Re: Black Holes and tidal forces

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

> SO the gradient changes faster for a smaller black
> hole...okay so how would I explain that to people who don't
> know what a gradient is...the change?

Yes.

You could contrive an example ,say the moon, where you calculate the
difference in the gravitational force between the head and feet.


Then do the same thing for a larger radius planet.  (you will have to use a
density for that planet so that the gravitational force at the surface is
the same as for the moon.)  And compare the difference in the gravitational
force between your head and your feet.

Naturally the force at the feet is the same (you contrived the example so
that is true.)  But the force at the head should be smaller for the moon
case.  Thus the difference (or change) is greater for the moon.  (larger
gradient).


> Tina
>
> Tina Fanetti
> Physics Instructor
> Western Iowa Technical Community College
> 4647 Stone Ave
> Sioux City IA 51102
> 712-274-8733 ext 1429
>
> > > > Roger Haar <haar@PHYSICS.ARIZONA.EDU> 11/27/01 02:37PM >>>
> Tina,
>         It is the gradient of the gravitational force
> that is important.  Consider going in feet first.
> For a small black hole the force on your feet is
> much larger that on your head.  Plus if you are in
> a circular orbit, the  orbital velocity and force
> associated with orbiting change quickly with
> radius for small holes.
>
> Thanks
> Roger Haar