Myth: When you toss something, air aside, it follows a parabolic path

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SUMMARY

The discussion centers on the trajectory of objects in motion, specifically addressing the claim that they follow a parabolic path. It is established that while a perfectly parabolic path is theoretically possible under uniform gravitational conditions, in practice, objects typically follow elliptical trajectories unless launched at escape velocity. The conversation references Newtonian physics and the implications of gravitational variations, emphasizing that real-world conditions often lead to deviations from idealized models. The participants agree that the distinction between parabolic and elliptical paths is minimal for practical scenarios, particularly with objects like baseballs.

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  • Understanding of Newtonian Physics and gravitational forces
  • Familiarity with conic sections, specifically parabolas and ellipses
  • Knowledge of escape velocity and its implications on orbital mechanics
  • Basic grasp of classical mechanics and motion equations
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  • Research the concept of escape velocity and its role in orbital mechanics
  • Study the differences between parabolic and elliptical trajectories in physics
  • Explore the effects of gravitational variations on object motion
  • Learn about the correspondence principle in physics and its applications
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Students of physics, educators in classical mechanics, and anyone interested in the nuances of motion and gravitational effects will benefit from this discussion.

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TL;DR
This gets into the same quibbling as many of those other "Math Myths" just posted. Even when atmospheric effect are avoided, the parabola will not precisely be the most common conic section.
Actually, the path can be precisely parabolic, but you will need quite an arm to get that result.
More commonly, the path will be a section of an ellipse - one that terminate when it hits the ground.
In general, it will be a segment of one of the conic sections.
 
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I believe you have to throw the object at escape velocity in order for its path to be parabolic, correct? Or has my Kerbal Space Program experience done me wrong here?
 
The parabola requires the assumption of uniform field. Then ##x\sim t## and ##z\sim t^2## so its a parabola. Hardly a myth.
 
@Drakkith Yes, given Newtonian Physics, escape velocity yields a parabola, less than that yields an ellipse, more than that yields a hyperbola.

Of course, even Newtonian is a myth - so a perfect parabola would be equally mythical.
 
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I believe it is time for a nap.
 
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.Scott said:
Summary:: This gets into the same quibbling as many of those other "Math Myths" just posted.
Just posted where?
 
.Scott said:
even Newtonian is a myth
Not at all. It's an approximation, but that doesn't make it a myth.

.Scott said:
a perfect parabola
Who is claiming a "perfect parabola"? Nobody in this thread, as far as I can tell.
 
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PeterDonis said:
Just posted where?
Never mind, I see the just posted threads in the General Math forum. However, this particular thread is not about math, it's about physics, so I am moving it to the Classical Physics forum.
 
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.Scott said:
Actually, the path can be precisely parabolic, but you will need quite an arm to get that result.
More commonly, the path will be a section of an ellipse - one that terminate when it hits the ground.
Is that just a guess or can you show that the path would always be more elliptic than parabolic?
 
  • #10
.Scott said:
Summary:: This gets into the same quibbling as many of those other "Math Myths" just posted. Even when atmospheric effect are avoided, the parabola will not precisely be the most common conic section.

Actually, the path can be precisely parabolic, but you will need quite an arm to get that result.
More commonly, the path will be a section of an ellipse - one that terminate when it hits the ground.
In general, it will be a segment of one of the conic sections.
If you object to someone considering the local gravitational field over short distance near the surface of the Earth to be constant (which produces a parabola), then they may object to your assuming the Earth is a perfect sphere without any surface irregularities, natural or human-made. And to ignoring the gravitational influence of the Moon, for example.

They might also object to your using Newtonian gravity, as opposed to GR.
 
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  • #11
The "myths" here could be called the correspondence principle in a broad sense.

Seems the Flat Earth Society is right after all. Locally.😎
 
  • #12
Any well behaved function can be approximated to arbitrary precision as a parabola over a small region. So the question isn’t if it is a parabola, it is how large of a parabola is it.
 
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  • #13
PeroK said:
They might also object to your using Newtonian gravity, as opposed to GR.
This is PF. We always use GR. Even for inclined planes problems.
 
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  • #14
Vanadium 50 said:
This is PF. We always use GR. Even for inclined planes problems.
we love to see it 👏😄
 
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  • #15
DrStupid said:
Is that just a guess or can you show that the path would always be more elliptic than parabolic?
If the object is launched at less than escape velocity, it will be in an elliptical orbit with the center of the Earth at one focus. That is what the OP is saying in what you quoted. Are you disputing that statement? If so, you are the one who needs to show your work, since the statement is a well known fact.
 
  • #16
PeterDonis said:
If the object is launched at less than escape velocity, it will be in an elliptical orbit with the center of the Earth at one focus. That is what the OP is saying in what you quoted. Are you disputing that statement? If so, you are the one who needs to show your work, since the statement is a well known fact.
It is a known fact for point masses or at least for spherical symmetric mass distributions, but not for objects in the local gravity field of Earth.
 
  • #17
Are we counting angels here? Arguing about the shape of the head of the pin? Ideological purity? Any calculation is either good enough or not good enough. They can always be made better. Kudos to @Vanadium 50
 
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  • #18
DrStupid said:
It is a known fact for point masses or at least for spherical symmetric mass distributions, but not for objects in the local gravity field of Earth.
If you mean that the Earth is not perfectly spherically symmetric, yes, that means that orbits are not exact ellipses, but it doesn't mean they are closer to parabolas than ellipses.
 
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  • #19
hutchphd said:
Are we counting angels here?

Yes.

The deviation between an ellipse and a parabola is something like a micron. That's less than the head of a pin by three orders of magnitude.
 
  • #20
Vanadium 50 said:
The deviation between an ellipse and a parabola is something like a micron.
For what initial conditions? Obviously the deviation will vary according to the initial conditions.
 
  • #21
PeterDonis said:
For what initial conditions?

A baseball a few meters in the air.
 
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  • #22
This sounds like a semantic argument, which tends not to be very productive.
 
  • #23
1. As mentioned, the difference between a parabola and an ellipse is tiny compared to the length scales involved. (Assuming we are talking about baseballs and not rockets)

2. "Parabola" and "ellipse" both are idealizations. Deviations from these approximations are large compared to the difference between parabola and ellipse. How do I know? The difference between parabola and ellipse is the same as the difference between constant gravity and 1/r2 gravity. And that is really hard to measure - you need a tower in the desert far away from anything to do that.

3. Physics is not about The Truth. It's about models, and comparing those models to data. Part of learning physics is learning which approximations to use when. The approximation that 1/r2 is constant over the scales of the problem is often a good approximation. We shouldn't be afraid to teach it because there are other approximations one could make, approximations that are often no better.
 
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  • #24
Ultimate quibbling.

Quantum mechanically, the baseball does not have a definite path.
The Earth rotates, so the path in a Earth fixed frame has deflections eastward or westward
The Earth's gravity departs from inverse square,
Relativitistic effects are also present

This can go on and on. I agree with the post above. Pay attention to the models.
 
  • #25
The OP has been sufficiently discussed. Thread closed.
 
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