I'm trying to calculate the angle (in degrees) between two latitude/longitude pairs, but with a twist. Most calculations I see use the Great Circle / bearing method, but this does not seem correct for my need.
I am sitting at the following coords (for the sake of argument): Point A: 38.89755 -76.96362
I would like to watch tomorrow's NASA / Orbital rocket launch at: Point B: 37.83832 -74.51674
The rocket will rise over 10deg altitude from my place, so I should be able to see it well.
How do I calculate the direction that I should look in, assuming the rocket flies straight up (I know it won't, but perhaps the math is easier)? I will be setting up a fair bit of astrophotography stuff and there are a few obstacles within a couple degrees of my target (126.83° according to Great Circle), so I would like to calculate it exactly to avoid them (I know my magnetic declination is currently -10.604, so I can add that later).
I understand that in many cases, a rocket launched straight up from Point B will never be visible from Point A since the paths never converge, so this makes the math slightly more frightening.
I've tried to calculate this in the past as well, where I grew up on one side of Lake Michigan and tried to imagine how tall a building on the other side (50 miles away) would need to be to be visible, and at what azimuth/altitude, but it seems to elude me.
The distance to the horizon is
$$d=r\tan\left(\sec^{-1}\left(1+\frac{h}{r}\right)\right)$$
where
$$d=\text{distance to horizon}$$ $$h=\text{height of the observer}$$ $$r=\text{radius of the sphere (aprox 6378137m for the Earth)}$$
http://www.wolframalpha.com/input/?i=distance+to+the+horizon
You need to calculate this twice - once for you and once for the rocket. If the sum of these is greater than the distance between you then you can see it (barring topography).