Kepler's Laws of Planetary Motion

Spec mapping: OCR H556 Module 5.4 — Gravitational fields (Kepler's three laws of planetary motion: orbits are ellipses with the Sun at one focus; equal areas swept in equal times; $T^{2} \propto r^{3}$ ; derivation of Kepler's third law for circular orbits from Newton's law of gravitation combined with centripetal acceleration; use of Kepler's third law to determine the mass of a central body from orbital observations). Refer to the official OCR H556 specification document for exact wording.

Two decades before Newton wrote down his law of gravitation, a German astronomer named Johannes Kepler distilled twenty years of his teacher Tycho Brahe's meticulous naked-eye observations of the night sky into three concise empirical laws describing planetary motion. The orbits Kepler deduced were not the perfect circles of Greek astronomy but slightly flattened ellipses. Sixty years later, Newton showed that all three of Kepler's laws follow as direct mathematical consequences of the inverse-square law of gravitation applied to a small body orbiting a much larger central mass.

This lesson states each of Kepler's laws, explains the physical content (angular momentum conservation for the second law, $T^{2} \propto r^{3}$ for the third), derives the third law for the special case of a circular orbit, applies it to the ISS, the Solar System, Phobos at Mars, and ratio problems, and lands the A* discriminators on (i) the focus-not-centre wording of the first law, (ii) the squaring of $T$ in the third law, (iii) the central-body dependence of the constant of proportionality $4\pi^{2}/(GM)$ , and (iv) the distinction between Kepler's kinematic description and Newton's dynamic explanation.

Kepler's First Law — the Law of Ellipses

The orbit of every planet is an ellipse, with the Sun at one focus.

An ellipse is a closed curve where the sum of distances from any point on the curve to two fixed points (the foci) is constant. If the two foci coincide, the ellipse degenerates to a circle (so a circle is a special ellipse with eccentricity zero). For planetary orbits, the Sun sits at one focus and the other focus is empty — there is nothing physical at the second focus, just a geometric reference point.

Most Solar-System planetary orbits are only slightly elliptical:

Body	Eccentricity $e$	Closeness to a circle
Venus	0.007	almost perfectly circular
Earth	0.017	1.7% departure from a circle
Mars	0.093	noticeable ellipticity — key to Kepler's discovery
Mercury	0.206	most elliptical major planet
Halley's comet	0.967	highly elongated

Mars was the crucial body for Kepler. Its eccentricity is just large enough that the orbit cannot be fitted with a single circle — but with twenty years of Tycho's precise positional measurements (accurate to about a minute of arc, the best naked-eye data ever assembled), Kepler could detect the systematic departure of Mars's path from a circle and recognise the ellipse.

Vocabulary

Term	Meaning
Perihelion	Point in the orbit closest to the Sun
Aphelion	Point farthest from the Sun
Semi-major axis $a$	Half the longest diameter — equals the average of perihelion and aphelion distances
Semi-minor axis $b$	Half the shortest diameter
Eccentricity $e$	Geometric "squashedness", $e = \sqrt{1 - b^{2}/a^{2}}$ ; $e = 0$ is a circle, $e \to 1$ is an extreme ellipse, $e > 1$ is a hyperbola (unbound)

A discriminator.* OCR mark schemes are strict about the wording. Saying "the orbit is a circle with the Sun at the centre" loses marks — the orbit is an ellipse with the Sun at a focus, even when the ellipse happens to be nearly circular. The centre of a circle and the focus of an ellipse are different geometric concepts and OCR examiners check that students know it. For OCR calculations you may treat the orbit as a circle (a fair approximation for most planets), but you must state the first law correctly.

Kepler's Second Law — the Law of Equal Areas

A line joining a planet and the Sun (the "radius vector") sweeps out equal areas in equal intervals of time.

This is a deeply geometric statement. If you draw the radius vector from Sun to planet and watch it sweep out a wedge-shaped region over (say) one week, the area of that wedge is the same no matter where in the orbit the planet is during that week — at perihelion (close to the Sun) or at aphelion (far from the Sun).

The kinematic consequence: the planet moves faster when close to the Sun (perihelion) and slower when far away (aphelion). At perihelion, a short arc must sweep out the same area as a much longer arc at aphelion, so the planet must traverse the short arc in less time, i.e. at higher speed. For the Earth, perihelion speed (3 January, ~30.3 km s⁻¹) is about 3% higher than aphelion speed (4 July, ~29.3 km s⁻¹).

The Modern Reading — Conservation of Angular Momentum

In modern language, Kepler's second law is a statement of conservation of angular momentum. The gravitational force on the planet is a central force — always directed from the planet straight back to the Sun, along the radius vector. A central force exerts zero torque about the Sun (since $\vec{\tau} = \vec{r} \times \vec{F}$ and $\vec{r}$ is parallel to $\vec{F}$ ). With zero torque, the planet's angular momentum $\vec{L} = m\vec{r} \times \vec{v}$ is conserved.

The connection to areas: the area swept out by the radius vector in time $dt$ is $dA = \tfrac{1}{2} r v_{\perp} \, dt$ where $v_{\perp}$ is the component of velocity perpendicular to $\vec{r}$ . Since $L = m r v_{\perp}$ , this gives

$\frac{dA}{dt} = \frac{L}{2m}$

— a constant whenever $L$ is conserved. Equal areas in equal times is exactly angular momentum conservation, written in geometric language. Kepler had no notion of $L$ in 1609 when he formulated this law; Newton recognised the equivalence later.

For a circular orbit, $r$ and $v_{\perp}$ are both constant, so the speed is constant and Kepler's second law reduces trivially to "the speed does not change". The rest of this lesson works in that simplification.

Kepler's Third Law — the Law of Periods

The square of the orbital period is directly proportional to the cube of the semi-major axis of the orbit.

In symbols, for any two bodies orbiting the same central mass:

$T^{2} \propto r^{3}$

With the constant of proportionality written out (derived below):

$T^{2} = \frac{4\pi^{2}}{GM} r^{3}$

where $M$ is the mass of the central body, $r$ is the semi-major axis (or, for a circular orbit, simply the orbital radius), $G$ is the universal gravitational constant, and $T$ is the orbital period.

This is the most quantitatively useful of Kepler's three laws and the one OCR examines most heavily.

Derivation for a Circular Orbit

This is a required derivation at OCR A-Level. Memorise the four lines below.

For a body in a circular orbit at radius $r$ moving at speed $v$ , gravity supplies the centripetal force:

$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}$

Cancel $m$ and rearrange for $v$ :

$v^{2} = \frac{GM}{r} \quad \Rightarrow \quad v = \sqrt{\frac{GM}{r}}$

The period is the orbital circumference divided by the speed:

$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi \sqrt{\frac{r^{3}}{GM}}$

Square both sides:

$T^{2} = \frac{4\pi^{2}}{GM} r^{3}$

— Kepler's third law for a circular orbit, derived in four lines from Newton's law of gravitation and Newton's second law. The result also holds for elliptical orbits (replacing $r$ with the semi-major axis $a$ ), with a proof that requires more advanced calculus.

A discriminator.* The constant of proportionality $4\pi^{2}/(GM)$ depends on the central body's mass $M$ . Kepler's third law only compares bodies orbiting the same central mass — Solar System planets share the constant $4\pi^{2}/(GM_{\odot})$ , but Earth's Moon (orbiting Earth) has a different constant $4\pi^{2}/(GM_{E})$ . Mixing the two — applying $T^{2}/r^{3}$ from the Solar System to a lunar orbit, say — is a guaranteed mark-loss.

Why the Form Matters

Two scaling consequences worth noting:

Doubling the orbital radius multiplies the period by $\sqrt{8} \approx 2.83$ . The Moon at twice its current distance would orbit Earth in 77 days instead of 27.3.
Quadrupling the central mass at fixed radius halves the period. Jupiter's moons orbit it much faster than Earth's Moon orbits Earth despite comparable radii, because Jupiter is 300+ times heavier.

The relation also goes the other way: if you observe $T$ and $r$ for any orbit, you can solve for $M$ — the central mass falls out. This is the single most-used technique in astronomy for measuring masses of distant bodies.

Putting the Solar System Through the Test

Using $M_{\odot} = 1.99 \times 10^{30}$ kg and Kepler's third law, we can predict each planet's period from its semi-major axis and compare with observation:

Planet	$a$ / $10^{11}$ m	$T$ predicted / years	$T$ observed / years
Mercury	0.579	0.241	0.241
Venus	1.082	0.615	0.615
Earth	1.496	1.000	1.000
Mars	2.279	1.881	1.881
Jupiter	7.783	11.86	11.86
Saturn	14.29	29.46	29.46
Uranus	28.71	84.01	84.01
Neptune	44.98	164.8	164.8

Agreement to 4 sf across all eight planets — a triumph for Kepler (who found the pattern from data) and for Newton (who explained why the pattern must hold). The ratio $T^{2}/a^{3}$ is the same for every Solar-System body to within high precision.

For Solar-System problems in convenient units (years and astronomical units, where 1 AU $= 1.496 \times 10^{11}$ m and 1 year $= 3.156 \times 10^{7}$ s), Kepler's third law takes the especially clean form

$T^{2}[\text{years}^{2}] = a^{3}[\text{AU}^{3}]$

— useful for quick sanity-checks but be careful to convert to SI before doing any calculation that mixes systems.

Worked Example 1 — Period of the ISS

The International Space Station orbits at 400 km altitude. Calculate its orbital period. Use $M_{E} = 5.97 \times 10^{24}$ kg and $R_{E} = 6.37 \times 10^{6}$ m.

Distance from the Earth's centre: $r = R_{E} + h = 6.37 \times 10^{6} + 0.40 \times 10^{6} = 6.77 \times 10^{6}$ m.

Using $T = 2\pi\sqrt{r^{3}/(GM)}$ :

$r^{3} = (6.77 \times 10^{6})^{3} = 3.10 \times 10^{20}$ m³.
$GM = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} = 3.98 \times 10^{14}$ m³ s⁻².
$r^{3}/(GM) = 3.10 \times 10^{20} / 3.98 \times 10^{14} = 7.79 \times 10^{5}$ s².

$T = 2\pi\sqrt{7.79 \times 10^{5}} = 2\pi \times 882.6 = 5547 \text{ s} \approx 92.5 \text{ min}$

The ISS completes one orbit just over 90 minutes — astronauts see 16 sunrises and sunsets every 24 hours. The figure matches the official mission spec to within rounding, and is independent of the station's mass (which cancelled at the centripetal-force step).

Worked Example 2 — Mass of the Sun from the Earth's Orbit

Earth's orbital period is 365.25 days, i.e. $T = 3.156 \times 10^{7}$ s. Its orbital semi-major axis is $1.496 \times 10^{11}$ m (1 AU). Use these to estimate the Sun's mass.

Rearranging Kepler's third law: $M = 4\pi^{2} r^{3} / (GT^{2})$ .

$r^{3} = (1.496 \times 10^{11})^{3} = 3.349 \times 10^{33}$ m³.
$T^{2} = (3.156 \times 10^{7})^{2} = 9.96 \times 10^{14}$ s².
$4\pi^{2} \approx 39.48$ .

$M_{\odot} = \frac{39.48 \times 3.349 \times 10^{33}}{6.67 \times 10^{-11} \times 9.96 \times 10^{14}} = \frac{1.322 \times 10^{35}}{6.64 \times 10^{4}} = 1.99 \times 10^{30} \text{ kg}$

— the accepted value. Kepler's third law applied to the Earth-Sun orbit gives the Sun's mass directly. The same method, applied to a moon orbiting a planet, gives the planet's mass; applied to a star orbiting a galactic centre, gives the central black-hole mass. This is the central-mass-from-orbit technique in astronomy.

Worked Example 3 — Ratio of Orbital Radii (No $G$ or $M$ Needed)

Jupiter's orbital period is 11.86 years; Mercury's is 0.241 years. Find the ratio of their orbital radii without using $G$ or $M_{\odot}$ .

For two bodies orbiting the same central mass, the $4\pi^{2}/(GM)$ constant cancels in a ratio:

$\frac{T_{J}^{2}}{T_{M}^{2}} = \frac{r_{J}^{3}}{r_{M}^{3}} \quad \Rightarrow \quad \left(\frac{T_{J}}{T_{M}}\right)^{2} = \left(\frac{r_{J}}{r_{M}}\right)^{3}$

Substituting:

$\left(\frac{11.86}{0.241}\right)^{2} = (49.21)^{2} = 2422 = \left(\frac{r_{J}}{r_{M}}\right)^{3}$

$\frac{r_{J}}{r_{M}} = 2422^{1/3} = 13.4$

Jupiter's orbit is 13.4× the radius of Mercury's — checking against actual values ( $r_{J} = 7.78 \times 10^{11}$ m, $r_{M} = 5.79 \times 10^{10}$ m gives $r_{J}/r_{M} = 13.4$ , agreement to 3 sf).

The technique is general: any ratio question with two orbits around the same central body can be answered using $(T_{1}/T_{2})^{2} = (r_{1}/r_{2})^{3}$ , without needing $G$ or $M$ .

Worked Example 4 — Phobos Reveals Mars's Mass

Phobos, the inner moon of Mars, orbits at $r = 9.38 \times 10^{6}$ m with period $T = 27\,540$ s (7 h 39 min). Calculate the mass of Mars.

$M = 4\pi^{2} r^{3} / (GT^{2})$ as before.

$r^{3} = (9.38 \times 10^{6})^{3} = 8.25 \times 10^{20}$ m³.
$T^{2} = (27\,540)^{2} = 7.58 \times 10^{8}$ s².

$M_{\text{Mars}} = \frac{39.48 \times 8.25 \times 10^{20}}{6.67 \times 10^{-11} \times 7.58 \times 10^{8}} = \frac{3.26 \times 10^{22}}{5.06 \times 10^{-2}} = 6.44 \times 10^{23} \text{ kg}$

Accepted value of Mars's mass: $6.42 \times 10^{23}$ kg — agreement to 0.3%. Identical method to Worked Example 2, applied to a different central body. Note the central-mass constant $4\pi^{2}/(GM_{\text{Mars}})$ is different from the Solar-System one $4\pi^{2}/(GM_{\odot})$ — never mix the two.

Visualising the Three Laws

graph TD
    N["Newton's law of gravitation<br/>F = GMm/r²<br/>(deeper principle)"]
    N --> K1["First law<br/>Orbits are ellipses with<br/>the Sun at one focus<br/>(geometry of bound orbits)"]
    N --> K2["Second law<br/>Equal areas in equal times<br/>(angular momentum conservation)"]
    N --> K3["Third law<br/>T² ∝ r³<br/>(scaling of bound orbits)"]
    K3 --> APP["Mass of central body from T, r"]
    K2 --> APP2["Faster at perihelion, slower at aphelion"]

Newton's law is the deeper physics; Kepler's laws are its geometric and kinematic consequences. Historically, the laws came in reverse order — Kepler's empirical fits came first (1609–1619), Newton's mathematical explanation came later (1687) — but pedagogically the modern presentation derives Kepler from Newton, not the other way around.

Kepler's Laws of Planetary Motion

Kepler's Laws of Planetary Motion

Kepler's First Law — the Law of Ellipses

Vocabulary

Kepler's Second Law — the Law of Equal Areas

The Modern Reading — Conservation of Angular Momentum

Kepler's Third Law — the Law of Periods

Derivation for a Circular Orbit

Why the Form Matters

Putting the Solar System Through the Test

Worked Example 1 — Period of the ISS

Worked Example 2 — Mass of the Sun from the Earth's Orbit

Worked Example 3 — Ratio of Orbital Radii (No $G$ or $M$ Needed)

Worked Example 4 — Phobos Reveals Mars's Mass

Visualising the Three Laws

Synoptic Links

More in Physics

Kepler's Laws of Planetary Motion

Kepler's Laws of Planetary Motion

Kepler's First Law — the Law of Ellipses

Vocabulary

Kepler's Second Law — the Law of Equal Areas

The Modern Reading — Conservation of Angular Momentum

Kepler's Third Law — the Law of Periods

Derivation for a Circular Orbit

Why the Form Matters

Putting the Solar System Through the Test

Worked Example 1 — Period of the ISS

Worked Example 2 — Mass of the Sun from the Earth's Orbit

Worked Example 3 — Ratio of Orbital Radii (No GGG or MMM Needed)

Worked Example 4 — Phobos Reveals Mars's Mass

Visualising the Three Laws

Synoptic Links

More in Physics

Worked Example 3 — Ratio of Orbital Radii (No $G$ or $M$ Needed)