While fishing off the coast of Florida, I noticed that eventually, I couldn’t see any land or any of the skyscrapers. I realize that this is in fact due to the curvature of the Earth, but I was told that I could only see 7 miles away due to this curvature. Is this true? If not, where did it come from?
It depends on how tall you are. One of the links in the above google search gives 1.17*sqrt(your height in feet)=distance to horizon in nautical miles. For an eye height of 6 foot that gives a horizon of about 2.8 nm or close to 3 statute miles. Not very far at all. Of course if you’re in an aeroplane at 35000 feet it’s more like 220 nautical miles.
The equation should be:
Earth circumference * acos(earth radius / (earth radius + observer’s height)) / 2PI
Or, more simply:
Earth radius * acos(earth radius / (earth radius + observer’s height))
Plugging some real values (in metres) in there:
6,378,136.6 * acos(6,378,136.6 / (6,378,136.6 + 1.8)) = 4,791.8 m, or about 3 miles.
But, of course, you have to account for the height above the Earth’s surface of whatever you’re looking at. For a 6ft tall person looking at a 6ft tall object, you can double it. Looking at a skyscraper from out at sea, you’ll be able to see it from much further away.
E.g. in clear conditions you can easily see across the Straits of Dover (about 20 miles) from the French beaches to the high cliffs of England.
If you read old books about naval adventures, this is why they were always climbing to the topmost eminence in the ship – a lookout 60 or 70 feet above deck could see the 60 or 70 foot mast of an opposing ship from considerably further away than I could see you standing on a vast open prairie, for instance.
I believe you can accomplish that easily enough by applying the equations given above twice (once for your height, once for the other object’s height) and adding the two numbers together.
From my house, I can see a mountain that’s almost 200 km away. At those distances, however, you probably also need to account for atmospheric refraction effects.
Yes, if what you’re looking for is tall, then you add your distance to the horizon plus their distance to the horizon.
By the way, the simpler formula for distance to the horizon is
distance to horizon = square root of (observer’s height * radius of the earth)
So if you’re four times as high as your buddy, you should see twice as far.
This is all theoretical of course, and doesn’t account for waves and refraction.
Was it recently? We’re going through a very hot, humid spell here right now (temp this morning at 5 am was 85 F with dew point at 79 F) and you seem to lose visibility due to this. Not sure exactly how this works from a science standpoint or if it’s just in my head. But cold dry air seems to allow a longer field of vision* than hot, humid air.
I’m sure some of the SDSAB geniuses will be in to correct me if I’m wrong.
*If that is the right term.
A geometric mean. 
That’s nothing. From my house, I can see a galaxy that’s almost 24 quintillion km away.
“Yes, Lord?”
“Peter…I can see my house from here!”
oh, man… 
<backing away from the inevitable lightening strike>
Not when it’s over the horizon you can’t! 
Sure I can! It’s just when it’s under the horizon that’s a problem.
When I was a kid and we’d take vacations in seaside hotels, my dad would make me do the math on how far we could see from our balcony on the whateverith floor.
What really drove me freaking nuts, though, is when I’d ask how much farther it was and he’d say, “Well, we just passed the 43 mile marker, and we’re going 65 miles an hour.” I thought everybody’s dad was a math asshole.
Do “Mathletes” grow up to be “Matholes”?
But can he see Russia from his house?