Solar System

Despite appearances from our home planet’s surface, the planets orbit the Sun on nearly circular, nearly coplanar paths. Kepler provided a concise description of this orderly motion (alongside an outdated theory of magnetic fibers), and Newton developed a plausible mechanism about a half century later.

Kepler’s Laws

In a feat of 16-17th century data science, decades of astronomical data were accurately described with three simple rules:

Each planet moves on an ellipse with the Sun at one focus. Most planetary orbits are only slightly elongated, so they look almost circular, but the offset matters for precise predictions. Drag the slider to stretch the orbit and watch the Sun sit at a focus — never the centre.

Sun (focus)

In units of years and astronomical units, the constant in the third law is exactly one for objects orbiting the Sun:

(Tyr)2=(aAU)3.\left(\frac{T}{\text{yr}}\right)^2 = \left(\frac{a}{\text{AU}}\right)^3.

This is why larger orbits take dramatically longer: Mercury laps the Sun every 0.240.24 years while Neptune needs about 165165. Closer to the Sun the gravitational attraction is stronger, demanding a higher orbital speed, and the path around is shorter too. Inner planets are both faster and have less ground to cover, so they lap the outer ones again and again.

Heliocentric Motion

A heliocentric model places the Sun at the center. This is a convenient reference frame for viewing the planets, as their primary motion is simply around the Sun. But many planets have moons, which are not solely orbiting the Sun. The Moon orbits Earth, while the Earth-Moon system orbits the Sun. So in a heliocentric view, the Moon’s path is the combination of both motions — a small loop riding along a big one.

The same two-frame thinking explains a famous puzzle. Throughout most of the year, an outer planet like Mars drifts slowly eastward against the stars. For a few weeks it appears to stop, reverse into retrograde, and then resume. Since Earth is on a faster inner orbit, we overtake Mars on the inside. From our point of view, Mars appears to slide backward against the distant background.

Problem Solving

Kepler's Third Law for a New Planet

Problem

A planet orbits the Sun at an average distance of a=4AUa = 4\,\text{AU}. How long is its year?

Use

(Tyr)2=(aAU)3.\left(\frac{T}{\text{yr}}\right)^2 = \left(\frac{a}{\text{AU}}\right)^3.

Solve

T2=43=64T=64=8years.T^2 = 4^3 = 64 \quad\Rightarrow\quad T = \sqrt{64} = 8\,\text{years}.

Check

A larger orbit means a longer year, and 88 years for 4AU4\,\text{AU} is reasonable — it falls between Mars (1.5AU1.5\,\text{AU}, 1.91.9 yr) and Jupiter (5.2AU5.2\,\text{AU}, 11.911.9 yr).

Jupiter's Distance in Kilometers

Problem

Jupiter is about 5.2AU5.2\,\text{AU} from the Sun. Estimate that distance in kilometers, using 1AU1.496×108km1\,\text{AU} \approx 1.496 \times 10^8\,\text{km}.

Solve

5.2×(1.496×108)7.78×108km.5.2 \times (1.496 \times 10^8) \approx 7.78 \times 10^8\,\text{km}.

Check

That is several times Earth’s distance from the Sun, which matches Jupiter’s place well out in the solar system — and shows why AU is the friendlier unit.

Orbits Checkpoint

Kepler's second law says a planet sweeps equal areas in equal times. What does this imply about its speed?