The shape of the Earth

The Earth is not a perfect sphere. It bulges at the equator, flattens at the poles, and its surface ripples with irregular mass distributions. For most mapping purposes you can ignore all of that. But knowing what you're ignoring — and when it matters — is the difference between a solid codebase and one that breaks on edge inputs.

Three models, three answers

The Earth has three commonly used shapes, in increasing accuracy:

Model	Description	Error vs. reality	When to use
Sphere	Perfect ball, radius ≈ 6,371 km	< 0.5%	Consumer apps, distance/bearing math
Ellipsoid	Flattened at poles (WGS84)	< 0.001%	GPS, surveying, aviation
Geoid	Lumpy gravitational equipotential	~0	Elevation referencing, geodesy

Each model gives a different answer to "where is this point?" — differing by hundreds of metres depending on location and model chosen.

The sphere

A sphere is the simplest model: a ball with a single radius $R$. It's wrong, but usefully wrong — for most consumer applications, the error is under 0.5%, far smaller than GPS receiver noise.

The most commonly used spherical radius is the mean radius of the WGS84 ellipsoid:

$$R = 6{,}371{,}008.8 \text{ m} \approx 6{,}371 \text{ km}$$

Rule of thumb: Use the sphere. Unless you're writing geodetic software, aviation systems, or need sub-metre accuracy, the sphere is the right choice. It's simpler, faster, and the errors are invisible in practice.

The ellipsoid (WGS84)

The Earth bulges at the equator because it rotates — centrifugal force pushes mass outward along the equatorial plane. The WGS84 ellipsoid (World Geodetic System 1984) captures this shape with two numbers:

$$a = 6{,}378{,}137.0 \text{ m} \quad \text{(equatorial radius)}$$ $$b = 6{,}356{,}752.314245 \text{ m} \quad \text{(polar radius)}$$

The difference $a - b \approx 21.4 \text{ km}$. The Earth is slightly fatter around the middle than it is tall.

The flattening $f$ quantifies this:

$$f = \frac{a - b}{a} = \frac{1}{298.257223563} \approx 0.003353$$

The geoid

The geoid is the shape that Earth's oceans would take if they were perfectly still and extended under the continents — a gravitational equipotential surface. It's neither a sphere nor a smooth ellipsoid; it bulges and dips based on variations in underground mass density.

The geoid matters for elevation. When a GPS device reports altitude, it reports ellipsoidal height (distance above the WGS84 ellipsoid). But when a topographic map says a mountain is 8,849 m tall, it's using orthometric height (height above the geoid — essentially above sea level). The difference, called geoid undulation, varies from about −107 m (Indian Ocean) to +85 m (near New Guinea).

The "good enough" sphere

For ~99% of app development, a sphere with radius $R = 6{,}371{,}008.8$ m gives errors under 0.5%. Use it.

When it matters:

• High-accuracy land surveying → use an ellipsoidal formula (Vincenty, Chapter 8)
• Aviation and aerospace → WGS84 ellipsoid required
• Elevation data → geoid corrections required

When it doesn't:

• Showing a map with pins
• Calculating "nearby" distances for a search radius
• Drawing routes and overlays
• Tile math (the projection already handles curvature — Chapter 5)

Worked example

Q: What is the circumference of the Earth at the equator using the spherical model?

$$C = 2\pi R = 2 \times \pi \times 6{,}371{,}008.8 \approx 40{,}030{,}228 \text{ m} \approx 40{,}030 \text{ km}$$

The "real" equatorial circumference (using WGS84's equatorial radius $a$) is:

$$C_{\text{equator}} = 2\pi a = 2 \times \pi \times 6{,}378{,}137.0 \approx 40{,}075{,}017 \text{ m}$$

The spherical estimate is ~45 km short — 0.1% error. About the distance from central London to Heathrow. For mapping work: invisible.

Key numbers to remember