16 Sep '07 23:46>
You can't do that. Gravity will compress the planet into a sphere, unless the laws of the universe will have significantly changed by then.
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.
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.
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?
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...
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?
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.
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.
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?
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.
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.