Distance to the celestial horizon Calculating the distance to the horizon, defined as the point at which a ship will vanish from sight because it's blocked by the curvature of the earth, is fairly simple.  But how about something a bit more complicated?
A satellite bearing a bright light is in geosynchronous orbit over point A on the Earth, on the equator.  A ship is on the equator, sailing due east.  At point B, the crew can see the satellite on the horizon.  How far apart are A and B?
Not a homework question or anything; I just thought of this scenario and realized I had no idea how to calculate it!
 A: If everything's on the equator, this is actually fairly simple to calculate.  We have two concentric circles: the surface of the Earth and geocentric orbit, both of which center upon the center of the Earth.  A bit of Googling turns up the radius of the Earth at 3,959 miles, and geocentric orbit over the equator as 26,199 miles from the center of the Earth.
From any point on the surface of the Earth, the center is subjectively straight down and the horizon is straight ahead, which means that the three points--point A' (the location of the satellite), point B, the location of the ship, and point C, the center of the Earth, can be described as the vertices of a right triangle, with the line from C to A' passing through A.  From here it's all calculator work.  First, we get the ratio of CB to CA':
3959 / 26199 = 0.1511126378869422
The cosine of an angle is defined as the length ratio between the adjacent non-hypotenuse side and the hypotenuse, which means that we just calculated the cosine of the arc distance between points A and B.  Taking the arc-cosine of that value returns approximately 81.31 (in degrees), which means that A and B are almost a quarter of the world away from one another.
Putting this value into the NOAA's latitude/longitude distance calculator returns 5614 standard miles, or 4879 nautical miles, which may be more appropriate since we're talking about a ship.
