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  4. Does a gravity calculation always work?
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Does a gravity calculation always work?

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Offline jeffreyH (OP)

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Does a gravity calculation always work?
« on: 15/02/2014 04:37:13 »
What about calculating the gravity of an object when the centre of gravity contains no mass? How can the forces be accurately determined as the field distribution does not contain mass at the centre?
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Offline JP

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Re: Does a gravity calculation always work?
« Reply #1 on: 15/02/2014 21:02:45 »
You can write Newton's law of universal gravitation out in integral form and do the calculation as long as you know the mass density of the object over space.
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Offline jeffreyH (OP)

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Re: Does a gravity calculation always work?
« Reply #2 on: 15/02/2014 22:34:14 »
Quote from: JP on 15/02/2014 21:02:45
You can write Newton's law of universal gravitation out in integral form and do the calculation as long as you know the mass density of the object over space.

Thanks for that. I wondered what type of answer people would give. I am hoping to do this for various shapes of mass.
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Offline Bored chemist

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Re: Does a gravity calculation always work?
« Reply #3 on: 16/02/2014 09:47:32 »
Are you talking about something like a donut?
There's no dough at the C of G, but the G of G is still there.
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Re: Does a gravity calculation always work?
« Reply #4 on: 17/02/2014 00:06:17 »
Quote from: Bored chemist on 16/02/2014 09:47:32
Are you talking about something like a donut?
There's no dough at the C of G, but the G of G is still there.

But the forces cannot be calculated in the same way as for a sphere where the centre of gravity does contain mass. The mass distribution has to be taken into account. I am looking into numerical relativity to find a solution.
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Offline JP

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Re: Does a gravity calculation always work?
« Reply #5 on: 17/02/2014 14:40:20 »
Quote from: jeffreyH on 17/02/2014 00:06:17
Quote from: Bored chemist on 16/02/2014 09:47:32
Are you talking about something like a donut?
There's no dough at the C of G, but the G of G is still there.

But the forces cannot be calculated in the same way as for a sphere where the centre of gravity does contain mass. The mass distribution has to be taken into account. I am looking into numerical relativity to find a solution.

There's nothing magical about when the center of mass contains mass.  It's the same calculation, but in some cases you can simplify the general formula.  For example, you could use numerical methods to compute the field from a general distribution of mass, but in the case of a non-rotating, spherically symmetric distribution, you get the Schwarzschild metric as a simple solution (outside the body).  You could also use more rigorous methods of numerical relativity to get the same answer, as the Schwarzschild solution is just a way of expressing the solution of the equations.
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Re: Does a gravity calculation always work?
« Reply #6 on: 17/02/2014 16:46:54 »
Just integrate F = GMm/r^2 over the shape and test position you have. Most standard symmetric integrals (shell, ring, torus, etc. ) are tabulated in textbooks, or you can resort to elemental analysis if you have ther time and inclination. It only gets difficult when the elements are separated and moving - i.e. astronomy.
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Offline jeffreyH (OP)

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Re: Does a gravity calculation always work?
« Reply #7 on: 18/02/2014 15:25:16 »
Quote from: alancalverd on 17/02/2014 16:46:54
Just integrate F = GMm/r^2 over the shape and test position you have. Most standard symmetric integrals (shell, ring, torus, etc. ) are tabulated in textbooks, or you can resort to elemental analysis if you have ther time and inclination. It only gets difficult when the elements are separated and moving - i.e. astronomy.

That is something I will include. Although I need to have small enough segments that feedback is negligible. The last plot of Ve from centre of gravity in the new theories post is a simplification and my thoughts are that I will get a logistic curve out from the centre that looks like the representation shown for an event horizon. Of course, though the field strength will be much lower, I feel the that shape will be the same. I may well disprove this assumption.
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