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UPDATE: Thanks to Shah, real-life rocket scientist from crookedindifference, for correcting me on the fact that GPS satellites are actually in Medium-Earth Orbit, not geosynchronous or geostationary orbit. Check out his post for more orbital knowledge.
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 They are actually in Medium Earth Orbit). 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. 
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UPDATE: Thanks to Shah, real-life rocket scientist from crookedindifference, for correcting me on the fact that GPS satellites are actually in Medium-Earth Orbit, not geosynchronous or geostationary orbit. Check out his post for more orbital knowledge.

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 They are actually in Medium Earth Orbit). 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. 

    • #science
    • #answer bag
    • #the-gentlemen-are-coming-by
    • #space
    • #orbit
    • #altitude
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    My only question upon reading this is: does it take into account the Earth’s rotation (as the launch would not allow the...
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  29. cats-pajamas reblogged this from crookedindifference and added:
    I just had a conversation (an admittedly less technical conversation, over Skype and wine) about this very thing a few...
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  31. bonedust reblogged this from crookedindifference and added:
    has anyone else gotten incredibly turned all by these intelligent men oozing such sexy science (that I by no means...
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  40. for-all-mankind reblogged this from crookedindifference and added:
    How fascinating indeed! that’s always something I’ve wondered, but I’m also curious how far up/down the East coast...
  41. thecosmosowl reblogged this from crookedindifference and added:
    Oh god i feel like i’m watchin a Science Episode of GunBuster.
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    mt for later
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I'm Joe Hanson, Ph.D. biologist and host/writer of PBS Digital Studios' It's Okay To Be Smart. Check out my "Episode Extras" here. There's a lot of amazing science out there. Let's go discover it together.

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