[extropy-chat] Juicy Classical Physics Problem InvolvingGravitational Potential

[Lee Corbin]

Stuart, our Avantguardian writes

> Now I have a puzzle for you guys: What is the maximum
> value of g, the acceleration of earth's gravitational
> field, and where is this maximum g located?

What a fine question, thanks!  Since I am totally unable to
think about anything else since having read this a while 
ago (and I've tried), then attend to it first I must. I also
appreciate your crafty

> Hint: Don't get stuck in two dimensions. :)

Not to spoil this for anyone, I'll give my answer at the
very bottom of this reply.

> --- Lee wrote:
>> Newton---but not I---was able to
> > calculate the spherical oblateness of
>> the Earth by carefully regarding this
>> notion of gravitational potential.
>> 
>> But my Newtonian calculation...is off
>> by a factor of two! Perhaps someone
>> here who really likes classical physics
>> can see what is going wrong.

> I did not actually do the calculation myself however
> there are two possible errors which may have led to
> the discrepancy. 
> 
> 1. Did you remember the factor (1/2) in your kinetic
> energy term? i.e. KE = (1/2)mv^2 for your particle.

Well, I think so.  Writing "w" for omega, I used the
value (wr)^2 divided by two, as you can check in
my original post.

> 2. The mass of the earth that contributes to the
> gravitational field at the poles is LESS than the mass
> contributing to the field at the equator. The reason
> for this is Gauss' Law in gravitation.

Maybe I can understand how this could be
true if by "gravitational field" you mean 
something involving potential. Because 
insofar as the *force* is concerned, that
surely cannot be right. At the poles, the
force on you is the perfect sphere beneath
your feet plus the "winged" material at the
equatorial bulge, that also contributes to
a "downward" pull.

> In other words, the mass of the earth that is included
> as M in GM/R should be the portion of the mass that is
> contained within a perfect sphere of radius R. If R is
> the polar radius this will be the total mass of the
> earth minus the mass of those parts of the earth that
> bulge out past the polar radius at the equator.

Remarkable.  Your term, "GM/R" is gravitational
potential, and I can only presume that I was in error
to think about the *force* in my previous paragraph.
[For God's sake I hope no one thinks that I'm being
sarcastic. I've been in trouble writing like this before.]

Hmm, so Gauss's Law applies to the non-spherical
Earth?  I suppose.  But I can't say I understand it.

> At any point, a given distance from the barycenter of
> an extended body, the only mass that contributes to
> the gravitational field at that point is that portion
> of the mass that is closer to the barycenter.

Well, that's a nice result about gravitational 
*potential*, I suppose.  I can't credit it for
a statement about forces though. I'll have to
think about Gauss' Law some more.

Now for your puzzle:

> Now I have a puzzle for you guys: What is the maximum
> value of g, the acceleration of earth's gravitational
> field, and where is this maximum g located?
> Hint: Don't get stuck in two dimensions. :)

Well, Stuart, I was going to write that it was maximal
at the poles, but then I thought of a neat way to prove
it to myself:  I would find out the value of g at the 
equator, then subtract off the centrifugal force. So I
tried to look up the two values of g, but instead ran
into http://forums.randi.org/archive/index.php/t-1104.html
(page down to Iconoclast's contribution) :

     Iconoclast 18th February 2003, 02:46 AM OK, here's
     (http://www.lns.cornell.edu/spr/1999-02/thrd3.html#0014
     446) a thread on the subject from the
     sci.physics.research newsgroup from a few years ago,
     the thread in question is entitled "Baltimore Sun Says
     Gravity Inside Sphere is Uncalculable" (the thread
     title is misleading, that's not what the article was
     implying). From this discussion it appears that:

     - For a homogeneous sphere, gravity is always at a
     maximum at a point on the surface, it falls off above
     and below that point. Gravity falls off faster as we
     move outwards from the surface than than it does as we
     move inwards.

     - It is believed that the earth has a much higher
     density at the core than at the surface, thus gravity
     on earth will increase for quite a few hundred miles as
     we descend. I didn't know either of those things.
 
So that wrecks it for me; I wouldn't have thought of the
variable density factor. It says elsewhere on that page
that someone heard in high school that it's maybe a
thousand miles beneath the pole.

Lee