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Doesn't the acceleration vector points towards the center of the Earth and not just downwards along an axis vector. I know that the acceleration vector's essentially acting downwards for small vertical and horizontal displacements but if the parametrization of projectile motion doesn't trace out a parabola, what is the shape of projectile motion?

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    $\begingroup$ Hint: all projectiles are essentially satellites just like the moon and the International Space Station. The reason a bullet out of a revolver does not stay up is because it's orbit intersects with the ground (I'm ignoring friction with air of course) $\endgroup$
    – slebetman
    Commented Feb 2, 2021 at 22:54
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    $\begingroup$ By this reasoning, we should also say that cows aren't spheres and air resistance isn't always negligible. $\endgroup$
    – Sandejo
    Commented Feb 2, 2021 at 23:56
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    $\begingroup$ Note that it's perfectly reasonable to work from a reference relative to the Earth's surface, just like how we can work from a rotating frame of reference when discussing centripetal/fugal forces. As far as I'm aware, the projectile motion example doesn't specify that it's working from a reference that separately accounts for the Earth's curvature. $\endgroup$
    – Flater
    Commented Feb 3, 2021 at 13:20
  • $\begingroup$ Technically they are all ellipses, but because the Earth is so large we assume the ground is perfectly flat and so the parabola is a spherical-cow approximation. $\endgroup$
    – KingLogic
    Commented Feb 3, 2021 at 22:17
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    $\begingroup$ Does this answer your question? Elliptical Trajectory, or Parabolic? $\endgroup$
    – Quillo
    Commented Jan 28, 2023 at 14:26

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Just as the motion of body around the earth is ellipse (1st Kepler law replacing sun by earth), so is the motion of a projectile. Notice that almost everything we deal is an approximation, the earth is not a massive perfectly rounded ball, and we are neglecting the air, so it is not a sin to consider the motion of the projectile as a parabola (at least in our everyday experience as, for example, throw a stone in the river)

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    $\begingroup$ "Better to calculate an approximation than nothing at all" - Albert Einstein 2020 $\endgroup$ Commented Feb 3, 2021 at 10:48
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    $\begingroup$ @AccidentalTaylorExpansion "Don't believe every quote you see on the internet" - Abraham Lincoln $\endgroup$
    – DKNguyen
    Commented Feb 3, 2021 at 16:54
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    $\begingroup$ @DKNguyen , I advise you to read that quote yourself because there weren't any internet during his lifetime. $\endgroup$
    – lee
    Commented Feb 4, 2021 at 16:06
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    $\begingroup$ @lee whoosh...and I advise you to determine whether Einstein has any quotes made last year. $\endgroup$
    – DKNguyen
    Commented Feb 4, 2021 at 16:31
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Assume Galilean relativity and Newtonian gravity. Neglect the drag due to the atmosphere. The gravitational field of the Earth is the same as the one produced by a point particle in its center (with the same mass, the usual $1/r^2$ gravitational force field). Now, you may know that a test particle in this $1/r^2$ force field of the Earth can have different orbits (closed or open, depending on the initial velocity and the initial position). Leave out open orbits, which means that you are shooting the projectile at infinity. All other orbits are ellipses. However, the Earth is not a point and has a finite radius: some of those ellipses (starting at the Earth surface) will intersect at later times the Earth surface again.

Why do you have parabolas in the "simple" setting you are describing? Because you can always approximate locally an ellipse with a parabola. See e.g. Can a very small portion of an ellipse be a parabola?

Edit: yes, in a spatially constant and uniform gravitational field, a parabola is exact. Other answers can be found in these "duplicate" posts: Elliptical Trajectory, or Parabolic? and Why does the Earth follow an elliptical trajectory rather than a parabolic one?

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    $\begingroup$ On a flat Earth, a parabola isn't an approximation, it's the exact solution. $\endgroup$
    – Mark
    Commented Feb 4, 2021 at 2:28
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    $\begingroup$ @Mark Is it, though? :) $\endgroup$ Commented Feb 4, 2021 at 5:52
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    $\begingroup$ In a vacuum, if the flat surface is of infinite horizontal extent, and there are no local density anomalies such as bodies of water... $\endgroup$
    – Ryan C
    Commented Feb 4, 2021 at 10:23
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    $\begingroup$ @FilipMilovanović, true, a flat Earth isn't sufficient. It needs to be an infinite featureless plain. $\endgroup$
    – Mark
    Commented Feb 4, 2021 at 20:52
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The other answers have already mentioned the approximations related to gravity, but there is more to that when discussing a projectile motion:

  • For example, one typically neglects friction, and if one were to include friction, the choice of the formula for the friction term would be an approximation for a particular situation (proportional to velocity or its square or cube), and the friction coefficient will be an approximation, since it actually depends on the shape of the projectile.
  • One could then also take into account that the projectile has finite size, and that it may rotate in 3 dimensions while flying.

One could go on further, as much as one's knowledge of the situation allows. The art of doing physics is to a large extent one's ability to develop models closer or less approximating real situations, capturing the most essential details, and solving them mathematically.

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Also the earth is rotating and is a non-inertial (accelerating) reference frame. For long-range projectiles the effect of the Coriolis force is important. Some of the earliest numerical simulations using computers were calculations of long-range projectile motion.

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It is an approximation, as everything in physics is an approximation based on mathematical/statistical modeling. And by definition, a model is an imperfect representation of reality. Models employ simplifying assumptions in order to make problems tractable so we can explain what is happening and hopefully make predictions. How good your explanation/predictions are depends on 1)how good the model is and 2)how close your situation is to the assumptions made by the model.

In the case of projectile motion, the standard equations work quite well if you throw a small dense object inside a windless room. It's a much different story if you throw a light big object during a windy day. But no matter what, the model will never exactly explain what is going on because reality never exactly matches the assumptions of a model.

Paraphrasing a quote often attributed to the statistician George Box "all models are wrong, some are useful". For projectile motion, you lose accuracy because the earth is not a perfect sphere with uniform gravitation, there's no such thing as a perfect vacuum , etc etc etc. Because of these things there is no exact shape that describes such motion, much like there is no shape that exactly describes orbital motion. Parabolas and ellipses are very close, but as you suspected they will always be approximations.

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Projectile motion is technically an approximation but it's better to think about it as a model that works perfectly at a certain scale or within a certain regime, as this is basically true in the way we understand the universe across all fields of science. Most of the time we're only concerned with making predictions in specific regimes which allows us to ignore many of the intricacies and complexities across scales or processes.

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