# Escape velocity of 4 bodies

This is a follow-up of question http://www.experts-exchange.com/Other/Math_Science/Q_26846773.html
The 4 bodies have equal mass and same initial speeds. Their initial position is in a square with their velocities shown here:
``````           Va
a -->
/\
Vb^  /  \
| /    \
|/      \
b        d
\      /|
\    / |
\  /  v Vd
\/
<-- c
Vc
``````
In the previous question, we learned that the perfect speed |Vp| required so that the 4 bodies rotate in a circle that inscribes the square, is determined by the equation:
f = G (m²)/(2*r)² + 2*G*(m²/S²)*cos 45 = m |Vp|²/r
where r is the radius of the inscribed circle and S is the length of the side of the square.

Now, if the initial speed is greater than |Vp|, then it may be possible that the bodies escape from their orbit. How do you determine the escape speed, |Ve|? If it is not possible to get a formula for |Ve|, then what numerical algorithm can be used to determine the value?
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Commented:
if you have exactly enough energy to escape to infinity, PE and KE would go to 0 at infinity
exactly 0 would be a parabolic orbit.
KE + PE sightly greater than 0 would be a hyperbolic orbit.
KE + PE less than 0 would be an elliptical orbit
KE = -PE/2 would be a circular orbit around a point mass.

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Commented:
Escape velocity is when kinetic energy equals gravitational potential energy
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Commented:
Anything over Vp will cause an escape.
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Commented:
Ozo, if it's in orbit already, then it needs to be greater than, not equal to, or it will remain in orbit.
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Commented:
One of us really should write that simulator.
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Commented:
Over circular orbit speed but less than escape velocity will be an elliptical orbit.
Generally escape velocity is sqrt(2) * circular orbit  velocity i.e. double the kinetic energy,
but I'm not sure if its the same when orbiting distributed masses instead of a point mass.
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Author Commented:
>> One of us really should write that simulator.
Well, I'm hoping for an analytic approach. And then use a simulator to verify the results.

>> Anything over Vp will cause an escape.
What are the equations to be used to show this?

>> Escape velocity is when kinetic energy equals gravitational potential energy
For one body, KE = 1/2 m v²

For PE and escape velocity, I found:
http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html
http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html#c2

I see that PE has a negative sign. So, when r -> oo, then PE -> 0. If the 4 bodies escape, then does this mean that they go to oo (where their PE increases to 0)? And, at oo, their KE is still > 0 (because if =0, then the bodies would fall back to each other)?
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Commented:
This is an interesting subject. The easiest way to get a grip on the subject is to look up “escape velocity” in the index of your (or another) general physics text book. But a summary can be given here.
The calculations of the previous problem are not relevant.
To escape the earth,  consider a simplified version of the problem. You have a point mass m on the surface of the earth (mass M and radius R) neglecting all friction.. We consider two energies. The potential energy of the system  ( or the PE of m in the gravitational presence of M). Consider the KE of m (in the frame of reference attached to M)
The gravitational PE of m is  PE = -GMm/R
The KE of m is 0.5 * m * v^2
The initial energy of m I          E = PE + KE = -GMm/R + 0.5 * m * v^2
The final energy when m has escaped is Ef = 0 + 0 becaise escape velocity is defined at the velocity necessary to just get m to the place where gravitational PE = 0 (hence KE there = ) too). Because the only energy involved is the initial KE (ie a long running rocket engine is not present) we have
0 = -GMm/R + 0.5 * m * v^2n   or 0.5 * m * v^2 = -GMm/R.
Hence v = (2GM/R)^0.5
Or putting in the values for G, M, and R       v = 11.2 km/s
\Note tha the value of m cancelles out so it does not make any difference wheather you have 3 or 4 or 256 small masses.
It does not make any difference whether the v is up or sideways because there is no friction and M soon becomes pointlike. (KE is not a vector)
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Author Commented:
Thanks all for your replies. I'll be back with my equation attempts (or else a plea).

>> The easiest way to get a grip on the subject is to look up “escape velocity”...
Right, aburr. I'm checking out some course videos to freshen up on G PE and escape.
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Author Commented:
Ok, per aburr's advice, I have reviewed 3 video course lectures on Work-Energy, KE, PE, Conservation of Energy, and Escape Velocity.

>> Generally escape velocity is sqrt(2) * circular orbit  velocity i.e. double the kinetic energy,
>> but I'm not sure if its the same when orbiting distributed masses instead of a point mass.

For the 4-body problem (in a square, so only need to consider R2):

- If you pick 3 of 4 bodies and consider their potential field, it is not just U(x,y), but rather U(x,y,t). But I don't think the t will complicate the escape velocity solution.

- My initial belief now is that the escape velocity for one body will be less than sqrt(2) * circular orbit  velocity since the CM of the other 3 bodies is not fixed, and is moving away from one body being affected by the CM of the other 3 bodies. I will see if I can figure out the escape velocity this weekend. If anyone else comes up with a formula for escape velocity, please post for comparison purposes (no solutions please - I don't want to be influenced by your approach).
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Commented:
Not necessarily the simplest way to do it but the most fundamental Consider just the four masses
Initial    sum of KE = 0               final      sum of KE = 0
Sum of PE = P                  sum of PE = 0
Therefore one must supply P worth of KE to the four masses.
Therefore one needs to find P which is the amount of work you can get out of the system whn you assembled it. Which is the amount of work you can get out of the system when you assemble it.
Energy with 0 masses      0
Energy with 1 masses       0
Energy with 2 masses      PE = G m M/R = (G * m^2)/S                  S = side of square
Energy with 3 masses       PE             (G * m^2)/S  + (G * m^2)/2r      (r = ½ the diameter
Energy with 4 masses      PE            (G * m^2)/S + (G * m^2)/S + (G * m^2)/2r = P
Thus KE required  = (1/2) * 4 * m * v^2
Now if the four masses are assembled on the surface of the earth, see my earlier post
Nothing has changed. (The additional PE of assembling the four masses is negligible compared to the PE due to the earth
HENCE
v = 11.2 km/s
ie the escape velocity for ANYTHING is the same. It is a function of the large mass from which you are escaping Just the conservation of energy
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Commented:
If the symmetry is preserved, the bodies will move in sync, which means that the force they feel will be proportional to their distance from the center of mass of the system, so their orbit behavior would be exactly the same as orbits around a single mass at the center of mass of the system.
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Author Commented:
aburr, I glanced at your latest post. From the length, you may have provided a full solution instead of an answer. You don't know how hard it is for me not to look at your details at this time. So far, I have resisted temptation. I'll tackle this problem and put down my answer and compare.

If anyone else tries this and gets a different answer, then please don't show the solution, just the answer.

Also, from the lectures, they like to relate escape velocity in terms of orbit velocity.
Ve = f( Vo )

I'll try to do that as well. If anyone wants to show their f( Vo ) without the derivation, please; that would be useful.

Thanks again!
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Commented:
Yes that was a full solution - sorry
in short your 4 masses will leave the earth if they  (or any one) has a velocity greater then
11.2 km/s
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Author Commented:
aburr, still not looking, so no problem. :)
Would you like to show f( Vo ) for this problem using symbols (e.g., like G, m if they don't cancel out) rather than numbers?
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Author Commented:
Hmm, I just realized something. The problem was stated with symbols, like r, S, m, Vp (which is what I'm now calling Vo, pure circular orbital velocity). So, I would expect that Ve is a function of some of these symbols rather than a fixed value of 11.2 km/s.
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Commented:
In my first post I said
v = (2GM/R)^0.5
where G is Newton's constant r is the radius of the earth and M is the mass of the earth.
I know, it says nothing about S or m becasuse they play no part in the escape velocity
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Commented:
Since all the distances in a symmetric system are proportional to the distance to the centre of mass, the relationship between orbital velocity and escape velocity will be the same for a four body system as for orbits around a single mass,
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Author Commented:
aburr, Ok, Sorrry. When I saw that post, I read the first line where you advised me to read to look up “escape velocity”, and I stopped reading your post, since I thought I needed to review the material first. Then I forgot to go back to reading your post again. Sorry again. I now see that:

"a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth. The surface velocity decreases with the cosine of the geographic latitude, so space launch facilities are often located as close to the equator as feasible"
http://en.wikipedia.org/wiki/Escape_velocity

I am hoping to derive Ve = f(r,m) for the 4-body problem in a universe in which there are just these 4 bodies, each having a mass, m, and initial radius, r, orbiting around the origin in a plane.

>> Since all the distances in a symmetric system are proportional to the distance to the centre of mass, the relationship between orbital velocity and escape velocity will be the same for a four body system as for orbits around a single mass
Sounds reasonable. I'll try to verify this. Then to get the actual Ve = f(m, r), then I guess one alternative approach would be to determine the effective virtual stationary single mass at the origin.
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Commented:
In my second post I derived the escape vel of just the 4 masses as
"Thus KE required  = (1/2) * 4 * m * v^2 =  (G * m^2)/S + (G * m^2)/S + (G * m^2)/2r = P"  (solve for v)
(now there are a few reasonable assumption here which if not made get you in trouble) (such as what frame of reference are you using, if there is nothing else in the universe, how do you know they are rotating. etc....)
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Author Commented:
>> In my second post I derived
Great! Thanks for summarizing the result. I'll look at your derivation after I try to work this out independently.

>> how do you know they are rotating
I don't know.

>> what frame of reference are you using
The frame of reference is the CM of the 4 sentient beings on their 4 respective bodies.
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Commented:
It is best to use an inertial (non-accelerating) such as the CM of the entire system.
Each of the sentient beings on their 4 respective bodies is in a non-inertial frame.
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Commented:
> "a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth."

That's because the Earth's equator is moving at 0.465 km/s relative to the CM of the Earth.
In both cases, escape velocity relative to the CM is the same from the same radius.

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Author Commented:
>> Each of the sentient beings on their 4 respective bodies is in a non-inertial frame.
Knew that. I originally had something like this  (in response to  how do you know they are rotating):

The frame of reference is the CM of the 4 respective bodies which are 4 sentient beings. Then the CM is at the origin (0,0,0). The 4 sentient bodies have lab equipment and can see each other. Possibly they will not know they are rotating, but they do have accelerometers which at least will let them know that there is some kind of acceleration (or some kind of gravity). Personally, I don't know they are accelerating since I don't exist is that universe.

So, I'm thinking that aburr is bringing up a good point (unrelated to this question, though), which is how could these bodies know they are rotating. Possibly they may think they are not moving at all, and attribute the centripetal acceleration to some other phenomena.

>> In both cases, escape velocity relative to the CM is the same from the same radius.

Knew that too. The link also said "On the surface of the Earth, the escape velocity is about 11.2 kilometers per second (~6.96 mi/s)"
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Commented:
>> So, I'm thinking that aburr is bringing up a good point (unrelated to this question, though), which is how could these bodies know they are rotating. Possibly they may think they are not moving at all, and attribute the centripetal acceleration to some other phenomena.

I dont think we needed distant stars to figure it out: Apples fall, so would planets.  Then again, I guess all the planetary and lunar movement we saw helped us figure it out.

I think tommy is right:
>>TommySzalapski:Anything over Vp will cause an escape.
>>TommySzalapski:Ozo, if it's in orbit already, then it [KE] needs to be greater than, not equal to, [PE] or it will remain in orbit.

Think of it like this:
the objects start out in a circular orbit such that thier speeding away from each other is perfectly matched by thier acceleration towards each other (KE=PE).  If we increase this velocity (increase momentum) just a tiny bit, they will no longer be accelerated back towards each other fast enough(KE>PE).  So they will move farther apart from each other(increase PE- liner function), which will decrease the  acceleration they feel towards back toward each other(the decrease in KE decreases exponentially). This will have a cyclyic effect and the particles will spiral away from each other.  The further they get, the straighter thier tragetories will become, because the acceleration will dimish.
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Commented:
> spiral
Only if you consider parabolas and hyperbolas to be spirals.

Also, there may be confusion about where zero potential energy is measured from.

If you start with (KE>PE) then increase PE and decrease in KE, you soon get back to (KE=PE)

Although I would call circular orbit conditions KE=-PE/2
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Commented:
> how could these bodies know they are rotating
see Foucault pendulum
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Author Commented:
Thanks again for all your inputs. As you know I reviewed three video lectures (two more than aburr advised). One of the lectures was about CM of a system of particles, and you know I asked a question about that. I was hoping to come up with a simplifying way to solve for the escape velocity, Ve. I have temporarily given up on trying to do that. I'll try to get back to that in another  question.

So, I went with what I learned about Work-Energy. I no longer think it is necessary to have the bodies spinning in a square to compute Ve (although they could be). The Ve computed by Work-Energy concept is the same whether their velocity is increased tangentially or whether moving radially away from the center of the square. I don't think the path to oo matters when computing Ve.

In OP we had the perfect orbit speed relationship shown as:
F = G (m²)/(2*r)² + 2*G*(m²/S²)*cos 45 = m |Vp|²/r

Now the force, F, is what I'm interested in. Suppose I have a massless ring and the 4 spherical bodies are place on the ring to form a square. The distance from the center of the bodies to the center of the square is R.

If an outward external radial force is simultaneously applied on the 4 bodies so as to move them slowly to oo, then I've done work, which is computed as follows:
``````        oo
W = §  F dr
R
``````
I compute how much work is needed to move the 4 bodies to oo, and then say that in absence of this external force, then to reach oo, a body needs enough KE to compensate for the work performed against gravity. So, set KE = W (and I believe that this is another way of expressing the KE/PE formulations given in some posts).

From this I find that (Ve/Vp)² = 2.   !!
We already know Vp, so now we know Ve.

From this formula, I now believe that if  Vp < V < sqrt(2) Vp, then the 4 bodies do not escape, but instead travel in some path (allegedly elliptical).

I went back to the lecture, and learned that the (Ve/Vp)² = 2 formula holds true even if you have one body (say a satellite) rotating around the Earth !!

I am surprised that the ratios are the same in both the 2-body and 4-body problems.

I will now be able to peek at your answers in more detail and get back to you tonight or tomorrow.
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Author Commented:
Of course, if you review the above solution, and take issue with it, then don't hesitate to mention the problem. I may not be able to answer, but hopefully we'll resolve the issue through teamwork.
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Author Commented:
So, now I wonder in the OP configuration, what the speeds should be for the bodies to move from a distance r to a max height of, say, 1.25r; and then how close to the center of the square will the bodies get.

>> see Foucault pendulum
I saw one as a kid at the science museum; but thanks, the wiki animation made me realize what is going on. If the 4-bodies are the 4 sentient beings and they were not spinning on an axis, as Earth and our Moon does (since they would be very dizzy otherwise), then would the Foucault pendulum still work to prove they are rotating around the square's center?
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Commented:
> (Ve/Vp)² = 2.
> the ratios are the same
The essential observation is that given that symmetry is preserved, all distances remain proportional to r, thus all forces remain proportional to 1/r²
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Commented:
> Earth and our Moon does (since they would be very dizzy otherwise)
Are sentient beings on the Earth and our Moon very dizzy?

> would the Foucault pendulum still work to prove they are rotating around the square's center
Yes, as a carefully constructed Foucault pendulum on Earth can work to prove we are rotating around the sun.
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Author Commented:
>> Are sentient beings on the Earth and our Moon very dizzy?
On some amusement rides where the radius is small and the speed is high, people come away dizzy.
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Author Commented:
aburr,
>> The calculations of the previous problem are not relevant.
I know there are usually multiple ways to solve a problem; and in this case I did start with your force formula.

I tried to solve for Ve from your equation in http:#35041583 , but I get a different answer than what I got using the Work - Energy concept in http:#35116178 . All I can conclude is that either only one of us is wrong, or both of us are wrong. Can you refine this conclusion?
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Commented:
> where the radius is small and the speed is high
I don't think those speeds would come from gravitational orbits.

By the way, the minimum orbital period around an object with the same density as the Earth
is the same as the minimal orbital period around the Earth.
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Commented:
http:#35041583 appears to assume that the other 3 bodies are fixed in place, whereas http:#35116178 assumes that all bodies move symetrically
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Commented:
> (allegedly elliptical)
The shape of the path is a more complicated computation, but if you will accept arguments from authority,
Kepler found observationally and Newton found analytically the shape of orbits in a 1/r² field.
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Author Commented:
So, thanks to the tight restrictions of symmetry w.r.t. initial position and velocity imposed on this problem, this boils down to a 1/r² field, and so we have an elliptical orbits if Vp < V < Ve. Actually, I am now guessing that we get an elliptical orbit if 0 << V < Ve. I made 0 << V so that the bodies don't get so close as to collide into a near degenerate orbit.

If Vp < V < Ve, then the bodies start moving away from each other (all the time maintaining a square). They reach some max distance from each other (TBD) and start moving closer to some min distance. Would that min distance be their initial radius, or is it possible that the ellipse's min distance from (0,0) is less than their initial distance? I am guessing now that the answer lies in just a two body analysis (where one is considered fixed, like Earth) given that the potential field of the 4-body case is just -K/r² and here K is just a different constant that that of the Earth potential field. If this is so, I can try to review more about elliptical orbits for the two body case and just apply the principle here.

I don't suppose there is an initial velocity where the 4 bodies could reach and maintain a higher orbit without additional impulse engines to change their direction at the desired orbit.
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Commented:
> Would that min distance be their initial radius
At and only at the apoapsis and periapsis of an elliptical orbit, the radial component of the velocity is 0.
If the body started with a radial velocity of 0, then it was at either the apoapsis and periapsis of its orbit.
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Commented:
then it was at either the apoapsis or periapsis of its orbit.
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Author Commented:
>> If the body started with a radial velocity of 0, then it was at either the apoapsis or periapsis of its orbit.
Ok, thanks much.

>> the minimum orbital period around an object with the same density as the Earth is the same as the minimal orbital period around the Earth.
Interesting tidbit which no doubt has some interesting ramifications.
Reviewed T, and found Te = 2pi Re^(3/2)/sqrt( G Me ), where e ~ Earth

Then I let R = k Re, and amazingly the k^(3/2) factor in computing T cancels out leaving T = Te. Interesting.

>> http:#35041583 appears to assume that the other 3 bodies are fixed in place
I'll have to think about this assertion in ~ 12 hours (or maybe aburr will chime in and confirm).
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Author Commented:
If any participating experts have any differing viewpoints, today would be a good day to make your views known. Thanks again for all your inputs.
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Commented:
ozo probably has the correct observation
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Author Commented:
When I listened to some arguments that were different from the results, I could not dispute them; they sounded reasonable. Only after working out the details did I come up with a result, which came as a surprise to me and definitely not intuitive. Thanks for your patience. I learned much from this thread.
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