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

• B
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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.

Drakkith
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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?

hutchphd
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The parabola requires the assumption of uniform field. Then ##x\sim t## and ##z\sim t^2## so its a parabola. Hardly a myth.

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@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.

weirdoguy
hutchphd
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I believe it is time for a nap.

phinds
PeterDonis
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Summary:: This gets into the same quibbling as many of those other "Math Myths" just posted.
Just posted where?

PeterDonis
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even Newtonian is a myth
Not at all. It's an approximation, but that doesn't make it a myth.

a perfect parabola
Who is claiming a "perfect parabola"? Nobody in this thread, as far as I can tell.

epenguin, Richard R Richard and Dale
PeterDonis
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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.

jim mcnamara
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?

PeroK
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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.

nasu
haushofer
The "myths" here could be called the correspondence principle in a broad sense.

Seems the Flat Earth Society is right after all. Locally.

Dale
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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.

Twigg and nasu
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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.

dRic2, jfmcghee, Twigg and 8 others
etotheipi
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This is PF. We always use GR. Even for inclined planes problems.
we love to see it

jedishrfu
PeterDonis
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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.

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.

hutchphd
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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

DaveC426913
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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.

DrStupid
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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.

PeterDonis
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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.

Staff Emeritus
For what initial conditions?

A baseball a few meters in the air.

Dale
This sounds like a semantic argument, which tends not to be very productive.

Staff Emeritus
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.

nasu, jim mcnamara, hutchphd and 2 others
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.

PeterDonis
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The OP has been sufficiently discussed. Thread closed.

jim mcnamara and Dale