I am curious as to research that calculates the shape of gravitational wells, and their limits, and affect on time, for different masses and spread of mass. The actual question is st the end.

For example: When the sun becomes a red giant the Earth's orbit is supposed to move out with the redistribution of mass, but what is the science behind this? Re-edit: Which I think was explained as the spreading density of the suns mass (it was a "TV" program. Another example, is it is said that the termination of the Sun's gravity field is 1.5 light years out. I am unaware of any distance studies to measure shape to specifically prove the theory.

The Question: Explanation of how the shape, time over distance, and extent of a gravitational well changes with the density of the mass, and the science behind this please? Re-edit: The curvature, research to verify theory, and actual simple graphical description of how the physical shape responds and changes based on contributing mass distribution. Say, does a more dense matter object cause the gravity well to more tightly curve to the surface of the matter, than to the surface of a cloud of gas of equal weight but magnitudes bigger. How does that look in physical shape over distance, how does the field terminate in shape. I'm interested in observational research on the profile. For instance does it continue the same decay equation or does it change/flatten out etc to a different equation at distance. This is more looking at verification/explanation of the conventional versus deviations. If we can only say so much at X distance from verified studies, that would be appreciated?

As we know there has been speculation based on deviations in observation of gravity on grander scales, such as across the galaxy. But I do not wish to go into those hypothesises, only the limits of what we have verified we know, which is a good starting point onto looking into this further.

  • $\begingroup$ en.m.wikipedia.org/wiki/Shell_theorem $\endgroup$ – safesphere Nov 25 '18 at 12:16
  • $\begingroup$ The Earth's orbit around the Sun will grow because the Sun loses mass as it becomes a red giant. It has nothing to do with the distribution of mass inside the Earth's orbit. $\endgroup$ – Rob Jeffries Nov 26 '18 at 8:03

The question is a bit open-ended, so this is more of an introduction to the topic than a total answer - since a full answer involves both potential theory and general relativity.

If we start with gravitational wells in classical mechanics, they have a simple form for point masses: $$U(r)=-\frac{GM}{r}$$ where $r$ is the distance from the point. They are defined everywhere but the point at $r=0$, and stretch to infinity (where the potential is zero).

If you have several point masses their potentials just add up: $$U(x)=-G\sum_i \frac{M_i}{d_i}$$ where $M_i$ is the mass of the $i$th particle and $d_i$ the distance to the point $x$ where the potential is measured.

One can analyse continuous mass distributions in the same way, by integrating contributions from infinitesimally small particles: $$U(x) = -G\int_{t\neq x} \frac{\rho(t)}{|x-t|} dt$$ where $\rho(t)$ is the mass density at the point $t$. The integral goes over all of space except $x=t$. This way you can (with some effort) calculate the gravitational field around spherical bodies, rings, planes etc.

One of the key results is the shell theorem of Newton: if you have a spherically symmetric mass distribution, the potential only depends on the distance to the center (it is also spherically symmetric). Also, there is no gravitational force from the mass shells outside the point you are measuring from: they still contribute potential, though -- at the center of Earth you would be weightless yet have a fairly low potetial. It does not matter how dense the body is, if you are outside it you will only feel a potential dependent on the total mass.

This also explains your secondary question about the Sun as a red giant. The reason Earth is expected to spiral out a bit is that the Sun will start losing mass though strong solar wind, not that the density will go down.

Further, one can find other ways of calculating potentials. Gauss law is very useful. It states that the gravitational force, when integrated around an arbitrary closed surface, is proportional to the mass inside: $\int\int_S g\cdot dA = -4\pi G M$. Note how it implies the shell theorem. One can differentiate this, and then arrive at Poisson's equation $$\nabla^2 U = 4\pi G\rho.$$ Potentials in empty space where $\rho=0$ follows Laplace's equation $$\nabla^2 U =0$$ -- they are the subject of potential theory and are sometimes called harmonic functions.

To describe their shapes one can make use of the fact that if $U_1$ and $U_2$ are two potentials that solve Laplace's equation, then $U_1+U_2$ also solves it. So one can add together simple potential solutions (just as in the point particle case) to build more complex ones. There is a vast mathematical theory on how to do this. For gravity usualy the important thing is expansions in multipole moments expressed in spherical harmonics. Since most mass distributions are roughly spherical blobs, these combinations usually produce accurate results with few terms. In particular they can easily handle the distortion from an oblate body (the potential is a bit squashed near the body), or more complex things such as the lumpiness of Earth. Once you have the potential you can for example calculate how it affects satelite orbits.

One of the important results is that higher-order harmonics decay faster with $r$ than $1/r$. They have less effect on the distant parts of the potential. A cubical planet will have a somewhat cube-like field near the surface, but further away it will be nearly like it was a sphere (see this paper, although they don't use the harmonic method).

Beyond this is general relativity. There is not exactly a gravitational potential in GR (the metric tensor is doing some of the job) although for weak fields one can approximate the classical one and calculate gravitational time dilation. There is also a counterpart to the shell theorem.

  • $\begingroup$ Anders, a wonderfully complete answer. Thank you. But I was seeking a more practically illustrative answer based on verified observation. I have not looked for years, but am not aware of much long distance research projects to prove theory. I have re-edited the question extensively to narrow down the direction of the question, and in reflection of a few example statements I have come across. $\endgroup$ – user213923 Nov 26 '18 at 6:28

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