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Measuring the whole Earth sounds like a job for satellites. About 2,200 years ago, Eratosthenes did it with a stick, a shadow and a distance between two cities. The method uses nothing beyond Class 9 geometry, and you can repeat it with a friend in another city. This post explains his reasoning, the parts that are uncertain, and a version for students.
Who was Eratosthenes?
According to the MacTutor biography of Eratosthenes from the University of St Andrews, he was born in 276 BC in Cyrene, in present-day Libya, and died in 194 BC in Alexandria. Around 240 BC he became the third librarian of the great library at Alexandria. He worked across many fields, which earned him the teasing nickname "Beta", the second letter of the Greek alphabet, because he was thought to fall just short of first place in each subject. Students today know him for the sieve of Eratosthenes for finding prime numbers.
His book On the Measurement of the Earth is lost. MacTutor explains that the details come through later writers, including Cleomedes, Theon of Smyrna and Strabo. That matters, because some numbers in the story are reconstructions.
How did he measure the Earth?
The method rests on two observations and one assumption, as summarised by MacTutor and by the American Physical Society's history column on Eratosthenes:
At Syene (modern Aswan), at noon on the summer solstice, the Sun was directly overhead. Vertical objects cast no shadow.
At the same time in Alexandria, to the north, a vertical stick cast a shadow. The Sun's rays made an angle of about 7.2° with the vertical, about one-fiftieth of a full circle.
The Sun is so far away that its rays reaching both cities are parallel.
If the rays are parallel, the 7.2° angle at Alexandria equals the angle at the Earth's centre between the two cities. That is the alternate-angles rule for parallel lines.
The arithmetic:
360° ÷ 7.2° = 50. The arc from Syene to Alexandria is one-fiftieth of the full circle.
The distance between the cities was taken as 5,000 stadia.
Circumference = 50 × 5,000 = 250,000 stadia.
MacTutor notes that the angle has also been reported as 7°12′. Since 12′ is 12/60 = 0.2 of a degree, that is the same 7.2°.
How accurate was his answer?
That depends on how long a stadium was, and nobody is sure. MacTutor reports two values scholars have proposed: 157.2 m and 166.7 m.
With 157.2 m: 250,000 × 157.2 m = 39,300,000 m, or 39,300 km.
With 166.7 m: 250,000 × 166.7 m = 41,675,000 m, or about 41,700 km.
For comparison, the NASA Earth fact sheet gives an equatorial radius of 6,378.137 km. The equatorial circumference is 2π × 6,378.137 ≈ 40,075 km. Either way, Eratosthenes was within a few per cent.
Be careful with stronger claims. MacTutor describes the historian Dennis Rawlins' argument that the only measurement Eratosthenes made himself was the angle at Alexandria, and that it was off by about 16′. Some popular accounts say he hired professional pacers to measure the distance; treat that as uncertain. And any error in the distance between the cities carries straight into the answer.
Can you repeat it at school?
Yes, and it is easiest at an equinox, around 20 March or 22 to 23 September. On those days the Sun is overhead at the equator at local noon, so the shadow angle from vertical at your school roughly equals your latitude.
Worked example with illustrative numbers. Two schools, roughly due north-south of each other and 1,000 km apart, each set up a vertical stick 1.00 m tall at local solar noon.
The southern school's shadow is 0.231 m. The angle from vertical is tan⁻¹(0.231 ÷ 1.00) ≈ 13.0°.
The northern school's shadow is 0.404 m. The angle is tan⁻¹(0.404) ≈ 22.0°.
Difference: 22.0° − 13.0° = 9.0°.
360° ÷ 9.0° = 40, so the circumference is 40 × 1,000 km = 40,000 km.
Common mistakes:
Using 12:00 on the clock. Local solar noon, when the shadow is shortest, can be well away from 12:00 IST, especially in the far east or west of India. Record the shortest shadow over half an hour.
A tilted stick. Check the stick with a plumb line, a stone on a string. A 1° tilt is a large error when your whole difference is 9°.
Cities far apart east-west. The method needs the north-south separation. Use distance along the meridian, not road distance.
Our Class 7 guide to the Earth, Moon and Sun explains why the Sun's height changes with latitude and season. For a different method, used centuries later from a mountain top, read our post on al-Biruni's measurement of the Earth's radius.
What can a student take from Eratosthenes?
Find the one assumption that unlocks the problem. Parallel sunlight turned an impossible measurement into a ratio.
Separate method from data. His method was exact; his data were rough. Exams test the method.
Quote uncertainty honestly. "Between 39,300 and 41,700 km" is a more accurate answer than a single number nobody can justify.
If the link between shadow angle and centre angle is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Try it: at the next equinox, measure your noon shadow, ask a friend in another city to do the same, and compute the Earth's circumference from your two angles.
