Curvature Of The Earth


From how far on a clear day, and assuming there are no obstructions, could you see from the top of the Eiffel tower?

Take the height of the Eiffel tower as 300 metres and the radius of the Earth as 6240 km.


We shall answer this question in two ways:

Let us begin by considering the distance from the top of the tower to furthest point on the surface of the earth (the tangental length).

Applying the Pythagorean Theorem, 6240.32 = d 2 + 62402. Hence d approximately 61.2 km.

However, this does not represent the distance along the curvature of the earth (the arc length). To do this we apply trigonometry to the right angle triangle above, cos θ = 6240/6240.3, thus θ approximately 0.562 degrees.

As the radius of the earth is 6240 km, the circumference is 2π times 6240 approximately 39187 km. So the arc length will be 0.562/360 times 39187 approximately 61.2 km, which is the same answer as the other method... coincidence?

Does the tangental length always equal the arc length?

What fraction of the earth is visible from the top of the Eiffel tower?

To simplify this problem, assume that the viewing range is a flat circle with a radius equal to the maximium viewing distance (61.2 km).

From how far away, on a clear day and with no obstructions, can the Eiffel tower be seen?

Note: This is not the same as the original question. Consider the following question first...

What is the maximum distance that two 2 metre tall people can see each other?

What is the typical range of a geostationary satellite?

Their orbit is about 35900 km (yep, that's right!) above the surface of the earth.

What fraction of the globe does a geostationary satellite typically cover?

WARNING: This problem is VERY difficult!

Problem ID: 44 (Apr 2001)     Difficulty: 2 Star

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