Over a 10 kilometer run, how high is the Earth's curvature at the mid- point?

Here's how I would solve this problem...I could be wrong though.

Radius of earth = 6378 kilometers.
The length of the arc on the surface is 10.000 km The angle, theta, it subtends is

                         10.0
              angle =   ------  x 57.295 degrees   = 0.0898  degrees
                          6378

The length of the chord connecting the end points of the arc is from some trigonometry

 2    2    2                             2
L  = R  sin (theta)  +( R - Rcos(theta) )

so that L = 9.9946 kilometers which means the chord is 5.3 meters shorter than the 10 kilometer distance along the surface. If we approximate the difference in the arc and the chord as two small triangles, the height of the arc above the mid point of the chord is just

 2                     2                   2
h    = (5002.66 meters)   - (5000.0 meters)

or 163 meters.

So, at the midpoint of the 10 kilometer arc, I would predict that the straight-line chord is 163 meters below this point inside the Earth.


Copyright 1997 Dr. Sten Odenwald
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