What is the ground distance covered by a plane travelling from west to east horizons?

Question

Assume the viewer is at sea level, there are no mountains or other obstructions.
The plane is travelling at 39,000 ft
The ground distance should be based upon when the plane appears on the west horizon and when it disappears on the east horizon.
It should be assumed the track of the plane is perfectly west/east over the viewer.

Imagine you are sitting in a boat surrounded by sea. To your west, you see a plane flying - just appearing at the horizon. The plane is flying at 39000 feet. You watch it for around half an hour - it flies over your head and eventually disappears over the eastern horizon. What is the distance on the ground, that the plane has travelled from, when you first saw it in the west to when you lost sight of it over the eastern horizon?

Answer 1

I assumed that you can see the plane when it's above the plane tangent to your location on Earth. Relative to the center of the Earth, you are at a radius R_e and the plane is at a radius R_p = R_e + 39000 ft. The plane is visible as it traces out an angle theta on each side of you. That angle theta is arccos(R_e/R_p), because that's what puts it at your height or greater in a cartesian plot. The ground path of the plane is equal to 2*R_e*theta, with theta in radians, from circle geometry. The radius of the Earth is 20905470 ft, so theta is .061 radians (or 3.5 degrees) and the ground path length is 2551937 ft, or about 483 miles. Your question made perfect sense and was a lot of fun to work out.

Answer 2

once you come and forth east you're going with the organic cutting-edge of the Earth's rotation because the Earth spins east on it really is axis. as in case you've been going west you would possibly want to be going adversarial to the organic air bypass and there might want to be more desirable resistance on the plane and accordingly taking more desirable time.

Answer 3

you're not making sense, dude.

try rephrasing your question