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What are Kepler's laws, and how did Newton explain them with gravity?

Kepler's three laws of planetary motion, the use of Kepler's third law in the form T squared over r cubed equals a constant, and Newton's law of universal gravitation.

A focused answer to Edexcel GCSE Astronomy statements 8.4 and 8.6 to 8.9, covering Kepler's three laws of planetary motion, how to use Kepler's third law in the form T squared over r cubed equals a constant (including how the constant depends on the central mass), and Newton's law of universal gravitation explaining Kepler's laws.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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What this dot point is asking
Kepler's three laws
Using Kepler's third law
Newton's law of universal gravitation
How Edexcel examines this
Try this

What this dot point is asking

Edexcel statements 8.4 and 8.6 to 8.9 want you to understand Kepler's three laws of planetary motion, to use Kepler's third law in the form $\dfrac{T^2}{r^3} = \text{constant}$ (and that the constant depends inversely on the central mass), and to know that Newton explained Kepler's laws using his law of universal gravitation (force proportional to the product of the masses and inversely proportional to the square of the separation).

Kepler's three laws

The first law fixes the shape (ellipse), the second the speed (equal areas, so faster when nearer), and the third the relationship between period and size of orbit. The second law explains why a planet or comet speeds up near the Sun, connecting to libration (Topic 2) and the Equation of Time (Topic 4). The third law is the one you calculate with.

Using Kepler's third law

This is the headline calculation of the topic. Choosing years and AU makes the constant $1$ , so $T^2 = r^3$ : cube the radius to get $T^2$ , then square-root for $T$ . The third law works for any system orbiting one central body (planets round the Sun, moons round a planet), but the constant differs between systems because, per statement 8.7, it depends inversely on the central mass: a more massive central body gives a smaller constant.

Newton's law of universal gravitation

You are not required to use the algebraic formula, only to state the law in words and know it underpins Kepler's laws. The Sun's gravity provides exactly the inward force that keeps each planet on its elliptical path, makes it sweep equal areas, and produces the period-radius relationship. Newton's achievement was to unify falling apples and orbiting planets under one law of gravity, turning Kepler's empirical patterns into consequences of a physical force.

Worked example: Kepler's third law in AU and years

A planet orbits the Sun at a mean radius of $9.0\,\text{AU}$ . Find its orbital period in years.

Step 1: Find the constant using the Earth's known orbit

Because the Earth orbits at $r = 1.0\,\text{AU}$ with $T = 1.0\,\text{year}$ , the constant in $\dfrac{T^2}{r^3} = \text{constant}$ is $\dfrac{1.0^2}{1.0^3} = 1$ . Working in years and AU always makes the constant equal to 1, so we can write simply $T^2 = r^3$ .

Step 2: Cube the orbital radius to find $T^2$

With $r = 9.0\,\text{AU}$ :

T^2 = r^3 = 9.0^3 = 729 \,\text{year}^2

Cubing the radius gives us $T^2$ directly, because the constant is 1.

Step 3: Take the square root to find the period

We have $T^2 = 729$ , so we take the square root to get $T$ . A common mistake is to stop at this step and give $729$ as the answer.

T = \sqrt{729} = 27\,\text{years}

Final answer: The orbital period is $27\,\text{years}$ (close to Saturn's actual period of about 29 years).

How Edexcel examines this

This is naked-eye Paper 1 content and the richest calculation point in the topic. Kepler's third law calculations are highly likely: in years and AU the constant is $1$ , so use $T^2 = r^3$ , cube the radius, then square-root for the period (or reverse it for a radius). Watch for the square root and consistent units. Kepler's laws are also tested by statement: the elliptical orbit with the Sun at a focus, equal areas in equal times (faster near the Sun), and the period-radius law, plus that the third-law constant depends inversely on the central mass. Newton's law is tested in words (proportional to the product of the masses, inversely proportional to the square of the distance) and by explaining that it produces Kepler's laws; the algebraic form is not required. Synoptic links run to stable orbits (previous dot point) and to the formation and stability of systems (Topic 12). The biggest errors are forgetting the square root and stating the force law incorrectly, so secure both.

Try this

Q1. State Kepler's third law in the form used at GCSE. [1 mark]

Cue. $\dfrac{T^2}{r^3} = \text{constant}$ (the square of the period over the cube of the radius is constant).

Q2. State how the gravitational force between two bodies depends on the distance between them. [1 mark]

Cue. It is inversely proportional to the square of the distance (it falls off as distance squared).

Exam-style practice questions

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

Edexcel 1AS0 20224 marksA planet orbits the Sun with a mean orbital radius of 4.0 AU. Using Kepler's third law T squared over r cubed = a constant, and that for the Earth T = 1.0 year at r = 1.0 AU, calculate the orbital period of the planet in years.

Show worked answer →

For the Earth, $\dfrac{T^2}{r^3} = \dfrac{1.0^2}{1.0^3} = 1$ (in units of years squared per AU cubed), so the constant is $1$ (1 mark). For the planet, $\dfrac{T^2}{r^3} = 1$ , so $T^2 = r^3 = 4.0^3 = 64$ (2 marks). Taking the square root, $T = \sqrt{64} = 8.0\,\text{years}$ (1 mark). Markers reward finding the constant as $1$ from the Earth's values, substituting $r = 4.0\,\text{AU}$ to get $T^2 = 64$ , and taking the square root to give $T = 8.0\,\text{years}$ . Using AU and years makes the constant equal to $1$ , which simplifies the calculation.

Edexcel 1AS0 20213 marksState Newton's law of universal gravitation in words, and explain how it accounts for Kepler's laws of planetary motion.

Show worked answer →

Newton's law of universal gravitation states that the gravitational force between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them (so the force grows with mass and falls off as distance squared) (2 marks). It accounts for Kepler's laws because the inward gravitational pull of the Sun on a planet provides exactly the force needed to hold the planet in an elliptical orbit, with the planet moving faster when closer (perihelion) and slower when further away (aphelion), and with the period-radius relationship of the third law following from the strength of that force (1 mark). Markers reward stating the proportionality to the product of the masses and the inverse square of the separation, and explaining that this gravitational force produces the orbits Kepler described. The algebraic form of the law is not required.

Related dot points

Sources & how we know this

Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Astronomy (1AS0) specification — Pearson (2017)