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Re: Saturday Morning Puzzle, Part 2



I wrote:

Hmm. I'd be interested in knowing how a distribution that
places most of the mass in one object at arbitrarily large
distances (not to mention in directions that cover a solid
angle of nearly 2*pi) from any given portion of the other
object could end up giving a larger net force than a
distribution that keeps all of the mass within a small
distance *and* a smaller range of solid angles.

To which, John Denker (apparently ignoring this argument) wrote:

Hint: Olber's paradox.

If that's not enough of a hint, read on.

It is enough; it helps me to understand the source of your error.

Consider a small patch of mass ("Moe") on one disk looking
across the gap at a similar small patch of mass ("Joe") on the
other disk. Assume the gap is tiny compared to the diameter of
the disks,

I already did.

and assume Moe and Joe are not too near the edge.

Granted. I made the same assumption which is well-justified in
the limits we are discussing.

Now move the disks apart, doubling the size of the gap.

O.K., but then Moe's and Joe's masses *both* increase by a factor
of two also since the problem stated that the separation is about
the same as the thickness of the two lamina. (I am assuming that
you mean for Moe and Joe to maintain the same physical area. You
aren't clear about this, but I perceive that implicit assumption
in what you say a little further on.)

The effect of Moe on Joe goes down by a factor of 2 squared.

No, it stays the same. Their increase in mass cancels the effect
of the increased distance.

But when Moe looks out at the other disk, he sees not just Joe
but Joe and three of Joe's friends in his field of view (the
field that used to be occupied by Joe alone).

Right, but only if Joe's *physical area* and that of his three
friends is independent of separation distance as I previously
assumed you meant above.

So there are four times as many effective sources.

Correct, but go on ... Each of them has twice as much mass also.

Result: the force on the two disks is constant, independent
of gap-size (except in the fringing field near the edge).

No. It's a little more complicated than that:

(1) The force on Moe due to Joe and his three companions is four
as much as the original force due to Joe alone. (Let's review:
The separation has doubled, Moe's mass has doubled, Joe's mass and
that of his three friends is eight times what Joe's mass used to
be.)

(2) Therefore, the net force on Moe's disk is twice what it used
to be. (Moe has half as many fellow disk citizens as he used to
have since all of them have doubled their masses. Four times the
force on each of half as many folks yields twice the net force.)

As I said in my original answer, the force is proportional to the
separation. I like the Moe and Joe method, but it's far easier to
get to the correct result (and less fraught with potential errors)
using Gauss' law to find the gravitational field due to one disk
and then simply multiply by the (fixed) mass of the other.

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm