multibody gravity question

[Phil Hobbs]

On 3/16/2015 6:57 PM, jurb6006@gmail.com wrote:

"Could you give an example for a couple masses in 3D or 2D space
for how to do this analytically? "

I made a crude drawing :

https://dl.dropboxusercontent.com/u/29948706/grav01.jpg

1 is with two bodies acting equally on a body centered between their
relative masses. Doe snot have to be equal masses or equal distances
but the are reciprocal. More mass requires more distance.

2 is with three bodies acting, but all on one plne. That is in 2D.
There could be an infinite number of bodies there as lond as they are
all on the same orbital plane. This would be PFR (Pretty Fucking
Rare)

3 is even more rare. The arrows top and bottom illustrate that the
ONLY way to do this in 3 dimensions is for the whole system to be
spinning. Otherwise the masses acting on the body in the middle will
collide due to their own gravitation toward each other.

Now, this does not mean you have a stable systme here. The way
gravity works, that object in the middle moves just a hair and that
gravitational effect is no longer equal and is higher from the object
toward which it has moved.

This is positive feedback in a DC system. Once it crashes it crashes.
No such system is likely to exist in the univers but them again it
culd3. I ust think that sooner or later, even if by external forces,
it will not be sustainable. Even without external forces for this to
sustain it would have to be so precisely balanced it ain't funny. You
want to run it forever that way ?? Then there is no room for error
whatsoever. And I mean NONE. Because onbe little iota of a cunt hair
osf a quark of an electron of even a thought will throw it off.
Remember we are not talking the quarterly report, we are talking
forever.

Maybe my input here is not worth a shit, but really to gwt
gravitational forces equal is not easy nor likely. If you are talking
in a theoretical space like in mathematics maybe, but in the real
uivers, of all the kagillions of galaxies and stars, I give it about
50/50 of such a system existing. The variables are piled up against
it.

The original statement of the problem is that the masses are fixed.
That's an easy problem.

Once you let them go, it becomes far harder.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Jamie M]

On 3/16/2015 3:27 PM, Phil Hobbs wrote:

On 3/16/2015 6:17 PM, Phil Hobbs wrote:
On 3/16/2015 5:53 PM, Jamie M wrote:
On 3/16/2015 2:35 PM, Phil Hobbs wrote:
On 03/16/2015 05:07 PM, Jamie M wrote:
On 3/16/2015 1:25 PM, Phil Hobbs wrote:
On 03/16/2015 03:53 PM, Jamie M wrote:
Hi,

For a multibody stationary system of uniform mass points
distributed in
2D or 3D space, is there a way to precisely find or estimate the
location in 2D or 3D space where the gravitational field strength is
lowest in a given bounded domain?

ie for 2 masses 1 meter apart, centered in a 3D space that is a
2meter
cube, the gravitational field strength will be lowest directly
between
the two masses.

Are there examples of a stationary system where there are multiple
point locations with the same minimum gravitational field?

I am wondering if there is a single formula to calculate the
location
of minimum field strength or if an algorithm is needed to narrow
down
where the lowest field strength is and recursively narrow down the
field location which could use a lot of computation as I want to
make
this calculation each time I add a new mass to the system.

cheers,
Jamie

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

Cheers

Phil Hobbs


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


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


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


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:

http://en.wikipedia.org/wiki/Gravitational_potential

http://upload.wikimedia.org/math/1/2/8/128f3420b7f96f902df75b94a0df99cd.png

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

 

"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? "

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

It is still the same though, the force of any body is proportional to its mass and to the inverse square of distance. If things are spinning it is possible 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 nature, 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 to assume a system spinning within a system that is spinning at a perpendicular angle. It really is the only way this can exist in the physical universe.

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 lift some of his stuff off of Freenet from CWRU.

And hey, I might be wrong. Fact is I stole my education. Of course there are holes in it you could drive a truck through, but from what I've seen these 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 fucked whom. Sop how did this question pop up ?

Just curious.

 

[Jamie M]

On 3/16/2015 4:53 PM, Phil Hobbs wrote:

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!

 

[Reinhardt Behm]

Jamie M wrote:

Hi,

For a multibody stationary system of uniform mass points distributed in
2D or 3D space, is there a way to precisely find or estimate the
location in 2D or 3D space where the gravitational field strength is
lowest in a given bounded domain?

ie for 2 masses 1 meter apart, centered in a 3D space that is a 2meter
cube, the gravitational field strength will be lowest directly between
the two masses.

Are there examples of a stationary system where there are multiple
point locations with the same minimum gravitational field?

Imagine a system with an infinite number of equal masses positioned along a
straight line and equally spaced. You will have an infinity number of point
with minimum g field.

I am wondering if there is a single formula to calculate the location
of minimum field strength or if an algorithm is needed to narrow down
where the lowest field strength is and recursively narrow down the
field location which could use a lot of computation as I want to make
this calculation each time I add a new mass to the system.

cheers,
Jamie

--
Reinhardt

 

[DecadentLinuxUserNumeroUn]

On Mon, 16 Mar 2015 16:53:12 -0700, Phil Hobbs wrote:

you can find the minimum analytically, but it gets messier as N
increases.

That's good. Splattered his brains all over the void between.

It was an uber ober cut.

 

[Jamie M]

On 3/16/2015 8:45 PM, Reinhardt Behm wrote:

Jamie M wrote:

Hi,

For a multibody stationary system of uniform mass points distributed in
2D or 3D space, is there a way to precisely find or estimate the
location in 2D or 3D space where the gravitational field strength is
lowest in a given bounded domain?

ie for 2 masses 1 meter apart, centered in a 3D space that is a 2meter
cube, the gravitational field strength will be lowest directly between
the two masses.

Are there examples of a stationary system where there are multiple
point locations with the same minimum gravitational field?

Imagine a system with an infinite number of equal masses positioned along a
straight line and equally spaced. You will have an infinity number of point
with minimum g field.


I am wondering if there is a single formula to calculate the location
of minimum field strength or if an algorithm is needed to narrow down
where the lowest field strength is and recursively narrow down the
field location which could use a lot of computation as I want to make
this calculation each time I add a new mass to the system.

cheers,
Jamie

Ya that's true but only if as you move a point down the line in one
direction there are still the same number of masses on either side of
the point, otherwise the gravitational force will be infinitesimally
larger in one direction.

cheers,
Jamie

 

[DecadentLinuxUserNumeroUn]

On Tue, 17 Mar 2015 13:17:52 -0400, Phil Hobbs wrote:

That's good. Splattered his brains all over the void between.

It was an uber ober cut.


Not so intended. I don't think that knowing calculus makes one that
special, but it's sure useful.

Cheers

Phil Hobbs

It was the "gets Messier" Messy-er is, one would suppose, the only way
to make that distinction in modern times, eh?

 

[DecadentLinuxUserNumeroUn]

On Tue, 17 Mar 2015 16:02:02 -0400, Phil Hobbs wrote:

On 03/17/2015 02:20 PM, DecadentLinuxUserNumeroUno wrote:
On Tue, 17 Mar 2015 13:17:52 -0400, Phil Hobbs wrote:

That's good. Splattered his brains all over the void between.

It was an uber ober cut.


Not so intended. I don't think that knowing calculus makes one that
special, but it's sure useful.

Cheers

Phil Hobbs


It was the "gets Messier" Messy-er is, one would suppose, the only
way
to make that distinction in modern times, eh?


I wasn't thinking of nebulae at the time.

Think about a reader's POV.

I am surprised now that you did not do that deliberately, and now you
know the real reason I chuckled at first.

Gets messier when one of them blows up... hehehe...

 

[Phil Hobbs]

On 03/17/2015 02:31 AM, DecadentLinuxUserNumeroUno wrote:

On Mon, 16 Mar 2015 16:53:12 -0700, Phil Hobbs wrote:

you can find the minimum analytically, but it gets messier as N
increases.


That's good. Splattered his brains all over the void between.

It was an uber ober cut.

Not so intended. I don't think that knowing calculus makes one that
special, but it's sure useful.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Phil Hobbs]

On 03/17/2015 02:20 PM, DecadentLinuxUserNumeroUno wrote:

On Tue, 17 Mar 2015 13:17:52 -0400, Phil Hobbs wrote:

That's good. Splattered his brains all over the void between.

It was an uber ober cut.


Not so intended. I don't think that knowing calculus makes one that
special, but it's sure useful.

Cheers

Phil Hobbs


It was the "gets Messier" Messy-er is, one would suppose, the only way
to make that distinction in modern times, eh?

I wasn't thinking of nebulae at the time.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Robert Baer]

Jamie M wrote:

On 3/16/2015 1:50 PM, Jack wrote:
Jamie M <jmorken@shaw.ca> wrote:

Hi,

For a multibody stationary system of uniform mass points distributed in
2D or 3D space, is there a way to precisely find or estimate the
location in 2D or 3D space where the gravitational field strength is
lowest in a given bounded domain?

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

ie for 2 masses 1 meter apart, centered in a 3D space that is a 2meter
cube, the gravitational field strength will be lowest directly between
the two masses.

also if they moves.

Are there examples of a stationary system where there are multiple
point locations with the same minimum gravitational field?

For Sun-Earth system they are called Lagrangian points.

Bye Jack


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

cheers,
Jamie

Umpteen years ago, (BC, that is Before Computers) an N-body system
was insoluble where n>=3.
Things have improved a bit since then.
To make computation a bit easier,start with the solution you do have.
and posit one added item with correct position but "zero" mass, and
slowly increase ts mass.
I guarantee that the masses will not be in a stable configuration.

 

[Robert Baer]

jurb6006@gmail.com wrote:

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 worried 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 this.

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

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

As such you will eventually fall somewhere. It might take 99 quadrillion years, 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 relation to each other.
Does a tree out in the middle of nowhere make a sound when it falls?

 

[Reinhardt Behm]

Jamie M wrote:

On 3/16/2015 8:45 PM, Reinhardt Behm wrote:
Jamie M wrote:

Hi,

For a multibody stationary system of uniform mass points distributed in
2D or 3D space, is there a way to precisely find or estimate the
location in 2D or 3D space where the gravitational field strength is
lowest in a given bounded domain?

ie for 2 masses 1 meter apart, centered in a 3D space that is a 2meter
cube, the gravitational field strength will be lowest directly between
the two masses.

Are there examples of a stationary system where there are multiple
point locations with the same minimum gravitational field?

Imagine a system with an infinite number of equal masses positioned along
a straight line and equally spaced. You will have an infinity number of
point with minimum g field.


I am wondering if there is a single formula to calculate the location
of minimum field strength or if an algorithm is needed to narrow down
where the lowest field strength is and recursively narrow down the
field location which could use a lot of computation as I want to make
this calculation each time I add a new mass to the system.

cheers,
Jamie


Ya that's true but only if as you move a point down the line in one
direction there are still the same number of masses on either side of
the point, otherwise the gravitational force will be infinitesimally
larger in one direction.

cheers,
Jamie

Therefor I said "_infinite_ number of equal masses".
--
Reinhardt

 

[Jamie M]

On 3/17/2015 7:57 PM, Robert Baer wrote:

Jamie M wrote:
On 3/16/2015 1:50 PM, Jack wrote:
Jamie M <jmorken@shaw.ca> wrote:

Hi,

For a multibody stationary system of uniform mass points distributed in
2D or 3D space, is there a way to precisely find or estimate the
location in 2D or 3D space where the gravitational field strength is
lowest in a given bounded domain?

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

ie for 2 masses 1 meter apart, centered in a 3D space that is a 2meter
cube, the gravitational field strength will be lowest directly between
the two masses.

also if they moves.

Are there examples of a stationary system where there are multiple
point locations with the same minimum gravitational field?

For Sun-Earth system they are called Lagrangian points.

Bye Jack


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

cheers,
Jamie

Umpteen years ago, (BC, that is Before Computers) an N-body system
was insoluble where n>=3.
Things have improved a bit since then.
To make computation a bit easier,start with the solution you do have.
and posit one added item with correct position but "zero" mass, and
slowly increase ts mass.
I guarantee that the masses will not be in a stable configuration.

Hi,

The masses are in fixed positions so it is a static gravitational field,
but new masses in fixed positions can be added.

cheers,
Jamie

 

[Phil Hobbs]

 Umpteen years ago, (BC, that is Before Computers) an N-body system
was insoluble where n>=3.

Nope, not so. Three-body _orbits_ are insoluble _in_closed_form_in_terms_of_elementary_functions. That's a far weaker statement. Even a Keplerian orbit needs elliptic integrals in general, and of course even sines and cosines are transcendental functions that can't be computed exactly except for special values.

Perturbation theory was finding new planets by the disturbance in the orbits of other planets in the 18th Century. It doesn't need to be exact, it just needs to be good enough.

Cheers

Phil Hobbs

 

[Martin Brown]

On 18/03/2015 02:05, Jamie M wrote:

Hi,

The masses are in fixed positions so it is a static gravitational field,
but new masses in fixed positions can be added.

cheers,
Jamie

In that case you can always contrive to insert the next mass in such a
position as to create a perfect location where all the forces sum to
zero. The fields are linearly superposable so if you pick a location and
work out the field there due to the current mass distribution you can
place your next mass so that it produces an equal and opposite force at
that location. And since you have somehow nailed them down so that they
don't move at all under their mutual gravity you are done.

In practice they do move and the dynamical solutions stability depends
critically on the mass ratios between participating masses. Various
online simulators will let you play with multibody gravitation and some
will even let you collide galaxies fairly realistically.

Equilateral triangle trying to rotate uniformly about its point of
symmetry is a good one to try for starting point and highly educational.

This was all supercomputer simulation stuff way back when it was first
developed but i7-3770 modern PCs are broadly comparable with 100x Cray
XMP if Intels benchmarks are to be believed (seems high to me).

--
Regards,
Martin Brown

 

[Martin Brown]

On 18/03/2015 11:55, Phil Hobbs wrote:

Umpteen years ago, (BC, that is Before Computers) an N-body system
was insoluble where n>=3.

Nope, not so. Three-body _orbits_ are insoluble _in_closed_form_in_terms_of_elementary_functions.
That's a far weaker statement. Even a Keplerian orbit needs elliptic
integrals in general, and of
course even sines and cosines are transcendental functions that can't
be computed exactly except for special values.

It was hard lines on the poor computors who had to do the calculations
back in the days when it was all pencil paper and log tables.

Perturbation theory was finding new planets by the disturbance in the orbits of other planets in the 18th Century.
It doesn't need to be exact, it just needs to be good enough.

Challis would have beaten Le Verrier and Galle to finding Neptune if he
had trusted the calculations given to him by Adams. It is known that
Galileo actually observed Neptune in the same field as Jupiter in 1612
and noted it moved but didn't see the importance at the time.

Nice history of who saw what and when online at St Andrews astronomy.

http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Neptune_and_Pluto.html

It is an interesting aside that although our solar system appears to be
as regular as clockwork a proof of its long term stability is still an
unsolved problem and very little progress has been made on Ovenden's
conjecture which is the most plausible theory underpinning "Bode's law".
(which is in fact a misnamed crude heuristic)

--
Regards,
Martin Brown

 

[Phil Hobbs]

On 3/18/2015 8:44 AM, Martin Brown wrote:

On 18/03/2015 11:55, Phil Hobbs wrote:

Umpteen years ago, (BC, that is Before Computers) an N-body system
was insoluble where n>=3.

Nope, not so. Three-body _orbits_ are insoluble
_in_closed_form_in_terms_of_elementary_functions.
That's a far weaker statement. Even a Keplerian orbit needs elliptic
integrals in general, and of
course even sines and cosines are transcendental functions that can't
be computed exactly except for special values.

It was hard lines on the poor computors who had to do the calculations
back in the days when it was all pencil paper and log tables.

Yup, that's for sure. When Shanks finished his 707-place computation of
pi (of which the last couple of hundred places were wrong), he said it
had taken him "three years of Sundays". (i.e. 21 years, for the folks
at home.)

Perturbation theory was finding new planets by the disturbance in the
orbits of other planets in the 18th Century.
It doesn't need to be exact, it just needs to be good enough.

Challis would have beaten Le Verrier and Galle to finding Neptune if he
had trusted the calculations given to him by Adams. It is known that
Galileo actually observed Neptune in the same field as Jupiter in 1612
and noted it moved but didn't see the importance at the time.

Nice history of who saw what and when online at St Andrews astronomy.

http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Neptune_and_Pluto.html

It is an interesting aside that although our solar system appears to be
as regular as clockwork a proof of its long term stability is still an
unsolved problem and very little progress has been made on Ovenden's
conjecture which is the most plausible theory underpinning "Bode's law".
(which is in fact a misnamed crude heuristic)

Michael Ovenden was my galactic dynamics prof at UBC. A good guy, who
unfortunately died pretty young.

Cheers

Phil Hobbs



--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Jack]

Martin Brown <|||newspam|||@nezumi.demon.co.uk> wrote:

This was all supercomputer simulation stuff way back when it was first
developed but i7-3770 modern PCs are broadly comparable with 100x Cray
XMP if Intels benchmarks are to be believed (seems high to me).

you get much more GFlops using the graphic card to do computation
(OpenCL, Cuda,...).

Bye Jack
--
Yoda of Borg am I! Assimilated shall you be! Futile resistance is, hmm?