multibody gravity question

[Jamie M]

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

 

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.

 

[DecadentLinuxUserNumeroUn]

On Mon, 16 Mar 2015 13:29:17 -0700, jurb6006 wrote:


> Years ago I as cavorting with a broad on a sex website.

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 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. "

Huh ?

If you mean what I think you mean, you ay be right. OK honey let's have some fun. I will buy you dinner, clothes, whatever. But keep your itts off my houses, keys, cars, bank cards, all of that. We be friends, not sistas. We are not joined at the hip.

the Chairman Of The Board, as he was called, had four Wives, but really you did not hear much about them. they were no Hillary to say the least. It was low key compared to today when ant Kardashian sneezes it is a major catastrophe.

The media is not the same now. Amd I couldn't care less. there are some people who I would like to meet and possibly even admire to an extent, but it is iunlikely they were ever on TV. No, Nixon. you know that motherfucker won the nmoney for his Presidential campaign in a high stakes poker game ?

Now that I can respect. the rest, well...

 

"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.

 

[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

 

[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

 

[George Herold]

On Monday, March 16, 2015 at 7:17:37 PM UTC-4, Phil Hobbs wrote:

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.
(dang, speed of light.)

George H.

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 03/16/2015 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

--
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]

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

 

[Jamie M]

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

 

[Jamie M]

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

 

[Phil Hobbs]

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

--
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

 

[Martin Brown]

On 16/03/2015 21:04, 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.

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.

--
Regards,
Martin Brown

 

[Jamie M]

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

 

[Jamie M]

On 3/16/2015 2:47 PM, Martin Brown wrote:

On 16/03/2015 21:04, 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.

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,

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

 

[Phil Hobbs]

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

--
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 3/16/2015 5:47 PM, Martin Brown wrote:

On 16/03/2015 21:04, 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.

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.

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

--
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 3/16/2015 6:21 PM, Phil Hobbs wrote:

On 3/16/2015 5:47 PM, Martin Brown wrote:
On 16/03/2015 21:04, 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.

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.


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


--
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 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

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Dr Philip C D Hobbs
Principal Consultant
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