We actually received the initial question well over a year ago, and the reason it has taken so long to publish is that no one wanted to do the math (British: maths). Chronos managed to sidestep the math with his very neat argument, and we all applauded in awe.
Ha! You’ve come to the right place! For a spherical planet (well, the Earth, really), Cecil covered the answer in his column, “What if you fell into a tube through the earth?” As I recall, there were at least a couple of other threads discussing this in more detail, if you want to search for them.
For a toroidal planet (frictionless, airless, etc), I suspect that the point where you turned around would depend on the aspect ratio of the torus (the “fatness” of the donut), and whether you’re talking about standing on the inside or the outside of the donut. For example, if you were on the outside of a skinny donut, you’d jump in the hole, fall through the rock, exit out the other side (The “inside” of the torus), turn around, fall back, and return to the point that you started from (conservation of energy). If the torus was substantially fatter, you’d fall through the rock, exit to the inside of the torus, continue through the center, enter the rock on the other side, keep going until you got to the outside surface (symmetry), turn around, and return.
Yeah, you could use the disk approach as per your modification there, Shadow Mint, but that still leaves three integrals, really, because integrating over dA for each disk is really two integrals. It’s often represented as one, as a shorthand, but that doesn’t really decrease the amount of work involved. You might also try dividing it up into a bundle of thin rings, and doing the other integrals afterwards… I’m not sure which would be easier.
So long as you are talking about being symmetrically located relative to the edges of the strip, you can simply ignore that and only consider a ring with no dimension out of plane of the ring. The non-radial forces all cancel out leaving only radial forces, which you might as well collapse into a ring with no width.
The basic argument should follow along the lines of symmetry. At the center, all forces cancel perfectly; it’s a metastable point. At the edge, most of the force is towards the center, since the mass that’s close to you on either side cancel each other out. The mass that’s directly opposite the center from you gets increasingly smaller for the purposes of determining radial force. T
As you get farther and farther away from the center, the mass that pulled you away start pulling you back toward the center, but since it’s a metastable point, the force must peter out before you get to the center.
This suggests that the behavior is like an inverse Lagrangian point, but ring-shaped. The maximum force occurs along a concentric ring.
TTFN
I like Zut’s intuative analysis.
Hey cronos - here’s something that I ‘learned’ a long time ago but I’m not so sure is true. In fact, I’m pretty sure it’s not true (by doing some Zut-style intuative analysis), but figured I would ask in this thread.
I was told that if you take a collection of discrete masses (stars perhaps in a cluster), find the center of gravity, and center there an imaginary sphere that is large enough to enclose the masses, then for any object on the outside of that sphere the system appears gravitationally the same as a single point source in the center with the same mass (ignoring GR).
True?
Not quite true, sford, but it looks like that might be a reference to the afore-mentioned Gauss’s Law. Gauss’s Law, in its gravitational form (it can also be applied to electric forces, or magnetic, or a few other more obscure applications) says that if you have a closed surface, any closed surface, and you add up the gravitational field over the entire surface, then the total you get will be directly proportional to the mass inside the surface. Now, if you’ve got the right sort of symmetry to your problem (usually spherical), then you know that the gravitational field is the same everywhere on your surface, so it would only depend on the total mass enclosed, not on the distribution.
However, if the mass distribution is not spherically symmetric, then you will be able to tell the difference, from just outside the distribution. For instance, satellites’ orbits are actually changed slightly by the fact that the Earth is flattened, rather than a perfect sphere. If you get far enough away, then any distribution of mass looks almost spherical, but right up next to the masses, it does make a difference.
All of this confusing stuff. Has anyone dealt with the fact that it depends on the donut?
Sorry, calc. makes me eyes glaze over, but…
Look, whether the side of the donut “overhead” attracts one toward the middle of the donut hole depends on the ratio of the thickness of the donut to the diameter of the hole–or the diameter of the whole, as the case may be. If the hole is a small area within a much larger body, the planet will collapse into it, and then no more hole. If we’re talking Larry Niven’s Ringworld, then the other side is too far away to have a noticeable gravitational pull.
So, the answer is, “It depends!”
And as I mentioned, you applauded inappropriately, because the argument isn’t correct.
Solving the definite integral is difficult because it leads to elliptical integrals. However, it can be solved numerically.
No matter what the size and proportions of the donut, it’ll still collapse on itself, if it’s not strong enough. Therefore, we’re implicitly assuming that it is strong enough. Given that, the proportions of the donut don’t have any effect on the direction of the force on a person on the inner rim.
The Ringworld isn’t actually relevant to the gravitational question at all, since it’s spinning: The primary force keeping folks’ feet on the ground there is centrifugal. (Yes, I know that centrifugal force is fictitious. I also know that it doesn’t matter that it is. :P)
<< Solving the definite integral is difficult because it leads to elliptical integrals. However, it can be solved numerically. >>
… Difficult AND tedious, which is why no one wanted to bother.
The argument I used in the report is admittedly not ironclad. I used that argument because I figured it would be the easiest to understand. The Gauss’s Law argument, however, is ironclad, and it gives the same answer.
Of course, both arguments give just the direction of the force, and not the magnitude. If we want the magnitude, then I don’t know how to avoid the integrals. Someone send me a paycheck for it, and I’ll tell you the magnitude-- but not before.
Although the Gauss’ Law argument is correct, doing the integration numerically is simple and that easily answers some of the other questions. I have tried this, and the force at the inner rim, of course, does pull away from the hole. On the surface, the force has a tangential component which tends to pull toward the inner rim, so this is where oceans would collect on a non-spinning donut planet. For the question about a hole drilled through and whether an object dropped through would ever reach the center of the main hole, zut is correct. If b/a>0.2736 the object will fall all the way through to the other side (b=radius of the circle which is revolved about an axis to define the torus and a=distance from the axis to the center of the revolved circle).
Since Chronos wouldn’t give the analytical solution unless paid, I felt obliged. There was a recent GQ asking if partial differential equations could be used for anything, and this problem is a good example. Instead of doing the integration, the partial differential equation for potential field could be solved. The expression I found for the potential is an infinte series, but just a couple of terms give a value within 1% of the numerical results, unless b/a is very small (a skinny torus). When b/a is small, the solution is well approximated by an elliptic integral expression. Here is the solution for potential field (normalized by G, density of the torus, and mass of an object in this field) around a torus:
Let c=sqrt(a[sup]2[/sup]-b[sup]2[/sup]) and H be defined from tanh(H)=sqrt(1-b[sup]2[/sup]/a[sup]2[/sup]). The normalized potential field is given by a sum over n=0 to infinity of terms with the form:
A[sub]n[/sub]cos(nt)*P(n-0.5,0,cosh(h))*sqrt(cosh(h)-cos(t))
Where h and t are functions of spatial location (the point where the field is being evaluated), and A[sub]n[/sub] is a coefficient dependent only on the geometry of the torus. The field is axisymmetric so, using cylindrical coordinates, depends on r and z. The variables h and t are defined by:
tanh(h)=2cr/(r[sup]2[/sup]+z[sup]2[/sup]+c[sup]2[/sup]) and cos(t)=cosh(h)-c/r*sinh(h)
P(n,m,x) is the associated Legendre Function of the first kind.
The first two terms of the series are usually sufficient for accuracy, and the first two A[sub]n[/sub] are closely approximated by:
A[sub]0[/sub]=-c[sup]2[/sup]pi[sup]2[/sup]sqrt(2)(11/8/(sinh(H))[sup]2[/sup]+3log(tanh(H))/4)
A[sub]1[/sub]=-5*c[sup]2[/sup]*pi[sup]2[/sup]sqrt(2)(1/8/(sinh(H))[sup]2[/sup]+log(tanh(H))/4)
I bow down before you, Manlob. Admins, we’ve got to send this guy a coffee mug, at least.
SPOOFE
Not unless you mean “exponentially” in a figurative sense.
My understanding is that the sphere is the basic shape, and all closed shapes are created by adding handles or Mobius strips. For instance, if you take a sphere, cut out a disk, and then replace the disk with a Mobius strip, you end up with a Klein bottle.
Exponent = 2 I believe.
“Exponential” means a constant raised to a variable, not a variable raised to a constant.
Well, you’ve just eradicated a tiny corner of ignorance Ryan. Correction noted!
I made a plot of the g-field of a ring (actually 200 point masses arranged in a ring). No surprises.
http://ogre.nu/images/gringi.jpg
The plane of the image is perpendicular to that of the ring.
The force is stronger where the equipotential curves are closer together.
The centripetal acceleration of an object moving in a circle is 4 pi^2 R / T^2, where R is the radius of the circle and T is the period.
I find it convenient to remember that one rpm gives one gee at 900 meters (oh all right, 894) and work up or down from that:
-
if you have the radius, divide it by 894m and take the square root to get the period in minutes;
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if you have the period in minutes, square it and multiply by 894m to get the radius.
From this we learn incidentally that a ringworld that spins in one day (1440 min), for one gee, has a radius of 1.854E9 meters, or 6.185 light-seconds, or 4.823 lunar orbits.
Dexter wrote:
> Actually, from a certain mathematical (algebraic topology)
> point of view, there aren’t any two dimensional surfaces
> other than a plane, a sphere, a torus, and a Klein bottle.
And projective plane: what you get if you glue opposite edges of a rectangle, with a half twist in each pair.
> Any homological or homotopic differences simply arise
> from adding handles.
Well then, you could leave the torus out of the list – as well as the plane, which is equivalent to a sphere with one point removed.
Four colors on a sphere,
six on a Klein bottle or projective plane,
seven on a torus.
See also http://www.math.niu.edu/~rusin/papers/known-math/97/chromatic.genus
(I dunno what happens to the coloring-number if you add handles to a non-orientable surface.)
This is great stuff, bronto. I hope you’re planning on sticking around and posting more? We can always use more in-depth, well-thought posts.