#1




Are freefall trajectories REALLY parabolas?
I understand that for everyday calculations where the curvature of the earth and air resistance are negligible, it makes sense to say that an object in freefall traces out a parabola (or straight line), but is it REALLY a parabola, or is it actually a small piece of an ellipse, since the direction which gravity pulls changes slightly with the horizontal motion of the object?

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#2




I think you are entirely correct. If you ignore air resistance, and assume the earth is flat, you get a parabola. This is what you normally do, because the error is normally exceptionally negligile.
If you ignore air resistance, and assume the object is attracted to the centre of the earth inversely proportional to the distance between them (which is what actually happens except for nonuniformities in the earth), you get an ellipse. If you try incorporate air resistance, it's a whole 'nother kettle of doves. 
#3




As mentioned by Shade, air resistance will mess up the parabola. I’m also pretty certain that the gravitational force must be constant with elevation in order to produce a perfect parabola. This is not the case in real life.

#4




So, sounds to me like there are three different factors that you have to simplify:
 air resistance. Nobody seems quite sure how to account for this in all circumstances  it's a wild card.  force of gravity changing in intensity with elevation/distance from the center of the earth.  force of gravity changing in direction slightly as you move sideways  this would be the flat earth versus round earth factor. So, if you keep air resistance out and change the other two to reflect reality more closely, your parabola turns into an ellipse segment. What if you only change one of the second two factors... either a round earth whose gravity is unaffected by distance, (supposedly,) or a flat earth that has a differential effect of gravity at different altitudes. Would a projectile trajectory describe a parabola or an ellipse under these conditions, or something different? 
#5




I would wager that, for most common projectiles (e.g. bullet from a gun), air resistance is by far the predominate factor that skews the neatntidy 2nd order equation.

#6




Somewhat offtopic here, I remember that in the old 'tank wars' computer game, wind speed was treated as a constant sideways acceleration vector to any projectile, no matter whether there was anything that would tend to shield it from the wind or not. Somewhat simplistic, but hey.
The interesting thing about that was that it didn't keep the trajectory arc from describing a parabola, at least with relatively low wind speeds. All that the wind factor would do was skew the axis of the parabola so that it was no longer straight up and down, (and sometimes affect how wide or narrow the parabola was, as well.) Basically, wind and gravity combined to create a single force acting on the projectile, at an angle, depending on how strong the wind was compared to gravity. 
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OP didn't mention the third factor, 'intensity of pull,' but any changes in that will also be negligible when the other two are, I would think. How far up above the surface of the earth do you need to get for gravitational pull to go down 2%, say?? 
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#11




Acceleration due to gravity is proportional to 1/r^{2}. The radius of the earth is r~6x10^{6}m. So for a change of 1% you want a height h so:
1/r^{2} * 0.99= 1/(r+h)^{2} 1.01 ~ (1+h/r)^{2} 1.01 ~ 1+(h/r)^{2} 0.01 ~ (h/r)^{2} h ~ 0.1 * r Or a height of about 600km. Did I get that right? If so that's really quite astonishingly negligible. And sideways is even less negligible, since drawing a diagram shows moving sideways is the same as moving up very little. So yes, a parabola is a very good approximation, ignoring air resistance. With air resistance, it might not fall at all, witness a sheet of paper 
#12




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In general, you can solve the secondorder differential equation d^{2}z/dt^{2} = f(z) for an arbitrary function f(z) to yield a pretty messy general result (via quadrature): t(z) = Integral( (2 Integral(f(z) dz) + c_{1})^{1/2} dz) + c_{2} where c_{1} and c_{2} are constants of integration and are determined by the initial position and velocity (or whatever other boundary conditions you care to impose.) Now, if f(z) = g for all z, then you can solve the integrals in closed form and the answer becomes t = Sqrt(2 g z + c_{1}) + c_{2} or, rearranging, z =  g (t  c_{2})^{2}/2 + c_{1}/2g which is certainly a parabola. However, this result depends pretty critically on the acceleration being a constant for all values of z, and for an arbitrary f(z) you won't get a parabola. Quote:

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MikeS's discussion of an "odd force of gravity" which goes completely to zero is inconsistent with any known model of gravity and therefore can be dismissed as unphysical. By comparison, av8rmike's hypothetical monkey planet is conceivable, but complicated and unlikely enough that one need not address it.
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Time travels in divers paces with divers persons. As You Like It, III:ii:328 
#15




So, not bothering with the math, what you guys are really saying is that the path of a projectile can be approximated very accurately by either a parabola or an elipse segment. Or, more to the point, parabola segments can be very accurately approximated by elipse segments. Right?

#16




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The way I see it, the 'flat earth gravity constant' picture is a simplification, a model of the world that misrepresents reality in several ways, but is useful because these misrepresentations represent either (a) variables that we cannot accurately quantify or predict, (b) factors that are small enough to be negligible over the scales of our daytoday existence, or (c) both of the above. Conceptually, I say that those two misrepresentations of gravity/earth are interrelated, but seperable. You can build a model where the surface is flat but the force of gravity decreases with 'height' in a similar fashion to the observable decrease of gravity with distance from the surface of the earth. You can build a model with a round planet whose gravity does not decrease at all no matter how far away you get from it. You can build a model, either round or flat, where gravity is 'on or off' based on a particular altitude threshold. NONE of these models are consistent with newton's gravitation. Neither is flatworld gravity constant. They're all interesting imaginary realms to play in, and to varying degrees interesting bases on which to set up computer simulations of flying objects and trajectories. The only unique feature about flatworld gravity constant is that it has a strong claim to being the simplest out of all of these models. If you want a set of parameters that reliably describe the real world, and that is consistent with Newton's gravitation, then you set up a model with a round earth and variable gravity. Depending on how closely you want it to model the real world, you set up air resistance, weather patterns, radar stations, air force interceptors... you get the idea. PS to jawdirk on preview: as the 'other end' of an ellipse gets further and further away, the section of it close to you comes to resemble a parabola more and more. A parabola can be considered the limit of an infinite series of ellipses. (I hope I got that right.) 
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You'll have to forgive a theoretical physicist for thinking about interesting problems that don't necessarily correspond to the real world, I guess. 
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Another way to look at it is that the further away from the slab you are, the larger a portion of the slab is significant. So even though the force from any little piece of the slab is less, the total force remains the same. And MikeS, I was perhaps being too hard on you. Yes, that's a perfectly valid force field, and could be realized electromagnetically, for instance, by an appropriate distribution of charged plates (as long as you don't mind your particle penetrating your plates). But gravity it's not. 
#19




The real question, of course, is: is it possible to build a monkeyshaped planet?
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Rigardu, kaj vi ekvidos. Look, and you will begin to see. 
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#21




Chrisk,
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. 
#22




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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. 
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Google "Gauss' law" for lots of web tutorials; you'll just have to mentally substitute g for E in most of them. 
#25




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. 
#26




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)
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[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 freefall 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? [AaCOP] 
#27




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 mediumlong 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. 
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