multibody gravity question

It's simple enough that you can do it analytically, and automate that.

Cheers

Phil Hobbs

Ha, I probably won't be of much help here but...

Years ago I as cavorting with a broad on a sex website. She wasn't too worr ied about thihngs because she is an Indiana state trooper and sharpshooter. Her 14 year old daughter had a question having to do with something like t his.

The question was when out in space will you fall. Well actually after getti ng understood about orbits and all that we are talking about if you are jus t out there, and "relatively" stationary", would you fall.

I had to answer yes. Because when you are out there, even though gravitatio nal force go by inverse square, one body's gravity will pull on you kore th an another. I almost want to go so far as to say that there is no place whe re all the gravitational forces are equal and cancel out.

As such you will eventually fall somewhere. It might take 99 quadrillion ye ars, but it has to happen eventually.

Even if you were to find a spot in the univers that has totally nulled out gravitational forces, that will change because everything it moving in rela tion to each other.

with 2 bodies yes (analitically), with 3 or more... you need to do some number crunching to solve the thing.

also if they moves.

For Sun-Earth system they are called Lagrangian points.

Bye Jack

The mere fact that you use the same moniker that Frank Sinatra used, pretty much says a lot about how lame your mindset must be when it comes to women.

The problem I'm working with has stationary masses only, but new masses are added one at a time.

cheers, Jamie

To do it analytically, I think I would just consider one mass at a time, and fill the field strength for grid points in 2D or 3D, then add the contribution from each subsequent mass, or is there a better method?

This wont give the exact point of minimum field strength, but it should be useful to give the grid point in 3D space of the lowest field strength (which is pretty much the same thing just less accurate)

cheers, Jamie

That's doing it numerically. Analytically, the potentials due to all the masses add, so you can trivially have an analytical expression for the total potential. Then you find the minimum numerically. (You can do that analytically too of course, but there can be quite a few local minima.)

Cheers

Phil Hobbs

You aren't making a lot of sense. Usually in multibody gravity simulations the whole object of the exercise is to compute how they will move under their mutual gravitational attraction.

Classic is to do a log2(N) trick where the closest together pairs are treated as the sum of their masses at the appropriate position applying conservation laws and so on upwards.

You might find the analytical book Celestial Encounters helpful to avoid reinventing the wheel Daicu & Holmes, Princeton 1996. Bit dated now but some things don't change - I expect a few more closed forms are known.

Warning it is not a particularly easy read. Advanced mathematics is assumed - I expect Phil H will enjoy it if he hasn't already seen it.

Hi,

Could you give an example for a couple masses in 3D or 2D space for how to do this analytically? Should I be combining all the masses and find the center of mass then that point should have no gravitational force, but it may not be the lowest field strength still, so really the point on the map furthest from the center of mass would have the lowest field strength? ie a part of the domain on an edge or corner of the domain will still have a gravitational force towards the center of mass, but may be a lower field intensity than the center of mass.

cheers, Jamie

Hi,

Thanks, for the application I am using the multibody gravity for, the masses do have fixed positions, but actually it is not a 3 dimensional problem, I have n-dimensional space of stationary objects! :)

cheers, Jamie

As I say, you just add up the gravitational potentials from each one. At a point *X* [vector, i.e. (x, y, z)], the potential is

phi(*X*) = G sum M_i/|*X* - *X_i*| , i=1 to N

where *X_i* is the CM postition of the i_th mass. If the masses are very far from spherical, you may need to use higher multipoles. Everything you want to calculate goes as some derivative of the potential.

Cheers

Phil Hobbs

Does sound fun.

Haven't done any of that stuff since I was an honours astronomy/physics undergrad, 1978-81. It was pretty well all analytical in those distant days--inner and outer Lindblad resonances, density waves, multipole perturbation theory, that sort of stuff. Fun. (Of course as an undergraduate you really just dip a toe into the real work of a field like that.)

Cheers

Phil Hobbs

Okay, there were a few on ABE for $5, so I took a flyer.

Thanks for the steer.

Cheers

Phil Hobbs

Forgot an overall minus sign. The gravitational potential is always negative. Force is minus the gradient of the potential.

Cheers

Phil Hobbs

Hi,

Thanks, all the masses are just point masses. Does this mean for each configuration of n stationary masses, I need to use calculus to find an equation that can solve for the point in space of lowest gravitational field intensity?

I found the formula you mention here:

I think that just solves for the potential at point x, but I would like to find the exact point with lowest field strength analytically.

cheers, Jamie

Jamie,

As I said, in any given situation, you can find the minimum analytically, but it gets messier as N increases.

Have you taken first year calculus? It's all the same stuff.

Cheers

Phil Hobbs

Jamie,

As I said, in any given situation, you can find the minimum analytically, but it gets messier as N increases.

Have you taken first year calculus? It's all the same stuff.

Cheers

Phil Hobbs

Thanks Phil! :)

OK, so in this instance it is either one point in time, or purely theoretic al. If the stuff ain't spinning it ain't happeming, unless of course you ar e talking about one point in time.

It is still the same though, the force of any body is proportional to its m ass and to the inverse square of distance. If things are spinning it is pos sible in this universe though as I said, would be extremely rare.

The spinning unitary system in figs 1 and 2 are less unlikely to occur in n ature, but that might not satisfy your putforth. If it is only true on one plane then the object in the middle cannot deviate from the plane witout an outside force. that means it is affected. So in any model you would have t o assume a system spinning within a system that is spinning at a perpendicu lar angle. It really is the only way this can exist in the physical univers e.

However, as my math dude has pointed out, spac is not alays space, which is on top of what I read from Professor Robert F. Heeter about velocity space . He was talking about particles in a reactor, velocity space is defined by relative volocity and vector. I didn't study under Mr. Heeter but I did li ft some of his stuff off of Freenet from CWRU.

And hey, I might be wrong. Fact is I stole my education. Of course there ar e holes in it you could drive a truck through, but from what I've seen thes e days, people with those crippling student loans are not much better.

It is good to see people wanting to discuss such things. In school or not, mosdt people want to talk about the party last Friday or who f***ed whom. S op how did this question pop up ?

Just curious.