 # Gravity of a non-spherical object

If the earth were more elliptical rather than so close to spherical, would those standing on the edge of the ellipse feel more gravitational pull than those standing near the center of the ellipse? How do you calculate the gravity of a non-spherical object?

It has been said that the gravitational force between 2 objects is proportional to the product of their masses and inversely proportional to the square of the distance between them (F=m1m2/r^2), the distance being measured from the center of mass of each objects. And force is a vector quantity so it has a magnitude and a direction and is additive. But most of what we know about gravity comes from large spherical objects (planets and stars) with a symmetrical mass distribution.

Now consider a binary star system. If you were feeling the net gravitational effect of a binary star system, you would feel more gravity when the two stars were in line with each other than you would feel when they were next to each other because the vectors would be in line with each other and one of the stars would be closer to you thereby magnifying its gravitational effect.

Now consider a non-spherical object like a galaxy or a nebula. If I were standing some distance away from a galaxy perpendicular to its axis (so the galaxy looks like a circle), I would feel a certain gravitational pull toward its center. But if that galaxy were to rotate 90 degrees without changing the location of its center (so now it looks like a long narrow ellipse), then I would feel a much greater gravitational force from the stars that rotated closer to me.

In both examples, the masses have not changed, nor have the distances between the observer and the centers of mass, but the gravitational force on the observer has changed.
So the question is, how do you calculate the gravity of a non-spherical object taking into account its shape and orientation, short of adding up the gravitational pull from each molecule or each atom? Has a formula been worked out for a thin disk or ellipse or a cube or any other 3-dimensional object other than a sphere? Do any objects exist close enough to us such that we can test these formulas or can we test them using smaller scale objects here on earth?
I’m sure these questions have been asked by much smarter people (such as yourself) and probably at least partially answered, but I cannot find them addressed in basic physics books. Can you help increase the understanding of gravity by the Teeming Millions? Thank you.

R.M., Seattle, WA

For the general case, you add up all the forces. For some simple shapes, it’s possible to determine an equation for the force - try googling “Gravitational force ellipsoid” or “gravitational force disk” for some examples.

Someone at the equator of Earth as it is now feels less gravity than someone at the North pole both because he is further away from the center of the Earth, and because of the effect of the Earth’s spin.

You can get a good approximation by assuming all the mass of the earth is concentrated at the center of mass. This works better the farther away you are, but still would work OK near the surface.

Newton and Maclaurin did most of the heavy lifting, some three hundred years ago (give or take.) Newton showed that spheres can be simplified as points. Maclaurin showed that rotating ellipses were only meta-stable: any disturbance tends to make them more spherical. He also showed that Saturn’s Rings couldn’t be solid, but had to be made up of particles.

Unfortunately, the calculus to integrate the forces for someone standing on the surface of an ellipsoid is…ugly. I think that “closed form” solutions do exist, but, really, you might as well just use numerical integration.

From orbit, an ellipsoid (like the earth) can be approximated as a sphere with a “belt.” Sort of like building a big long square barn all the way around the equator. This is the first approximation that was used by the early ICBM targeters. Nowadays, they have mapped out the mountains and mass-concentrations to incredible detail, and numerical integration is the only way to go.

The approximation you get by assuming the mass is all at the centroid is only an approximation, and if the question is “how does the force differ from that of a mass at the centroid” then this approximation is merely a statement that you’re going to ignore the difference.

You can use the integral calculus to calculate the force exactly for shapes you can define mathematically, if you can find a solution to the integral. I have done this for an ellipsoid, but it was a lot of work and the fanciest math I have ever done. Today I would use a numerical approach.

Without going through the calculations, you can easily show that for most non-spherical shapes the local gravitational field will generally not point to the center of your figure, and won’t be normal to the surface (except at certain points of symmetry – in the case of your ellipsoid, it’ll point to the center when you’re at the extremes, but not in general). i pointed this out in our discussions about cubical earths. The fact that the gravitation vector isn’t normal to the surface is a Big deal – the surface material will tend to “flow” to even out things so that the vector IS normal. That’s how we ended up with a spherical earth in the first place. (Actually, you have to take into account the centrifugal fictitious force and the nonuniform mass distribution into account as well. That determines the direction of the effective force of gravity, which determines the local normal). So a weird-shaped Earth probably won’t be very stable, and will tend to collapse to a spherical form.

You may be interested in http://en.wikipedia.org/wiki/Gauss's_law_for_gravity which, with cleverly defined surfaces, can simplify calculations. In particular,

The gravitational flux through any closed surface is proportional to the enclosed mass.

Cecil has also written about walking on a cubic planet

Which was very interesting.

From the individual “stars that rotated closer,” you would certainly feel a greater attraction, but not from the galaxy as a whole. Taking all the mass of the galaxy as if it were concentrated at a point in the center (which is a reasonable approximation at sufficiently great distances from a distributed mass), it is highly unlikely you would feel a “much greater” force due to “stars that rotated closer”.

From a “reasonable” distance away from the center of the galaxy, along its axis of rotation, all the stars are within a fairly narrow range of distances from you, and the summation of their individual gravities would establish the baseline force under discussion.

At the same distance from the center of the galaxy, only this time, aligned with its flattened “disk” shape, many stars would now be very much closer to you, and their contribution to the summation of stellar gravities would be multiplied, due to the “inverse square law,” whereby the force of gravity increases as the distance decreases, by the square of the difference (i.e., at half the distance, the force of gravity is four times stronger.)

BUT—and here is the kicker—an equal number of stars will be very much farther from you, and their contribution will be lessened by an equivalent amount.

The net effect of these two situations is that the force of gravity will be roughly the same whether you are hovering over the center or at the edge of the disk. Of course, inhomogeneities in the composition of the galaxy or other random fluctuations might distort the true equivalence, but not sufficient for you to feel a "much greater gravitational force."

I don’t think this applies, because gravitational force depends on the square of distance. For example, if you have 2 x 1 unit masses a distance of 2 apart, and you’re 2 units from the center, you could have:
mass (2 distance) mass (1 distance) you

or
mass
(2 distance)----------------(you)
mass

and the forces would be different. They would appear the same if you’re sufficiently far, though.

Thanks for the insightful responses. The links were very informative and the discussions pointed out some common misconceptions. I’m glad I posed the question to the Teeming Millions. There are clearly some very educated people out there.

R.M. Seattle, WA