Kepler's third law with real planets

Curriculum links

Quoted word for word from the official documents.

Step by step

The caption is what the class reads on screen; the talking points are in your notes drawer (N) when you present. Steps marked Pupils only appear only in the pupil lesson.

1. 01

   Orbits at true scale

   Presented and for pupils

   Caption

   These are the planets' real orbits at true scale, seen from above. The further a planet is from the Sun, the longer its year.

   Talking points

   • JPL's elements give semi-major axes of 5.203 au for Jupiter, 9.537 au for Saturn, 19.19 au for Uranus and 30.07 au for Neptune. JPL's sidereal orbital periods for the four are 11.86, 29.45, 84.02 and 164.79 years.
   • NASA: half of an ellipse's longest axis is its semi-major axis, and Kepler's third law says the squares of the planets' periods are proportional to the cubes of their semi-major axes.
   • At this distance the four inner orbits crowd round the Sun in the middle of the picture. The next step moves in on them.

   Sources 1 2 4Present from this step →

2. 02

   T² ∝ a³

   Presented and for pupils

   Caption

   Kepler's third law: a planet's period squared is proportional to its semi-major axis cubed. With T in years and a in astronomical units, T²/a³ comes out as 1.

   Talking points

   • From JPL's semi-major axes and sidereal periods (years of 365.25 days), T²/a³ is 1.0000 for Mercury, Venus, Earth and Mars, 0.9991 for Jupiter, 0.9998 for Saturn, 0.9990 for Uranus and 0.9988 for Neptune: the same to within 0.12 per cent, while a ranges over a factor of 78.
   • Earth's entry in JPL's table is the Earth-Moon barycentre, the point the two orbit together.
   • JPL says its elements are "simply the result of being adjusted for a best fit", valid from 1800 to 2050. The viewer moves the planets on this same table, so the orbits on screen keep to the law.

   Ask the class

   Jupiter's semi-major axis is 5.203 au. Use T² = a³ (T in years, a in au) to predict its orbital period in years.

   Answer11.87 years. Answers from 11.82 to 11.92 are marked right.

   WhyT = a^(3/2) = √(5.203³) = √140.85 = 11.87 years. JPL gives Jupiter's sidereal period as 11.86 years.

   Sources 1 2 3Present from this step →

3. 03

   A straight line on log axes

   Presented and for pupils

   Caption

   If T² ∝ a³, then lg T = 1.5 lg a + a constant. A graph of lg T against lg a is a straight line.

   Talking points

   • AQA's specification suggests logarithmic plots of T against r for given data (MS 3.11).
   • With JPL's values: Mercury lg a = −0.412 and lg T = −0.618; Neptune lg a = 1.478 and lg T = 2.217. The gradient between them is 2.835 ÷ 1.890 = 1.500.
   • A least-squares line through all eight planets has a gradient of 1.4999 and passes through lg T = 0 at lg a = 0 (to four decimal places), because T = 1 year at a = 1 au.

   Ask the class

   What is the gradient of a graph of lg T against lg a for the planets?

   Answer1.5. Answers from 1.48 to 1.52 are marked right.

   WhyT² = ka³ gives 2 lg T = 3 lg a + lg k, so lg T = 1.5 lg a + 0.5 lg k: the gradient is 3/2 = 1.5. Mercury and Neptune give (2.217 + 0.618) ÷ (1.478 + 0.412) = 1.500.

   Sources 1 2 9Present from this step →

4. 04

   Where the law comes from

   Presented and for pupils

   Caption

   Gravity provides the centripetal force: GMm/r² = mv²/r. With v = 2πr/T this gives T² = (4π²/GM) r³.

   Talking points

   • The planet's mass m cancels, so the constant 4π²/GM depends only on the Sun's mass M. Every planet of the Sun shares it.
   • OCR's specification expects learners to derive this equation from first principles. The derivation takes a circular orbit of radius r; Kepler's law, as NASA states it, uses the semi-major axis of the ellipse.
   • NASA: Kepler's laws "were instrumental in Isaac Newton deriving his theory of universal gravitation, which explains the unknown force behind Kepler's Third Law."

   Ask the class

   Planet B's orbit has 4 times the semi-major axis of planet A's. How many times longer is planet B's year?

   1. A2
   2. B4
   3. C8Right answer
   4. D16

   WhyT ∝ a^(3/2), so T_B ÷ T_A = 4^(3/2) = (√4)³ = 2³ = 8.

   Sources 4 10Present from this step →

5. 05

   Weighing the Sun: GM

   Presented and for pupils

   Caption

   Rearranged, the law measures the Sun: GM = 4π²a³/T². Earth's orbit is enough to do it.

   Talking points

   • NASA's Earth Fact Sheet gives a semi-major axis of 149.598 million km and a sidereal orbit period of 365.256 days, which is 3.15581 × 10⁷ s.
   • With those figures GM = 4π² × (1.49598 × 10¹¹)³ ÷ (3.15581 × 10⁷)² = 1.3271 × 10²⁰ m³ s⁻². JPL's heliocentric gravitational constant is 1.32712440041 × 10²⁰ m³ s⁻².
   • JPL quotes the Sun's GM to many more figures than NIST can give G: NIST's relative standard uncertainty in G is 2.2 × 10⁻⁵, so the Sun's mass on its own is known to about five figures.

   Ask the class

   Earth's orbit has a = 1.496 × 10¹¹ m and T = 3.156 × 10⁷ s. Calculate GM for the Sun, in units of 10²⁰ m³ s⁻².

   Answer1.327 × 10²⁰ m³ s⁻². Answers from 1.317 to 1.337 are marked right.

   WhyGM = 4π²a³/T² = 39.48 × (1.496 × 10¹¹)³ ÷ (3.156 × 10⁷)² = 39.48 × 3.348 × 10³³ ÷ 9.960 × 10¹⁴ = 1.327 × 10²⁰ m³ s⁻². JPL gives 1.32712 × 10²⁰.

   Sources 3 5 8Present from this step →

6. 06

   The Sun's mass

   Presented and for pupils

   Caption

   Dividing GM by the gravitational constant G gives the mass of the Sun.

   Talking points

   • CODATA 2022, from NIST: G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻².
   • NASA's Sun Fact Sheet gives the Sun's mass as 1,988,400 × 10²⁴ kg, which is 1.9884 × 10³⁰ kg, and 332,900 times Earth's.

   Ask the class

   Using GM = 1.327 × 10²⁰ m³ s⁻² and G = 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻², find the Sun's mass in units of 10³⁰ kg.

   Answer1.988 × 10³⁰ kg. Answers from 1.978 to 1.998 are marked right.

   WhyM = GM ÷ G = 1.327 × 10²⁰ ÷ 6.674 × 10⁻¹¹ = 1.988 × 10³⁰ kg, which matches NASA's 1.9884 × 10³⁰ kg.

   Sources 7 8Present from this step →

7. 07

   Time Mercury yourself

   Pupils only

   Caption

   Your turn. The clock runs at one day a second. Time one orbit of Mercury against a fixed direction on your screen.

   Talking points

   • NASA's Mercury Fact Sheet gives a sidereal orbit period of 87.969 days. JPL's 0.2408467 years of 365.25 days is the same, 87.969 days.
   • The camera stays fixed in space in this step, so Mercury's return to the same direction from the Sun on the screen is one orbit against the stars, its sidereal period.
   • Answers from 83 to 93 days are marked right.

   Pupil question

   How many days did your one orbit of Mercury take?

   Answer88 days. Answers from 83 to 93 are marked right.

   WhyNASA gives Mercury's sidereal period as 87.969 days. Reading each date to the nearest day gives an answer within a day or two of 88.

   Pupil task

   Choose a point Mercury passes, such as straight to the right of the Sun on your screen. Pause when Mercury is there and note the date. Play, and pause again when Mercury is back at the same point. Leave the view where it is while you time.

   HintAt one day a second an orbit takes about a minute and a half. Count the days between the two dates, then type the answer in the box above.

   The viewer checks the task as the pupil works and says when it is done.

   Sources 2 3 6Open this step as a pupil →

8. 08

   From period to distance

   Presented and for pupils

   Caption

   The law works both ways: a planet's period gives its distance from the Sun.

   Talking points

   • NASA's Mercury Fact Sheet gives a semi-major axis of 57.909 million km, and 57.909 ÷ 149.598 = 0.3871 au.
   • The same rule gives Neptune's distance from its period: 164.79^(2/3) = 30.06 au, against JPL's 30.07 au.

   Ask the class

   Mercury's period is 0.2408 years. Use T² = a³ to find its semi-major axis in au.

   Answer0.387 au. Answers from 0.384 to 0.39 are marked right.

   Whya = T^(2/3) = 0.2408^(2/3) = 0.387 au. NASA gives 57.909 million km, and 57.909 ÷ 149.598 = 0.387 au.

   Sources 1 2 5 6Present from this step →

Sources

Every fact in this lesson comes from these sources, and each step lists the ones it uses.

1. Approximate Positions of the Planets (Keplerian elements, table 1, 1800 to 2050), JPL Solar System Dynamics
2. Planetary Physical Parameters, JPL Solar System Dynamics
3. Astrodynamic Parameters, JPL Solar System Dynamics
4. Orbits and Kepler's Laws, NASA Science
5. Earth Fact Sheet, NASA NSSDCA
6. Mercury Fact Sheet, NASA NSSDCA
7. Sun Fact Sheet, NASA NSSDCA
8. Newtonian constant of gravitation, CODATA 2022, NIST
9. AQA AS and A-level Physics specification (7407, 7408), version 1.3
10. OCR A Level Physics A (H556) specification, version 3.0