How to calculate the drop as one moves in a straight line from the surface of the earth

Question

I am trying to come up with a formula that will calculate the drop as one moves in a straight line from earth. In other words, if one travels in a straight line from the surface of the earth, to what distance will the surface fall away as a function of the distance traveled? It will have to involve the formula for calculating the length of a chord of a circle. Put in geometric terms, the question would be: "What is the distance from a line tangent to a circle to the surface of the circle, the distance being perpendicular to the tangent?" The formula for calculating the length of a chord is

l = 2$\sqrt{r^2 – d^2}$

where r is the radius of the circle and d is the perpendicular distance of the chord from the center of the circle. The radius of the earth is 3,959 miles. How would one come up with the formula that would do this?

Answer 1

It is not totally clear what you are asking but looking at this:

then

$$a = \sqrt{r^2+d^2}-r$$ $$b = r-\sqrt{r^2-d^2}$$ though $b$ is only real when $d \le r$.

When $d$ is much smaller than $r$ then both $a$ and $b$ are about $\dfrac{d^2}{2r}$.

Answer 2

If I understood what you are asking, you want a formula to find $x$, given $R$ the radius of the Earth and $d$ the distance traveled.

First notice that $$x=R-R\cos \theta$$ and since $\theta=d/R$ the formula is $$x=R(1-\cos \frac{d}{R}).$$

For example, if you travel $d=200km$ : $$x=\sim 3 km. $$

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Notice that this does not mean that every time you travel $200km$ you 'fall' down of $3km$, this is only true when you travel along a maximal cricle (called geodesic). In the other cases the falling distance will be smaller. This is because you can't talk about straight line on the surface of earth, since a straight line lies on a plane and there is no plane contained in the surface of a sphere.