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Is it possible to calculate the radius of Earth based on two height measurements, and knowing the distance between the observers? If so: how?

I have the following situation: enter image description here

Here I can use the Pythagorean theorem to find R = ($a^2/2s$)-$s/2$ but what if the observers are on 2 different heights? (Like the picture but you have s1 and s2 instead) How can I then calculate the radius of the Earth?

Tore
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  • Here's a MathJax tutorial :) – Shaun Apr 27 '17 at 13:27
  • The theorem on the power of a point (https://en.wikipedia.org/wiki/Power_of_a_point) may be useful: given a fixed circle and a point $P$, take any line that intersects the circle, and let $A$ and $B$ be the two intersection points. Then the product of $PA$ and $PB$, called the power of $P$, is always the same. (If you choose a tangent line that intersects the circle at just one point $T$, then the power of the point is $(PT)^2$.) – Connor Harris Apr 27 '17 at 13:42

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The problem is impossible if we relax our assumption (implicit in your diagram) that the line between the two observers is tangent to the Earth. Let's reverse our perspective and suppose that the Earth's radius is $R$, the observers have height $h_1$ and $h_2$, and that our task is to fix the two observers so that they're a distance $d$ apart. By the cosine formula for triangles, we can put them an angle $\theta$ apart such that $d^2 = (R+h_1)^2 + (R+h_2)^2 - 2(R+h_1)(R+h_2) \cos \theta$, or $$\theta = \arccos \frac{(R+h_1)^2 + (R+h_2)^2 - d^2}{(R+h_1)(R+h_2)}$$ and as long as $h_1, h_2, R, d$ are chosen such that this expression is in the domain of $\arccos$, this can be satisfied. This means that any three measurements $h_1, h_2, d$ can work for a range of potential $R$s, so they can't pick out any specific one.

If the line between the two observers is tangent to the Earth, though, then by the power-of-a-point theorem, the distance from observer $i=1, 2$ to the Earth along this tangent line is $\sqrt{h_i (h_i + 2R)}$, and the length of the tangent line $d$ is the sum of these. The problem thus becomes solving $d = \sqrt{h_1(h_1+2R)} + \sqrt{h_2(h_2 + 2R)}$, and this has a unique solution because $\sqrt{h_1(h_1+2R)} + \sqrt{h_2(h_2 + 2R)}$ is a monotonically increasing function of $R$.

Connor Harris
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    Thank you very much for your answer! So I guess it is impossible to solve my problem then. I found somewhere that someone used R = 2a/(sqrt (2h1)+sqrt (2h2)). But it didn't show the explanation behind it. – Tore Apr 27 '17 at 14:40
  • I'll add that the problem could be solved if we had some other bit of information, such as the angle between the line through both observers and the vertical line from one observer to the ground. But with just the three quantities, it's not possible. – Connor Harris Apr 27 '17 at 15:01
  • Strange, but when I work in reverse (like you purposed) it seems that the formula I found fits very good to the result. Or is the math wrong? It seems that it is based on $d= {\sqrt R} * {\sqrt (2h1)} + {\sqrt (2h2)}$ If this gives a good estimate then I only need to know d, h1 and h2 to calculate R. – Tore Apr 27 '17 at 15:15
  • Your formula for $d$ works if $h_i \ll R$, because then $\sqrt{h_i (h_i + 2R)} \approx \sqrt{2h_i R}$. – Connor Harris Apr 27 '17 at 15:38
  • Thank you so much for your help!! And $hi$ will be around 3-5 meters so I guess this means I can use that formula. It will not be exact, but it will do. – Tore Apr 27 '17 at 15:53