Center of mass for 2 dimensional objects ?

[Externet]

Hi all.

Imagine two extremely large 'two-dimensional' objects -say like a brutally large flat sheet of paper- where their thicknesses is irrelevant in relation to thousands of square miles.

 

Would their centers of mass still be a point ?

 

- If no, what shape would it be ?

 

- If yes, would bringing those two bodies together make their intergravitational attraction nearly infinite, impossible to pull them apart ?

[timo]

Center-of-mass points are points by definition. Bringing such sheets sufficiently close to another (and ignoring that at some point the atomic structure will come into play) can cause arbitrarily large attractive forces. However, it seems to me as if you think that the gravitational attraction between two objects is always a center-of-mass-to-center-of-mass one. That is not the case in general.

 

Minor comment: I don't think that "nearly infinite" is a term that makes sense, given that "infinite" roughly means "larger than any real number" and "larger than nearly any real number" and "nearly larger than any real number" both make no sense.

[Externet]

Thanks.

Can you expand when the center-to-center of masses is not the distance for calculation of attractive force ?

[Schrödinger's hat]

Thanks.

Can you expand when the center-to-center of masses is not the distance for calculation of attractive force ?

 

Any time your objects are of significant size compared to the distance between them, you need to consider that the mass isn't all at a single point. Otherwise you will get an inaccurate result.

 

There's a special case for spherically symmetric things where the maths works out the same as having a single point at the centre of mass.

 

This will work as long as you are outside the spherical object, Inside a hollow spherical shell a [math]\frac{1}{r^2}[/math] force (like gravity) goes to zero, So for solid spherical objects you can consider the mass that is deeper inside the object than you as a single point in the middle.

Edited September 29, 2011 by Schrödinger's hat

[Externet]

OK. Thanks.

 

A massive torus or pipe section will have its centre of mass point at the void.

 

(Something like this just to be pictorial ----> http://www.emcocontrols.com/files/emcobilleder/flow/vrf.jpg )

 

A bowling ball will have its centre of mass point at ~its geometrical centre.

 

Both being spherically symmetric.

 

Fitting the ball loose inside the torus would allow both centres of mass points sharing the same location; d=0

Their attraction would be infinite; but it is not.

 

=======================

 

The International Space Station ISS has a centre of mass point somewhere within its guts, Say it happens to be at a void location. Placing a bowling ball (or any object centre of mass there would make it 'unmovable' by d=0 )

 

But not even a hint of that behavior happens.

 

=======================

 

Is the equation [ math] F = G frac{m_1 m_2}{d^2} [ /math] missing some factor in the denominator ? Like 0.00001 + d^2 instead

Edited September 29, 2011 by Externet

[Schrödinger's hat]

OK. Thanks.

 

A massive torus or pipe section will have its centre of mass point at the void.

 

(Something like this just to be pictorial ----> http://www.emcocontr...er/flow/vrf.jpg )

 

A bowling ball will have its centre of mass point at ~its geometrical centre.

 

Both being spherically symmetric.

 

Fitting the ball loose inside the torus would allow both centres of mass points sharing the same location; d=0

Their attraction would be infinite; but it is not.

 

=======================

 

The International Space Station ISS has a centre of mass point somewhere within its guts, Say it happens to be at a void location. Placing a bowling ball (or any object centre of mass there would make it 'unmovable' by d=0 )

 

But not even a hint of that behavior happens.

 

=======================

 

Is the equation [ math] F = G frac{m_1 m_2}{d^2} [ /math] missing some factor in the denominator ? Like 0.00001 + d^2 instead

 

Perhaps this:

There's a special case for spherically symmetric things where the maths works out the same as having a single point at the centre of mass.

 

This will work as long as you are outside the spherical object, Inside a hollow spherical shell a [math]\frac{1}{r^2}[/math] force (like gravity) goes to zero,

Was a little unclear.

I was saying that the [math]\frac{GMm}{r^2}[/math] formula works if and only if you are outside an object which is spherically symmetric (it will be approximately correct for things that are far enough away that they are approximately spherical, but never exact). If you are inside a spherically symmetric thing the force goes to zero (if earth were hollow and you went anywhere inside you'd feel no force. If you dig down a bit you only feel the gravity from the mass that is at lower altitude than you).

 

 

A torus is not spherically symmetric.You can rotate it one way and it's indistinguishable (you don't know whether or not it has been rotated, because it looks the same), but if you rotate it the other way you can tell the difference (a doughnut lying flat on the table is distinct from a doughnut on its edge).

As a result the [math]\frac{GMm}{r^2}[/math] formula isn't going to work.

 

 

Not only that, but any formula which just depends on r isn't going to work, you need two parameters to describe the field (because of circular symmetry, for something with no simplifying symmetry you need three parameters).

 

That being said, your intuition about the 0.0001 + r^2 is pretty good. I can't be bothered working it out for a thick torus right now right now, but the formula you get in situations like that if you restrict your location to the axis of symmetry (the middle of the torus, which you could spin it around and it wouldn't look any different) then you get something a lot like what you said

IIRC for a torus that is very thin compared to its radius (imagine a thin ring rather than a doughnut), it takes the form: [math]\frac{kr}{(r^2+R^2)^{\frac{3}{2}}}[/math]

Where R is the radius of the torus.

Which looks a lot like [math]\frac{k}{r^2}[/math] at long distances, but decreases as you get close to r=0.

 

If you go off the symmetry axis it gets more complicated.

 

For a thick torus it'll most likely look much the same, but probably have some logarithms in it.

 

This is about as far as I can go without introducing some more advanced maths. Are you familiar with vectors and/or integration perchance?

Edited October 1, 2011 by Schrödinger's hat

[Externet]

Thanks.

University vectorial analysis and calculus were 40 years ago; so I prefer to say am not familiar now. Understand your post; had to be some detail as you exposed were the formula in text books for celestial bodies and at extreme fine reality differ.

 

What would be your opinion, on what factors would be the reason [ to the specific case on the ISS and another object in it sharing both the same center of masses point with no 'effect' of external gravity ] not showing the masses attraction ?

[Schrödinger's hat]

What would be your opinion, on what factors would be the reason not showing the masses attraction ?

Not 100% sure I interpreted your question correctly, but I'll have a go anyway.

 

The object in the middle of the space station is attracted by it.

It's just that the attractions are all in different directions. If the object is at the centre of mass¹ of the space station, they will all cancel out. It's being pulled in all sorts of different directions, but the net result is it's not being pulled at all.

 

When the object is outside the space station, and a long way away, all of the different bits of the space station are pulling in the same direction, and they're pulling from roughly the same distance away so you can treat it like a point at its centre of mass.

 

Then, to confuse the issue we have big spherical things like planets.

You can use 1/r^2 for these because -- when you add up all the little pulls from different distances and directions in a sphere -- you get exactly the same result as if you were being pulled by a point at the planet's centre of mass. This only works from outside the sphere.

You can think of this as a coincidence². Ie. that the same formula for points just happened to work for spheres.

 

 

 

¹This is not true for non-spherical systems, but -- without getting into detailed explanations involving calculus -- it's close enough.

 

²It's not really, it stems from the force being spherically symmetric. It's the same reason the formula is 1/r^2 in the first place.

[matty]

Even a sheet of paper is a 3D object.

 

It has not just a given length and width but depth, thickness, as well in order even to exist.

Edited October 19, 2011 by matty