes

T^{2} = a^{3} / (M / M_{⊙})

when the planet's mass is small, which for the Sun is simply

T^{2} = a^{3}

Show the derivation

In a short time $d t$ the planet moves a distance $v d t$ . The area swept is a thin triangle with its point at the Sun, a base of length $r$ along the line to the planet, and a height equal to the part of the movement at right angles to that line, $v_{⟂} d t$ . Its area is $d A = \frac{1}{2} r v_{⟂} d t$ , and since the angular momentum is $L = m r v_{⟂}$ , this is $d A = (L / 2 m) d t$ . When $L$ is constant, equal times give equal areas.

For the third law, the area of the whole ellipse, $π a b$ , is swept in one period, so $T = 2 π a b m / L$ . Newton's solution of the orbit, which Section 3.3 describes, gives $L^{2} = G M m^{2} a (1 - e^{2}) = G M m^{2} b^{2} / a$ . Putting this in gives $T^{2} = 4 π^{2} a^{3} / G M$ , and replacing $M$ by $M + m$ allows for the motion of the central body (Section 3.7).

At perihelion and aphelion the planet moves at right angles to the line from the Sun, so the second law gives $r_{p} v_{p} = r_{a} v_{a}$ . The speeds at the two ends are in the inverse ratio of the distances.

ResultSpeeds at the closest and furthest points(3.3)

\frac{v_{p}}{v_{a}} = \frac{r_{a}}{r_{p}} = \frac{1 + e}{1 - e}

$v_{p}, v_{a}$	the speeds at the pericentre and at the apocentre
$r_{p}, r_{a}$	the distances of those points from the central body

The rule holds only at the two ends of the major axis, where the velocity is at right angles to the line from the central body. Elsewhere only the part of the velocity at right angles to that line enters the second law.

Earth moves 1.034 times as fast in January as in July. This is why the Sun takes 7 days longer to go from the March equinox to the September equinox than to come back, since that half of the orbit includes the aphelion and the slow part of the year.

Example 3.1The orbit of Halley's Comet

Halley's Comet moves on an orbit with semi-major axis 17.83 au and eccentricity 0.967. How close does it come to the Sun, how far away does it go, what is its period, and how much faster does it move at perihelion than at aphelion?

Solution

By (3.1), the perihelion distance is $17.83 \times (1 - 0.967) = 0.59$ au, inside the orbit of Venus, and the aphelion distance is $17.83 \times 1.967 = 35.1$ au, beyond Neptune. The comet's mass is tiny, so by (3.2) in years and astronomical units,

T = a^{3 / 2} = {17.83}^{3 / 2} = 75.3 years .

By (3.3), the ratio of the speeds is $1.967 / 0.033 = 60$ .

AnswerHalley's Comet comes within 0.59 au of the Sun and goes out to 35.1 au, returns every 75.3 years, and moves about 60 times as fast at perihelion as at aphelion. It spends most of its time far out, which is why it is seen for only a few months in each return.

3.1.3Weighing with Kepler's third law

The third law contains the mass of the central body, so an orbit weighs it. This is how we know the mass of the Sun from Earth's year, the mass of every planet that has a moon, and the masses of stars in binary systems, which Section 3.7 and Chapter 8 take further. Solved for the mass, the law reads $M = 4 π^{2} a^{3} / G T^{2}$ when the orbiting body is light.

Example 3.2Weighing Jupiter with Iofinish it

Io, the innermost of Jupiter's four large moons, orbits at 421 700 km from the centre of Jupiter with a period of 1.769 days. Take Io's mass as small. What is the mass of Jupiter?

The start

The period in seconds is $1.769 \times 86 400 = 152 800$ s, and the semi-major axis is $4.217 \times 10^{8}$ m. Solved for the mass, (3.2) gives $M = 4 π^{2} a^{3} / G T^{2}$ .

Your stepNow find

M

× 10²⁷ kg

Common errorsSection 3.1

The Sun is at a focus of the ellipse, and the centre of the ellipse is empty. The distance from the Sun to the perihelion is $a (1 - e)$ , and the distance from the centre to the perihelion is $a$ .
The form $T^{2} = a^{3}$ holds only with $T$ in years, $a$ in astronomical units and a central mass of one Sun. For a moon round a planet, or a planet round another star, the full form of (3.2) is needed, with the mass of the central body.
The second law is about areas. Near perihelion the planet covers a long arc in a given time and near aphelion a short one, while the areas swept in the two times are the same.
The ratio $v_{p} / v_{a} = r_{a} / r_{p}$ holds only at the two ends of the major axis. At any other point the velocity is at an angle to the line from the Sun, and only its part at right angles to that line enters the second law.

ExercisesSection 3.16 questions

3.1.1
Mars orbits the Sun with a semi-major axis of 1.524 au. What is its period?
years
3.1.2
At which point of its orbit does a planet move fastest?
3.1.3
A satellite is placed on a transfer orbit with its perigee 300 km and its apogee 35 786 km above Earth's surface. Earth's radius is 6371 km. What are the semi-major axis and the eccentricity of the orbit? Give the eccentricity as the answer.
3.1.4
What is the period of the transfer orbit in the last exercise? Take $G M$ for Earth as $3.986 \times 10^{14}$ m³ s⁻².
hours
3.1.5
A comet passes perihelion at 1.0 au from the Sun moving at 40.0 km/s, and its aphelion is at 9.0 au. How fast does it move at aphelion?
km/s
3.1.6
A planet orbits a star of half the Sun's mass with a semi-major axis of 4 au. What is its period?
years

Olympiad problemsPast IOAA problems, restated1 problem

2007A comet's period and the area it sweeps4 points

Adapted from IOAA 2007 (Chiang Mai), theory problems 1.6 and 1.7. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

A periodic comet goes round the Sun. At its furthest it is 31.5 au from the Sun and at its closest 0.5 au.

(a) What is the period of the comet? [2 points]

(b) What area, in square astronomical units per year, does the line joining the comet to the Sun sweep out? [2 points]

I solved it without the solution

AnswerThe period is 64 years, and the line from the Sun sweeps out 3.12 au² every year. The eccentricity is $31 / 32 = 0.969$ , close to that of Halley's Comet.