The whole difference is between a very large, but still finite, flat plate and an infinite one.
A flat steel plate 1E10km on a side by 1E3km thick would have gravity that decresed with height above the surface and which had a sideways vector unless you were hovering over the center.
A flat steel plate of infinite extent by 2 molecules thick would have gravity that doesn’t decrease with height. The thicker it gets, the more the gravity.
Infinity is a tough concept to really grok, as opposed to just compute with, or worse yet, merely bandy about.
Okay… I can understand that the intuitive notions I have from a lifetime of dealing with finite objects can break down when they get to infinities. But that doesn’t mean that I’m able to see why what you say should be, even with infinities, and frankly I’m not ready to take your word for it either.
Damn, infinity makes my brain hurt. I had started making this huge and insightful argument about cutting up the infinite plane into segements one mile square, when I realized that I was setting myself up for the infinite inkeeper’s dilemma and that I shouldn’t try any further on that tack. (The man running a hotel with an infinite number of rooms, each numbered with a natural number, is all booked up when an infinite number of new guests arrive - how does he find rooms for everybody??
He bumps up every current guest into the room with double their old room number, thus moving all current guests into rooms with even room numbers and freeing up the odd rooms for the new incoming guests.)
Any chance you can point me to somewhere that I can try getting into the real meat of this stuff? I’m a math geek, as I’ve already said on this thread, I’m pretty good with my limits and my calculus even.
Would there not be an area close to the centre of the plate where the gravity was effectively indistinguishable from (i.e. within an arbitrarily-small error of, and you get to pick the error) that of an infinite plate?
The argument Chronos made above is a simple application of (the integral form of) Gauss’ Law, which is commonly discussed in EM textbooks (because of its applications to electrostatics) but which holds for other inverse-square forces like Newtonian gravity as well. It’s also discussed (in math contexts) in multivariate calculus courses in relation to divergenceless vector fields. (The Newtonian gravitational field in free space, and the electrostatic field in a charge-free region, are divergenceless.)
Google “Gauss’ law” for lots of web tutorials; you’ll just have to mentally substitute g for E in most of them.
A simple explanation of why an infinitly wide 2d plate with finite thickness produces a gravity field only dependent up altitude above the plate:
Lets use something similar to how you started and divide the plate up into 1 m x 1 m plates with you floating above the center of one of them. Abviously, the plate below you will pull you straight down. The plate to your immediate right will pull you down and to the right and the one to your left will pull you down and to your left. Looking at the vectors we see that the right and left components of these vectors cancel leaving only the sum of the two downward pulls. This is less than twice the pull from the plate below you as the right and left plates are both farther away from you, and only a portion of their gravitational attraction was pointed down.
Now do the plates immediately in front and behind you with the same effect. Due to symmetry we see the total force is straight down and the contribution of outer panels drops off quickly (by 1/[(x^2+h^2)^3/2] where x is the horizontal distance to the panel considered and h is your vertical distance aboe the panel). When this effect is integrated over the 2d plate from negative infinity to positive infinity in both directions we come up with standard Newtonian gravity dependent upon height alone and decreasing by 1/h^2.
Not for the first time, I’ve posted what I thought was a very simple question, with what I hoped was a very simple answer, and got flooded with math I won’t begin to understand for a few more years. (I’m only halfway through high school right now)
I’m not sure if I’m understanding you correctly. Let me try to rephrase the OP:
[Attempt at Clearer OP]
If we ignore the fact the earth is curved, and assume that gravity pulls in the same direction no matter what, we get a parabola. However, if we are extremely realistic, and acknowledge that gravity’s magnitude and direction both change (very slightly) as an object moves, its path is actually an ellipse. So, neglecting air resistance, but not neglecting anything else, any object in free-fall falls in an ellipse. Sure, that ellipse may sometimes look almost exactly like a parabola, but it’s actually not.
Is all that correct, or is there some kind of mistake?
TJdude – You got it. I think your written OP left one one point, but I’m pretty sure you do understand it.
If we assume that gravity pulls in the same direction and the same magnitude everywhere, then (in the absence of other forces such as air resistance) the path is a parabola.
With gravity as a point source (so the direction is always towards the point), getting weaker with distance as inverse square, then a projectile’s path is an ellipse. (And by the way, a perfect sphere acts just like a point source, gravitationally. ) That’s why a planet’s orbit around the Sun is an ellipse.
But for most cases along the lines of throwing baseballs, catapulting pianos, and up to medium-long range artillery, the difference between the ellipse and parabola is very, very small, possibly less than the width of the projectile. And of course air resistance is big, big, huge gigantic in comparison. So using the parabola assumption is well justified.
On the other hand, for ICBMs, the parabola assumption is not well justified. Most of an ICBM’s projectile is in space, so we actually can ignore air resistance, but since the range is significant compared to the size of the Earth, you have to treat it as an ellipse.
[QUOTE=Sunspace]
The real question, of course, is: is it possible to build a monkey-shaped planet? [ul][li]How big can a monkey-shaped planet get before it succumbs to sphericality?[]How is the monkey posed? (This will affect what orbits are possible close to the planet.)[]Is the monkey obese? (This will affect the location of the oceans.)[]Is there vulcansim? Plate tectonics? an atmosphere?[]And most importantly, does the monkey have wings and an attitude of faithful service? (I know I’d like to see a monkey butler 6000 km tall, and since like any planet it will be in orbit somewhere, it will by default be flying.) [/ul] :)[/li][/QUOTE]
I propose we build a space elevator out of carbon nano-fiber “barrel o’ monkey” monkeys, arms all linked together. When extrapolated out to infinity, the monkey equations reduce to a spherical monkey approximation, and solve the planet-sized monkey problem.
When I do it I get that the force is proportional to h/(x[sup]2[/sup]+h[sup]2[/sup])[sup]3/2[/sup]. Integrating x over the whole plane, the result is constant–it does not depend on h. Thus, brute force verifies Chronos’s elegant explanation of why you do not get Newtonian dependence on height.
