MHB Simple Law of Cosines Question

[Xarzu]

Simple Law of Cosines Question

Huge suspension bridges are build with the Earth's roundness in mind. The two towers are plumb line straight up and down and yet, because of their colossal size, they are a bit further apart at their tops than they are at their base. So, how can we calculate what this difference would be?

Here is the input data:

If we know the Earth's radius; the location of the base of the towers above sea level; The distance (from the center of each base of the tower) between the bases; and the height of the tower, how would we calculate the distance from the tops of the towers.

I made this video to explain what I am talking about but I want to have the mathematical formula to predict the distance differences between the tops of the towers compared to the base:

[YOUTUBE]8NuNga3Bpns[/YOUTUBE]

I have seen a similar question answered once using something called "the law of cosines" where, if you know an angle and the length of two vectors, you can calculate the distances between the two vectors? I hope that helps and gives us a clue.

 

[MarkFL]

Suppose we consider the Earth to be a sphere with radius $r_E$, and our two towers have their bases at the same distance from the center of the Earth, Let $h_1$ be the height of the first tower and $h_2$ be the height of the second tower.

If we know the distance $a$ along the surface of the Earth between the two towers, then the angle between them is:

$$\theta=\frac{a}{r_E}$$

And so the straight-line distance between the bases of the towers is:

$$d_B=\sqrt{2r_E^2\left(1-\cos(\theta)\right)}$$

And the straight-line distance between the tops of the towers is:

$$d_T=\sqrt{\left(r_E+h_1\right)^2+\left(r_E+h_2\right)^2-2\left(r_E+h_1\right)\left(r_E+h_2\right)\cos(\theta)}$$

Does that help?

 

[I like Serena]

Since the radius of the Earth is absurdly big with respect to anything we're doing, we can neglect any differences between arcs and straight lines. In other words, there's no real need for cosines or such.

Suppose $a$ is the distance at the base, $h$ is the common height of the 2 towers, and $R$ is the radius of earth.
Then the corresponding angle $\theta$ is:
$$\theta= \frac aR$$
The arc length at the top is then $(R+h)\theta$.
And the difference in distance between top and bottom is:
$$\Delta s = (R+h)\theta - R\theta = h\theta = \frac hR a$$

As an example, suppose $h=600\text{ m}$, $a=1\text{ km}$, and $R=6000\text{ km}$.
Then:
$$\Delta s = \frac {0.6}{6000}\cdot 1\text{ km}=0.1\text{ m}$$
That is, if the towers are as high as the highest tower in the world, we have 10 centimeters over a distance of a kilometer.