Chronology |
Current Month |
Current Thread |
Current Date |

[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |

*From*: Peter Schoch <pschoch@fandm.edu>*Date*: Thu, 9 Oct 2014 05:38:03 -0400

Thank you to everyone for the response -- and for wading through my muddled

question.

John -- you have every right to be confused by the question as I muddled

the students arguments with my own query. My defense is that I am finding

this years class quite challenging -- while they are extremely bright, they

won't believe anything I say if it is not in the textbook. I could say

2+2=4 and they would ask me where that is stated in the book. I am finding

it to be very vexing.

On Thu, Sep 25, 2014 at 1:15 PM, John Denker <jsd@av8n.com> wrote:

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

center.

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.

===================

Tangential remarks:

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

are confused!

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

https://www.av8n.com/physics/weight.htm

_______________________________________________

Forum for Physics Educators

Phys-l@www.phys-l.org

http://www.phys-l.org/mailman/listinfo/phys-l

**Follow-Ups**:**[Phys-L] believing everything they read, or not***From:*John Denker <jsd@av8n.com>

- Prev by Date:
**[Phys-L] American Association of Chemistry Teachers launched (Sept. 11, 2014)** - Next by Date:
**[Phys-L] some climate concepts, numbers, and references** - Previous by thread:
**[Phys-L] American Association of Chemistry Teachers launched (Sept. 11, 2014)** - Next by thread:
**[Phys-L] believing everything they read, or not** - Index(es):