30
$\begingroup$

Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.

$\endgroup$
53
$\begingroup$

The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.

On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.

At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.

$\endgroup$
  • 4
    $\begingroup$ For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction. $\endgroup$ – NLambert Apr 1 '19 at 1:48
  • 3
    $\begingroup$ "This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.) $\endgroup$ – luk32 Apr 1 '19 at 8:42
  • $\begingroup$ For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant $\endgroup$ – slebetman Apr 2 '19 at 4:52
6
$\begingroup$

One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.

A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.

As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.

$\endgroup$
  • 3
    $\begingroup$ To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality. $\endgroup$ – Paul Sinclair Apr 1 '19 at 16:31
  • $\begingroup$ Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you $\endgroup$ – user56930 Apr 1 '19 at 19:47
  • $\begingroup$ What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to. $\endgroup$ – Arthur Apr 2 '19 at 12:11
  • $\begingroup$ To revisit, the problem of going from the mathematics of ellipse to the math of parabola did not come up when landing on the moon (in the movie Hidden Figures) but did come up when landing on the earth. Perhaps the gravitational force of the larger, more massive earth is no longer functioning as a point source. Would this distributed gravitational pull on earth produce a parabola? $\endgroup$ – user56930 Jan 15 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.