This was a super-fun question to research. For one thing, it made me go back and re-learn a lot of basic geometry that would have the average 9th grader laughing in my face. I guess that’s not so fun, on second thought. But I did get to learn a lot about satellite orbits, which are really fascinating.
On with the question.
When thinking about how high something would have to be launched to be visible from the opposite coast of the United States, we are really just talking about a more advanced “horizon problem”. Essentially it looks like this:
If you were standing along the line segment h, with your head as the O, your line of sight to the horizon would be the line d. Your actual distance to the horizon would be the arc s, though. When talking about viewing the horizon while standing on the beach, those are almost exactly the same, so the math is easy (horizon = 3 miles for an average person at sea level). But when you are talking about distances like, say, Los Angeles to New York, it gets a bit more complicated.
Let’s say you’re gonna fire a rocket from the Santa Monica Pier and your friend is looking west from Battery Park, waiting to see it. How high will it have to go? Using the equations found here, we can work backwards to figure out the height h above Los Angeles that something will have to attain in order to be seen from NYC.
First we start by assuming that the Earth is a perfect sphere, which it is not. The average radius of this pretend-Earth is 6,378 km (that’s R up above). The land distance from NYC to LA is 3,932 km, give or take a few, since Google Maps insists on using roads (that’s s above). So using my Geometron 5000™ machine, I can solve for h, the height above LA to be seen along the line d from NYC.
Your friend in NYC would see your rocket when it reached 1,440 km above Earth. For comparison, most man-made satellites above Earth are less than 1,000 km up, including the ISS and almost every manned space mission. It’s about the same height as the Globalstar communications satellites, but much, much less than things like GPS satellites (which are in geosynchronous orbit at 35,786 km above sea level). Here’s a diagram of where most things orbiting Earth reside (our pretend rocket would be just outside the inner “ball”):
As for how big it would have to be to be seen? Well, a lot of that depends on how bright it is, how much background light there is, etc. Assuming it was super-reflective, not leaving a trail of fire, your friend was viewing with the naked eye and NYC had the most severe blackout in history … well, this one’s really hard. The human eye has the ability to see a 1.2 meter object at a 1 km distance according to Wikipedia, so along the line d above (4,521 km), the object would need to be 3,768 m wide by my calculations. That’s pretty big. If I screwed that last one up, let me know.
This is awesome and interesting since I work on unmanned spacecraft propulsion systems for a living (including previous work on the GPS program, but with most my experience in the design and launch/mission control of geosynchronous satellites).
There is one change I would make to this however. The GPS satellites operate in MEO or Medium Earth Orbit, which is still much higher than the astronauts go but well below GEO (geostationary orbit). A geosynchronous orbit is an orbit around the Earth with an orbital period of one sidereal day (approximately 23 hours 56 minutes and 4 seconds), matching the Earth’s sidereal rotation period. The sidereal revolution period of the GPS satellites is 11 hours 58 minutes (approximately half a sidereal day). The GPS constellation is made up of 24 satellites (21 plus 3 active spares) located in six orbital planes in almost circular orbits with an altitude of about 20,200 km above the surface of the Earth (lower than the listed 35,786 km for geostationary orbits).
Here is a diagram to help.
has anyone else gotten incredibly turned all by these intelligent men oozing such sexy science (that I by no means understand) all over this post?
I can’t be the only one…
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