Newton's Enigma?

Question

This question troubled Newton for many years. A little mass, "m," is a certain distance from the center of a globular cluster of masses and there is a certain force of gravity, due to the cluster of masses, on the little mass that pulls it toward the center of the cluster. Now consider the situation whereby neither the little mass nor the center of the cluster moves, but the cluster uniformly expands. As a result of this expansion some parts of the globular cluster are closer to "m" and some are farther from "m." After the expansion, the forces of gravity of the cluster on the little mass "m" will:

a) increase
b) decrease
c) remain unchanged

Answer 1

I probably miss the wisdom of Newton since for the foolish me the answer is quiet 'simple'.

The cluster of masses is far away and the radius of the cluster r is very small relative to the distance R between our little mass and the center of the cluster

Case 1. R>>r

Just like any other force between two masses. Force is along R

Case 2 R= r

The density of the cluster is significantly decreased. The total force is decreased as we are getting components of F 'radial ' to R. How much exactly requires a little bit of formulation

Case 2 R<
Our mass m will be almost at the center of the cluster and experience very little force or no force at all.

Oh well...
"Fools run where wise men fear to tread. " :-)

Answer 2

b) decrease
The cluster expands and I assume that it expands beyond the small mass. Before the expansion had started the mass was out of the cluster. So the whole of it attracted m towards itself.
We can assume the whole mass of the cluster to be situated at it's center of gravitational attraction, from where the force has a definite value. Even as the cluster expands but has not yet crossed m, this value of the force is retained because the center and the mass m are where they were, and as has been pointed out, even if some portions are coming close to the mass, some are moving away therefore I am assuming the complete mass to be situated at the center.
Hence, for some time, it remains unchanged. but as the cluster crosses the mass m, some of it's masses are 'on the other side' of m. As I type this I happen to recall the saying 'United we stand, divided we fall.' Till the masses were united, i e. they were on the same side of m, they all attracted it towards the center, but as they got divided, those on the other side of m started pulling it away from the center.
Let's draw a sphere with it's center at the center to touch the mass m. If we overlook the remaining cluster, what is the difference, between the sphere we have now and the cluster as it was when it just touched the mass m, except their mass. The hollow shell that remains has it's forces divided and thus it 's contribution, if there is any, to the force may by no ways make it larger than as it was when the cluster was in the situation of just touching the mass m.
And since the mass of the sphere is smaller than that of the cluster, therefore it's attraction towards the center is also smaller than the original cluster.
[EDIT: I saw Remo Aviron 's answer and came to know about the shell theorem, if it is correct, then the outer shell would exert no force immaterial of the position of the mass m. Therefore according to me the force would definitely decrease. But in the case, where, the cluster does not cross the mass, the force remains constant. ]
Clearly I have assumed the cluster to be spherical.
Hence the answer, according tome is b) decrease.

Answer 3

Of the above, Edward's answer is the best. But I think it can be stated much more definitely and straight-forwardly:

Assume that the cluster has radius R_c, and that the mass of the cluster is uniformly spread throughout the spherical region. Then the mass density is
p_c = M_c/(4πR_c^3/3).

a) The distance of the mass m is r, and r > R_c:
In this case, all the mass M attracts m. It is very well known (first shown by Newton himself) that the total effect of the attraction is the same as if the entire mass M were located at a point at the distance r. This is true ONLY because the mass distribution is spherically symmetric. Therefore:
for r > R_c, F(r) = - GMm/r^2

b) The distance of the mass r = R_c:
Just set r = R_c, and you get the right result by continuity:
for r = R_c, F(r) = - GMm/R_c^2

c) The distance of the mass r > R_c:
In this case, we consider the mass M in two parts:
- The part which lies at radial distance from the center of the cluster GREATER than r: When we consider all of this together, its net force on m CANCELS OUT. In other words, if a mass is inside a spherical shell, the net attraction it feels from that shell is 0. This is another aspect of Newton's theorem above. So all the mass at distance from the center of the cluster creates no attraction!
- The part which lies at radial distance from the center of the cluster LESS than r: this has the same attraction as discussed in part b). But in this case, the relevant spherical radius is not R_c, but r itself; and the relevant mass is not M, but that part of the globular cluster's mass which is within radius r.
So: In the answer for part b) change:
"M" => p_c*4π*r^3/3 = M*(r/R_c)^3
"R_c" => r
and we get
F(r) = -GmM*(r/R_c)^3 /r^2
= -(GMm/R_c^3)*r

So, in sum, the final answer is:
- for r > R_c:
F(r) = - GMm/r^2

- for r < R_c:
F(r) = -(GMm/R_c^3)*r

Note that:
- the two expressions match when r = R_c
- F(r) => 0 as r => 0

Now to answer the question:
- Suppose that initially, r > R_c. Then as R_c increases, there is no change in F provided r is still greater than R_c: the attraction depends only on r.
However, as soon as R_c reaches and exceeds r, we are the region where F is proportional to r/R_c^3. So now as R_c increases, the gravitational attraction will decrease.

- If initially, R_c were already greater than r, then the gravitational attraction would start to decrease immediately, according to r/R_c^3.

So the best "single" answer of those presented to you is "b)", although there could be grace period during which there is no decrease, until R_c catches up to r.

Answer 4

The easy answer is C.

This assumes that the cluster is fairly uniform around its cg (i.e. spherical or something close--density can decrease from center to edge of cluster so long as it does so uniformly), and that none of the mass actually expands past our little mass.

It all relates back to the inverse square law and the integration over the volume of a sphere. [Edit: and more specifically the Shell theorem, see below]

Now, off for a break while I think of all of the ways to get around what seems the obvious answer........... .......... ...........

.......... ............. ....... I'm back. I type in my answer and discover that Kirchwey and Edward beat me by 6 minutes and said some of the same things -- spent too much time typing. So I have to add something, neither of them accounted for the fact that the expansion of the cluster takes energy and energy means mass. How does this effect the mass and gravity of the cluster? Not much, but even a little is something to ponder. ....... ........ ........... But looking at Edward's answer, he is stumped at r=R. This the Shell Theorem. The answer to Newton's Enigma. Its at http://en.wikipedia.org/wiki/Shell_theorem and I used it in my answer to Dr. H's previous question. Much more to ponder. Hasta Manana ....................

......................Manana

Final Answer remains "C" for the reason stated above, particularly the Newton's shell theorem.

The shell theorem is what governs both the interaction near the radius of a large body and once you get inside a large body. Intuitively, it does not quite make sense, how could gravity act, when near the surface of a large object, as if all the object's mass was located at the cg of the large object? I can see why Newton was stumped. (The corallaries about the hollow shell and about ignoring mass farther away from the cg than the observer follow directly out of having the all the mass act as if it were located at the cg).

But the devil is always in the details. Now for the caveats (and I'll be the first to admit that, at least the last two caveats, are just playful attempts to circumvent the obvious answer):

1. I think either I or all of your other readers sans Kirchwey misinterpreted the question. I read it that "m" was outside the cluster (based on the general description and (i) the statement that m did not change in relationship to the cg of the cluster as it expanded -- which implies that it was not in or part of the cluster, and (ii) the statement that certain masses in the cluster got closer to m, which they would not have if m were part of the cluster), but several others did not. My original answer, above, stated specifically that I was assuming that none of the cluster mass actually expands past the little mass, which I think is the correct assumption.

If the little mass is within the cluster the answer, per the shell theorem, is B: As I discussed in answering your last question (using the hypothetical of a hollow earth), it is only the mass closer to the cg than the little mass that counts in determining the acceleration of gravity and, under this scenario, there would be less mass in the volume occupying the radial space between the little mass and the cg.

2. Kirchwey did a good job at pointing out that local anamolies have a much more pronounced effect on the little mass as the cluster expands. This is much like orbiting the earth, the closer you get, the more pronounced the gravity anamolies. But the cluster would be far less uniform than earth's surface. The secondary effects of the local anamolies could overwhelm, depending on location, the primary effect that the acceleration of gravity remains unchanged. And as Kirchwey points out, this could lead to an answer of a, b, or c, depending on location. However I would think that the average acceleration of gravity for all possible little ms would remain unchanged.

3. An expanded cluster, all things being equal, would be ever so slightly more massive than the shrunken cluster (again answer b). This is because you would be pulling stars out from the gravitational well. But the question is where does the energy necessary to cause this expansion come from? Maybe the guy who wrote the problem? Nah. Dark energy, maybe, but I'm not going there. Real life it has to come from some place and at a price.

Easiest solution (but not quite realistic) is it came from the cluster itself in some sorta adiabatic expansion. This would keep the overall mass the same, although you would have to convert some of the mass of the cluster into energy to accomplish this expansion. (I know, this sounds contradictory, you have to burn some mass, to push out some other mass to a point of greater potential energy, and overall mass will remain the same).

Most likely though is that whatever energy source caused the expansion was inefficient resulting in energy (i.e. mass) escaping the cluster and resulting in a slightly smaller acceleration of gravity. (This should be a miniscule effect)

4. Lastly and most interestingly, the expansion of the cluster towards the little mass will increase the gravitational force on the little mass while it is expanding.

Think about it. Gravity is limited to the speed of light. The little mass, because it is closer to one edge of the cluster, will see that edge expanding towards it many, many years before it sees stars on the far edge expanding in the other direction. Gravity, of course, will mirror what the little mass sees. It will percieve the cluster's cg as being closer to it than, in fact, it is. This means during the expansions (or at least the period of time the little mass sees the cluster expanding), the little mass will experience an increase (i.e., answer a) in gravity. ... I realize this should be de minimus, but it is definitely cool!

-- Dr. H, I'd appreciate your thoughts on these last two points. They seem logical, but from what little I know of gr, and that knowlege being quite stale, I cannot count on nieve intuition being correct. Thx Remo

Answer 5

Really, none of the above. It may increase or remain unchanged. I'm not sure if it can decrease. On average, and at sufficient distance, it's analogous to approaching a solid, spherical body of the same radius and average density; i.e., unchanged. But with a cluster you can be very close to the CG of one of its component bodies while remaining at a 1-radius distance from the composite CG. The inverse square law can have a large effect here, producing more g than you would experience on the surface of the equivalent solid body if the ratio of material density to average density is large enough. Basically the 2-body approximation fails and it's a multibody situation. Whether the result is an increase or a decrease in g depends on particulars of the conditions.