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On 09/25/2014 06:56 AM, Peter Schoch asked:
am I right in being able to tell him to just find the CM of thethe
portion "below" the chord, and the CM "above" the chord, and just find
effects due to both of them at each point along the chord?
I suspect I don't understand the question.
According to the famous "shell theorem":
a) Outside a uniform spherical shell of mass M,
the δg created by the shell is the same as the
δg created by a point mass M concentrated at the
b) Inside the shell, the δg is zero.
I'm confused because surely you already knew that.
It's needed in order to do the drop-along-a-diameter
exercise. Also it's easy to prove. It's Theorem XXXI
in the _Principia_.
It's true point by point, so it doesn't matter whether
you are moving along a diameter, or a chord, or an
arbitrary curved path. This should be obvious since
every point lies on "some" diameter.
There are probably dozens of AJP articles that mention
and/or use this theorem, but they're not going to tell
you anything you didn't already know.
If that's not the answer you were looking for, please
ask a more specific question.
Except in cases where the details don't matter, it is
better to call it the "gravitational acceleration"
(if that's what you mean) rather than «gravity».
That's because «gravity» can refer to a lot of things,
including the gravitational potential, the gravitational
acceleration, the tidal stress, et cetera.
Actually even the notion of "the" gravitational acceleration
is ambiguous. Virtually every textbook I've ever seen
gets this wrong, using the term one way in one chapter
and another way in other chapters. No wonder students
This is why I wrote δg rather than plain g above.
You can make the problem go away by (a) assuming
a non-rotating planet and (b) choosing the obvious
"preferred" frame ... but this is (a) unphysical
and (b) risky at best.
For details on this, see
Forum for Physics Educators