Kepler's laws of planetary motion explained: HSC Physics Module 5

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.

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.

First law

Every planet orbits the Sun in an ellipse, with the Sun at one focus (not the centre).

Second law

A line from the planet to the Sun sweeps out equal areas in equal times, so the planet moves fastest at perihelion and slowest at aphelion.

Third law

The square of the orbital period is proportional to the cube of the orbital radius (semi-major axis): $T^2 \propto r^3$.

Marks: one for each law stated correctly (ellipse and focus; equal areas and the speed consequence; $T^2 \propto r^3$).

Use $v = \dfrac{2\pi r}{T}$, the distance travelled in one orbit divided by the period.

$v = \dfrac{2\pi (8.00 \times 10^6)}{6.96 \times 10^3} = \dfrac{5.03 \times 10^7}{6.96 \times 10^3} = 7.22 \times 10^3\ \text{m s}^{-1}$.

Marks: one for the correct formula $v = 2\pi r / T$, one for substitution with units, one for the answer to three significant figures with the correct unit.

For bodies orbiting the Sun, Kepler's Third Law in AU-and-year units simplifies to $T^2 = a^3$ (because Earth, with $a = 1\ \text{AU}$ and $T = 1\ \text{yr}$, fixes the constant to $1$).

$T^2 = (3.20)^3 = 32.8$, so $T = \sqrt{32.8} = 5.73\ \text{years}$.

Marks: one for using $T^2 = a^3$ in AU/year units, one for $T^2 = 32.8$, one for $T = 5.73$ years.

(a) Kepler's Third Law gives $\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM}$, so $T^2 = \left(\dfrac{4\pi^2}{GM}\right) r^3$. For satellites of the same central mass $M$, this is a straight line through the origin with gradient $\dfrac{4\pi^2}{GM}$.

(b) Gradient $= \dfrac{\Delta T^2}{\Delta r^3} = \dfrac{(1.549 - 0.335) \times 10^9}{(1.56 - 0.34) \times 10^{22}} = \dfrac{1.214 \times 10^9}{1.22 \times 10^{22}} = 9.95 \times 10^{-14}\ \text{s}^2\text{m}^{-3}$.

(c) The gradient equals $\dfrac{4\pi^2}{GM}$, so $M = \dfrac{4\pi^2}{G \times \text{gradient}} = \dfrac{4\pi^2}{(6.67 \times 10^{-11})(9.95 \times 10^{-14})} = 5.95 \times 10^{24}\ \text{kg}$, close to Earth's known mass.

Marks: one for identifying that same-$M$ orbits give $T^2 \propto r^3$ (line through the origin), one for a correctly computed gradient with units $\text{s}^2\text{m}^{-3}$, one for equating the gradient to $4\pi^2/(GM)$, one for $M = 5.95 \times 10^{24}$ kg with the unit.

Both planets orbit the same central mass (the Sun), so $\dfrac{T_{\text{Mars}}^2}{r_{\text{Mars}}^3} = \dfrac{T_{\text{Earth}}^2}{r_{\text{Earth}}^3}$, giving $T_{\text{Mars}} = T_{\text{Earth}} \left(\dfrac{r_{\text{Mars}}}{r_{\text{Earth}}}\right)^{3/2}$.

$\dfrac{r_{\text{Mars}}}{r_{\text{Earth}}} = \dfrac{2.279 \times 10^{11}}{1.496 \times 10^{11}} = 1.523$.

$T_{\text{Mars}} = (3.156 \times 10^7)(1.523)^{3/2} = (3.156 \times 10^7)(1.880) = 5.93 \times 10^7\ \text{s}$.

Converting to years: $T_{\text{Mars}} = \dfrac{5.93 \times 10^7}{3.156 \times 10^7} = 1.88\ \text{years}$.

Marks: one for setting up the ratio $T_{\text{Mars}}^2/r_{\text{Mars}}^3 = T_{\text{Earth}}^2/r_{\text{Earth}}^3$ (no $G$ or $M$ needed), one for the radius ratio, one for $T_{\text{Mars}} = 5.93 \times 10^7$ s, one for converting correctly to $1.88$ years.

Band-6 plan. (1) State Law 1 (ellipse, Sun at one focus) so distance to the Sun genuinely varies. (2) State Law 2 (equal areas in equal times) and derive the speed-distance relationship from it. (3) Give the underlying physical reason (zero torque from a central force, hence conservation of angular momentum). (4) Describe the full-orbit behaviour (perihelion to aphelion and back) and tie it together.

Model answer. Kepler's First Law states that a planet's orbit is an ellipse with the Sun at one focus, not the centre. Because the Sun is off-centre, the planet's distance from the Sun genuinely changes over the orbit, reaching a minimum at perihelion and a maximum at aphelion, rather than staying constant as it would for a circular orbit centred on the Sun.

Kepler's Second Law states that the line joining the planet to the Sun sweeps out equal areas in equal times. The area swept in a short time $\Delta t$ is approximately $\Delta A \approx \tfrac{1}{2} r v \Delta t$, so a constant rate $\Delta A/\Delta t$ requires the product $rv$ to be constant. Where $r$ is small (near perihelion) $v$ must be large, and where $r$ is large (near aphelion) $v$ must be small.

The physical reason is that gravity is a central force: it acts entirely along the line joining the planet to the Sun, so it exerts zero torque about the Sun. With no external torque, the planet's angular momentum $L = mvr$ (for the component of velocity perpendicular to $r$) is conserved throughout the orbit. Kepler's Second Law is exactly this conservation law expressed geometrically as equal areas.

Putting the two laws together: as the planet moves from aphelion toward perihelion along the ellipse, $r$ decreases and conservation of angular momentum forces $v$ to increase correspondingly; after perihelion, as $r$ increases again toward aphelion, $v$ decreases back to its slowest value. The planet therefore speeds up and slows down cyclically once per orbit, fastest at perihelion and slowest at aphelion, with no change to the total energy of the orbit (the trade-off is between kinetic and gravitational potential energy).

Marker's note: the top band explicitly links the ellipse (First Law) to the varying $r$, derives $rv = \text{constant}$ from the equal-areas statement, and names conservation of angular momentum (zero torque from a central force) as the physical cause - not just "the law says so". A response that only restates the two laws without connecting them to angular momentum caps in the middle band.

Band-6 plan. Thesis: Kepler's Third Law can determine the star's mass, but only if $T$ and $r$ are both known in absolute (not just relative) units, and only under the circular-orbit approximation. Argue (1) the derivation and what it needs, (2) what real exoplanet observations actually measure and the gap this creates, (3) the limits (mass of the planet, eccentricity), then judge.

Model answer. Kepler's Third Law, $T^2/r^3 = 4\pi^2/(GM)$, was derived by equating the gravitational force to the centripetal force for a circular orbit, $GMm/r^2 = mv^2/r$, and substituting $v = 2\pi r/T$; the orbiting mass $m$ cancels, so the ratio $T^2/r^3$ depends only on the central mass $M$ and the constant $G$. In principle, therefore, measuring an exoplanet's orbital period $T$ and its orbital radius $r$ and rearranging to $M = 4\pi^2 r^3/(GT^2)$ gives the star's mass directly, with no other information required.

In practice this is genuinely useful: the transit method gives the period $T$ precisely (the time between dips in the star's brightness), and if the star's mass is otherwise estimated (from its spectral type), $r$ can instead be found from Kepler's Third Law - this is exactly how astronomers report exoplanet orbital radii. Used the other way around, if $r$ can be independently measured (for example from radial-velocity amplitude combined with an assumed inclination), the same equation returns $M$.

However, the method has real limits. Kepler's Third Law as derived here assumes a circular orbit; real exoplanet orbits can be eccentric, in which case $r$ must be replaced by the semi-major axis $a$, which is harder to measure directly from a transit light curve alone. The derivation also assumes the planet's mass is negligible compared to the star's, which is an excellent approximation for planets but would need correction for a binary-star system. Finally, $T$ alone (from transits) does not give $r$ without additional information (such as an independently known stellar mass or radial-velocity data), so period and radius are not both trivially available from a single observation technique.

Weighing this, Kepler's Third Law is the essential relationship connecting mass, period and radius, and is genuinely used to extract stellar or planetary masses in exoplanet astronomy - but it cannot be applied from period data alone; it must be combined with at least one other independent measurement (spectral-type mass estimate, radial velocity, or an assumed near-circular orbit) to fully solve for $M$.

Marker's note: the top band states the derivation and its assumption (circular orbit, $m \ll M$), correctly identifies that period alone is insufficient without a second independent quantity, and reaches an explicit evaluative judgement (useful but not sufficient alone). A response that only restates the formula and says "yes it works" without addressing the circular-orbit/eccentricity limitation or the need for a second measurement caps below the top band.