Circumference of Earth — Value, Formula, and How It Was Found

Book A Free Math Class

Circumference of Earth — Value, Formula, and How It Was Found

#Geometry

TL;DR

Earth's circumference is about 40,075 km around the equator and 40,008 km around the poles, found from C=2πrC = 2\pi rC=2πr using the planet's radius. This guide gives the modern values, the circumference formula, and how Eratosthenes measured it in 240 BCE using only shadow angles and a known distance — landing within about 1% of today's figure

BT

Bhanzu Team Last updated on July 13, 20269 min read

What Is the Circumference of Earth?

The circumference of Earth is the total distance around the planet — the length of a complete loop along its surface. Because Earth is very slightly flattened at the poles (an oblate spheroid, not a perfect sphere), the distance depends on the path you take:

The two differ by only about 67 km, which is why a single round-number value of roughly 40,000 km is often quoted. These figures come from precise modern surveying and satellite measurement, as catalogued by sources such as Wikipedia's record of Earth's circumference. The radius and circumference are tied together by the same relationship that holds for any circle, which is where the formula comes in.

How Is The Circumference of Earth Calculated Today?

The modern approach is short: measure Earth's radius (from satellites and geodetic surveys), then apply the circle formula. The mean equatorial radius is about 6,378 km, and the circumference is 2π2\pi2π times that. The hard part is measuring the radius accurately — the formula itself is the same one used for a coin.

The Formula for Earth's Circumference

Earth's circumference uses the standard circle formula. Treating the planet as a sphere of radius rrr (or diameter d=2rd = 2rd=2r):

C=2\pi r=\pi dC = 2\pi r = \pi dC=2πr=πd

This is worth understanding rather than memorising. The number π\piπ (about 3.14159) is defined as the ratio of any circle's circumference to its diameter — so circumference divided by diameter always equals π\piπ, which rearranges to C=πdC = \pi dC=πd. Since the diameter of a circle is twice the radius, C=2\pi rC = 2\pi rC=2πr as well.

Variable key: CCC is the circumference (distance around); rrr is the radius (centre to surface); d=2rd = 2rd=2r is the diameter (full width through the centre); π≈3.14159\pi \approx 3.14159π≈3.14159 is the fixed circle ratio. All distances use the same unit; the answer comes out in that unit.

How Eratosthenes Measured Earth's Circumference

The historical method is the best worked example the formula has, so it is worth following step by step. Eratosthenes of Cyrene (c. 276–194 BCE), the chief librarian at Alexandria, built his estimate on three observations:

  1. At Syene (modern Aswan), at noon on the summer solstice, the Sun was directly overhead — a vertical well lit all the way to the bottom and a rod cast no shadow.

  2. At Alexandria, due north, a vertical rod at the same moment cast a shadow whose angle was about 7.2° from vertical.

  3. The two cities were about 5,000 stadia apart (roughly 800 km) along the same north–south line.

Because the Sun is so distant, its rays arrive essentially parallel. The 7.2° shadow angle at Alexandria therefore equals the central angle between the two cities measured at Earth's centre — a result that comes straight from the alternate-angles rule for parallel lines crossed by a transversal.

That 7.2° is exactly 150\frac{1}{50}501​ of a full 360° turn:

360°7.2°=50\frac{360°}{7.2°} = 507.2°360°​=50

So the distance between the cities is 150\frac{1}{50}501​ of the whole way around. Multiply up:

C=50×800 km=40,000 kmC = 50 \times 800 \text{ km} = 40{,}000 \text{ km}C=50×800km=40,000km

Eratosthenes' figure, often cited as 252,000 stadia, lands close to 39,000–40,000 km depending on the exact length of his "stadion" — within about 1% of the true value, achieved with shadows and arithmetic. A century or so later Posidonius (c. 135–51 BCE) made an independent estimate using the star Canopus and reached a comparable figure; both are recorded in the historical sources above.

How Accurate Was Eratosthenes' Measurement?

Remarkably accurate — his figure lands within roughly 1% of the modern value of about 40,075 km, depending on the exact length of the stadion he used. For a measurement made with a vertical rod, a shadow, and the distance between two cities, an error of around 1% is extraordinary, and it stood as the best estimate for centuries.

Examples of Circumference of Earth

The examples move from a direct formula application to the Eratosthenes-style reasoning. Each states its units and the value of π\piπ used.

Example 1

Find Earth's equatorial circumference from its equatorial radius, r≈6378r \approx 6378r≈6378 km. Use π≈3.14159\pi \approx 3.14159π≈3.14159.

C=2πr=2×3.14159×6378≈40,074 kmC = 2\pi r = 2 \times 3.14159 \times 6378 \approx 40{,}074 \text{ km}C=2πr=2×3.14159×6378≈40,074 km

This matches the surveyed value of about 40,075 km.

Example 2

A student is told Earth's diameter is about 12,742 km and computes the circumference as C=πrC = \pi rC=πr, getting about 20,015 km. Find the slip and the correct value.

The student used radius and the diameter formula together. A quick check kills the answer: 20,015 km is less than the diameter doubled, yet a circle's circumference is always more than three times its diameter. Something is mismatched.

The error is mixing C=πdC = \pi dC=πd (which needs the diameter) with the radius. Using the diameter correctly:

C=πd=3.14159×12,742≈40,030 kmC = \pi d = 3.14159 \times 12{,}742 \approx 40{,}030 \text{ km}C=πd=3.14159×12,742≈40,030 km

Or with the radius, C=2πrC = 2\pi rC=2πr where r=6371r = 6371r=6371 km gives the same answer. Both forms agree; mixing one form's input with the other does not.

Example 3

Eratosthenes measured a 7.2° shadow angle and a 5,000-stadia (about 800 km) distance between cities. Find the circumference.

The 7.2° is 150\frac{1}{50}501​ of 360°, so the city distance is 150\frac{1}{50}501​ of the circumference:

C=50×800=40,000 kmC = 50 \times 800 = 40{,}000 \text{ km}C=50×800=40,000 km

Example 4

Suppose two cities on the same meridian are 555 km apart and the Sun's shadow angle differs by 5° between them. Estimate Earth's circumference.

The 5° is 5360\frac{5}{360}3605​ of a full turn, so:

C=360°5°×555=72×555=39,960 kmC = \frac{360°}{5°} \times 555 = 72 \times 555 = 39{,}960 \text{ km}C=5°360°​×555=72×555=39,960 km

Close to the true value, using the same logic Eratosthenes used.

Example 5

How far does a point on the equator travel in one full rotation of Earth (one day)? Use C≈40,075C \approx 40{,}075C≈40,075 km.

A point on the equator traces the full equatorial circle once per day, so it travels the whole circumference:

Distance=40,075 km in 24 hours\text{Distance} = 40{,}075 \text{ km in 24 hours}Distance=40,075km in 24 hours

That is a speed of about 40,07524≈1,670\frac{40{,}075}{24} \approx 1{,}6702440,075​≈1,670 km/h — you are moving that fast right now, standing still.

Example 6

A plane flies a great-circle route once around Earth over the poles at an average 900 km/h. Roughly how long does the flight take? Use the polar circumference ≈40,008\approx 40{,}008≈40,008 km.

Time=Cspeed=40,008900≈44.5 hours\text{Time} = \frac{C}{\text{speed}} = \frac{40{,}008}{900} \approx 44.5 \text{ hours}Time=speedC​=90040,008​≈44.5 hours

So a non-stop polar circumnavigation at that speed takes a little under two days of flying.

Why Measuring the Planet Still Matters

Knowing Earth's size is the foundation under navigation, mapping, and space travel.

The lasting lesson sits underneath all of it: Eratosthenes showed that a careful measurement of a small thing, plus the right geometric relationship, measures the largest thing in reach. He never saw the whole Earth, yet he sized it — the original proof that geometry lets you reason far beyond what you can see.

Where Earth-Circumference Problems Go Wrong

Mistake 1: Mixing the radius and diameter forms

Where it slips in: Choosing between C=2\pi rC = 2\pi rC=2πr and C=\pi dC = \pi dC=πd when the problem gives one of radius or diameter.

Don't do this: Put the radius into C=\pi dC = \pi dC=πd (or the diameter into C=2\pi rC = 2\pi rC=2πr) — it halves or doubles the answer.

The correct way: Match the formula to what you are given: radius goes with 2πr2\pi r2πr, diameter with πd\pi dπd. The first-instinct error is grabbing whichever formula is remembered and feeding it the wrong length; the check is that circumference is always a bit over three times the diameter.

Mistake 2: Treating Earth as a perfect sphere when precision matters

Where it slips in: Problems that distinguish equatorial from polar circumference.

Don't do this: Report a single circumference as if every great-circle loop were identical.

The correct way: Earth bulges at the equator, so the equatorial loop (about 40,075 km) is longer than the polar loop (about 40,008 km). For a rough estimate, 40,000 km is fine; when the question asks for equatorial versus polar, keep them apart. The memorizer who learned only "40,000 km" can't answer which loop is longer.

Mistake 3: Confusing the shadow angle with the wrong central angle

Where it slips in: Eratosthenes-style problems, when relating the measured shadow to Earth's central angle.

Don't do this: Assume the city distance is some random fraction of the circumference instead of the fraction the angle dictates.

The correct way: Because the Sun's rays are parallel, the shadow angle equals the central angle between the cities, so the distance is angle360°\frac{\text{angle}}{360°}360°angle​ of the circumference. The second-guesser who measured the angle correctly but distrusts the parallel-rays step is exactly who stumbles here — the parallel rays are what make the two angles equal.

Conclusion

Practice and Next Steps

Work through these problems to solidify your understanding, then check each against the formula above.

  1. Find Earth's circumference from a mean radius of 6,371 km (π≈3.14159\pi \approx 3.14159π≈3.14159).

  2. Two cities on the same meridian are 1,000 km apart with a 9° shadow-angle difference. Estimate the circumference.

  3. At the equatorial circumference of 40,075 km, how fast (in km/h) does an equatorial point move during one 24-hour rotation?

To explore circles, π\piπ, and measurement with a teacher who builds each idea from the ground up, explore Bhanzu's geometry tutor, our middle school math tutor, or math classes online. Want a live Bhanzu trainer to walk through more circle and circumference problems? Book a free demo class.