Kepler's laws

Law of Orbits. Every planet moves around the Sun in an elliptical orbit, with the Sun at one focus of the ellipse — not at the centre. This replaced the older, wrong assumption of perfectly circular orbits centred on the Sun, an assumption that had made planetary tables slightly but stubbornly wrong for centuries.

Law of Areas. The radius vector joining a planet to the Sun sweeps out equal areas in equal times. A planet therefore moves faster near the Sun (perihelion) and slower far from it (aphelion). This is a direct consequence of angular momentum conservation about the Sun: gravity is a central force, always pointing along the Sun–planet line, so it exerts zero torque about the Sun and angular momentum stays fixed.

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

where: $T$ is the orbital period and $a$ is the semi-major axis of the elliptical orbit. For a circular orbit the semi-major axis is just the radius, $a=r$, so $T^2\propto r^3$ — the form this section derives and uses throughout.

Where circular-orbit gravity fits in. For a circular orbit, gravity itself supplies the centripetal force keeping the planet on its path:

where: $G$ is the gravitational constant, $M_s$ is the Sun's mass, $m$ is the planet's mass, $r$ is the orbital radius, $v$ is the orbital speed and $\omega=2\pi/T$ is the angular velocity. This single balance — gravity equals centripetal requirement — is what the derivation below turns into Kepler's third law.

Why the historical order matters. Kepler reached his three laws around 1609–1619 by fitting geometry to observation alone; he had no notion of a force of gravity. Newton, working roughly seventy years later, showed that all three laws fall out of one assumption — an inverse-square, centrally directed attractive force between the Sun and each planet. Reducing three independently discovered empirical rules to consequences of a single universal law, one that also explains falling apples and ocean tides, is one of the most celebrated triumphs of classical mechanics. It is precisely this connection — Kepler's observational law of periods meeting Newton's law of gravitation through the centripetal-force condition — that the circular-orbit proof below reproduces.

Planet A orbits the Sun at radius $r_A = 1.0\times10^{11}\ \text{m}$ with period $T_A = 1.0\ \text{year}$. Planet B orbits at radius $r_B = 4.0\times10^{11}\ \text{m}$ (both orbits approximately circular). Find the period of Planet B.

Step 1. Apply Kepler's third law between the two planets: $\dfrac{T_B^2}{T_A^2} = \left(\dfrac{r_B}{r_A}\right)^3$.

Step 2. Compute the radius ratio: $\dfrac{r_B}{r_A} = \dfrac{4.0\times10^{11}}{1.0\times10^{11}} = 4$, so $\dfrac{T_B^2}{T_A^2} = 4^3 = 64$.

Step 3. Solve for $T_B$: $T_B = T_A\sqrt{64} = 1.0\times8 = \mathbf{8.0\ years}$. Planet B, orbiting four times farther out, takes eight years to complete one orbit — consistent with $T\propto r^{3/2}$.

Competitive stretch

Using $T^2 = \dfrac{4\pi^2}{GM_s}r^3$ with Earth's data ($r=1.5\times10^{11}\ \text{m}$, $T=3.15\times10^7\ \text{s}$, $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$):

Step 1. Rearrange for the central mass: $M_s = \dfrac{4\pi^2r^3}{GT^2}$.

Step 2. Compute the pieces separately: $r^3 = (1.5\times10^{11})^3 = 3.375\times10^{33}\ \text{m}^3$, $T^2 = (3.15\times10^7)^2 = 9.9225\times10^{14}\ \text{s}^2$, and $4\pi^2\approx39.48$.

Step 3. Substitute and divide: $M_s = \dfrac{39.48\times3.375\times10^{33}}{(6.67\times10^{-11})(9.9225\times10^{14})} \approx \mathbf{2.01\times10^{30}\ kg}$ — matching the known solar mass and showing how the third law lets us "weigh" the Sun from Earth's orbit alone, without ever sending a probe there.

• Applying $T^2\propto r^3$ using orbital radius for elliptical orbits — strictly the semi-major axis $a$ belongs in the formula; the syllabus derivation only covers circular orbits, where $a=r$.
• Misreading the law of areas as constant speed — speed varies (faster near the Sun, slower far away); it is the rate of area sweep that stays constant.
• Forgetting the constant $4\pi^2/GM_s$ depends on the central mass — it differs for planets orbiting the Sun versus satellites orbiting Earth.
• Attempting a general elliptical-orbit derivation of the third law — explicitly outside the syllabus; only the circular-orbit proof is required.
• Confusing the law of orbits (shape: an ellipse) with the law of periods (timing: $T^2\propto r^3$) — they answer different questions and are often mixed up in rapid-fire exam recall.
• Assuming the third law's proportionality constant is universal — it is $4\pi^2/GM_s$, tied to the Sun's mass; a satellite orbiting Earth obeys the same form of law but with a different constant, $4\pi^2/GM_E$.

Ask the tutor: Want a step-by-step walkthrough of how angular momentum conservation forces the law of areas, or more practice problems comparing periods and radii across different orbits?