If you believe Earth is flat, you probably reject the results of Eratosthenes’s shadow experiment. While the Sun was directly over one city, he measured the length of a pole’s shadow in a city 800 km away. From this, he calculated Earth’s radius with surprising accuracy—all based on the assumption that Earth is round.
But let’s bring some healthy April Fools’ Day skepticism to this assumption. What if Earth was, in fact, flat?
If this were the case, how far from Earth would the Sun need to be to explain Eratosthenes’ measurement?
Before we waltz into the shadows, congratulations to this month’s winner Susan Callister who found the Sun’s distance to be about $8000\ \text{km}.$ Also recognized is solver Johan Basberg who suggested we get rid of the shadows by burying the Sun $2\ \text{m}$ below Earth’s surface… invalid, but thought-provoking nonetheless.
To get some intuition for the situation, let’s redraw the diagram with the heights a bit embellished:
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Drawing the rays of light as they’d come from the sun, we can see a pair of similar triangles traced out.
The first is the one formed by the shadow, and the pole in Alexandria. The second is formed by the span from the left edge of the shadow to the city of Syrene, and the span from the city of Syrene to the Sun itself.
Because these triangles are similar (their corresponding angles are equal), the ratio of their sides will be equal too. So, forming the ratio of base to height in both triangles, we get
$$ \frac{0.1\ \text{m}}{1\ \text{m}} = \frac{0.1\ \text{m} + 800\ \text{km}}{1\ \text{m} + d}. $$
Since $800\ \text{km}$ is much bigger than $0.1\ \text{m}$ and, presumably, the distance to the Sun is still much greater than a meter we can ignore these amounts and the ratios reduce to
$$ \frac{0.1\ \text{m}}{1\ \text{m}} \approx \frac{800\ \text{km}}{d}, $$
or $d \approx 8000\ \text{km}.$