Earth's Curvature in a swimming pool

[A. BETTIK]

I'd have to post this over in Esther's Follies if I tried to actually measure such in a swimming pool, though I suspect it can be detected given the right equipment.

What I did see at the pool today was a kid squirting a foam water gun/cannon into the air. These guns hold a fair amount of water and take a second or two to empty, so a 5-10 foot high arc of water developed and ended over about a two second period before collapsing back to the pool surface.

The classical beauty of the parabola reminded me that even for just a few seconds these drops were following the very faintest pattern of an orbit. To be sure, to actually achieve an orbit would require an elevation in altitude and a boost of about 18,000 mph if I remember correctly.

A two second segment of a low earth orbit as such would be subjected to near the same amount of gravity or bending in path as the water gun. A two second path of a space shuttle would have as much drop as that stream of water from the water gun (albeit by summing the up and down distances).

Dividing 18,000 mph by 60 minutes and then 60 seconds gives about 5 miles per second of low earth orbit path.

Knowing that then allowed me to conclude (perhaps correctly) that the earth curves from Austin to Round Rock about as much as a kid's water gun can squirt a stream of water into the air and back.

 

[Perham1]

Knowing that then allowed me to conclude (perhaps correctly) that the earth curves from Austin to Round Rock about as much as a kid's water gun can squirt a stream of water into the air and back.

I'm not a physics expert by any stretch of the imagination, but how do you account for the expulsive power of the water gun? Suppose gun A can eject the water to a height of 10 feet and gun B to a height of 30 feet? How does this show the curve of the earth? Is the curve 10 feet? 30 feet? Neither?

I don't think the "up and down" is the way to do this. I've heard that if you shoot a rifle (.22, say) (holding the gun level, ground is also level) and drop a bullet at the same time, both bullets hit the ground at the same time. This seems to be more of a gravity thing. I have no idea what it has to do with measuring the curvature of the earth.

Interesting post. Thanks for making it. Coincidentally, I'm re-reading a book on Richard Feynman. I don't understand any of the physics there, either.

 

[Bluepies]

Things in orbit are technically in free-fall, still affected by the Earth's pull of gravity. One way to think of an object in orbit is that it's continuously falling, but its horizontal velocity is so great that the Earth's curvature falls away from the object at the same rate that the object actually falls.

I'm at work and don't have time to see if the OP's numbers (or methods) check out, but at first glance, it looks right, in theory.

Just FYI, I used to teach high school physics. Not exactly a physics genius, but I do know a thing or two.

 

[texas_ex2000]

If there was no atmosphere to drag you down, and you had enough power to get you to orbital velocity, you could orbit the earth at sea level (okay maybe a little higher to account for mountains and such).

 

[Lake_Travis_Horn]

Perham's statement about the bullets is correct, if you neglect any effect of the air (giving the shot bullet lift, not slowing it down). The both start out with a vertical velocity of zero, although the shot bullet has a substantial horizontal velocity. Gravity acts the same way on both bullets, accelerating them downward at the same rate so they'll hit the ground at the same time.

 

[Statalyzer]

In reply to:

---

---