Last night I watched an ISS flyover from my backyard - it passed almost directly overhead, reaching a maximum height of something like 88 degrees - which made me wonder:
Say the ISS makes a theoretical pass from my viewpoint from the horizon to directly 90 degrees overhead to the opposite horizon.
At the points at which it crosses the horizon for me (“rising” and “setting”), it would be directly overhead potential observers at other locations - the question is, how far away from me would they be?
Oh! At first, I thought this was just “how far away is the horizon.” But not quite!
Basically, you draw two circles: one is the earth, and the other is the ISS orbit. You draw a tangent line across the inner circle to the outer circle, and then drop verticals from those intersection points to the earth’s surface.
All very easy to do graphically. Using geometry and equations and stuff is harder…
Let O be the center of the Earth, which I’ll assume is spherical. Let U be your location on Earth. Let I be the location of the ISS when it hits the horizon. Then, triangle IOU is a right triangle where:
side OU = the radius of the Earth (3959 miles)
side IO = OU + the height of the ISS’s orbit (3959 miles + 249 miles = 4208 miles)
angle IOU in radians = the distance we want in Earth radii.
Breaking out the simple trigonometry, angle IOU = arccos(OU/IO) ≈ 0.3457… radians. Multiplying that by the radius of the Earth ≈ 1369 miles, a lot further than I would have guessed.
This is true, the adding together of horizons. The map of the satellite’s horizon is also known as its footprint so you can get a pretty good idea by googling for ISS footprint.