NSWPhysicsSyllabus dot point

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)

Q: 2022 HSC HSC: Use 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.)

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.

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 OpusReviewed by Better Tuition Academy6 min answerUpdated 2026-05-18

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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: $\left(T_1 / T_2\right)^2 = \left(r_1 / r_2\right)^3$ .

Worked example with numbers

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.

Convert: $T = 27.3 \times 86400 = 2.36 \times 10^6$ s.

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

For a satellite at $r = 4.22 \times 10^7$ m:

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

$T = \sqrt{7.40 \times 10^9} = 8.60 \times 10^4$ s, or about $23.9$ hours.

This is the period of a geostationary satellite, 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.

In one sentence

Kepler's three laws state that orbits are elliptical with the Sun at one focus (1st), the radial line sweeps equal areas in equal times (2nd, conservation of angular momentum), and $T^2 / r^3 = 4 \pi^2 / (G M)$ for every body orbiting the same central mass (3rd, derivable from Newton's law of gravitation).

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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 $r$ :

IMATH_3
$r = \left(\frac{G M T^2}{4 \pi^2}\right)^{1/3}$ .

Substituting:

IMATH_5
IMATH_6
IMATH_7
$r^3 = 7.52 \times 10^{22}$ 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.