Why is there no DM halo around the solar system?

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  • #1
Buzz Bloom
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If anyone knows a reference that discusses this question, please post a citation.

Naively it seems that any central mass would attract DM from the mass's general vicinity so that an approximately spherically symmetric accumulation would develop over time about the central mass. This seems to be accepted as the explanation of why the velocities of stars (or gas) within spiral galaxies do not exhibit a Newtonian pattern. Since the planets of our solar system DO have a Newtonian pattern, there must not be any such DM halo. But, why not?
 

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  • #2
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It's there, it's just not very big. The distribution of DM seems fairly uniform within the galaxy, but that density is so low that in places where mass is fairly densely concentrated (like a solar system) it's gravity is negligible. DM's gravity comes from the fact that there is so much empty space in the galaxy, and it's got just as much DM as the solar system.

Think about it like this: a cloud in the sky can weigh over a million pounds. That's a million pounds of water, which is a very large amount, but when the cloud comes down to earth and you run through it as fog, you barely notice it's mass.

Also, matter conglomerates because matter sticks to other matter. Atoms collide with each other and lose some of their energy to friction, slowing them down and allowing fast clouds to cool and condense. Dark matter does not seem to interact with anything, including itself, so there is nothing to slow it down, disperse it's energy, and make it form dense areas, only clouds the size of galaxies.
 
  • #3
Janus
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It's not that the Solar system doesn't contain DM, it's a matter of the relativity density of the solar system with respect to a galaxy. The density of DM in the vicinity of the Sun is ~5e-13 kg per cubic km. So let's take the volume inside the radius of Pluto's orbit and work out how much DM would be contained in that volume. It works out to be ~5e17 kg, or something like the mass of a small asteroid. Since the Sun has a mass of 2e30 kg, its is no wonder that the DM in the solar system has no noticeable effect on the planet's orbits.

Now, lets calculate the amount of DM at the same density, enclosed in the volume inside a radius equal to the Sun's distance from the center of the Galaxy. We are ~27,000 light years from the center so that is ~2.6e17 km, for a volume of ~7e52 cubic km, which give us a DM mass of 19 billion suns. Since the total estimated mass of the stars in our galaxy is ~64 billion solar masses, this amounts to a good fraction of the mass effecting the Sun's orbit around the galaxy, and we would expect it to have a significant effect.
 
  • #4
Buzz Bloom
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Hi Janus and newjerseyrunner:

Thank you both for your posts. I have a sense that your explanations regarding the thread's question missed an important aspect of the question, but I am not an expert in this subject, so I may be wrong.

any central mass would attract DM from the mass's general vicinity so that an approximately spherically symmetric accumulation would develop over time about the central mass.
I think that making a calculation based the current average density of DM in the universe,
ρDM = ρc x ΩDM,​
fails to consider that a central mass of Baryonic matter of mass M would pull all DM within some distance D towards the mass's center of gravity, which over a period of time will concentrate the DM into densities much greater than ρDM. One can get a rough estimate of D by considering the distance at which the escape velocity equals the Hubble velocity:
√(2 G M / D) = h D → D = (2 G M / h2)1/3
Then the "captured" DM mass would be
MDM = (4/3) π D3 ρDM

I know this is an oversimplification, since one needs to consider how H and ρc change over time. However, I think it will be OK as a very rough approximation.

I suggest doing two calculations, one for the solar system, and one for some spiral galaxy (perhaps the Milky Way) where a reasonable estimate of the mass of its baryonic matter is known. My guess is that the ratio, MDM/M, of of "captured" DM mass to each system's original baryonic mass will be comparable. If my guess is right, then my "Why" question would remain a mystery.

My skills at making such calculations are marginal at best, but I plan to do what I can, although I expect it will take me at least several days.

Regards,
Buzz
 
  • #5
Jonathan Scott
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Capturing DM is not like capturing baryonic matter, because of the lack of non-gravitational interaction or anything like "friction" (apart from gravitational tidal effects). It isn't easy to understand how galaxies can capture it, and the DM passing through the solar system would probably barely notice it.
 
  • #6
Buzz Bloom
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Capturing DM is not like capturing baryonic matter
Hi Jonathan:

Thanks for your post.

I understand that there are EM interactions among baryonic particles that cause their kinetic energy reduction due to photons created in the interactions radiating the energy away. (See https://www.physicsforums.com/threads/the-dark-sky-ahead.836784 post #7.) However, if we assume that DM is like dust, imagine a sphere of DM of non-rotating dust with uniform density ρ and radius R in a locally otherwise empty universe. That is, assume an empty shell of outer radius Rv and inner radius R, and Rv >> R.

If we think of this DM sphere in terms of radially symmetric shells of zero thickness separated from each other by dR, the shell at radius D is attracted towards the center by a mass
(1) M(D) = (4/3) π D3 ρ.​
The acceleration towards the center acting on this shell is
(2) A(D) = G M(D) / D2 - h2 D (1+(3/2) Ωmass).​
This assumes that Ωradiation << 1. From this we calculate that
(3) A(R) = 0 for R = (G M(R) / (h2 D (1+(3/2) Ωmass))1/3
Let this value of R be called Rlimit, and assume that the sphere radius
(4) R < Rlimit.​

I am not sure that the following assumption is correct, but for this discussion, I am assuming it anyway to keep the math simple. Assume that as the nested shells move towards the center, that the shells do not pass each other. If we assume that the sphere has uniform density at time T0, then, one can calculate where each shell will be at a later time T > T0.

My guess is that a current DM halo around a galaxy will be more centrally concentrated than a halo around the solar system since the galaxy's halo has been contracting for about 13 Gy, while the solar system halo has been contracting for only 5 Gy. However, I am doubtful that this will be an explanation for why no halo is detectable around the solar system.

Regards,
Buzz
 
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  • #7
phinds
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My guess is that a current DM halo around a galaxy will be more centrally concentrated than a halo around the solar system since the galaxy's halo has been contracting for about 13 Gy, while the solar system halo has been contracting for only 5 Gy. However, I am doubtful that this will be an explanation for why no halo is detectable around the solar system.
Bad guess, as it turns out. DM spends more time in the halo around a galaxy than in the center. This is a direct result of the fact that it only interacts gravitationally and so is not slowed as it passes through the center and out to the other side and back again.
 
  • #8
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Dark matter is affected by the galaxy far less than the galaxy is affected by the dark matter. I don't like thinking about the universe as a bunch of galaxies with DM halos around them. I prefer to think of the universe as clouds of dark matter that have little galaxies bobbing around inside of them.

Here is a nice visual:
https://qph.is.quoracdn.net/main-qimg-1ad964633d3fb229579f9a1f9d8ecf8b?convert_to_webp=true [Broken]
In this analogy, the galaxy is the bubble, the DM halo is the water. The DM (water) is affected by the movement of the galaxy (bubble) inside, but who's really in charge?
 
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  • #9
Buzz Bloom
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I now see I made a math error.
(1) M(D) = (4/3) π D3 ρ.​
The acceleration towards the center acting on this shell is
(2) A(D) = G M(D) / D2 - h2 D (1+(3/2) Ωmass).
(3) A(R) = 0 for R = (G M(R) / (h2 D (1+(3/2) Ωmass))1/3
Let this value of R be called Rlimit, and assume that the sphere radius​

Substituting (1) into this (2) gives
(5) A(D) = G (4/3) π D3 ρ / D2 - h2 D (1+(3/2) Ωmass).
= (4/3) π G ρ D = h2 D (1+(3/2) Ωmass).​
Therefore (5) with A(D) = 0 cannot be solved for D.

I need to think about what this means. Any suggestions?

ADDED
I must have made a mistake in calculating the formula for the acceleration effect of the expanding universe.
A = h2 D (1+(3/2) Ωmass)​
I will look for that mistake.

Regards,
Buzz
 
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  • #10
Buzz Bloom
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Bad guess, as it turns out. DM spends more time in the halo around a galaxy than in the center. This is a direct result of the fact that it only interacts gravitationally and so is not slowed as it passes through the center and out to the other side and back again.
Hi phinds:

Thank you for the excellent point in your post. I have never considered that before.

If I understand your concept correctly, my assumption about the "dust" with no angular momentum slowly falling inward failed to consider that over time each dust particle would acquire sufficient radial velocity to become dynamically stable with a repeated motion that passes through the center and comes out the other side.

I will now need to think about the dynamics that brings that stability about, and its effect on the shell model.

I still don't understand why this phenomenon would lead to our solar system having no halo.

Regards,
Buzz
 
  • #11
DaveC426913
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Hi phinds:

Thank you for the excellent point in your post. I have never considered that before.

If I understand your concept correctly, my assumption about the "dust" with no angular momentum slowly falling inward failed to consider that over time each dust particle would acquire sufficient radial velocity to become dynamically stable with a repeated motion that passes through the center and comes out the other side.

I will now need to think about the dynamics that brings that stability about, and its effect on the shell model.

I still don't understand why this phenomenon would lead to our solar system having no halo.

Regards,
Buzz
You're over-thinking it.

Each particle starts with a random amount of angular momentum. It neither gains nor loses it as it gravitates inward. It simply orbits the CoM of the galaxy. Since it does not interact with nay bodies in it orbit, there is no need to apply some 'stability' factor.
 
  • #12
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Think about a bunch of marbles on a sheet, they bend the sheet very slightly, you've seen this analogy before:
91964-004-30C6274D.gif


Now give those marbles random directions and velocities, they start hitting each other. click clack click clack. When two marbles hit each other head on, going 5mph relative to each other they hit and make a big CLICK, and then move away from each other it slightly less than 5mph of total velocity. Why? Because the collision caused them to lose some energy, it got converted to sound and heat. Now when you have a lot of marbles that get close to each other, there is a lot of collisions and they quickly disperse outwards, but they've lost enough energy that they now can't overcome the mutual pull of all of the other marbles, so they start falling back towards the center of mass, so eventually, you'll get clumps or marbles that have very little velocity relative to each other.

Now imagine those marbles can't hit each other, they just go straight through each other. They're still affected by the curve in the sheet caused by the other particles, but they don't collide. How would it act?
 
  • #13
Buzz Bloom
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You're over-thinking it.
Hi Dave:

Thanks for your post.

I wanted to avoid including angular momentum because it complicated calculations, and because I think it doesn't effect the overall conclusion about the difference in the DM in the vicinity of a galaxy or our solar system. If I ignore the value changes with time of h and dh/dt, in order to calculate M(R) I will still need to consider what the fraction of the time a particle with a given maximum distance D from the central mass spends at a distance d in the range [R,R+dR].

Regards,
Buzz
 
  • #14
Buzz Bloom
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Now imagine those marbles can't hit each other, they just go straight through each other. They're still affected by the curve in the sheet caused by the other particles, but they don't collide. How would it act?
Hi newjerseyrunner:

Thanks for your posts.

I think my posts #10 and #13 responding to phinds and Dave answer the quoted question you asked. If I am still missing something, please let me know.

Regards,
Buzz
 
  • #15
phinds
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If I understand your concept correctly, my assumption about the "dust" with no angular momentum slowly falling inward failed to consider that over time each dust particle would acquire sufficient radial velocity to become dynamically stable with a repeated motion that passes through the center and comes out the other side.
exactly

I still don't understand why this phenomenon would lead to our solar system having no halo.
most likely because the DM just passes through the solar system on the way to the center the same as it does through the galaxy center
 
  • #16
Buzz Bloom
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most likely because the DM just passes through the solar system on the way to the center the same as it does through the galaxy center
Hi phinds:

Thanks for your post.

This quote confuses me.

The gravitational acceleration by the maximum total estimated mass of the Milky Way (3 x 1042 kg), assumed to all be at the galaxy's center, on any location in the solar system (distance 2.57 x 1021 m) is 3 x 10-11 m/s2. The gravitational acceleration by the Sun's mass (2 x 1030 kg) on Pluto (semi-major axis 1.5 x 1011 m) is 6 x10-3 m/s2. That is, the Sun's gravitational influence on the solar system is greater than the Milky Way's gravity on the solar system by more than a factor of 108.

Therefore, it seems most likely that the Sun's influence on local DM would gravitationally capture that DM rather than that DM passing by the solar system on it's way to or from the center of the Milky Way.

Regards,
Buzz
 
  • #17
phinds
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Hi phinds:

Thanks for your post.

This quote confuses me.

The gravitational acceleration by the maximum total estimated mass of the Milky Way (3 x 1042 kg), assumed to all be at the galaxy's center, on any location in the solar system (distance 2.57 x 1021 m) is 3 x 10-11 m/s2. The gravitational acceleration by the Sun's mass (2 x 1030 kg) on Pluto (semi-major axis 1.5 x 1011 m) is 6 x10-3 m/s2. That is, the Sun's gravitational influence on the solar system is greater than the Milky Way's gravity on the solar system by more than a factor of 108.

Therefore, it seems most likely that the Sun's influence on local DM would gravitationally capture that DM rather than that DM passing by the solar system on it's way to or from the center of the Milky Way.

Regards,
Buzz
Your logic/math seems sound to me. I wasn't sure, which is why I said "most likely". Must be something we're both missing. I think it probably has to do with the speed of the DM "particles" as they pass through the solar system. They are probably traveling faster than the escape velocity of the solar system.
 
  • #18
Buzz Bloom
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I think it probably has to do with the speed of the DM "particles" as they pass through the solar system. They are probably traveling faster than the escape velocity of the solar system.
Hi phinds:

Thanks for your post.

My guess is that your suggestion above won't work either. I will have to some math (a slow process for me) to check it out. This is what I plan.

Consider an extended line through the Sun and the center of the Milky Way. Assume a DM particle P on this line with a maximal distance Dm from the center of the Milky Way, DM equal to some multiple of the radius of the Milky Way. Calculate P's velocity when it passes near the Sun. At some point, distance D from the Sun, this velocity will be less than the escape velocity at the radius D.

Now, consider all possible radial paths of a particle P through the center of the Milky Way. If P were to pass within a distance D of the Sun, it would be captured. I could possibly then calculate what fraction of such particles moving from a distance Dm through the Milky Way's center would pass within D of the Sun. This would then be an approximate estimate of the fraction of the Milky Way's DM that would be captured by the Sun.

Regards,
Buzz
 
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  • #19
Buzz Bloom
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I have been thinking hard about the scenario from my previous post. I have decided that it might produce an interesting result, but I am now thinking it will not be very relevant for reasons I will discuss later. Here are some beginning results.

I decided to chose for Dm the distance from the Milky Way (MW) center, PMW, at which the escape velocity equals the Hubble velocity.
Vesc = √(2 G MMW / Dm)
Vh = h Dm
G = MMW = 6.69 x 10-11 m3 kg-1 s-2
MMW = 1.4 x 1042 kg
h-1 = 5.45 x 1017 s
Dm = (2 G MMW / h2)3/2 = 3.82 x 1022 m​
I assumed all of the MW mass is at the galaxy's center, and at time t = 0, a DM particle is at distance y0 = Dm with zero velocity. The particle falls radially toward the center on a radial line that passes through the sun. The time t at which the particle is at distance y is given by:
eq-t(y).png
where μ = G MMW.
(See https://en.wikipedia.org/wiki/Free_fall Inverse-square law gravitational field.)

The time at which the test particle reaches PMW is
t(0) = 8.56 x 1017 s = 27.2 Gy.​

The distance between the sun and PMW is
ysun = 2.57 x 1021 m.​
The time at which the test particle reaches the sun is
t(ysun) = 1.37 x 1017 s = 4.34 Gy.​

I will add in a later post calculations for the velocity of the test particle at y = 0 and y = ysun.

My reasons for expecting the results this scenario to not be very relevant are the following.
I think that the way DM in the MW halo actually passes by the sun is not radially. Also the time frame for the particle to fall is too long for the scenario to have much relevance.

I think a scenario closer to reality is to consider that the DM halo is (1) more-or-less on the average not moving, and (2) it has a density ρ(y) that varies with the distance y from PMW. Also, at the time the sun and it's system was being formed, (3) the baryonic material that was to become the solar system was passing through the halo DM at an approximate radial distance ysun with the DM density being ρDM(ysun), and (4) this proto-solar system material having in it's orbit with respect to PMW a velocity through the DM. This velocity can be compared with the MW's escape velocity at ysun.

It may be that (a) the orbital velocity of the sun about PMW is too fast for the sun to accumulate enough DM around it to effect planet orbits, and (b) without the sun being able to capture DM, the ρDM(ysun) density might be too small to effect the orbits of the sun's planets. I think this scenario would be more relevant than the one in my previous post for answering the thread's question.

Any comments or questions?
 
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  • #20
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I think the assumption that DM moves in a similar speed to the matter is not correct. I was under the impression that DM is traveling every which way. It only interacts gravitationally, so it'd be difficult for it to transfer energy around and all end up the same way, only the tidal force of gravity would even slightly pull on the DM.

You meant that you would expect the DM near the solar system to travel more or less together, sort of like how the other stars near us are all traveling more or less the same direction and speed?
 
  • #21
phinds
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As for your statement:
We assume that the DM material in the MW halo has more-or-less approximately zero velocity component in the plane of the sun's orbit
I'm not clear on what you mean, but it doesn't not sound right. DM moves in and out of the center of the galaxy in the plane of the suns orbit, so clearly cannot have zero velocity in that plane.

Oh ... are you talking about the DM halo of the solar system (if there even is one?). The total amount of dark matter in the solar system has been estimated to be about the same as a modest-sized asteroid. Negligible in terms of having any gravitational effect regardless of its kinematics.
 
  • #22
Buzz Bloom
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Hi @newjerseyrunner, @Janus, @Jonathan Scott, @phinds, @DaveC426913:

Thanks to all of you for helping me better understand the question I asked to start this thread. I think I have now learned enough to understand a plausible reason why the solar system might have acquired some DM which might affect orbits in the solar system. I hope you will look at my explanation below, and respond with comments agreeing or disagreeing with my analysis.

First a summary:
The sun has a (near) circular orbit around the center of the Milky Way (MW). Assumptions:
(a) I think that most of the DM material in the MW halo has approximately zero velocity component in the plane of the sun's orbit, VDM. For the purpose of this analysis, I assume VDM = 0.
(b) I assume sufficient DM exists near the sun's orbit so that the DM that is close enough to the sun to be captured by the sun's gravity will be able to affect solar orbits.​
I calculate the distance from the sun, Desc, at which its escape velocity Vesc equals the sun's velocity, Vsun, in it's orbit. The only DM that can be captured by the sun's gravitation must come within the distance D from the sun's center as the sun moves through the DM. Below I show the calculation of D.

The answer I got is D = 5.04 x 106 km. This is less than to the radius of Mercury's orbit, 5.7 x1010 km. (I would much appreciate someone checking my math.)

Calculation
(1) Vsun = 8.28 x 108 km/h = 2.3 x 105 m/s
(2) G = 6.67 x 10-11 m3 s-2 kg-1
(3) Msun = 1.99 x 1030 kg
(4) Vesc2 = 2 G Msun / D = 2.66 x 1020 m3/s2
(5) Desc = Vsun2 / 2 G Msun = 5.04 x 109 m​
This calculation (if I did it correctly) depends on several assumptions.
The (a) assumption has positive and negative possible affects. If VDM > 0 then only some of the DM closer to the sun than Desc would be captured. On the other hand, some of the DM further than Desc from the sun would also be captured.

Given the result of the calculation, I think that the (b) assumption may need further investigation, but for now I need to think about the implications of the small value of Desc.
ADDED
I just thought of another possibility that might reduce the effectiveness of the captured DM to affect orbits. Each captured particle would have an elliptical orbit with its perihelion being the distance it passed closest to the sun at the velocity Vsun. The particle's aphelion is likely to be be most of the time much further away from the sun, where it would have less (or maybe negligible) effect on orbits. I need to look into this possibility also.

Regards,
Buzz
 
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  • #23
Buzz Bloom
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OK. I made a big error about the semimajor axis of Pluto's orbit!! I just corrected it in my previous post.
It is 5.87 x109 km.
The value of Desc = 5.04 x 109 m = 5.04 x 106 km
is less than Mercury's orbit of 57 x 106 km.

Back to the drawing boards.
 
  • #24
Buzz Bloom
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I think the assumption that DM moves in a similar speed to the matter is not correct.
Hi @newjerseyrunner:

I am not sure what in particular I said that suggested I was assuming "DM moves in a similar speed to the matter". Can you quote from my text so I can explain what I intended to say?

Regards,
Buzz
 
  • #25
Buzz Bloom
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I'm not clear on what you mean, but it doesn't not sound right. DM moves in and out of the center of the galaxy in the plane of the suns orbit, so clearly cannot have zero velocity in that plane.
Hi @phinds:

Thanks for your post.

I assume that the DM has all kinds of different orbits with respect to the center of the MW. Relative to the MW, let
Πsun be the plane of the sun's orbit, and
ΠDM. be the plane of a DM particle's orbit.​
Also, let
VDM be the speed of the DM particle.​
Let
θ be the angle between Πsun and ΠDM.​
When θ is close to π/2, the DM particle's velocity has only a small component in Πsun. For every DM orbit with angle θ, its speed component in Πsun is
VDM.sun = VDM sin θ.
Let VDM.sun be the velocity of the particle relative the sun's motion.​
Half the time, the sign of the velocity of the particle relative the sun's motion will be positive and half negative.
(1) When it is positive, the velocity that is relevant to the particle's being captured (that is to be compared with Vsun), is
Vsun - VDM.sun.​
This means that value of Resc for this particle will be greater than for the case when θ = π/2.

(2) When it is positive, the velocity that is relevant is
Vsun + VDM.sun.​
This means This means that value of Resc for this particle will be less than for the case when θ = π/2.​
My guess is that these two case more-or-less will balance out. That is what I means by
The (a) assumption has positive and negative possible affects. If VDM > 0 then only some of the DM closer to the sun than Desc would be captured. On the other hand, some of the DM further than Desc from the sun would also be captured.
I hope this adequately clarifies my intention. Please let me know if I misunderstood your post.

Regards,
Buzz
 

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