Built on Facts

Elliptical Arguments

Here, straight from the Wikipedia article, is a lovely picture of a basketball in a free-flight trajectory.

i-561201805a894940d18cb97d73579279-flight.png

You probably expect a parabolic trajectory, and we do get pretty close. There are some deviations. The resistance of the atmosphere is the largest, and the rotation of the ball will itself result in aerodynamic effects that distort the flight of the ball from its idealized trajectory.

But in fact even in a perfect vacuum with no external forces but gravity we still won’t get a parabola. We’ll get a section of an ellipse.

Why? Newton’s laws tell us that if you’re in the gravitational field of a spherically symmetric object like the earth, it’s mathematically identical to a situation with an orbit about a point mass located at the center of that object. Without air resistance or other perturbations, as far as that basketball is concerned it’s in a long, thin, very eccentric orbit about the center of the earth. In a very, very exaggerated visualization of the effect the actual trajectory will look like the lower one here and the parabolic trajectory is the upper one (arbitrary units):

i-5a20b43cbc8bc47ad36d989e10cd1512-ellipse.png

Let’s try to put some numbers to how big this effect will be. The shape of an ellipse is characterized by its eccentricity. An ellipse with an eccentricity of 0 is a circle, and it gradually gets more and more squashed as it approaches the maximum possible value of 1. At 1 and beyond it’s no longer a closed curve; it’s a parabola or hyperbola. The equation for the eccentricity of an ellipse in central force motion is

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Plugging in the values for E, L and k, I find that for a normal gravitational potential the eccentricity is:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

That’s after dropping terms in the fourth or higher powers of v, but for velocities lower than the kilometer per second range this is a fine approximation. Plugging in a test figure of v = 10 m/s, I see that we get an eccentricity of about e = 0.9999984. Pretty eccentric. And since e = 1 corresponds to a parabola exactly, it follows that the trajectory we see will in fact be a parabola to a very close approximation. How close? My rough order-of-magnitude estimate with these numbers is that the deviation of the ellipse from the parabola will be nanometers at typical basketball trajectories. Almost every other effect from the moon to local gravitational variations will probably swamp that so it’s probably permanently beyond the capabilities of tabletop experiments.

Still, interesting to think about!

[Exam report from last week: These are estimates as they haven't been passed back yet, but I think my classical mechanics exam went very well, but my E&M II exam was probably pretty sketchy. I can live with that though, there's always the final.]

Comments

  1. #1 Chip Gentry
    November 17, 2008

    I once tried to convince an news science reporter that the NASA training plane “Vomet Comet” trajectory was an ellipse that is very nearly a parabola instead of a true parabola. I never got him to concede.

    Thanks for this presentation, which is much clearer than the one I used with the reporter.

    Chip G.

  2. #2 Uncle Al
    November 17, 2008

    http://www.mazepath.com/uncleal/tettet2.png
    When approximation fails… (x,y,z),

    1.146388 0.4067684 0.3201515
    0.04216192 1.348825 0.1133035
    1.043776 1.151007 -0.9389371
    1.407833 1.849292 0.2978272

    4.817326 0.4826342 0.06582354
    4.714714 1.226873 -1.193265
    4.453269 -0.2156502 -1.170941
    5.81894 0.2848162 -0.9864171

    Calculate the centers of mass gravitation, (4)(4)/r^2. Calculate all (8)(8-1)/2 = 28 unique handshakes and sum their forces parallel to the centers of mass line, respecting direction. The answers are different.

    Theory derived from convenience is wrong, but is it good enough? The Equivalence Principle is convenient. Perform a parity Eotvos experiment, left-handed versus right-handed single crystal quartz test masses. Self-similar opposite chirality mass distributions do not average – they sum. GR might be insufficiently correct.

  3. #3 Mark Dow
    November 17, 2008

    It seems like you should mention the approximations used in typical calculus or classical mechanics class that lead to this discrepancy:

    - The direction of gravitational force (toward the center of the earth) changes as the ball moves horizontally.

    - The magnitude of the gravitational force drops with height, proportional to the square of the separation.

  4. #4 Tim Gaede
    November 18, 2008

    #1 (Chip Gentry)

    Yep. It’s an ellipse alright. High school physics courses have a derivation of the parabolic trajectory based on the assumption of uniform gravity. Such a derivation is within the grasp of novice physics students and emphasizes the concept of vector components. A derivation of the elliptical shape is presumably far more difficult and as Matt Springer pointed out, deviates only marginally from a parabola for short distances. The eccentricity for the aircraft seems to be about 0.9996, which would not deviate from a parabola by more than few feet over the 25 seconds of simulated weightlessness.

    However, I think that recognizing the trajectory of the “vomit comet” as being an ellipse is important because it is essentially doing what orbiting craft do. The net force on the airplane is due to gravity as thrust cancels out lift & drag.

    One more misconception about the training plane is that it needs to be going down to simulate weightlessness.

  5. #5 Tim Gaede
    November 18, 2008

    Hey Matt,

    You forgot to take the square root:

    sqrt(0.9999984) = 0.9999992

  6. #6 Mermer
    March 25, 2009

    Calculate the centers of mass gravitation, (4)(4)/r^2. Calculate all (8)(8-1)/2 = 28 unique handshakes and sum their forces parallel to the centers of mass line, respecting direction. The answers are different.

    Theory derived from convenience is wrong, but is it good enough? The Equivalence Principle is convenient. Perform a parity Eotvos experiment, left-handed versus right-handed single crystal quartz test masses. Self-similar opposite chirality mass distributions do not average – they sum. GR might be insufficiently correct.

  7. #7 Weslo
    August 26, 2010

    Thanks for the post! I play basketball all the time, and I had no idea it wouldn’t be perfectly parabolic.

  8. #8 Proform
    October 18, 2010

    Oh that’s just neat. Love it.

  9. #9 Jeff Smythe
    January 18, 2011

    I totally agree with all that. But of course I don’t have a PhD or anything of the sort – I’m wondering what the use of this knowledge would be but for space travel?

    You mention that the trajectory varies in nanometers…Fun experiment, but practically a NASA issue.

  10. #10 Jim B.
    January 18, 2011

    How funny! I remember studying this in Trig YEARS ago, and we were told it was always a parabolic arc.

    (Can I get you to convince my professors I was right all along? lol)

  11. #11 Bones J.
    March 15, 2011

    Funny this post takes me back to college. I studied fitness and physics and I was looking to reinvent arc training but now I just work at a gym.

  12. #12 ball valves
    March 27, 2011

    The solid truth about this blog article is, that it spoke to me deeply. Thank you for sharing your thoughts and concerns.