It's a parabola in a uniform gravity field, an ellipse in a circular gravity field coming from a point mass.
So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density, but it isn't. In reality it's a potato like mass blob that's approximated by what geodesists call the "geoid". So in order of approximations the path of a ball thrown on earth is a parabola -> ellipse -> numerical integration of 6-dof initial conditions and spherical harmonics approximation of the earth gravity field.
> So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass.
This turns out to be not pedantic but very important if you're guiding an ICBM. And when landing on the Moon, Apollo had to deal with irregularities in the Moon's gravity due to mass concentrations, called mascons.
(If you're interested in missile guidance, take a look at the book Inventing Accuracy. Among other things, it discusses some of the efforts to map the Earth's gravity field to increase missile accuracy for Trident and Minuteman missiles. I knew a physicist who worked on this.)
(Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull.)
It wouldn't be an ellipse even if Earth were a point mass. The gravity of the Moon and Sun, the gravitational lumpiness in the sky, has the same effect as the gravitational lumpiness underground. The combined result may be no closer to an ellipse than it is to a parabola.
But on low earth orbit, the L2 term of earth's oblateness dominates by an order of magnitude compared to the moon and the sun, and the rest of the planets are negligible.
Source is a table in the first chapters of Fundamentals of Astrodynamics.
Spherical mass can be replaced with a point mass. Earth however is not spherical (biggest deviation is polar flattening). And even then, Einstein's model, unlike Newton's, says it's not an ellipse even for a point mass.
So, moral of the story - we have to be not too pedantic.
> Spherical mass can be replaced with a point mass
I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
In Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance.
Proving this is a classic problem in undergraduate physics.
Thanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
The way the result works is that if you are a meter underground, it is equivalent to standing on top of a sphere with the top meter of mass removed.
Or equivalently, if you are inside a shell (sphere outside, hollow sphere inside) then the gravitational effect is zero. This can be explained by analogy with light which also follows the inverse square law--changing your position inside a hollow sphere does not change the fact that you see the sphere all around you. Same reason that there is no electrical field inside a hollow conductive object.
Check out gauss law for electromagnetism and gravity. It says that total flux through closed surface is proportional to strength of field sources inside the surface. Flux from outside sources cancels out.
You can use this law to see what's the gravity fields as you move under ground and in other very symmetrical cases.
>What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
It doesn't break it at all. The meter above you can be treated as a hollow shell, which surprisingly has zero net pull, and the solid sphere below can be treated as a point mass just as before.
Just remember these two facts, each provable with a simple integral calculation, usually done in high school physics or freshman college physics: a uniform sphere has the same gravitational pull on an object as a point mass at it's center, and the net gravitational pull on an object inside a spherical shell is zero.
This all works under perfect spheres, uniform (at the spherical shell level at least) density... There are other cases it works, but this simple case is the basic idea.
And that brings fluid dynamics into the picture, which produces discrepancies orders of magnitude greater than the parabolic/elliptical/etc. distinction!
> if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center
If the mass is spherically symmetric, this will not be the case; all of the Newtonian forces from the masses further away from the center than you are will cancel out. This is called the "shell theorem", and it turns out to hold even in General Relativity.
I don’t think this is correct. I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface. A perfect sphere of uniform density would not meet the shell theorem assumptions.
> I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface.
No, it's stronger than that. It says that any spherically symmetric distribution of matter outside a certain radius exerts no "gravitational force" on anything inside that radius, whether there is matter inside that radius or not.
What if my distance to one point on the shell was zero? Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?
> What if my distance to one point on the shell was zero?
Then you are not inside the shell, you are on it. The shell theorem only applies if you are inside the shell.
> Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?
In General Relativity, matter is not viewed as point particles. It is viewed as a continuum, described by the stress-energy tensor. This is one of those cases where the difference shows up.
I think how gravity works at really short distances like 10^(-50)m is still a mystery as we don't have a consistent quantum theory of gravity that works there.
But as you get close to stuff, say two atoms, the electrostatic and other forces are far greater than the gravitational ones.
Any point on the shell, whether the shell has zero thickness or not, has zero mass, but a finite mass density is assigned to it. Any finite mass is spread out, so you cannot have zero distance to enough of its points to feel infinite force.
You and pdonis are correct but you are saying different things. Inside a perfect solid sphere, you would of course feel force everywhere except at the center, but this force would only be due to the mass that is nearer to the center than you.
The intuition we came up with when we had to solve this issue in undergrad physics was interesting.
For each point of any given distance, you can find a disk of points at the plane of the same distance whose gravitational pull will be equivalent to a single point at their center.
You do this for all distances x, and you will find an equivalent rod going from the attracted objected to the center of the sphere, of a non-uniform but symmetrical density.
We find that the density is linearly correlated to the area of the corresponding section, and we then take the closer half of the rod, multiply the density by 1/r^2, and find that it is constant!
For the far half of the rod, we cannot do this, so instead we divide it into two halves of equal pull, then divide those to halves and so on, and find that it this reduces to the equivalent of a point.
Now that we have two points in-line with the object, we find the point with the same total mass that exerts and equivalent force, and lo and behold it is at the center.
I'm sure there is a much simpler way to intuit it, though :)
> "It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density"
Would it? Even of uniform density, the mass would not the at the core, only the center of mass would be. Most of the mass is actually closer to the surface. I'm no physicist, but I'd imagine that would have quite some impact[0] on any trajectory that crosses the surface.
And for any ball thrown at human speeds, I'd expect Earth's gravity would be much closer to a uniform gravitational field than one from a point mass.
[0] Pun not originally intended but I'm quite happy with it now.
In relativity, there is no such thing as a "uniform gravity field", if by that you mean a field where the "acceleration due to gravity" is the same everywhere. The closest you can come is the "gravity field" inside a rocket accelerating in a straight line in empty space, where the acceleration felt by the crew is constant. That kind of "gravity field" has an "acceleration due to gravity" that decreases linearly with height.
Look up the Bell Spaceship Paradox. In relativity, two spatially separated objects (or two ends of a spatially extended single object) that have the same proper acceleration in the same direction do not stay at rest relative to each other; they move apart, as seen in each of their own frames. This is different from the behavior predicted by Newtonian mechanics. In order to have the two objects (or two ends of a single extended object) remain at rest relative to each other, the one in front has to have a smaller proper acceleration than the one in back. (Rindler coordinates are often used to describe this case.)
Based on what I've seen over the years I've been around here, I'm quite certain that's a primary motivator for a non-trivial number of commenters here.
> it's never an ellipse because the Earth is not a point mass.
At least classically, a sphere is indistinguishable (gravitationally) from a point mass while you're outside it. The earth is pretty sphere-ish, locally speaking.
There are very few stable orbits close to the lunar surface. Basically a couple of polar orbits with very specific parameters, and that's it.
The rest get so perturbed by gravitational anomalies that they fall out of orbit after a few months or years--faster than low Earth orbit where there is atmospheric drag!
Yeah, the moon's gravitational field is quite lumpy compared to Earth's, due to its smaller size and the way it is believed to have formed. Plus you can orbit much closer to the surface due to the lack of atmosphere, so the lumps are steeper.
And a ball thrown in the air follows a parabola, locally speaking. At least, you're better off correcting for air resistance before you sub in the ellipse equation.
So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density, but it isn't. In reality it's a potato like mass blob that's approximated by what geodesists call the "geoid". So in order of approximations the path of a ball thrown on earth is a parabola -> ellipse -> numerical integration of 6-dof initial conditions and spherical harmonics approximation of the earth gravity field.