Derivation of the Drop due to the curve of the Earth
This is a "supporting slide" and is not a part of the course. It is included here to show the derivation, and justify the use of, the Ground Drop formula in the simulations. If you know some geometry and trigonometry, you might enjoy having a look.


We want to derive a formula that calculates the drop (d) of the surface of the Earth as a function of the horizonal distance (x).
Note that x is the "x-component" distance referenced in the simulations of the laws of motion.

Notice that the red lines form a right triangle with known dimensions. Applying the Pythagorean equation we get: x² + (r - d)² = r²

Using the binomial expansion of the form (a + b)² = a² + 2ab + b² and expanding the term in parantheses we get: x² + r² - 2rd + d² = r²

Subtracting r² from both sides of the equation we simpify to: x² + d² - 2rd = 0

We want the function in the form of d = f(x). x is the independent variable and d the dependent.
Accordingly we reorder the terms to set it up as a quadratic equation: d² - 2rd + x² = 0

The solutions for equations in the form \(ax^2 + bx + c = 0\) is the quadratic equation \( x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

In our case a = 1, b = -2r, and c = x² and the solutions are \( d = {-(-2r) \pm \sqrt{(-2r)^2-4x^2} \over 2}\) = \( {2r \pm \sqrt{4r^2-4x^2} \over 2}\) = \( {2r \pm 2\sqrt{r^2-x^2} \over 2}\)

Finally divide out the 2 and choose the negative solutuion: \(d = {r - \sqrt{r^2-x^2} }\)

We choose the negative because the limits fit the model. d = 0 when x = 0 and d = r when x = r.
When x > r, d becomes imaginary because there is a square root of a negative term. This makes sense because it lies outside the model.

This equation will work on the circular cross section of any spherical body but for the Earth in particular, r = appr 4000 miles = and so \(d = {4000 - \sqrt{16000000 - x^2} }\) in miles:

Drop Calculator: x:  (miles)   d:     (inches)