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Natural Science Forum / Physics / General Physics / July 2008



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gravitational attraction of a suspended thin cylinder

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Prime Mover - 13 Jul 2008 15:54 GMT
Hello all physicists!

What would be the Earth's gravitational attraction on a cylinder in
the configuration below:

         .
       .      .
     .      .
   .      .                       <------- side view of the thin
cylinder
 .      .
      .

- - - - - - - - - - - - - - - - -  (Earth's surface)

You have to imagine that the cylinder is sustained by something up to
a certain height over the Earth's surface. What that "thing" that is
supporting the cylinder is does not matter.

I'm curious on how to take into account the inclination of the
cylinder as well as the heigh above the Earth's surface.

Thank you all.
Uncle Al - 13 Jul 2008 20:05 GMT
> Hello all physicists!
>
[quoted text clipped - 19 lines]
>
> Thank you all.

a = GM/r^2  M is the Earth's mass, r is the radius to its center of
gravity.  Integrate dr for large /_\r above the surface.

The test mass' center of mass is accelerated toward the Earth's center
of mass (with centripetal correction to center of gravity) by 1 gee.
For a non-local mass distribution a quadrupole tidal force rotates its
largst moment of inertia axis until it is aligned parallel to gee.

This is one reason why ISS FUBAR is such a loathsome piece of sh.t.
In free fall its long axis would point toward the Earth's center.
NASA wants it flying tangent to the surface for photo ops.  Two real
world consequences ensue:

  1) ISS FUBAR constantly burns outs steering reaction wheels, and

  2) Despite ISS FUBAR being the size of a half-dozen 747s, the total
true free fall volume within ISS FUBAR is about four basketball's
worth.

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Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

 
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