Astrophysics  ·  18 October 2026

Kepler's Laws of Planetary Motion Explained: Ellipses, Equal Areas and the Harmony of the Worlds

Prague, 1601. Tycho Brahe dies, and Johannes Kepler inherits two things: the most accurate planetary observations anyone had ever made, and the job of explaining Mars. He tries for years to make Mars run on circles. He gets close — his best circular model matches Tycho's positions to within eight arcminutes, about a quarter of the Moon's width in the sky, better than any astronomer before him had managed. And he refuses to accept it. "Since the divine goodness has given us in Tycho Brahe a most careful observer, from whose observations the error of 8′ is shown… these eight minutes alone have led the way to the reformation of the whole of astronomy." Two thousand years of circles ended on a discrepancy most people would have rounded away.

There are three things on this page to push on, one for each of the laws that came out of that refusal. Two pins and a loop of string that draw a real planet's orbit, so you can see where the Sun actually sits. A stamp pad that lets you lay filled wedges on an orbit and read the same number off wedges of wildly different shape. And a set of scales that weighs Jupiter, then Earth, then the black hole at the centre of the Milky Way, from nothing more than a clock. Each law is also a tool. Somebody used each of them to find out something that could never have been measured directly, and those stories are here too, with their dates attached.

What Are Kepler's Three Laws of Planetary Motion?

Stated plainly, the three laws say this about every planet going round the Sun:

  1. The law of ellipses. Each planet moves on an ellipse, and the Sun is at one focus of that ellipse — not at its centre.
  2. The law of equal areas. A line drawn from the Sun to the planet sweeps out equal areas in equal times. The planet moves fastest when it is nearest the Sun and slowest when it is furthest away.
  3. The harmonic law. The square of a planet's orbital period is proportional to the cube of its average distance from the Sun: T2∝a3T^2 \propto a^3.

The first two appeared together in Astronomia Nova in 1609, after the fight with Mars. The third came nine years later: Kepler recorded that the idea arrived on 15 May 1618, and he published it in Harmonices Mundi in 1619. He never numbered them, and he never called them laws — the "three laws" framing is how later astronomers tidied up a body of work that, in Kepler's own books, is tangled up with harmonies, solids and a great deal of arithmetic. The numbering does not even follow the order of discovery. The second law came first.

What they replaced was the circle. Plato had asked for the heavens to be explained by uniform circular motion, and for two thousand years astronomers did as they were asked. Ptolemy stacked circles on circles — epicycles — until the model tracked the planets. Copernicus moved the Sun to the middle, which was the great step, and then kept the circles, so his system needed epicycles too. Kepler's laws are the moment someone let the data choose the curve.

Kepler's First Law: Why Are Orbits Ellipses and Not Circles?

Mars was the right planet to fight with, and Kepler was lucky to be handed it. Of the planets Tycho could track well, Mars has the most eccentric orbit, e=0.093e = 0.093, which is exactly enough for a circle to fail by eight arcminutes and not enough for the failure to be obvious. Venus would have hidden the truth; Mercury is too hard to see.

The war with Mars: the 1604 egg and the 1605 ellipse

Kepler's first non-circular attempt, in 1604, was an egg — an oval fatter at one end. It did not fit. In 1605 he tried the ellipse, the curve the Greeks had studied as a slice through a cone, and Mars fell into place. He had, in fact, already computed numbers that were the ellipse without recognising them, and he wrote about the moment with some embarrassment: the answer had been on his desk for a while.

What the law actually claims is worth getting exact, because textbook diagrams mislead. It does not say orbits are noticeably squashed. It says the Sun sits at a focus — one of the two special points of an ellipse — rather than at the centre. For Earth those are very different statements. Earth's orbit is 0.014% shorter across than along; no drawing on a screen can show the difference from a circle. But the Sun sits 2.5 million kilometres from the centre of that orbit, 3.6 times its own radius off to one side. Kepler's first law is a law about where the Sun is.

An ellipse is what you get from two pins and a loop of string. Push the pins into a board, loop the string round them, hold it taut with a pencil and go round: the pencil traces an ellipse, because every point it reaches has the same total distance to the two pins — the string's length, 2a2a. The two pins are the foci. Move them together and you get a circle. Pull them apart and the curve flattens. The distance from the centre to each focus is c=aec = ae, where ee is the eccentricity: 0 for a circle, close to 1 for a curve stretched almost to a line. The half-width of the ellipse is b=a1−e2b = a\sqrt{1 - e^2}, the nearest point to the Sun (perihelion) is a(1−e)a(1-e) and the furthest (aphelion) is a(1+e)a(1+e).

Try it yourself

  1. It loads with Earth, and the pencil is already going round — one lap in about four seconds. Watch the two string segments trade length as it goes; the legend under the canvas reads "string length 2a = 2.000 AU" and it never changes. The banner reads "Earth: the ellipse is 0.014% shorter across than along — no eye could tell it from a circle. What Kepler saw is that the Sun sits 2.5 million km off centre." Look for the faint cross at the centre of the ellipse: the amber Sun pin is to the right of it, and the ledger's "Sun off-centre (ae)" row puts that at "2.5 million km (3.6 R☉)".
  2. Switch "Circle centred on the Sun" to Shown. A dashed circle labelled "circle centred on the Sun" appears — the wrong model, with the Sun in the middle — and it is the same shape as the orbit, displaced sideways. At the far left a short caliper labelled "ae = 2.5 million km" measures the gap between the two curves, drawn in true proportion: a couple of pixels. That gap is the whole content of the first law for Earth.
  3. Choose Mars in the Orbit menu. The pencil redraws and the banner reads "Mars: visibly off-centre, barely squashed — the Sun sits 21.3 million km from the centre, yet the ellipse is only 0.44% shorter across than along." The perihelion label reads 1.381 AU and the aphelion label 1.666 AU. This is the orbit Kepler spent years on: still indistinguishable from a circle by shape, but with the Sun far enough off centre that eight arcminutes gave it away.
  4. Choose Halley's comet. Now the string trick makes a cigar: "Halley's comet: the orbit is 3.9× longer than it is wide, and the Sun sits 2.58 billion km from its centre — closer to the near end than Venus is to the Sun." The legend says the string is 35.66 AU long and that the Sun is drawn oversize (at true scale it would be a fraction of a pixel here). Turn the ghost circle on: it has nothing to do with this orbit at all.
  5. Drag the Eccentricity slider and the menu flips to "Custom (a = 1 AU)". At e = 0.500 the banner reads "Custom orbit (a = 1 AU): the orbit is 1.2× longer than it is wide, and the Sun sits 74.8 million km from its centre — closer to the near end than Venus is to the Sun." Push it to 0.970 and the pins sit almost at the ends of the string, 4.1× longer than wide. Then drag it back to 0: both pins collapse onto the centre cross, the offset reads 0 km, and you have the one case in which "the Sun is at the centre" is true.
Earth: the ellipse is 0.014% shorter across than along — no eye could tell it from a circle. What Kepler saw is that the Sun sits 2.5 million km off centre.
The pins are the foci and the string is 2a long — this construction is not an analogy for an ellipse, it is the definition of one, and every point the pencil reaches has distances to the two pins that sum to exactly 2a. All major axes are drawn horizontal: the orbit’s orientation in space is not shown, nor are inclination and precession. The Sun disc is never drawn smaller than 3 px, which for the orbit shown is 1.0× its true size at this scale — everything else, including the ghost circle and the ae gap, is in true proportion.
Orbit
Eccentricity e
0.02
e = 0.0167 now. Dragging switches the orbit to Custom (a = 1 AU); the ellipse updates live.
Circle centred on the Sun
The pencil goes once round in about four seconds. Choosing a new orbit redraws it; the slider reshapes the finished curve in place.
Eccentricity e0.0167
Semi-major axis a1.000 AU
b/a0.99986
Sun off-centre (ae)2.5 million km (3.6 R☉)
Perihelion a(1 − e)0.983 AU
Aphelion a(1 + e)1.017 AU
String length 2a2.000 AU
Period √a³1.00 years

The ghost circle is the picture most people carry: Sun in the middle, planet going round. For Earth it is almost right — the two curves differ by 2.5 million kilometres over a 300-million-kilometre orbit — and that "almost" is why it took a Tycho to measure the difference and a Kepler to believe it. The eccentricity slider shows the other half of the story. Nothing about the pins-and-string construction changes as you drag it; the orbit is the same kind of curve at e=0.017e = 0.017 and at e=0.967e = 0.967. Circles are not a different law. They are the special case where the two pins happen to coincide.

How eccentric are real orbits?

The column that matters in this table is not b/ab/a, which is nearly 1 for everything but the comet. It is the Sun's offset from the centre.

Bodya (AU)eb/aSun off-centre (ae)Perihelion (AU)Aphelion (AU)Period
Mercury0.38710.20560.9786411.9 million km0.3080.46788 days
Venus0.72330.00680.999980.74 million km0.7180.728225 days
Earth1.00000.01670.999862.50 million km (3.6 R☉)0.9831.0171.00 year
Mars1.52370.09340.9956321.3 million km1.3811.6661.88 years
Jupiter5.20260.04890.9988038.1 million km4.9485.45711.9 years
Saturn9.55490.05650.9984080.8 million km9.01510.0929.5 years
Pluto39.480.24880.968551.47 billion km29.6649.30248 years
Halley's comet17.830.9670.254782.58 billion km0.58835.0775.3 years

Venus is the roundest orbit in the Solar System, and even Mercury and Pluto — the eccentric ones — are only 2% and 3% shorter across than along. Draw any planet's orbit to scale on a sheet of paper and you will draw a circle. What you cannot do is put the Sun in the middle of it. Halley's comet is the case where the shape finally shows: an orbit nearly four times longer than it is wide, with the Sun tucked 2.58 billion kilometres from the centre — so far toward one end that at perihelion the comet passes inside the orbit of Venus.

Why the seasons aren't caused by distance

Earth is at perihelion around 3 January and at aphelion in early July. So in January the whole planet is 3.3% closer to the Sun than in July, and receives about 7% more sunlight per square metre, since intensity falls with the square of distance. The northern hemisphere has winter then anyway. Distance cannot be what drives the seasons, because the two hemispheres have opposite seasons at the same distance; the cause is the tilt of Earth's axis, which decides how high the Sun climbs and how long it stays up. The 7% is real and it does show up, faintly: southern summers happen at perihelion and are a little more intense than northern ones, an effect that the far larger ocean area in the south mostly washes out. Kepler's first law puts a number on a distance effect that exists and then tells you it is too small to be the seasons.

Kepler's Second Law: Why Do Planets Speed Up Near the Sun?

Kepler found this one first, in 1602, before he had the ellipse. He was trying to compute where Mars should be at a given time, and he guessed that a planet's speed varies inversely with its distance from the Sun — his "distance law". That guess is exactly right at perihelion and aphelion and slightly wrong everywhere else, and in wrestling it into a form he could calculate with he arrived at something better: the line from the Sun to the planet sweeps out area at a constant rate. A planet near the Sun covers a lot of arc in a month; far from the Sun it covers less; but the thin, long wedge it sweeps out when distant has the same area as the short, fat one it sweeps when near.

Underneath, the second law is the same thing that makes a skater spin faster when she pulls her arms in: conservation of angular momentum. Gravity pulls the planet straight toward the Sun, so it exerts no torque about the Sun, and the planet's angular momentum L=mr2θ˙L = m r^2 \dot\theta about the Sun cannot change. The rate at which the Sun–planet line sweeps area is

dAdt=12 r2θ˙=L2m\frac{dA}{dt} = \tfrac{1}{2}\, r^2 \dot\theta = \frac{L}{2m}

which is constant because LL is. Pull the planet in to a smaller rr and θ˙\dot\theta has to rise to compensate. Kepler had no concept of angular momentum, and no force law; he found a conservation law by fitting positions, eighty-five years before Newton could say why it held.

Try it yourself

  1. It loads with Earth going round the Sun to scale, one full orbit every twelve seconds, and the banner's live line changes as it goes: "Earth · 0.983 AU from the Sun · 30 km/s" as it passes perihelion at the right-hand end, and about "1.02 AU from the Sun · 29 km/s" at the far left. Under that it says "Lay two stamps — one near perihelion, one near aphelion — and the banner compares them." The slider reads "Slice = 30.4 days" and the ledger's "Area per slice" row reads 0.2618 AU².
  2. Press Stamp as the body passes perihelion. A coloured wedge grows from the Sun to the body for one slice, then freezes with "① 0.2618 AU²" printed in it. Wait half a lap and press Stamp again near aphelion. Pressed exactly at the two ends, the banner reads "Stamp 1 near perihelion: 0.983 → 0.986 AU from the Sun, 0.53 AU of arc at 30 km/s average. Stamp 2 near aphelion: 0.52 AU of arc at 29 km/s. Both sweep 0.2618 AU² — equal areas in equal times." Your arcs and speeds will differ a little depending on when you press; the two areas never will.
  3. Switch Seasons to "Show equinoxes & solstices". Four lines through the Sun appear, labelled with their 2026 dates — 20 Mar, 21 Jun, 23 Sep, 21 Dec — with perihelion marked "3 Jan", and the caption reads "Mar → Sep 186 d · Sep → Mar 179 d". The banner turns amber and adds "March equinox → September equinox 186.4 d; September → March 178.8 d. Earth passes aphelion in early July, moving slowest, so northern spring + summer is the long half."
  4. Switch Body to Halley's comet. The stamps clear, the slider now reads "Slice = 6.3 years", and a faint dashed ring appears around the Sun — Saturn's orbit, for scale. The comet whips through perihelion at "Halley's comet · 0.588 AU from the Sun · 54 km/s" and then crawls; the banner notes "Halley spends 92.9% of its 75.3-year period beyond Saturn’s orbit (r > 9.55 AU)." The ledger reads 54.46 km/s at perihelion and 0.914 km/s at aphelion.
  5. Stamp as the comet leaves perihelion and the wedge fans out to 16.4 AU in a single 6.3-year slice. Stamp again out near aphelion and you get a sliver. The banner: "Stamp 1 near perihelion: 0.588 → 16.4 AU from the Sun, 18 AU of arc at 13 km/s average. Stamp 2 near aphelion: barely 1.3 AU of arc at 1.0 km/s. Both sweep 21.20 AU² — equal areas in equal times." Eighteen AU of arc against one, and the same area.
  6. Switch to Mercury — "Mercury · 0.308 AU from the Sun · 59 km/s" at perihelion, "Slice = 7.3 days", 0.03839 AU² per slice — and drag the Slice slider to its maximum of 48 slices per orbit: "Slice = 1.8 days", 0.009598 AU² each. Lay six quick stamps round the orbit and compare their widths; Clear stamps wipes them. The ledger's speeds, 58.98 and 38.86 km/s, are in the ratio 1.52 — exactly (1 + e)/(1 − e) for e = 0.2056.
Earth · 0.983 AU from the Sun · 0.0 km/s
Lay two stamps — one near perihelion, one near aphelion — and the banner compares them.
Time-lapse pacing: one full orbit takes 12 s of wall clock whatever the body, with mean anomaly advancing linearly, so the speed-up at perihelion is physically faithful but the absolute rate is not. Sector areas are exact from mean anomaly — (M₂ − M₁)/2π × πab — while arc lengths are summed along a 200-segment polyline, good to about 0.1%. Speeds are vis-viva about 1 M☉ with no planetary perturbations. Earth’s perihelion date drifts by a day or two from year to year, and the equinox and solstice dates shown are the 2026 values. Halley’s comet passed aphelion on 9 December 2023 at 35.14 AU (JPL); its next perihelion is 28 July 2061.
Body
Time-slice
12 per orbit
One slice is 1/12 of the period. Every stamp sweeps 0.2618 AU² regardless of where it is laid.
Stamp near perihelion, then again near aphelion, and compare the two numbers. Up to six stamps.
Seasons
BodyEarth
Period1.00 years
Slice1/12 · 30.4 days
Area per slice0.2618 AU²
Area rate3.141 AU²/yr
Speed at perihelion30.29 km/s
Speed at aphelion29.29 km/s
Stamps laid0 / 6

For Earth the two wedges are nearly the same shape, because Earth's orbit is nearly a circle; the law is doing very little work, and the speeds at perihelion and aphelion, 30.29 and 29.29 km/s, differ by 3%. For Halley the law is doing everything. The comet covers eighteen astronomical units of arc in the six years after perihelion and little more than one in the six years around aphelion, and the two wedges have identical areas to four significant figures. Nothing in the stamp pad computes the areas from the wedges; they come straight from the constant sweep rate, πab/T\pi ab / T, which for Earth is 3.141 AU² per year and for Halley 3.380.

Why is northern summer a week longer than winter?

Count the days between the equinoxes. From the March equinox to the September equinox is 186.4 days; from September back round to March is 178.8. Northern spring plus summer is seven and a half days longer than autumn plus winter, and the second law is the reason. Earth passes aphelion in early July, where it moves slowest, so it dawdles through the half of its orbit that holds northern summer; it passes perihelion around 3 January, moving fastest, and hurries through the winter half. Southern hemisphere readers get the short summer. The Seasons overlay above draws the four quarter-points where they really fall on the orbit, with perihelion 13° past the December solstice, and the two numbers in the caption are read straight off those lines — no calendar required.

Where is Halley's Comet right now?

Halley's comet last came through perihelion on 9 February 1986, at 0.588 AU from the Sun and moving at 54.5 km/s. It then spent nearly 38 years climbing away, slowing the whole time, until on 9 December 2023 it reached aphelion at 35.14 AU — most of the way out to Pluto's average distance — moving at 0.91 km/s, sixty times slower than at perihelion, and turned round. It is now falling back. It will cross Saturn's orbit inbound in late 2058 and reach perihelion again on 28 July 2061. Because the sweep rate is constant, the fraction of the period the comet spends at any distance can be read off the orbit's geometry: 92.9% of its 75.3-year lap is spent beyond Saturn's orbit, and the whole of the spectacular part — the months it spends inside the orbit of Mars, tail blazing — is the rest. If you were born after 1986 you have never seen it, and the second law says why: for most of a human lifetime it is a dim rock crawling round the far end of a very long ellipse.

Kepler's Third Law: How Do We Weigh Planets, Stars and Black Holes?

The third law arrived on 15 May 1618, by Kepler's own account, and the timing could hardly have been worse. The Thirty Years' War broke out in Prague that same month and would run for the rest of his life. Within two years his mother Katharina was arrested for witchcraft in Württemberg — in 1620 — and Kepler spent much of the next year conducting her legal defence; she was acquitted in 1621. Harmonices Mundi, published in 1619 in the middle of all this, is the book in which the law appears, and it is mostly about something else. Kepler wanted the planets' orbits to make music — literal harmonies, ratios of speeds that corresponded to musical intervals — and he spent most of the book on that. The power law he found on the way is one sentence in it.

What does Kepler's third law say, and what did Newton add?

Take a planet's orbital period TT and its semi-major axis aa (the average of its perihelion and aphelion distances). Then T2/a3T^2 / a^3 is the same number for every planet. Measure TT in years and aa in astronomical units and that number is 1:

T2=a3(T in years, a in AU)T^2 = a^3 \qquad (T \text{ in years},\ a \text{ in AU})

Jupiter at 5.2 AU takes 5.23=11.9\sqrt{5.2^3} = 11.9 years; Mars at 1.524 AU takes 1.88. It is a strange fact, and Kepler could only report it. Newton, in the Principia of 1687, explained it — and in explaining it he changed what it was for. If the Sun pulls with a force GMm/r2GMm/r^2, then a body in a circular orbit of radius rr must have GMm/r2=mv2/rGMm/r^2 = m v^2 / r with v=2πr/Tv = 2\pi r / T, and rearranging gives

T2=4π2GM a3T^2 = \frac{4\pi^2}{GM}\, a^3

which holds for ellipses too, with aa the semi-major axis (the gravitation post derives this step by step). Kepler's constant is 4π2/GM4\pi^2 / GM: it depends on the mass of the thing being orbited and nothing else. That is why every planet shares the same ratio — they all orbit the same Sun — and it is why the law became the most useful equation in astronomy. Rearrange it and you have

M=4π2a3G T2M = \frac{4\pi^2 a^3}{G\,T^2}

Measure how far something orbits and how long it takes, and you have weighed whatever it is going round. You do not need to touch it, land on it or know what it is made of. Jupiter is weighed by its moons, the Sun by its planets, the Milky Way's central black hole by a single star, and a planet no telescope can see is sized up from the shadow it casts. Every one of those is the same equation. The interactive below is that equation as a pair of scales: pick a system, guess the mass of the thing at the centre, and a ghost body laps at the period your guess predicts while the real one laps at the period actually observed. When they run in step, you have weighed it.

Try it yourself

  1. It loads with Io going round Jupiter, and the slider parked at −1 dex — a tenth of the true mass. The caption under the slider reads "1.90 × 10²⁶ kg · 31.8 Earths" and the banner reads "Your 1.90 × 10²⁶ kg would give Io a 5.59-day month; Jupiter actually turns it round in 1.769 days. Heavier." Under it: "T_yours / T_obs = 3.16× — the ghost falls behind by 68% of a lap every observed lap." The two clocks beneath Jupiter read "Observed 1.769 d" and "Yours 5.594 d". Watch the hollow amber ring fall further behind the solid body every lap — too little mass, too slow an orbit.
  2. Drag the slider up and watch the ring. Press Sync whenever you want the two bodies back together so the drift builds from zero. Overshoot to +0.50 dex and the banner flips: "Your 6.00 × 10²⁷ kg would give Io a 23.9-hour month; Jupiter actually turns it round in 1.769 days. Lighter." — the ghost now "pulls ahead by 78% of a lap every observed lap". You have bracketed Jupiter.
  3. Bring the slider to 0.00 dex — the badge accepts anything within ±2% — and the banner turns teal: "Weighed: Jupiter = 1.90 × 10²⁷ kg — 318 Earths, from one moon's clock." A pill appears on the canvas reading "Weighed: Jupiter = 1.90 × 10²⁷ kg". If the slider will not land, press "Snap" — it sets the mass to the Kepler-III answer and re-syncs. Then choose Callisto → Jupiter and snap again: the same 1.90 × 10²⁷ kg from a moon four and a half times further out (the ledger's Orbit radius a reads 0.01259 AU) whose month is 16.69 days. Two clocks, one answer.
  4. Choose Moon → Earth and snap. "Weighed: Earth = 6.03 × 10²⁴ kg — 1.01 Earths, from one moon's clock. That is 1% over Earth's 5.97 × 10²⁴ kg — the Moon is heavy enough that you have weighed Earth and Moon together." The law weighs the pair, and the Moon is 1.2% of Earth. Then Geostationary satellite → Earth: "Observed 23 h 56 min", and snapping gives "Weighed: Earth = 5.97 × 10²⁴ kg — 1.00 Earths, from one satellite's clock." — the satellite is light enough to weigh Earth alone.
  5. Choose S2 → Sagittarius A*. The orbit becomes the long thin loop the star really follows, e = 0.885, with the black hole at one end, and S2 visibly whips through pericentre. At −1 dex: "Your 8.46 × 10³⁵ kg would give S2 a 50.8-year lap; Sagittarius A* actually turns it round in 16.05 years. Heavier." Snap: "Weighed: Sagittarius A* = 8.46 × 10³⁶ kg — 4.25 million M☉, from one star's clock. 4.25 million Suns inside a region no wider than three times Pluto's orbit — a black hole or nothing."
  6. Finish with TRAPPIST-1e → TRAPPIST-1, a planet no one has ever seen directly: "Observed 6.100 d", and on snapping, "Weighed: TRAPPIST-1 = 1.78 × 10²⁹ kg — 0.0897 M☉, from one planet's clock." A star a little under a tenth of the Sun's mass, weighed by the timing of a dip in its light.
Your 1.90 × 10²⁶ kg would give Io a 5.59-day month; Jupiter actually turns it round in 1.769 days. Heavier.
T_yours / T_obs = 3.16× — the ghost falls behind by 68% of a lap every observed lap.
Every orbit but S2’s is drawn as a circle; the real ones are ellipses of small eccentricity, and the third law depends on the semi-major axis a alone, so the weighing is unaffected. S2 is drawn to its true e = 0.885 with Sagittarius A* at the focus, and its speed varies along the loop as the second law demands. Periods are sidereal (orbit-to-orbit against the stars), and one lap of the real body is eight seconds of your time whatever the system. The “true” mass is what a and T give through M = 4π²a³/GT² with G = 6.674 × 10⁻¹¹ N m² kg⁻²; the Moon’s answer is deliberately left 1% above Earth’s catalogue mass because the law weighs the pair, and the Moon is 1.2% of Earth. S2’s 4.25 million M☉ sits 1% under the GRAVITY collaboration’s 4.30 million because their fit is relativistic. “Snap to observed period” is the accessibility path to the badge for anyone who cannot chase a drifting ring.
System
Your mass guess
-1 dex
1.90 × 10²⁶ kg · 31.8 Earths
Drag until the hollow ghost stops drifting off the solid body. Within ±2% counts.
Sync puts the ghost back on the real body so you can watch the drift build from zero. Snap sets your mass to the Kepler-III answer.
Orbit radius a421 700 km
Observed period1.769 d
Your mass1.90 × 10²⁶ kg
Predicted period5.594 d
T_yours / T_obs3.16×
In familiar units31.8 Earths

The mass the badge reports is not looked up; it is what aa and TT give through the equation, which is why the Moon comes out 1% heavy — the law returns Earth plus Moon — and why Sagittarius A* comes out at 4.25 million solar masses rather than the 4.30 million the GRAVITY collaboration publishes from a fit that includes general relativity. Those are not errors. They are the equation telling you exactly what it can and cannot see from a single orbit.

How was the Milky Way's black hole weighed?

Two teams, one led by Reinhard Genzel in Germany and one by Andrea Ghez in California, began tracking individual stars around the centre of the Milky Way — Genzel's from 1992, Ghez's from 1995 — using infrared cameras to see through the dust and adaptive optics to sharpen the images. One star, S2, turned out to be on a 16.05-year orbit with a semi-major axis of about 1,030 AU and an eccentricity of 0.885 — Halley-like in shape, though larger than the whole Solar System. Its pericentre passage in May 2018 brought it to 120 AU from whatever sits at the centre, moving at 7,650 km/s, two and a half percent of the speed of light. Put aa and TT into Kepler's third law and out comes 4.25 million solar masses. That much mass inside a sphere no bigger than S2's closest approach — 120 AU, about three times Pluto's average distance from the Sun — cannot be a cluster of ordinary stars; no cluster that dense could survive. It is a black hole or nothing, and the 2018 pericentre passage also showed the gravitational redshift and the orbital precession that general relativity predicts for a black hole of exactly that mass. Genzel and Ghez shared the 2020 Nobel Prize in Physics for it, with Roger Penrose. The measurement rests on knowing the distance to the Galactic Centre, 8.25 kiloparsecs, to turn an angle on the sky into a length — the same chain of distance measurements that begins with parallax.

How do astronomers size up exoplanets they cannot see?

NASA's Kepler space telescope, named for him, flew from 2009 to 2018 and watched hundreds of thousands of stars for ones that dimmed by a fraction of a percent at regular intervals — a planet crossing in front. It found more than 2,600 confirmed planets that way. The transit tells you almost nothing directly: a dip, and how often the dip repeats. But "how often" is an orbital period, and if you know the star's mass — from its colour and brightness — the third law turns that period into an orbital radius, and the radius sets how much starlight the planet receives and so how warm it is. TRAPPIST-1e is the example everyone uses: it transits every 6.10 days, its star has 0.09 of the Sun's mass, and a=GMT2/4π23a = \sqrt[3]{GMT^2/4\pi^2} puts it at 0.0293 AU — thirteen times closer to its star than Mercury is to the Sun, which around so faint a star is the habitable zone. Nobody has seen the planet. Its distance from its star is known to three figures from a clock and Kepler's law.

Why do geostationary satellites sit at 35,786 km?

Turn the law round once more. Fix the mass — Earth's, 5.972×10245.972 \times 10^{24} kg — and fix the period to one sidereal day, 86,164 seconds, the time Earth takes to turn once relative to the stars. The law then permits exactly one orbital radius: 42,164 km from Earth's centre, which is about 35,800 km above the equator. A satellite there orbits once as Earth turns once and hangs over the same spot on the ground, which is why every satellite dish on every house points at the same fixed patch of sky. The number is not a design choice. It is the only altitude the equation allows for that period, and every geostationary satellite ever launched sits in the same ring. GPS satellites orbit at half the period, 11 h 58 min, and the same arithmetic puts them at 26,560 km from the centre, about 20,200 km up. The worked example below does the geostationary case in full. The commonly quoted 35,786 km is measured above Earth's equatorial radius of 6,378 km — a geostationary satellite sits over the equator — while subtracting the mean radius of 6,371 km, as an exam paper does, gives 35,790 km. The 7 km gap is Earth's bulge, not the orbit.

Where Kepler's third law breaks — and what that told us

In 1970 Vera Rubin and Kent Ford published rotation measurements of the Andromeda galaxy, using a new spectrograph to clock the speed of stars and gas at different distances from its centre. Kepler's third law made a clear prediction. Almost all of Andromeda's visible mass sits in its bright central bulge, so stars far out should orbit slowly, the way Pluto crawls where Mercury sprints — speed falling as 1/r1/\sqrt{r}. The stars did not. Out to the edge of the visible disc and beyond, the orbital speed stayed flat: stars in the outskirts were going as fast as those near the middle. The equation had not failed. If vv does not fall with rr, then M(r)M(r), the mass inside radius rr, must keep growing in proportion to rr, long after the visible stars have run out. There had to be mass out there that gave off no light — dark matter, most of the galaxy by weight — and the case for it was built on trusting Kepler's law more than the telescope. A law that had spent 350 years being confirmed became a discovery instrument the moment it appeared to break.

How Did Newton Explain Kepler's Laws?

In August 1684 Edmond Halley travelled to Cambridge with a question. He and Robert Hooke and Christopher Wren had been arguing about it in London: if the Sun attracted the planets with a force that fell off as the inverse square of distance, what path would a planet follow? Halley put it to Newton. Newton said an ellipse. Halley asked how he knew, and Newton said he had calculated it — and then could not find the calculation. He promised to redo it and send it on. The nine pages that arrived that November, De motu corporum in gyrum, grew over the next three years into the Principia, published in 1687, in which all three of Kepler's laws fall out of a single force law. An inverse-square attraction gives closed orbits that are conic sections with the Sun at a focus — the first law. Any central force gives equal areas in equal times, because it exerts no torque — the second. And the inverse-square strength specifically gives T2∝a3T^2 \propto a^3 with a constant set by the central mass — the third. Kepler's three separate rules turned out to be three faces of one equation.

Halley then did something with it that Kepler could not have. He collected records of bright comets and noticed that the ones of 1531, 1607 and 1682 had near-identical orbits and came round at 75-to-76-year intervals. Treating them as one object on a long Keplerian ellipse, and allowing for the tug of Jupiter, he predicted its return for the end of 1758. He died in 1742. The comet was picked up on Christmas night 1758, sixteen years after his death, and it has carried his name since — the first object ever shown to obey Kepler's laws before anyone had seen it do so. What happens to a body that arrives with more speed than any ellipse can hold — the open-ended orbits Newton's equation also allows — is the escape velocity post's subject.

How Are Kepler's Laws Examined?

In the Leaving Certificate, Kepler's third law sits in Strand 1.5 (the gravitational field) of the specification first examined in 2027, and the question shapes carry over from the old syllabus's Higher Level material: derive T2=4π2r3/GMT^2 = 4\pi^2 r^3 / GM for a circular orbit by equating gravitational force to centripetal force, then use it — the period of a satellite at a given height, the height of a geostationary orbit, or the mass of a planet from a moon's period and radius. At A-Level the same derivation appears, usually with the centripetal force mv2/rmv^2/r or mω2rm\omega^2 r written out first and a log–log plot of TT against rr (gradient 3/2) as the data-handling question. AP Physics 1 wants the third law understood as a consequence of Newton's gravitation, plus the qualitative content of the first two: where the Sun is, and where the planet is fastest. JEE examines all three quantitatively — eccentricity and the perihelion/aphelion distances a(1∓e)a(1 \mp e), the area law as constant LL and vprp=varav_p r_p = v_a r_a, and third-law numericals across different central bodies.

Three habits save marks. Use the semi-major axis, not the perihelion or aphelion distance, in the third law — and for a circular orbit that is just the radius. Keep TT and aa in the same unit system as your constant: with GG in SI you need metres and seconds, but between two bodies orbiting the same primary the constant cancels and T12/T22=a13/a23T_1^2 / T_2^2 = a_1^3 / a_2^3 works in any units at all, years and AU included. And remember that the mass in the equation is the mass of the body being orbited, never the orbiting one — a satellite's own mass appears on both sides and cancels, which is why a screw and a space station at the same altitude have the same period.

Worked Examples for Physics Exams

Worked Example

Example 1 — Orbital period of Mars from its semi-major axis

Mars orbits the Sun with a semi-major axis of 1.5237 AU. Find its orbital period in years and in days.

About the Sun, with TT in years and aa in AU, Kepler's third law reads T2=a3T^2 = a^3:

T2=1.52373=3.5375,T=3.5375=1.881 yearsT^2 = 1.5237^3 = 3.5375, \qquad T = \sqrt{3.5375} = 1.881 \text{ years}

T=1.881×365.25=687 daysT = 1.881 \times 365.25 = 687 \text{ days}

No constant, no GG, no mass of the Sun — those are all inside the fact that the ratio equals 1 for anything orbiting the Sun, once the units are years and AU. Earth calibrates it: 1 AU, 1 year.

Check it in the pins-and-string construction above: choose Mars and the ledger's Period row reads 1.88 years.

Worked Example

Example 2 — Altitude of a geostationary satellite

A satellite is to stay above a fixed point on the equator. Earth's mass is 5.972×10245.972 \times 10^{24} kg, its radius 6,371 km, and one sidereal day is 86,164 s. Take G=6.674×10−11G = 6.674 \times 10^{-11} N m² kg⁻². Find the orbital radius and the altitude.

Rearrange T2=4π2r3/GMT^2 = 4\pi^2 r^3 / GM for rr:

r3=GM T24π2=(6.674×10−11)(5.972×1024)(86 164)24π2r^3 = \frac{GM\,T^2}{4\pi^2} = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(86\,164)^2}{4\pi^2}

GM=3.986×1014 m3 s−2,T2=7.424×109 s2GM = 3.986 \times 10^{14} \text{ m}^3\text{ s}^{-2}, \qquad T^2 = 7.424 \times 10^{9} \text{ s}^2

r3=3.986×1014×7.424×10939.48=7.495×1022 m3,r=4.216×107 m=42 163 kmr^3 = \frac{3.986 \times 10^{14} \times 7.424 \times 10^{9}}{39.48} = 7.495 \times 10^{22} \text{ m}^3, \qquad r = 4.216 \times 10^{7} \text{ m} = 42\,163 \text{ km}

The altitude is the radius less Earth's radius:

h=42 163−6 371=35 792 km≈35 790 kmh = 42\,163 - 6\,371 = 35\,792 \text{ km} \approx 35\,790 \text{ km}

The figure usually quoted, 35,786 km, subtracts Earth's equatorial radius of 6,378 km instead of the mean radius used here — a geostationary satellite sits over the equator, where Earth is 7 km fatter. The orbital radius is the same either way, and the period is the only input that matters — the satellite's own mass never appears.

Check it in the scales above: choose Geostationary satellite → Earth. The ledger reads Orbit radius a 42 164 km, Observed period 23 h 56 min, and snapping to the observed period weighs Earth at 5.97 × 10²⁴ kg.

Worked Example

Example 3 — Mass of Jupiter from the orbit of Io

Io orbits Jupiter at a mean distance of 421,700 km with a period of 1.7691 days. Find Jupiter's mass.

Convert to SI first: a=4.217×108a = 4.217 \times 10^{8} m and T=1.7691×86 400=1.5285×105T = 1.7691 \times 86\,400 = 1.5285 \times 10^{5} s. Then

M=4π2a3G T2=39.48×(4.217×108)3(6.674×10−11)(1.5285×105)2M = \frac{4\pi^2 a^3}{G\,T^2} = \frac{39.48 \times (4.217 \times 10^{8})^3}{(6.674 \times 10^{-11})(1.5285 \times 10^{5})^2}

a3=7.499×1025 m3,T2=2.336×1010 s2a^3 = 7.499 \times 10^{25} \text{ m}^3, \qquad T^2 = 2.336 \times 10^{10} \text{ s}^2

M=2.961×10271.559=1.90×1027 kgM = \frac{2.961 \times 10^{27}}{1.559} = 1.90 \times 10^{27} \text{ kg}

That is 318 Earth masses, and it agrees with the catalogue value of 1.898×10271.898 \times 10^{27} kg to three figures. Callisto, at 1,882,709 km and 16.689 days, gives the same answer — a3/T2a^3/T^2 is 2.396×10162.396 \times 10^{16} km³/day² for every one of Jupiter's moons, which is the third law in a single number.

Check it in the scales above: Io → Jupiter, slider to 0.00 dex — or Snap — and the banner reads "Weighed: Jupiter = 1.90 × 10²⁷ kg — 318 Earths, from one moon's clock."

Worked Example

Example 4 — Halley's comet: speed at perihelion and aphelion from the area law

Halley's comet has eccentricity 0.967 and reaches 54.5 km/s at perihelion. Find the ratio of its perihelion speed to its aphelion speed, and its speed at aphelion.

At perihelion and aphelion the velocity is perpendicular to the Sun–comet line, so angular momentum conservation — Kepler's second law — gives directly

m rpvp=m rava⇒vpva=rarp=a(1+e)a(1−e)=1+e1−em\,r_p v_p = m\,r_a v_a \quad\Rightarrow\quad \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{a(1+e)}{a(1-e)} = \frac{1+e}{1-e}

vpva=1.9670.033=59.6\frac{v_p}{v_a} = \frac{1.967}{0.033} = 59.6

va=54.559.6=0.91 km/sv_a = \frac{54.5}{59.6} = 0.91 \text{ km/s}

The semi-major axis cancelled — the ratio depends on eccentricity alone. For Earth, e=0.0167e = 0.0167 gives a ratio of 1.034, a 3% difference; for Mercury, 1.52. Notice that the ratio of distances is also 59.6 (35.07 AU against 0.588 AU): the comet is sixty times further away and sixty times slower, which is Kepler's original "distance law" holding exactly at the two ends of the orbit.

Check it in the stamp pad above: choose Halley's comet and the ledger reads Speed at perihelion 54.46 km/s, Speed at aphelion 0.914 km/s.

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