Earth orbits the Sun at r = 1.50 × 10^11 m with period T = 3.156 × 10^7 s. Calculate the Sun's mass using Kepler's third law. [3 marks]

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Inquiry Question 3: How does the force of gravity determine the motion of planets and satellites?

Investigate the relationship of Kepler's Laws of Planetary Motion to the forces acting on, and the total energy of, planets in circular and non-circular orbits using v = 2 pi r / T and T^2 / r^3 = 4 pi^2 / (G M)

A focused answer to the HSC Physics Module 5 dot point on Kepler's three laws. Elliptical orbits, equal areas in equal times, the period-radius relationship, the derivation from Newton's laws, and the worked geostationary-satellite example.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-23

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to state Kepler's three laws of planetary motion, derive the third law from Newton's Law of Universal Gravitation for circular orbits, and apply $T^2 / r^3 = 4 \pi^2 / (G M)$ to calculate orbital periods, radii, and speeds. You also need to explain the physical meaning of each law in plain English.

The answer

Johannes Kepler stated three empirical laws of planetary motion (1609-1619) based on Tycho Brahe's observations. Newton later showed they follow from his law of universal gravitation.

Kepler's First Law (the law of ellipses)

Every planet orbits the Sun in an ellipse, with the Sun at one focus.

A circle is a special case of an ellipse where the two foci coincide. Most planetary orbits in the solar system are very nearly circular, but Mercury and Pluto have noticeably elliptical orbits.

Kepler's Second Law (equal areas in equal times)

A line drawn from a planet to the Sun sweeps out equal areas in equal times.

This means planets move faster when closer to the Sun (perihelion) and slower when farther away (aphelion). The law is a geometric expression of the conservation of angular momentum, $L = m v r$ , valid because gravity always acts along the line between the planet and the Sun (zero torque about the Sun).

Kepler's Third Law (the harmonic law)

The square of the orbital period is proportional to the cube of the semi-major axis:

T^2 \propto r^3

For orbits around a central mass $M$ :

\frac{T^2}{r^3} = \frac{4 \pi^2}{G M}

The ratio $T^2 / r^3$ is the same for every body orbiting the same central mass.

Derivation for circular orbits

For a circular orbit, gravity provides the centripetal force:

\frac{G M m}{r^2} = \frac{m v^2}{r}

Using $v = 2 \pi r / T$ :

\frac{G M m}{r^2} = \frac{m}{r} \left(\frac{2 \pi r}{T}\right)^2 = \frac{4 \pi^2 m r}{T^2}

Rearranging:

\boxed{\frac{T^2}{r^3} = \frac{4 \pi^2}{G M}}

This is Newton's derivation. Note that $m$ (the mass of the orbiting body) cancels, so the relationship depends only on the central mass $M$ .

Implications

All satellites of Earth obey the same $T^2 / r^3$ ratio. Knowing one orbit fixes the constant.
A higher orbit (larger $r$ ) has a longer period: geostationary satellites orbit at about $42200$ km from Earth's centre.
Comparing orbits of different planets around the Sun: $\left(T_1 / T_2\right)^2 = \left(r_1 / r_2\right)^3$ .

Using the Moon to predict a satellite's orbital period

The Moon orbits Earth with period $T = 27.3$ days at radius $r = 3.84 \times 10^8$ m. Calculate $T^2 / r^3$ and use it to predict the orbital period of a satellite at $r = 4.22 \times 10^7$ m.

Step 1: Convert the period to SI units

Kepler's Third Law requires consistent SI units throughout. Convert days to seconds:

T = 27.3 \times 86400 = 2.36 \times 10^6 \text{ s}

Step 2: Compute the Kepler constant from the Moon's orbit

The ratio $T^2 / r^3$ is the same for every body orbiting Earth, so computing it from the Moon gives us the constant we need. This works because the central mass $M_E$ is the same in every case:

\frac{T^2}{r^3} = \frac{(2.36 \times 10^6)^2}{(3.84 \times 10^8)^3} = \frac{5.57 \times 10^{12}}{5.66 \times 10^{25}} = 9.84 \times 10^{-14} \text{ s}^2 \text{m}^{-3}

Step 3: Apply the constant to the new orbit

Because the satellite also orbits Earth, it shares the same $T^2 / r^3$ ratio. Rearranging gives $T^2 = 9.84 \times 10^{-14} \times r^3$ , so for $r = 4.22 \times 10^7$ m:

T^2 = 9.84 \times 10^{-14} \times (4.22 \times 10^7)^3 = 9.84 \times 10^{-14} \times 7.52 \times 10^{22} = 7.40 \times 10^9 \text{ s}^2

Step 4: Take the square root to find the period

T = \sqrt{7.40 \times 10^9} = 8.60 \times 10^4 \text{ s} \approx 23.9 \text{ hours}

Final answer: The orbital period is $8.60 \times 10^4$ s, or about $23.9$ hours. This is the period of a geostationary satellite, which orbits once per sidereal day as expected.

Try it: Kepler's third law calculator - solve $T^2/r^3 = 4\pi^2/(GM)$ either way around, with planet presets for the central body.

Common traps

Quoting Kepler's First Law as "circular orbits." It is elliptical orbits, with the Sun at one focus. Circles are a special case.

Forgetting that $m$ cancels: The Third Law constant $4 \pi^2 / (G M)$ depends only on the central body's mass, not the orbiting body's.
Mixing units of $T$ and $r$: Use SI units throughout: seconds and metres. Days and kilometres need conversion first.
Applying Kepler's Third Law across different central bodies: The constant $T^2 / r^3$ is the same only for orbits around the same central mass. Earth satellites and Sun-orbiting planets have different constants.
Confusing semi-major axis with radius: For circular orbits they are the same; for elliptical orbits the semi-major axis is half the longest diameter and is the value used in Kepler's Third Law.

Examples in context

Example 1. Confirming Kepler-3 with Jupiter's moons through a Mt Stromlo telescope. ANU Mt Stromlo Observatory students nightly observe Jupiter's four Galilean moons. For Io: orbital radius $r = 4.22 \times 10^8 \text{ m}$ , period $T = 1.769$ days $= 1.528 \times 10^5 \text{ s}$ . Kepler-3 says $T^2 / r^3 = 4 \pi^2 / (G M_J)$ . Computing $T^2 / r^3 = (1.528 \times 10^5)^2 / (4.22 \times 10^8)^3 = 3.10 \times 10^{-16} \text{ s}^2 \text{ m}^{-3}$ . This yields $M_J = 4 \pi^2 / (G \times 3.10 \times 10^{-16}) = 1.90 \times 10^{27} \text{ kg}$ . The same ratio holds for Europa, Ganymede and Callisto - confirming Kepler's third law and giving Jupiter's mass directly.

Example 2. Halley's Comet's elliptical orbit observed from Parkes. Halley's Comet has perihelion $r_p = 0.586 \text{ AU} = 8.77 \times 10^{10} \text{ m}$ and aphelion $r_a = 35.1 \text{ AU} = 5.25 \times 10^{12} \text{ m}$ . The semi-major axis is $a = (r_p + r_a)/2 = 2.67 \times 10^{12} \text{ m} = 17.8 \text{ AU}$ . From Kepler-3, $T = \sqrt{a^3 \cdot 4\pi^2 / (G M_{\odot})} = \sqrt{(2.67 \times 10^{12})^3 \times 4\pi^2 / (6.674 \times 10^{-11} \times 1.99 \times 10^{30})} = 2.37 \times 10^9 \text{ s} = 75.0$ years. Parkes radio tracking confirmed Halley's 1986 perihelion to within minutes, matching the prediction made by Edmond Halley in 1705.

Try this

Q1. State Kepler's three laws of planetary motion. [3 marks]

Cue. (1) Elliptical orbits, Sun at one focus. (2) Equal areas in equal times. (3) $T^2 \propto r^3$ . One mark each.

Q2. Earth orbits the Sun at $r = 1.50 \times 10^{11} \text{ m}$ with period $T = 3.156 \times 10^7 \text{ s}$ . Calculate the Sun's mass using Kepler's third law. [3 marks]

Cue. $M_{\odot} = 4 \pi^2 r^3 / (G T^2) = 4 \pi^2 \times (1.5 \times 10^{11})^3 / (6.674 \times 10^{-11} \times (3.156 \times 10^7)^2) = 2.00 \times 10^{30} \text{ kg}$ .

Q3. A new asteroid is discovered with semi-major axis $a = 2.80 \text{ AU}$ . (a) Calculate its orbital period in years. (b) Explain why a more eccentric orbit with the same $a$ has the same period. (c) Sketch a labelled diagram showing how the asteroid's speed varies between perihelion and aphelion (Kepler-2). [2+2+2 marks]

Cue. (a) $T = a^{3/2}$ yr $= 2.8^{1.5} = 4.68 \text{ yr}$ . (b) Kepler-3 depends only on $a$ , not eccentricity. (c) Faster near perihelion, slower at aphelion; equal areas swept.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC5 marksUse Kepler's Third Law to derive the orbital radius of a geostationary satellite around Earth. (Mass of Earth = 5.97 x 10^24 kg, period T = 86400 s, G = 6.67 x 10^-11 N m^2/kg^2.)

Show worked answer →

Kepler's Third Law for orbits around a body of mass $M$ :

$\frac{T^2}{r^3} = \frac{4 \pi^2}{G M}$ .

Rearranging for $r$ :

$r^3 = \frac{G M T^2}{4 \pi^2}$
$r = \left(\frac{G M T^2}{4 \pi^2}\right)^{1/3}$ .

Substituting:

$r^3 = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86400)^2}{4 \pi^2}$
$r^3 = \frac{3.98 \times 10^{14} \times 7.46 \times 10^9}{39.48}$
$r^3 = \frac{2.97 \times 10^{24}}{39.48}$
$r^3 = 7.52 \times 10^{22}$ m $^3$ .

$r = (7.52 \times 10^{22})^{1/3} = 4.22 \times 10^7$ m, or about $42200$ km from Earth's centre (around $35800$ km altitude).

Markers reward the explicit derivation, the use of $T = 86400$ s (one sidereal day in seconds), and the final answer with units. Bonus credit for identifying that this orbit is in the equatorial plane.

2017 HSC3 marksExplain how Kepler's Second Law (equal areas in equal times) implies that a planet moves faster when it is closer to the Sun.

Show worked answer →

Kepler's Second Law states that a line joining a planet to the Sun sweeps out equal areas in equal times. The area swept in a small time $\Delta t$ is approximately a triangle with base $v \Delta t$ and height $r$ , so:

Area $= \frac{1}{2} r v \Delta t$ (approximately).

For the area per unit time to be constant, the product $r v$ must be constant. When the planet is closer to the Sun (smaller $r$ ), its speed $v$ must be larger; when farther away (larger $r$ ), its speed must be smaller.

This is a consequence of the conservation of angular momentum: with no external torque (gravity acts along the line to the Sun), angular momentum $L = m r v$ is conserved.

Markers reward the geometric interpretation of the area sweep, the conclusion that $r v$ is constant, and a reference to angular momentum conservation.