|
Post by Progenitor A on Apr 3, 2011 9:15:29 GMT 1
Baffling discussion on another board
Imagine a massive boomerang-shaped mass in space.
Its CofG will fall outside the boomerang and withinn the concave curve of the mass
A very smalll mass approaches this object
Is it possible for the smaller mass to orbit thi CofG whilst staying within the concave arc of the boomerang-shaped object?
|
|
|
Post by eamonnshute on Apr 3, 2011 10:38:35 GMT 1
No.
Consider a similar but simpler situation, with two fixed masses, and their COG in the middle. A particle which is nearer one mass will be pulled towards it and away from the COG, so it cannot orbit the COG (or centre of mass, being pedantic).
|
|
|
Post by carnyx on Apr 3, 2011 11:00:03 GMT 1
@nm
I suspect you could initially rule out any solution where movement away from the nominal C of G in a hollow body produces an increase in the attractive force, as opposed to a decrease. In other words the force would be proportional to minus 1/D^2 ...... a kind of anti-grav effect!
I reckon that in such 'negative' gravity zones, the counterbalancing centrifugal force would require a matching increase in orbital velocity with increasing radius, which would imply a decrease in mass or a reduction in velocity ( and so a decrease in energy) to preserve momentum... which does not seem possible at first glance.
I suppose a map of the contours of the attractive force could help in visualising object paths that may produce an arithmetic balancing centrifugal force at all points, and whose varying velocity would preserve momentum
Without a whole lot of detailed work, I intuit that there may be just one such path, and it would describe a hyperbola, .. a one-time pass-by .... and so not an orbit. It would need to be relatively small object so as not to have an appreciable 'negative' gravity gradient across it's internals, and therefore it woud need to be a small mass with a consequently high velocity to achieve the right momentum value.
If I am right, then all orbiting bodies will eventually crash, because of the 'negative' gravity gradient zone around the concavity. So, pretty soon it get filled in .. which may be by why stellar objects tend to be round?
So to cut it short ... my answer to the question is 'no'
|
|
|
Post by Progenitor A on Apr 3, 2011 19:09:39 GMT 1
No. Consider a similar but simpler situation, with two fixed masses, and their COG in the middle. A particle which is nearer one mass will be pulled towards it and away from the COG, so it cannot orbit the COG (or centre of mass, being pedantic). I agree with your conclusion but do not see the appplicability of a two-mass model as there is only one mass in the example given - the boomerang Of course in your example an orbit can be established around the CofG in a plane perpendcular to a line joining the two masses and passing through the CofG
|
|
|
Post by Progenitor A on Apr 3, 2011 19:14:52 GMT 1
@nm I suspect you could initially rule out any solution where movement away from the nominal C of G in a hollow body produces an increase in the attractive force, as opposed to a decrease. In other words the force would be proportional to minus 1/D^2 ...... a kind of anti-grav effect! I reckon that in such 'negative' gravity zones, the counterbalancing centrifugal force would require a matching increase in orbital velocity with increasing radius, which would imply a decrease in mass or a reduction in velocity ( and so a decrease in energy) to preserve momentum... which does not seem possible at first glance. I suppose a map of the contours of the attractive force could help in visualising object paths that may produce an arithmetic balancing centrifugal force at all points, and whose varying velocity would preserve momentum Without a whole lot of detailed work, I intuit that there may be just one such path, and it would describe a hyperbola, .. a one-time pass-by .... and so not an orbit. It would need to be relatively small object so as not to have an appreciable 'negative' gravity gradient across it's internals, and therefore it woud need to be a small mass with a consequently high velocity to achieve the right momentum value. If I am right, then all orbiting bodies will eventually crash, because of the 'negative' gravity gradient zone around the concavity. So, pretty soon it get filled in .. which may be by why stellar objects tend to be round? So to cut it short ... my answer to the question is 'no' I agree with your conclusion, and oddly, the feasibility of having a small mass rotate around the CofG was 'explained' in terms of 'negative gravity' where the gravitational force lines went in the direction from the mass of the boomerang to the CofG, in other words the mas of the boomerang provides a repulsive force I think that that is wrong and agree with your conclusion
|
|
|
Post by speakertoanimals on Apr 4, 2011 16:39:28 GMT 1
Not quite -- it lyes within the convex hull.
ANd it ISN'T a single mass, in the sense that basic mechanics deals with POINT masses, hence a single object with a continuous mass distribution needs to be viewed as the limit of an infinite number of point masses.
The point about orbiting around the centre of gravity/mass (the two not quite the same, but never mind that now!), is that Newtonian gravity says that for an arbitrary distribution of mass, if you are FAR enough away, the net effect is the same as a single mass placed at the centre of mass.
There ARE higher order terms (such as gravitational quadrupole, but these usually only important for things like artificial sateliites).
Hence if you are far away, you orbit the centre of mass, but this is only an approximation that holds when the separation of the masses becomes small compared to how far away you are. This is important, else why should Newton have been able to get away with treating the gravitational effect of the earth or the sun as the gravitational effect of a single point mass?
But once you are WITHIN the set of masses, this distant approximation no longer any good. Hence within a spherically symmetric shell of matter, we have ZERO gravitational effect.
Why does the centre of mass matter? Because if we have the individual masses attracting each other, the centre of mass is the point about which the forces between the bodies have no effect -- the centre of mass moves along at constant speed, no matter how much the individual masses may be orbiting each other.
Now if we think about a test mass inside the curve of the boomerang, can it orbit the centre of mass?
1) Of course not! the net gravitational effect being towards the centre of mass ONLY holds if you are at large distances compared to the size of the gravitating body.
2) Of course not! If inside the boomerange curve, then OBVIOUSLY all particles of the boomerang can only ever pull outwards, if there is a net effect. Since an orbit requires an inwards force, then orbiting inside the curve is obviously impossible. Which is why you couldn't have orbiting moons within a hollow earth.....................
No 'repulsion' required, each particle of the boomerang only ever attracts the test body.
|
|
|
Post by Progenitor A on Apr 4, 2011 17:26:54 GMT 1
Not quite -- it insights within the convex hull. Nothing can lye within a convex hull ! ANd it ISN'T a single mass, in the sense that basic mechanics deals with POINT masses, hence a single object with a continuous mass distribution needs to be viewed as the limit of an infinite number of point masses. The point about orbiting around the centre of gravity/mass (the two not quite the same, but never mind that now!), is that Newtonian gravity says that for an arbitrary distribution of mass, if you are FAR enough away, the net effect is the same as a single mass placed at the centre of mass. Yes we know that There ARE higher order terms (such as gravitational quadrupole, but these usually only important for things like artificial sateliites). Hence if you are far away, you orbit the centre of mass, but this is only an approximation that holds when the separation of the masses becomes small compared to how far away you are. This is important, else why should Newton have been able to get away with treating the gravitational effect of the earth or the sun as the gravitational effect of a single point mass? Indeed why! But once you are WITHIN the set of masses, this distant approximation no longer any good. Hence within a spherically symmetric shell of matter, we have ZERO gravitational effect. Two uncorrelated facts there and indeed the second is entirely irrelevant Why does the centre of mass matter? Because if we have the individual masses attracting each other, the centre of mass is the point about which the forces between the bodies have no effect -- the centre of mass moves along at constant speed, no matter how much the individual masses may be orbiting each other. Yes indeed Now if we think about a test mass inside the curve of the boomerang, can it orbit the centre of mass? 1) Of course not! the net gravitational effect being towards the centre of mass ONLY holds if you are at large distances compared to the size of the gravitating body. Huh ! Only one question was asked, but thank the Lord, at last the irrelevant waffle stops! 2) Of course not! If inside the boomerange curve, then OBVIOUSLY all particles of the boomerang can only ever pull outwards, if there is a net effect. Since an orbit requires an inwards force, then orbiting inside the curve is obviously impossible. Which is why you couldn't have orbiting moons within a hollow earth..................... No 'repulsion' required, each particle of the boomerang only ever attracts the test body. Who said anything about repulsion? In fact, in between all your irrrelevaicies and garbled English, you are quite right, but you must be the most awful teacher on God's earth But that apart, well done!
|
|
|
Post by Progenitor A on Apr 4, 2011 19:40:09 GMT 1
You said CONCAVE (wrong word). We have the convex hull of the points, the points themselves lye within this, as does the centre of mass. As may your test mass. Why you want to claim otherwise, god knows. Or do you think it is okay to use the totally wrong word, and let it pass without correction............................... you can give the impression that thet effect of ANY distribution of mass is the same as a spherically-symmetric distribution. Which isn't the case. Quadrupole isn't actually that weird, many people will have heard of electric or magnetic dipoles, for instance, and interesting if they ponder WHY what appears here is the quadrupole, not the dipole (hint what constitues an electric dipoles?). No it isn't! Indeed, it is key to seeing the short answer to the what happens if you drill a hole problem. But hey, you just ignore it, mix up concave and convex, and make up stuff the way you seem to prefer to! Except I'm NOT teaching here, I thought this was a discussion board! Except again, you don't seem to care about having a discussion. Repulsion WAS mentioned, in fact you mentioned it yourself:
|
|