Projectile motion while accounting for planetary curvature?

Question

On my last Physics test, there was a problem concerning a cannon firing a shell at 1200 m/s at 50 degrees from the horizontal. I was to find the horizontal distance the shell traveled. What I did is I broke the 1200 m/s into its vertical and horizontal components. Then I used those numbers to figure out how long the shell would be in the air, and used that to figure out how far away the shell would be when it landed. I got the question right.

But, I'm pretty sure there's a serious difference in the answer when you stop assuming that the earth is flat. What I'd like you nifty physics experts to do is take into account the curvature of the earth and the lessened strength of gravity when far away for the earth's surface (you can continue to ignore air resistance). Did the shell clear the earth's pull entirely? Is it in orbit? Did it finally hit the other side of the earth?

What's the real answer to my test question?

Answer 1

Doing a really rough calculation, with a lot of simplifying assumptions, I come up with:

2.2 additional seconds in flight
2.1 additional kilometers traveled (horizontally)
5% maximum reduction in gravity

If you want to see the spreadsheet I used to make these rough calculations, send me a note.

Answer 2

You're right!!!

One of the first concepts of rocketry was firing a cannon in a level shot from a mountain top with such a speed that it fell towards the ground at the same rate that the ground curved away from it. This model doesn't even require figuring out the different gravity since it is always at the same altitude.

In artillery, you get another process. The earth is spinning. If you fire straight up, the shell will not land on you, unless you're at a pole. When it leaves the gun, the shell has the same radial velocity as the gun, but as it rises, that velocity does not cover the same angle as the gun since it is farther from the center of the Earth than the gun.

Also, If you aim at a target to the north or south, it is at a different latitude, and so the target is moving at a different speed than the gun, so you would have to correct for that.

Good question. Great question. Keep asking, Keep learning. Who knows -- you might have a career at JPL. I threw in some reading material.

Answer 3

I've got a feeling that unless you're trying to actually hit something, the difference is negligible due to curvature. Although if the poster above me is right, that's about a 10% increase in the original distance of approximately 18km. (I took the same test, and the actual numbers are a bit fuzzy.)

Answer 4

You need calculus to handle those extra complications accurately, but I can say for sure that 1200 m/s is way too slow for escaping or even reaching orbit. Orbital velocity is about 8000 m/s and escape velocity is over 11000 m/s.

Answer 5

Go right to the source and ask the military. They have real world experience with these factors as well as wind, humidity, barometric pressure, altitude, projectile materials and spin, angular momentum...probably a hundred other variables, some that are undoubtedly classified.