Total eclipse

Classical Mechanics Level 2

On Monday, August 21, mid-afternoon there will be a solar eclipse visible in the United States. The path of totality ranges from West Coast to East Coast.

If all of the US population were to go to see the total eclipse, what would be the average population density along the path of totality, approximately?

Note

No precise answer is sought; only an order-of-magnitude estimate.

150 people/sq mi 15 000 people/sq mi 15 people/sq mi 1500 people/sq mi 150 000 people/sq mi

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Arjen Vreugdenhil
Aug 18, 2017

The United States is several thousands miles wide. The width of the path of a total eclipse is less than 100 miles. The population of the US runs in the hundred millions.

Therefore we find pop. density ∼ population length ⋅ width ∼ 3 × 1 0 8 people 1 0 2 mi ⋅ 3 × 1 0 3 mi = 1 0 3 people/sq mi . \text{pop. density} \sim \frac{\text{population}}{\text{length}\cdot\text{width}} \sim \frac{3\times10^8\ \text{people}}{10^2\ \text{mi}\cdot 3\times10^3\ \text{mi}} = 10^3\ \text{people/sq mi}.

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