Back to the on-screen lesson ·

Orbits and Kepler's third law

Farther planets have longer paths and move more slowly, so with distance in AU and period in years, the period squared equals the distance cubed; around a heavier star orbits are quicker.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to explain why outer planets have long years, and use Kepler's third law to find a period from a distance and a distance from a period.

2. What you already know

You know that gravity holds the planets in orbit around the Sun and that it weakens with the square of the distance. You know that Earth takes one year to go around the Sun, and that the outer planets are farther away. This lesson finds the exact rule connecting a planet's distance to the length of its year, and uses it to predict orbits anywhere.

3. Words for this lesson

TermWhat it means
Astronomical unit (AU)Earth's average distance from the Sun, about 150 million km or 93 million miles.
Orbital periodThe time a body takes to go once around its orbit.
EllipseA stretched circle; planets move in slightly elliptical orbits.
Kepler's third lawFor orbits around the Sun, the period in years squared equals the distance in AU cubed.
CubeA number multiplied by itself three times, such as 4 × 4 × 4 = 64.

4. Period squared equals distance cubed

Measure an orbit's average distance from the Sun $a$ in astronomical units and its period $P$ in years. Then for every planet, asteroid and comet:

$$P^2 = a^3.$$

  1. Cube the distance.
  2. Take the square root to get the period.

At $4$ AU: $4^3 = 64 = 8^2$, so the period is $8$ years. At $9$ AU: $9^3 = 729 = 27^2$, so $27$ years. A planet farther out takes much longer, because its path is longer and it moves more slowly. Around a star $M$ times the Sun's mass, the rule becomes $P^2 = a^3 / M$.

Orbital period in years against average distance from the Sun in astronomical units, with the curve where the period squared equals the distance cubed. Mercury at 0.39 AU takes 0.24 years, Venus at 0.72 AU 0.62 years, Earth at 1 AU 1 year, Mars at 1.52 AU 1.88 years and Jupiter at 5.2 AU 11.86 years. All five sit on the curve. The curve bends upward: five times farther takes more than eleven times longer, because a farther planet has a longer path and also moves more slowly along it.
Orbital period in years against average distance from the Sun in astronomical units, with the curve where the period squared equals the distance cubed. Mercury at 0.39 AU takes 0.24 years, Venus at 0.72 AU 0.62 years, Earth at 1 AU 1 year, Mars at 1.52 AU 1.88 years and Jupiter at 5.2 AU 11.86 years. All five sit on the curve. The curve bends upward: five times farther takes more than eleven times longer, because a farther planet has a longer path and also moves more slowly along it.

Another way: picture

Imagine runners on circular tracks of different sizes around a field, where the runners on the outer tracks are also more tired and jog more slowly. The outer runners take far longer to finish a lap, both because their track is longer and because they are slower.

Another way: steps

  1. Measure distance in AU and period in years.
  2. Cube the distance.
  3. Square-root the result for the period.
  4. To go backward, square the period and cube-root it.
  5. Around a heavier star, divide the cube by its mass first.

5. Kepler's search for a pattern

Johannes Kepler spent years working with the careful planet positions measured by the Danish astronomer Tycho Brahe. In 1609 he showed that planets move in ellipses with the Sun at one focus, and that they speed up when nearer the Sun. In 1619 he found a third rule, connecting the size of each orbit to its period.

Kepler had no explanation for why the rule held. Seventy years later Isaac Newton showed that it follows directly from gravity weakening with the square of distance. The third law is therefore not just a pattern in the numbers but a test of gravity itself, and it has passed that test for every planet, moon and satellite measured since.

6. Choosing convenient units

The law takes its simplest form when distance is measured in astronomical units and time in years, because then Earth's own orbit gives $1^2 = 1^3$. One AU is Earth's average distance from the Sun, about $150$ million km or $93$ million miles.

In these units, you can compare any orbit around the Sun directly with Earth's. Mars averages $1.52$ AU from the Sun; Jupiter $5.2$ AU; Neptune about $30$ AU. A comet that swings far out past Neptune might average $100$ AU, and its period would be $\sqrt{100^3} = \sqrt{1{,}000{,}000} = 1{,}000$ years.

7. Working the law forward

To find a period from a distance, cube the distance and take the square root. For an asteroid at $4$ AU: $4^3 = 4 \times 4 \times 4 = 64$, and $\sqrt{64} = 8$, so it takes $8$ years. Distances that are perfect squares, such as $4$, $9$, $16$ and $25$, give whole-number periods: $8$, $27$, $64$ and $125$ years.

For real planets the numbers are not so tidy, and a calculator is needed. For Mars, $1.52^3 \approx 3.51$ and $\sqrt{3.51} \approx 1.87$ years, very close to the measured $1.88$ years. The chart shows the real planets sitting right on the curve.

8. Working the law backward

To find a distance from a period, square the period and take the cube root. Halley's Comet returns about every $76$ years. Then $76^2 = 5{,}776$, and the cube root of $5{,}776$ is about $17.9$, so its average distance from the Sun is about $18$ AU, between Uranus and Neptune.

Periods that are perfect cubes give tidy answers: a period of $8$ years means a distance of $4$ AU, because $8^2 = 64 = 4^3$. This backward use is how astronomers find the sizes of orbits they cannot measure directly, from timing alone.

9. Why farther orbits are slower

If every planet moved at Earth's speed, a planet twice as far would simply take twice as long, because its path would be twice as long. But the Sun's pull is four times weaker at twice the distance, and a weaker pull can only hold a slower planet in orbit; a fast planet would fly off.

Earth moves along its orbit at about $30$ km every second. Jupiter, $5.2$ times farther out, moves at only about $13$ km/s, and Neptune at about $5.4$ km/s. Longer paths at lower speeds make the outer planets' years very long: Neptune has gone around the Sun only about once since it was discovered in 1846.

10. Heavier stars, faster orbits

Kepler's law in its simple form works only for orbits around the Sun. Around a more massive star, gravity is stronger at every distance, so a planet at the same distance must move faster to stay in orbit, and its year is shorter. The rule becomes $P^2 = a^3 / M$, where $M$ is the star's mass in units of the Sun's mass.

Astronomers use this the other way around. By timing a planet around another star and measuring its distance, they can weigh the star. The same idea, applied to the stars orbiting the center of our galaxy, revealed a black hole about four million times the mass of the Sun.

11. Checking an answer

Three checks catch most errors. A period in years should be larger than the distance in AU for any orbit beyond Earth's, and smaller for any orbit inside it: $P = a\sqrt{a}$. If your period for $9$ AU is $9$ years or $3$ years, you forgot the cube. Square your period and cube your distance; the two should match. And a farther orbit must always have a longer period.

12. What the model leaves out

The law treats the planet's mass as tiny compared with the Sun's, which is very nearly true even for Jupiter, at about a thousandth of the Sun's mass. It also uses the average distance of an elliptical orbit, which is the right measure for the period.

The law says nothing about the planet's own spin, its size or what it is made of. A pebble and a planet at the same distance have the same period. That surprising fact, that mass does not matter for the orbiting body, is the same reason a feather and a hammer fall together on the airless Moon.

The Sun at the center and the orbits of Mercury, Venus, Earth and Mars drawn to scale, with distances in astronomical units: 0.39, 0.72, 1 and 1.52. Each planet moves along its orbit at its real relative speed, so Mercury laps the Sun more than four times while Earth goes around once, and Mars takes almost two Earth years. Mercury's and Mars's orbits are visibly stretched ellipses, and each planet speeds up on the side of its orbit nearest the Sun. The planets and the Sun are drawn far larger than scale so they can be seen.
The Sun at the center and the orbits of Mercury, Venus, Earth and Mars drawn to scale, with distances in astronomical units: 0.39, 0.72, 1 and 1.52. Each planet moves along its orbit at its real relative speed, so Mercury laps the Sun more than four times while Earth goes around once, and Mars takes almost two Earth years. Mercury's and Mars's orbits are visibly stretched ellipses, and each planet speeds up on the side of its orbit nearest the Sun. The planets and the Sun are drawn far larger than scale so they can be seen.

13. In the world: when will the comet come back?

Edmond Halley used Newton's laws in 1705 to show that bright comets seen in 1531, 1607 and 1682 were the same object returning about every $76$ years. He predicted it would return in 1758. He died before then, but the comet appeared on schedule, and it has carried his name ever since.

Kepler's third law turns that period into an orbit size: $76^2 = 5{,}776$, and its cube root is about $17.9$ AU. Halley's Comet swings in closer than Venus at its nearest and out beyond Neptune at its farthest, averaging about $18$ AU.

Its last visit was in 1986, when spacecraft flew past it, and it is due back in 2061. Astronomers at NASA's Jet Propulsion Laboratory in Pasadena track thousands of comets and asteroids the same way, using their periods to predict exactly where they will be.

14. In the world: weighing a distant star

Since 1995, astronomers have found thousands of planets orbiting other stars. NASA's Kepler space telescope, named after Johannes Kepler, watched about $150{,}000$ stars for the tiny dimming that happens when a planet passes in front of its star, and timed how often each planet came around.

With a planet's period and the star's mass, the third law gives the size of its orbit, which tells astronomers whether the planet is close enough to its star to be scorching hot, or at a distance where water could be liquid. A planet around a star four times the Sun's mass, at $4$ AU, would orbit in $4$ years rather than the $8$ years it would take around the Sun.

The same law, applied to stars whipping around the center of the Milky Way, weighed the black hole there at about four million Suns, work that shared the 2020 Nobel Prize in Physics.

15. Twice as far, twice as long

It seems natural that a planet twice as far would take twice as long to go around, but that assumes every planet moves at the same speed. Farther planets feel a weaker pull and move more slowly, so their years are much longer: twice as far takes about 2.8 times as long.

Another mix-up is between a planet's spin, which gives its day, and its orbit, which gives its year. Kepler's law is about the orbit.

16. A period from a distance

  1. An asteroid orbits at $9$ AU. Cube the distance.

    $9^3 = 729$

    Nine times nine times nine.

  2. Set the period squared equal to it.

    $P^2 = 729$

    Kepler's third law.

  3. Take the square root.

    $P = 27\ \text{years}$

    Since 27 squared is 729.

  4. Compare with the distance.

    $27 > 9$

    Farther orbits are also slower.

17. A distance from a period

  1. A comet returns every $64$ years. Square the period.

    $64^2 = 4096$

    Kepler's law gives the period squared.

  2. Set the distance cubed equal to it.

    $a^3 = 4096$

    Period squared equals distance cubed.

  3. Take the cube root.

    $a = 16\ \text{AU}$

    Since 16 cubed is 4096.

  4. Place it in the solar system.

    $\text{between Saturn and Uranus}$

    Saturn is near 9.6 AU and Uranus near 19 AU.

  5. Check by cubing.

    $16 \times 16 \times 16 = 4096$

    The law holds.

18. A planet around a heavier star

  1. A planet orbits $4$ AU from a star $4$ times the Sun's mass. Cube the distance.

    $4^3 = 64$

    Distance in AU, cubed.

  2. Divide by the star's mass.

    $64 \div 4 = 16$

    A heavier star pulls harder.

  3. Take the square root.

    $P = 4\ \text{years}$

    The period squared is 16.

  4. Compare with the same orbit around the Sun.

    $\sqrt{64} = 8\ \text{years}$

    The Sun's lighter pull gives a slower orbit.

  5. Find how many times faster.

    $8 \div 4 = 2$

    Twice as fast around the heavier star.

  6. Say how astronomers use this.

    $\text{weigh the star}$

    Timing an orbit reveals the central mass.

19. Your turn: what is the period of an orbit at $16$ AU?

  1. Cube the distance.

    $16^3 = 4096$

    The period squared.

  2. Take the square root.

    $\sqrt{4096}$

    Undo the square.

  3. Your turn: work this step out. Its working is at the end of the packet.

    State the period.

20. Guided practice

Jupiter is about 5 times as far from the Sun as Earth, but its year is about 12 Earth years, not 5. Why?

21. Guided practice

Complete the worked solution: a dwarf planet orbits the Sun at an average distance of $16$ AU. Find its period in years.

  1. Cube the distance.

    $\text{distance}^3 =$ c

    This is the period squared.

  2. Take the square root.

    $\text{period} =$ p

    The number whose square is the cube.

  3. Compare with Earth's year.

    $\text{much longer than one year}$

    A longer path traveled more slowly.

22. Guided practice

Match each orbit's average distance from the Sun to its period, using the period squared equals the distance cubed.

1 year8 years27 years64 years
1 AU
4 AU
9 AU
16 AU

23. Practice

An asteroid orbits the Sun at an average distance of $9$ AU. Fill in the distance cubed, the period squared and the period in years.

value
distance cubed
period squared
period (years)

24. Practice

A planet orbits a star $16$ times as massive as the Sun. Around that star, the period in years squared equals the distance in AU cubed divided by $16$. Write the period squared, $P^2$, as a function of the distance $a$ in AU.

Answer:

25. Practice

A comet goes around the Sun once every $27$ years. What is its average distance from the Sun, in AU?

Answer: AU from the Sun on average

26. Somewhere new

Uranus orbits the Sun at an average distance of $19.2$ AU. Using Kepler's third law and a calculator, about how many Earth years does it take to go around once? Give three significant figures.

Answer: Earth years per orbit

27. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

28. Test question

A planet orbits a star $9$ times as massive as the Sun. Around that star, the period in years squared equals the distance in AU cubed divided by $9$. Write the period squared, $P^2$, as a function of the distance $a$ in AU.

Answer:

29. What you can do now

You can use Kepler's third law. Find the period of an orbit at 4 AU, and explain why it is more than 4 years.

Working for the steps left to you

19. Your turn: what is the period of an orbit at $16$ AU?, step 3

$64\ \text{years}$

Since 64 squared is 4096.