Originally posted by HolyTI don't, but I've seen it used for gravity as well.
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.
Originally posted by wormwoodPerhaps you can use Gauss to prove that a spherical homogenous body can be treated as a gravitational point in the center of the sphere. But any shaped object with any mass distribution inside, no...?
I don't, but I've seen it used for gravity as well.
Even Earth is not completely spherical. It's flatter at the poles. And this alone is a reason to compensate the orbits of the satelites continuasly, especially the low polar orbits. (Asides a lot of other disturbances that the orbit is not completely stable.)
Earth is not even homogenous in mass distrubution, but as long it is symmetrical it's not a problem. But even the level of sea is not a mathematical ellipsoid, but now we ar plunging to deep of disturbances of satellite orbits.
In the mathematical sphere case, you can use Gauss, but not with other shapes and mass distributions.
I would love you to do the experiment. Compare the result between a numerical calculation and a Gauss calculation of a horse shoe body in the size and mass of that of Earth. If a difference can be explained with anything other than accumulative roundings errors, then I'm right in this. If you're right I will eat my hat (baked of pie doe and fruit filling).
Originally posted by FabianFnasYou can use Gauss' theorem for any closed surface. And in physics the importance of Gauss's theorem reveals itself everytime the potential goes as 1/r. Just like teh potential and electrostatic potential.
In the mathematical sphere case, you can use Gauss, but not with other shapes and mass distributions.
Originally posted by adam warlockIf the math differ from the real world - what is right and what is wrong?
You can use Gauss' theorem for any closed surface. And in physics the importance of Gauss's theorem reveals itself everytime the potential goes as 1/r. Just like teh potential and electrostatic potential.
I cannot think of any stable orbit around a horse shoe formd planet, if not veeeery wiiiide.
Originally posted by FabianFnasBut just because you can't think of them it doesn't mean they can't exist. What Gauss's law for potentials that goes as 1/r says is if you're out of the bounds of the body causing the potential you might as well consider it a point particle. Now I don't have much time but later I'lltry to explain this better.
If the math differ from the real world - what is right and what is wrong?
I cannot think of any stable orbit around a horse shoe formd planet, if not veeeery wiiiide.
Originally posted by adam warlockNo, don't explain.
But just because you can't think of them it doesn't mean they can't exist. What Gauss's law for potentials that goes as 1/r says is if you're out of the bounds of the body causing the potential you might as well consider it a point particle. Now I don't have much time but later I'lltry to explain this better.
Make the math for a square planet and show it instead. Perhaps then you'll understand the difference bettween the math and the real world.
Originally posted by FabianFnasMaybe you should study some real physics then. Your views are very innocent you know? All you've done is to show you don't understand Gauss's law, and the relation between physics, mathematics and the real world.
No, don't explain.
Make the math for a square planet and show it instead. Perhaps then you'll understand the difference bettween the math and the real world.
Originally posted by adam warlockI think you do simplifications which lead to a far too simplified result.
Maybe you should study some real physics then. Your views are very innocent you know? All you've done is to show you don't understand Gauss's law, and the relation between physics, mathematics and the real world.
But never mind, there are no qubic planets so it has nothing to do with reality anyway. It's just hypothetical reasoning.
I stop here. Thank you for an interesting discussion.
Originally posted by sonhouseMinimum safe orbit is 13 000 km as measured from the center of the planet.
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?
Shape of the orbit is a square, because its a square planet, duh!
(just kidding, the shape can be an ellipse/circle)
1 edit
Originally posted by FabianFnasyep, I believe I've seen it only used for spherical objects. don't know if it would work for other shapes (I have my doubts), but that's the only case I've actually seen it used.
Perhaps you can use Gauss to prove that a spherical homogenous body can be treated as a gravitational point in the center of the sphere. But any shaped object with any mass distribution inside, no...?
Originally posted by FabianFnasyou just find a symmetry plane, and orbit the midpoint of the horse shoe cross section on that plane. whether the orbit will be a circle or not, depends solely on the shape of the shoe cross section. all lateral forces cancel each other out on the symmetry plane.
I cannot think of any stable orbit around a horse shoe formd planet, if not veeeery wiiiide.
Originally posted by wormwoodYou can use for objects that have some kind of symmetry. For some symmetries it is more helpful than others but still it can be used. The most efficient symmetries are the spherical and cylindrical.
yep, I believe I've seen it only used for spherical objects. don't know if it would work for other shapes (I have my doubts), but that's the only case I've actually seen it used.
1 edit
Originally posted by preachingforjesus"You all like to make things coplicated.
You all like to make things coplicated.
so long as the orbit remains outside of the body all you need is Keplers laws of planetary motion.
satelites around the earth are more complicated because of the moon having a gravitational feild to deal with.
so long as the orbit remains outside of the body all you need is Keplers laws of planetary motion."
... and those laws dies only work with two pointlike bodies.
"satelites around the earth are more complicated because of the moon having a gravitational feild to deal with."
... and polar flattening, and sundwind, and atmospheric drag in low orbits, and, and, and, ... Therefore things are complicated and need more attention than simple laws. Especially when the bodies can't be approximated by a pointlike source of gravitation.
Originally posted by FabianFnasI would agree with Fabian's points here. A square mass can only be approximated as a point when you are very far away as compared to the size of the side of the square. If you are close, then of course the points are going to throw off your orbit as you pass near to them. The points or edges will deflect your orbit away from a circle or an ellipse as you pass near to them. There is surely an equipotential surface you could travel along and always feel the same pull, but whether this could be maintained without artificial means, I'm not sure.
[b]"You all like to make things coplicated.
so long as the orbit remains outside of the body all you need is Keplers laws of planetary motion."
... and those laws dies only work with two pointlike bodies.
"satelites around the earth are more complicated because of the moon having a gravitational feild to deal with."
... and polar flat ...[text shortened]... aws. Especially when the bodies can't be approximated by a pointlike source of gravitation.[/b]
Gauss's law is not being applied properly to this situation as stated so far.