# 4 bodies moving in a circle treated as two systems

This is a follow-up of question http://www.experts-exchange.com/Other/Math_Science/Q_26846773.html
The 4 bodies have equal mass (m) and same initial speeds. Three of the bodies, known as B, C, and D will be treated as one system. The fourth body, A, is the second system. My goal is to derive an effective single point entity that represents the B-C-D system in order to compute the force on A due to the B-C-D system.

Their initial configuration is a square (whose diagonals have a length 2r), and their velocities are shown here:
``````           Va
A -->
..
Vb^  .  .
| .    .
|.      .
B        D
\      /|
\    / |
\  /  v Vd
\/
<-- C
Vc
``````

Assumption:
In order to determine the force acting on A due to the B-C-D system, I thought I could treat the B-C-D system as an effective single entity having mass equal to 3m and whose location is the CM of B-C-D.

If this is an incorrect assumption, could you provide me with the correct way to derive an effective single entity that acts on A.

Using this assumption, I determined the CM to be 1/3 the distance on the line from C to the other system, A. (As expected, the CM of the triangle is on a bisector of the angle BCD.) Since ABCD is a square, angle BCD = 90 degrees). Since the diagonal length is 2r, then the distance from A to the CM is (2/3)(2r) = (4/3)r.

But when I compute the force on A by this effective single 3m entity at the CM, the  force on A is different than the force in the previous related question. I am looking for consistency.

Maybe arithmetic or algebra is wrong, or maybe my assumption is wrong.
LVL 33
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Commented:
>> My assumption was that I could take the distance from r_A to r_cm and then plug it into the gravitational force formula to compute F_external.

This is NOT true if you define CM as
>>CM defined as a position vector, r_cm, is determined by:

>>Mt * r_cm = sum( m_i r_i )

Consider this example to see why:
Particle b and c are 1 unit above particle a on the y axis,
But they are one million units to the right and left of particle a, on the x axis.
The CM is 1 mm above particle a (large force), but this is NOT equivalent to the forces exerted by particles that are a million units away (small force).

I suggest you define CM as follows:
(F and D are vector quatities. sorry I used D as the distance between particles, rather than r- since r was used in the initial question a bit differently)

Sum the forces acting on particle A to get Ftotal, the force on A on the “effective”particle
F=Gmm/D^2
|F|=Gmm/|D|^2
Ft=Fb+Fc+Fd
Note: |Ft|  !=  |Fb|+|Fc|+|Fd|
Arbitrarily(?) assign mass to effective particle
Mt=3m
Now to determine position of the “effective” particle
Ft=GmMt/Dat^2
Plug in Ft from above
Solve for Dat
or more generally:
Sqrt(Mt/m(SUM:0->i(1/Dai^2))=Dat
more specifically:
Sqrt(Mt/m(1/(-r,r)^2+ 1/(0,2r)^2+ 1/(r,r)^2))=Dat

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Commented:
You can't replace the three corner masses by the center of mass. for gravitation calculations.

CM is a linear calculation.  Gravity is a inverse square.

Think about a small test mass at the center of your square.
All the forces are near balanced, not much going on.

Replace the corners with 4M at the center, and the test mass dives in.
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Author Commented:
When you consider all the atoms of the Earth each applying a gravitational force on a body (say, a satellite around the Earth), I thought we simplify the calculations by just taking the CM of the Earth (near the center of the Earth), and then say the entire mass of the Earth is at that point. Then the force on a body of mass m due to the Earth (if outside the boundaries of the Earth) is |Fg| = G Me m/r^2 where Me is the mass of the Earth, and r is the distance from the body to the Earth's CM.
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Commented:
>> I thought we simplify the calculations by just taking the CM of the Earth (near the center of the Earth), and then say the entire mass of the Earth is at that point.

You can only do this if the matter is even distributed spherically around that point. In fact the earth is not spherical nor even, as the necessity for the continual correction of satelitte orbits shows.
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Commented:
The Earth is close enough to being spherically symmetrical to be reasonably approximated as a point for purposes of problems like this.
The B-C-D system is not spherically symmetrical.
However, under the assumption that the A-B-C-D system starts and stays radially symmetrical,
we would know that any movement of A will be matched by movements of B-C-D that maintain the symmetry,
and all distances will adjust proportionally.
Under these assumptions, the direction of the force on any body in the A-B-C-D system will always be directly in line with the CM of the A-B-C-D system, and the magnitude of the force will be proportional to the inverse square of the bodys distance to the CM of the A-B-C-D system.
So to compute the equivalent mass at the CM of the A-B-C-D system, you just need to find the force on any body in the A-B-C-D system, and figure out what mass at the CM of the system will produce an equivalent force.
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Author Commented:
My (erroneous?) assumption in the OP was based on F = dP/dt, where P is total linear momentum of a set of particles. In absense of an external force, the total P remains constant. I thought that if there were a single external body exerting a force on the set of particles, then the system's CM (having a point mass of the total mass of the system) would accelerate via F = dP/dt = M_cm * a_cm. Then I thought that the reaction force on the single external body would just be -dP/dt = -M_cm * a_cm.

I haven't completed the review. I know something is wrong.

What I was hoping to do in my next question was to allow the initial speeds to vary from V_orbit and then using a simpler calculation to determine the orbits min/max distance from the center of the square, as well as to compute the escape velocity (in my existing open question).
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Author Commented:
From my review:
-----------------------------------------------------------------------------------
Have a system of particles
r_i   = position vector of particle i
m_i = mass of particle i
Mt   = sum( m_i ) = total mass of system of particles
v_i  = velocity of particle i; v_i = r_i'
a_i  = acceleration of particle i; a_i = v_i'

Pt   = total linear momentum of system of particles = sum( m_i v_i )

CM defined as a position vector, r_cm, is determined by:

Mt * r_cm = sum( m_i r_i )

first derivative:
Mt v_cm = sum( m_i v_i ) = Pt

2nd derivative:
Mt a_cm = Pt'

Pt' = F_external (since internal forces cancel out due to action/reaction)

F_external = Mt  a_cm

-----------------------------------------------------------------------------------

My understanding is that the above notes do not require any kind of spherical symmetry.

So, body A (which is external to the BCD system) is applying a force to the CM (r_cm) of BCD which results in an an acceleration of BCD's CM (r_cm) as follows:
F_external = Mt  a_cm = 3m  a_cm

I thought I could treat BCD as a single body of mass 3m which is concentrated at r_cm. And then the force of BCD acting on A would just be the reaction force,  -F_external.

My assumption was that I could take the distance from r_A to r_cm and then plug it into the gravitational force formula to compute F_external.

Which step(s) is/are wrong?
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Commented:
>>Pt   = total linear momentum of system of particles = sum( m_i v_i )

They are all moving in different directions and thus you cannot sum them thus.

I'm not happy about the approach, since it seems to assume a lot.

Take fior starters the force acting on a mass, m, at the corner of a sqaue of side a, where the other three corners have the same mass. Assuming that the corner (top left) has the coordinates x,y then the force from the top right mass is Gm²/(x-a)² since the y coordinate remains constant. Similary for the bottom left we have Gm²/(y-a)². The bottom right gives Gm²/[(x-a)²+(y-a)²] Adding these terms together then gives with Newtons Law a differential equation to determine the motion, which, in this case, assumes that the masses remain at the same distance relative to each other. As you will see the DE is not simple. Perhaps a formulation in spherical coords would be better - does E-E allow the symbol theta? I'll try it  ¿ ¿
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Author Commented:
Firstly, I guess EE doesn't yet like theta (but there are plans in the future to bring theta into the fold), but it loves beta ( ß ), and also, what I'll call lamda ( £ ) (which is great for eigenvalues), as well as mu ( µ ),  and upper/lower phi ( Ø or ø ).

>> then the force from the top right mass is Gm²/(x-a)²

In the previous related question noted in the OP, I learned how to compute the net force on body A from the other 3 bodies, B, C, and D. See
http://www.experts-exchange.com/Other/Math_Science/Q_26846773.html?cid=2035#a35011879

>> They are all moving in different directions and thus you cannot sum them thus.
Here is a reference which also suggests that it is OK to  sum the individual P_i's to get a total Pt and then the internal forces cancel out leaving only the external force to the system.
http://en.wikipedia.org/wiki/Momentum#Linear_momentum_of_a_system_of_particles
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Commented:

>>The key physics principle is that the masses will move in a circle IF there is a centripetal force applied.

Applies to a stationary mass at the center with a mass some point futher off AND that the orbit is in fact stationary, ie: the inward force causes the circulating mass to "fall" into an orbital path.

This principle DOES NOT apply to a multi-bodied situation, whose general situation is unsolvable.

There is an implicit assumption that the circular orbit is indeed possible, and what I would like to see is a proof of that from first principles.

>>>> They are all moving in different directions and thus you cannot sum them thus

Perhaps I should have written: What good does that do? Since one is interested in the actual motion of an individual particle and not the "center of mass".

The question, as asked, is at what rotational velocity will the system be unstable - ie: one mass will "fly off". This precludes the stability of the arrangement in the first place. If one could obtain a solution to the first problem (which you think has been solved) then a perturbation will solves the second.

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Author Commented:
>> Applies to a stationary mass at the center
In previous related question, there was no stationary mass at the center; and yet, we got a solution. If you want to read the entire previous link and can show that the solution is incorrect, then I will reopen the question. After the lengthy discussion (some of which was due to some terminology issue), I became convinced that the solution was OK.

>> whose general situation is unsolvable
I think that the since there was symmetry in the previous question, this symmetry resulted in some easier equations so a solution was found.

>> What good does that do? Since one is interested in the actual motion of an individual particle and not the "center of mass".
I am interested in learning about clusters of particles and the relation of their CM to external forces (and whatever else is related to this topic).

BTW - stability was covered in the previous question.
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Commented:
>> I became convinced that the solution was OK.

Well, I',m not, for the simple reason that aburr's derivation is based on results of a two body system.

But to answer the question, if you take aburr's formular for the force, which I get to be (Gm²/4r²) * (1+2*sqrt(2)) and equate this to a mass X at the center acting on a mass m at one of the corners the force at the corner will be GXm/r² and hence X will be m*(1+2sqrt(2))/4. This has to concur with the result from statics which I'll do later.
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Author Commented:
>> mass X at the center acting on a mass
Actually, in this question, I am trying to focus just on CM of the BCD bodies acting on A and vice-versa. This question is not concerned about equivalency possibilities at the center of the square.

Why do you say that aburr's derivation is based on results of a two body system? Can you pinpoint an error in the reasoning?

BTW - from aburr's solution, I had done the algebra and came up with the force on each of the four bodies as (Gm²/4r²) * (1+2*sqrt(2)) , so we are in agreement w.r.t. the net force (and its direction is radial - pointing to the center of the square, as aburr also pointed out).
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Commented:
You can place four identical masses at the corners of a square, find the net forces
on each mass due to gravity.

For each mass, these forces are directed toward the center of the square, which is
also the center of mass for the system.  If you give the masses the proper
transverse velocity they will rotate in stable circular orbits.

You can not move three of the masses, (B C and D) to their center of mass, and
expect  at the system to behave the same way.  You have to move each mass  closer to A, so the force on A goes up substantially.  But you already found this out.

You CAN replace B, C, and D by a single mass M so that for force on A and the
center of mass (and the orbit of A) remain the unchanged.

There are two equations:

F  = (Gm²/4r²) * (1+2*sqrt(2))  = (Gm²M/(r+x)²)      and    xM = r*m

I solved nummerically to find        M = 2.091*m     and    x = r/2.091
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Commented:
For the record :-

"The key physics principle is that the masses will move in a circle IF there is a centripetal force applied. F = (m * v^2)/r ."

Well, in ideal conditions yes, but in practice the orbits are ellipses, due to the fact that they are gravitating masses and not balls on the end of strings. The equation is the result of a two body analysis. The only argument which I have seen regarding your four body system is the principle of symmetry and equivalence, which may or may not apply.
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Commented:
Solving numerically, the force on A at radius r from the CM of the A-B-C-D system will be equivalent to the force from a mass of approximately
2.091281547*m
at a distance of approximately
0.478175691*r from the CM of the A-B-C-D system
This is only true because we know that the movements of A will be symmetrically mirrored by movements of B-C-D
Any deviation from perfect radial symmetry will grow to destroy that neat relationship.
Also, it is only from exactly at the position of A that the equivalence will hold.
(the equivalence for a point mass at the center of a spherically symmetrical mass distribution will hold
through all space outside the sphere)
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Author Commented:
FYI - I am getting close (I hope) to closing the 4-body escape velocity question (in case you are interested):
http://rdsrc.us/Tw9Zoa

Once closed, I'll be able to go through your responses. Thanks again.
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Author Commented:
Now that the other two previous questions are closed, I'll try to address comments made here as I absorb them.

>> Well, in ideal conditions yes, but in practice the orbits are ellipses, due to the fact that they are gravitating masses and not balls on the end of strings.

>> Well, I'm not, for the simple reason that aburr's derivation is based on results of a two body system.

These comments would have been better placed in one of the previous related questions.

We actually did cover elliptical paths in both questions. We computed the speed necessary to go in a perfect circle (assuming perfect spherical masses perfectly placed in a square with same specified rules about velocity). If the speed in the OP is less than Vp (i.e., the perfect speed that produces a circle) and large enough so the objects don't fall in so close to each other as to collide, then the path is an elliptical path where the bodies forever fall inward towards the square's center and then back to its initial position. We also determined that if Vp < V < Ve, (i.e., Ve = escape speed) then we also get an elliptical orbit. If interested, see:
http://rdsrc.us/V8QUd2   and   http://rdsrc.us/KGbqGq

Each body applies a force on the other bodies. One body experiences a force from 3 bodies, so I don't think it really matters how the centripetal force comes to be to produce the orbit.

I think I understand your practical concerns about how realistic this scenario could occur. But for those two questions, I am please that we were able to come up with theoretical results that I now believe are consistent with Newtonian physics.
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Author Commented:
@d-glitch,
Thanks for the response.
>> You CAN replace B, C, and D by a single mass M so that for force on A and the center of mass (and the orbit of A) remain the unchanged.
Sorry, got confused with the underlined part. I first thought you were saying that we can compute an effective mass, M, at the CM of BCD which produces the same force as experienced by A; namely,
F  = (Gm²/4r²) * (1+2*sqrt(2))

In fact, up to that point, I thought this was the same conclusion that ozo made in http:#35078371 :
So to compute the equivalent mass at the CM of the A-B-C-D system, you just need to find the force on any body in the A-B-C-D system, and figure out what mass at the CM of the system will produce an equivalent force.

But then you said
>> There are two equations:
F  = (Gm²/4r²) * (1+2*sqrt(2))  = (Gm²M/(r+x)²)      and    xM = r*m
and you found that:
x = r/2.091
But in the OP, I wrote:
>> the distance from A to the CM is (2/3)(2r) = (4/3)r

So, if my CM is correct, then I have to assume that your x is not related to the CM of BCD. Then, what does your x represent?

Also, could you please elaborate on the meaning of your 2nd equation: xM = r*m
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Commented:
My first post (and the first post to this question) said that you can not move B, C, and D to their Center of Mass
and expect their effect on A to stay the same.  This should be obvious:

B and D move   from d = 1.414*r   to  1.333*r
C moves from d = 2.000*r   to  1.333*r

All the masses move closer and all the forces are aligned.  The net force on A increases substantially.

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

You can do as ozo suggested, and find a mass at the CM that will generate the same force on A.
This mass will be smaller than 3m,  and the center of mass of the system will be different.
This two body system will not have the same stable orbital period or radius as the four body system.

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

You can do as I suggested in a later post, and find a mass that will generate the same force on A and leave
the CM for the system unchanged.

The position of this mass is not at the CM of BCD.  The orbital behavior of A in the new system is unchanged.

Your 4-body system and my 2-body system are the only configurations that give the same orbital behavior for A.

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

The equation     xM = r*m   describes a two body system with the CM at zero.

This mass will be smaller than 3m,  and the center of mass of the system will be different..
This two body system will not have the same stable orbital period or radius.

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Commented:
> I have to assume that your x is not related to the CM of BCD. Then, what does your x represent?
distance to the CM of ABCD
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Author Commented:
@ozo,

From d-glitch's explanation, my guess is that it represents a distance away from CM of 4-body system radially opposite from A. It's meaning, I think, is that if you replace BCD with M at x, then we end up with a binary system whose CM is still the original CM of the 4-body system. I will be working out 2-body problem keeping  each at same angular velocity, and see if this interpretation is correct.

I will also try to verify his statement, "This two body system will not have the same stable orbital period or radius as the four body system."

In your first post, you talk about replacing an equivalent mass at CM of the ABCD system, which is the center of the square of the original ABCD system. Were you saying that when doing this, we should assume that the new effective mass is fixed?

@all,

I think you are all saying my assumption is wrong. After a review, I posted http:#35093808 to explain the reason for my (wrong) assumption. I still don't have a good handle on the explanation in http:#35093808 and CM of a set of bodies. Maybe the OP question is not the best, and maybe I will have to try a new question to focus on http:#35093808
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Commented:
> "This two body system will not have the same stable orbital period or radius as the four body system."
"A" will have the same orbits with the same periods as the four body system, but unlike the 4 body system,
the two body system would be stable in that small perturbations would cause small changes in the orbit.

> replacing an equivalent mass at CM of the ABCD system, which is the center of the square of the original ABCD system. Were you saying that when doing this, we should assume that the new effective mass is fixed?

In that context the effective mass would have been fixed.
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Author Commented:
If the particles represent stars, then I am truly illuminated by this discussion.
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Author Commented:
Thanks all for this enlightening subject.
I liked Korbus response by setting up extreme configuration to make his point.

Happy holidays to all of you!
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EngineerCommented:
Can somebody explain what is going on?????

I get an expert alert about half an hour ago  that this question has been asked which matched my expert alert. The entire question is spelled in the email. But when I open this question I find that it was asked several months ago and is already accepted, probably today.
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Author Commented:
@ssaqibh,
That is strange. You should hit the Request Attention button and ask a moderator.
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