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

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?

• When you say perpendicular distance, is this distance perpendicular to the surface, or to the tangent? For any positive distance, those two are not exactly the same. Jan 3, 2016 at 7:44
• I mean the distance perpendicular to the tangent in the case of the tangent, and perpendicular to the chord in the case of the chord.
– Paul
Jan 3, 2016 at 7:45
• The question has been updated accordingly, thanks.
– Paul
Jan 3, 2016 at 8:08
• A 'straight' line doesn't exists on the surface of the earth, is it ok to consider a geodesic? Jan 3, 2016 at 8:36
• I don't understand the question.
– Paul
Jan 3, 2016 at 8:56

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 $\dfrac{d^2}{2r}$.

• Ok, thanks for this. So, would you know how to reformulate this if one is given a radius, which in this case for the earth would be 3,959 miles? In other words, after substituting 3,959 for r, how could the formula be simplified?
– Paul
Jan 3, 2016 at 22:08
• I have seen a formula b = 8in. * d ^2, where d is the distance in miles. This seems to check out with my results on a graphing calculator, etc., but I am wondering how it is derived.
– Paul
Jan 3, 2016 at 22:24
• $\dfrac{d^2}{2r}$ would give about $0.0001263 \,d^2$ miles if $d$ is measured in miles. Since there are $63360$ inches in a mile, you might think of that as about $8.002\, d^2$ inches if $d$ is measured in miles. Jan 3, 2016 at 23:55
• Forgive my ignorance, how does it happen that if d is much smaller the r you get the above formula, and what are the steps to get to it?
– Paul
Jan 4, 2016 at 1:53
• If $d$ is much less than $r$ then $r - \sqrt{r^2-d^2} \approx r - \sqrt{r^2-d^2+ \frac{d^4}{4r^2}} = r - \sqrt{\left(r^2- \frac{d^2}{2r}\right)^2} = \frac{d^2}{2r}$. Similarly $\sqrt{r^2+d^2} -r \approx \sqrt{r^2+d^2+ \frac{d^4}{4r^2}}-r = \sqrt{\left(r^2+\frac{d^2}{2r}\right)^2} -r = \frac{d^2}{2r}$ Jan 4, 2016 at 8:17

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 $d=200km$ : $$x=\sim 3 km.$$

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.

• For $d$ much less than $R$ you too have $x \approx \frac{d^2}{2R}$, as with my results. Jan 4, 2016 at 8:18
• Interesting and thanks for this, although what I was trying to ask was what Henry answered, i.e. not the drop for the distance traveled over the surface of the earth, but the drop for the distance traveled if one travels through space in a straight line from earth.
– Paul
Jan 6, 2016 at 1:39