# Are free-fall 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 free-fall 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?

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.

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.

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?

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 neat-n-tidy 2nd order equation.

Somewhat off-topic 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.

If you are neglecting curvature of the earth, why does the direction “gravity pulls” change? You’re changing the assumptions mid-problem. With a “flat Earth” assumption, gravity is always directed downwards, thus the solution to the 2nd order differential equation is a parabola.

Well said.

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??

Hmmmm. (struggles for a frame of reference.) So, about 1/3 of the height required for stable orbit above the earth, or 7 times as high as Mt Everest?

Acceleration due to gravity is proportional to 1/r[sup]2[/sup]. The radius of the earth is r~6x10[sup]6[/sup]m. So for a change of 1% you want a height h so:

1/r[sup]2[/sup] * 0.99= 1/(r+h)[sup]2[/sup]
1.01 ~ (1+h/r)[sup]2[/sup]
1.01 ~ 1+(h/r)[sup]2[/sup]
0.01 ~ (h/r)[sup]2[/sup]
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

Actually, no. Suppose you’re on a planar planet with an odd force of gravity that goes exactly to zero when you get above a certain height h. If you manage to fire your projectile to it gets up to that height h, it’ll no longer be experiencing any forces and continue in a straight line. Its trajectory was, however, curving beforehand, and I don’t know of any parabolas that start out curvy and go straight at the end.

In general, you can solve the second-order differential equation d[sup]2[/sup]z/dt[sup]2[/sup] = -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[sub]1[/sub])[sup]-1/2[/sup] dz) + c[sub]2[/sub]

where c[sub]1[/sub] and c[sub]2[/sub] 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[sub]1[/sub]) + c[sub]2[/sub]

or, rearranging,

z = - g (t - c[sub]2[/sub])[sup]2[/sup]/2 + c[sub]1[/sub]/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.

Not quite. By the binomial approximation (1 + h/r)[sup]2[/sup] ~ 1 + 2h/r, so h/r ~ 0.005 and the height you’re looking for is about 30 km. Still pretty high, though.

And suppose that our planet was monkey-shaped. I wasn’t trying to provide an answer that would cover 100% of possibilities, I was just pointing out a discrepancy in the OP’s problem statement that made it self-contradictory.

Actually, your factors 2 and 3 are just different aspects of the same thing. If one approximation is good, then the other will be as well. Even if you look at the extreme (and unphysical) case where those aren’t approximations, they’re still tied together. If you had a “world” consisting of a slab of matter extending infinitely in two dimensions, but finite in the third, (a completely “flat Earth”), then you would also have a gravity independent of height above that slab. Likewise, to have gravity truly unchanging with height, your world would have to be such an infinite slab.

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.

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?

I don’t see that either of those statements is accurate according to Newton’s law of gravitation. All of this is dependent on how thick the ‘slab’ is (and what’s underneath it of course,) but gravity would still diminish with distance from mass.

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 day-to-day existence, or © 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 flat-world 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 flat-world 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.)

Granted, but motion of a particle in an arbitrary force field is a perfectly well-posed physical problem, and if the force field has enough symmetry (as in this case) it’s perfectly soluble. I merely posited this example to show that just because things are translationally symmetric doesn’t necessarily mean you get parabolas, which seemed to me to be a misconception on av8rmike’s part. (Although now I see that wasn’t what he meant - mea culpa.)

You’ll have to forgive a theoretical physicist for thinking about interesting problems that don’t necessarily correspond to the real world, I guess.

Actually no. The easiest way to see this is by drawing field lines (a technique most often used with electromagnetism, but equally applicable to gravity). If I have an infinite slab, then it’s completely symmetric about any given point. So a field line drawn from that point must go straight out from the slab, normal to the surface. And just as electric field lines can only end at charges or at infinity, so too gravitational field lines can only end at masses or at infinity. Since the lines are all parallel and never ending, the density of lines is constant, and the density of lines is proportional to the field strength, so the field strength is constant. The same occurs for the corresponding electrostatic problem, where you have an infinite slab of charge (a reasonable approximation for the plates of a simple capacitor).

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.

The real question, of course, is: is it possible to build a monkey-shaped planet? [ul][]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]

Hmmmm… well, I’m willing to admit I might be wrong – but as a self-professed math nerd, it’s not making any sense to me yet.