Earth Curvature and Refraction, Explained Simply

Title: Earth Curvature and Refraction, Explained Simply Category: physics

Earth’s curvature describes how far the surface falls away from a straight, horizontal tangent as you look across it. Atmospheric refraction changes what you see: because air density usually increases closer to the ground, light often curves slightly downward through the atmosphere. That makes distant ground or objects appear higher than they would along a perfectly straight light path.

The calculations below give a useful everyday estimate. They assume a roughly spherical Earth and ordinary atmospheric conditions; unusual temperature layers can make refraction much stronger, weaker, or even reverse its usual effect.

How much does Earth curve away?

Let $d$ be the horizontal distance along the surface and $R$ Earth’s radius, about $6{,}371$ kilometers. The surface’s vertical drop below a tangent at the starting point is, exactly,

$$R\left(1-\cos\left(\frac{d}{R}\right)\right).$$

For distances small compared with Earth’s radius, this is well approximated by

$$C \approx \frac{d^2}{2R}.$$

The distance must use the same units as $R$. In convenient units, the result is about 0.0785 meters per kilometer squared, or about 7.85 centimeters per kilometer squared. Thus, over 10 km the geometric drop is about 7.85 m.

A familiar imperial approximation is about 8 inches per mile squared. In both versions, the distance is squared: doubling the distance makes the drop four times as large.

What refraction changes

For a simple estimate, surveyors represent the usual downward bending of a light ray with a refraction coefficient $k$. A commonly used value is about $0.13$, though conventions and atmospheric conditions vary. The ray follows part of Earth’s curve, so the apparent curvature drop is approximately

$$C_{\text{apparent}} \approx (1-k)\frac{d^2}{2R}.$$

With $k=0.13$, refraction reduces the geometric drop by about 13%. For 10 km, the estimate becomes

$$C_{\text{apparent}} \approx 0.87(7.85\text{ m}) \approx 6.83\text{ m}.$$

This is why curvature-only and curvature-plus-refraction calculations give different answers. Some surveying references use $k=1/7$ instead of $0.13$; those are close approximations, not exactly the same value.

Worked example: estimating the horizon

An object is visible if its height puts it above the observer’s apparent horizon. For a height $h$ in meters, the approximate distance to the horizon in kilometers is

$$d \approx 3.57\sqrt{h}$$

using geometry alone. With the typical refraction estimate $k=0.13$, it becomes about

$$d \approx \frac{3.57}{\sqrt{1-0.13}}\sqrt{h} \approx 3.83\sqrt{h}.$$

Suppose your eyes are 2 m above the water and a ship’s highest point is 100 m above it. Their horizon distances are approximately

$$3.83\sqrt{2} \approx 5.4\text{ km}, \qquad 3.83\sqrt{100}=38.3\text{ km}.$$

Adding them gives a maximum visibility estimate of about 43.7 km in typical conditions. This assumes an unobstructed view and ignores waves, haze, and the ship’s shape. With no refraction, the same calculation gives about 40.7 km.

Why the estimate can be wrong

Refraction depends on how temperature and air density change with height. Under ordinary conditions, the ray bends down a little. A strong temperature inversion can bend it much more, producing looming or mirages; other conditions can bend it less. A single coefficient cannot describe every sightline or weather pattern.

The horizon formulas also treat the observer and target as points at known heights above a smooth spherical surface. Terrain, buildings, waves, and the height of the observer’s eyes can change what is visible. For precise surveying or unusual atmospheric conditions, use measured refraction data or a specialized model.

Common mistakes

• Forgetting to square the distance: curvature drop grows with $d^2$, not $d$.
• Mixing units: use meters for both $d$ and $R$, or kilometers for both, before calculating.
• Subtracting the wrong amount: the typical refraction correction is about 13% of the curvature drop, not 13% of the distance.
• Treating the rule as exact: the geometric approximation is reliable for ordinary distances, but atmospheric refraction is variable.
• Confusing a tangent with a sightline: a straight tangent is a geometric reference. Light traveling through air follows a curved path when refracted.