Originally posted by AThousandYoungI think, au contrare, that the transition to a spheric shape will be quite quick.
Eventually, but it could take a while.
If we suppose that the qubic Earth have the same composition as our present Earth, which is more or less a drop of molten magma with a thin crust of solid rock, there is no reason to believe that this thin crust will have the strength to avoid turning into a sphere. The forces in progress will be of that extend that it's energy will melt the remaining rock into being molten.
1 edit
Originally posted by sonhouseDo you trully understand the difficulty of the problem? Orbits around spheres are easy to solve cause there symmetries all over the place. Now you have a problem with regular (cartesian coordinates) mixed with polar coordinates. You'll have 1/Sqrt( x²+y²+z² ) all over the place. They usually don't like to be integrated. But it's a good problem, gonna try to solve it.
A thousand years from now, we have interstellar spacecraft, a team finds an artifact the size of a planet but cube-shaped, 15,000 Km across one face and it masses the same as the earth, 6 E24 Kg.
There is no atmosphere. What is the lowest orbit you can safely acheive around such an object? It is homogenous density wise. What is the shape of the orbit?
Originally posted by wormwoodif you make an approximation of the cube to a point like structure. The problem is gravity is not constant along the orbit. And the problem strongly depends on initial conditions. (initial vector velocity)... i think...
the lowest orbit is just above zero 🙂 , or h_lowest = sqrt( 7500^2 + 7500^2 ) = 10 607km from the center of the cube, and the shape of the orbit is a superellipse I believe.
1 edit
Gauss's Law allows us to treat any closed surface, with any distribution of charge within that surface, as a point charge, in terms of the effect of the electric field on charged bodies outside that surface. And guess what? The same math gives the same result for mass and gravity. The location of the point to use as the center, for a cube of uniform density, is, simply, the geometrical center of the cube.
So the orbital problem boils down to one of geometry. Now we have to figure out what the questioner meant by "lowest orbit." 2 possibilities come to mind, but there are others: the orbit which will carry a satellite closest to the surface at any point (1) edges included, (2) only considering the center of any face. Counting corners only seems like a stretch. "Lowest" could also mean (a) closest to the center on average or (b) closest to the center at any point. It turns out that (1) and (a) can be satisfied by the same answer and (2) and (b) can be satisfied by the same answer.
In case (1), you could have any elliptical orbit which just grazes only 2 or all 4 of the edges, but a circular orbit which touches all 4 edges would be lower on average for the rest of the orbit, which is not one of my conditions above but seems closer to the spirit of the original question, and gives us something definitive to shoot for. Draw a square (the cross-section of the center of the cube), then circumscribe a circle around it. This circular orbit would have a radius of half of the diagonal of one face, or square root of 2 * 15,000 km * 1/2, or about 10,606.6 km.
For case (2), draw the square again, then imagine an elliptical orbit whose minor axis touches the top and bottom edges of the square (the top and bottom faces of the cube) and whose major axis goes to infinity. Since the orbit is infinite in size, it may as well be a hyperbolic orbit or a parabolic orbit. Along the minor axis, the orbit would have a periapsis (closest point of approach to the center) of 1/2 * 15,000 km, or 7,500 km (it would skim the surface).
What do I win?
Originally posted by HolyTI don't think you win at all.
Gauss's Law allows us to treat any closed surface, with any distribution of charge within that surface, as a point charge, in terms of the effect of the electric field on charged bodies outside that surface. And guess what? The same math gives the same result for mass and gravity. The location of the point to use as the center, for a cube of uniform density ...[text shortened]... the center) of 1/2 * 15,000 km, or 7,500 km (it would skim the surface).
What do I win?
Gauss' law is not applicable in this case.
1 edit
Originally posted by serigadothe gravity is constant on an orbit, the orbit just isn't circular nor an ellipse.
if you make an approximation of the cube to a point like structure. The problem is gravity is not constant along the orbit. And the problem strongly depends on initial conditions. (initial vector velocity)... i think...
think of gravitational potential around the cube. there are closed equipotential sheats around the cube, on which the gravitational pull is constant by definition. any of those sheats will act as a stable orbit at a certain initial velocity, and the lowest possible is the one which stays just above the ground over the midpoint of an edge.
I think?
Originally posted by wormwoodI agree. The centre of the cube is the centre of gravity, and the equipotential surfaces take the form of sperical shells extending outward from that centre. All circular orbits would lie on these shells, and the smallest orbit would be the circle that just misses grazing the midpoint of the side of the cube, as wormwood stated.
the gravity is constant on an orbit, the orbit just isn't circular nor an ellipse.
think of gravitational potential around the cube. there are closed equipotential sheats around the cube, on which the gravitational pull is constant by definition. any of those sheats will act as a stable orbit at a certain initial velocity, and the lowest possible is the one which stays just above the ground over the midpoint of an edge.
I think?
1 edit
Originally posted by PBE6the shells can't be spherical, as the gravitational potential at any point is the sum of the potential caused by every point of the cube. which means the gravitational pull is greater (well, now that I think it might be smaller as well, but it certainly isn't equal) at the tips of the cube than in the middle of the edges. which makes the equipotential surfaces sort of rounded cubes.
I agree. The centre of the cube is the centre of gravity, and the equipotential surfaces take the form of sperical shells extending outward from that centre. All circular orbits would lie on these shells, and the smallest orbit would be the circle that just misses grazing the midpoint of the side of the cube, as wormwood stated.
my hunch is they'll be the shape of 3d superellipsoids, as I gather the shape needs to be one than can be smoothly interpolated from cube to a sphere (which is the shape of the surface at infinite distance). but as my math is 'a bit' rusty, I couldn't figure out an analytical solution to it.
Originally posted by wormwoodIf a shape has uniform density, then its centre of mass lies at the centroid. For a cube, the centroid is just the centre. Any object orbiting the cube would be acted on as if all the mass were centred at the centre of mass.
the shells can't be spherical, as the gravitational potential at any point is the sum of the potential caused by every point of the cube. which means the gravitational pull is greater at the tips of the cube than in the middle of the edges. which makes the equipotential surfaces sort of rounded cubes. my hunch is they'll be the shape of 3d superellipsoids, ...[text shortened]... distance). but as my math is a bit rusty, I couldn't figure out an analytical solution to it.
http://en.wikipedia.org/wiki/Centre_of_mass
Why wouldn't the equipotential shells be spherical then?
1 edit
Originally posted by PBE6the center of mass is in the center, but it has nothing to do with the shape of the gravitational field, except in the case of a perfectly symmetrical ball or if the distance is far enough for the effects of the non-uniformity of the gravitational field of a non-uniform mass to be negligible. in others words, you don't have to take into account the gravity of mt. everest calculating an orbit around earth, because earth is so much bigger. but were the mt. everest the size of the moon, you'd definitely need to calculate its effect.
If a shape has uniform density, then its centre of mass lies at the centroid. For a cube, the centroid is just the centre. Any object orbiting the cube would be acted on as if all the mass were centred at the centre of mass.
http://en.wikipedia.org/wiki/Centre_of_mass
Why wouldn't the equipotential shells be spherical then?
the corners of the cube are like huge mountains, and will have a gravitational influence of their own. the closer you are to one of the 'mountains', the more it'll pull you. all the other parts of the cube are farther away, so their pull will diminish. that's why the field is not spherical, because the 'mountains' increase the gravity locally, compared to a perfectly spherical mass.
2 edits
Originally posted by FabianFnasGauss's Law does indeed apply, but I think I applied it a little wrong. Gauss's Law applies to closed surfaces (the cube is a closed surface) containing force-producing particles (masses in the case of gravity, charged particles in the case of electrical forces). But I think you have to enclose the cube in a spherical Gaussian surface to use the point-charge (point-mass) simplification.
Gauss' law is not applicable in this case.
By enclosing the cube inside a sphere (a "spherical Gaussian surface" ), everything else I said should be correct, up to the point of talking about the actual orbits. The sphere would touch the corners of the cube. The smallest circular orbit would then skim the surface of the Gaussian sphere at a distance from the center equal to the distance from a corner to the center. Using the Pythagorean Theorem twice, this distance is calculated to be 1.5 times half the length of one side of the cube, or 11,250 km.
The case of the infinitely long ellipse still works, since for such a huge orbit, even one that is close to the body for a very short time, the cube's gravity can be simplified to a point mass, even if it's inside the Gaussian sphere for a very short time at a very high speed.
Originally posted by HolyTI still don't think so.
Gauss's Law does indeed apply, but I think I applied it a little wrong. Gauss's Law applies to closed surfaces (the cube is a closed surface) containing force-producing particles (masses in the case of gravity, charged particles in the case of electrical forces). But I think you have to enclose the cube in a spherical Gaussian surface to use the point-charge ...[text shortened]... mass, even if it's inside the Gaussian sphere for a very short time at a very high speed.
If you have a sphere you can approximate this body to a pointlike source of gravitation. Any circular or elliptic orbit will be closed and therefore stable. You can't to this with a a qube formed object.
If you still think it is possible, then it is possible with any kind of shape. Like a dumb-bell shaped object, Like a long and thin rod, Like a tetraedic yet irregular shape, etc, or why not a gigantic coathanger... Every shape that is closed should then be possible to apply Gauss' law...? I don't think so.
I say like this. (We talk about a two body system, don't we?) A sphere is the only shape that has stable orbits in any inclination. Some other shapes can have some stable orbits. Some shapes can have semi-stable orbits. But of course if the orbits are wide enough you can approximate any body to a point like source of gravitation.
The homoqene sphere is the only body that can have stable orbits in any inclination. A qube like body - no.
If you think about where the source of gravitation lies in a body, you surely agree with me.
2 edits
Does it apply to any shape? Yes! That's the beauty of Gauss's theorem. You just have to enclose the shape within a (imaginary) sphere for it to work.
Anyone who has a better understanding of Gauss's theorem, particularly as it applies to mass and gravity, please jump in and educate us! Thanks.