Re: Black Holes and tidal forces

[Jack Uretsky <jlu@HEP.ANL.GOV>]

        Yeah, but how do you explain it to people who, probably like most
of your students, don't have much of a grasp of algebra - or the graph of
a monomial function?


On Tue, 27 Nov 2001, Rondo JEFFERY wrote:

> One way to think of it is that the force decreases with the square of
> the distance from the center (1/r^2), whereas the tidal force decreases
> with the cube of the distance from the center (1/r^3).  That is why the
> tidal force from the moon on the oceans is larger (by about a factor or
> 2) than the tidal force from the sun.  The sun obviously has a much
> greater mass than the moon, but the r-cubed factor drops the tidal force
> greatly.  So, if the forces are the same, then the tidal force will be
> greater for the object with the smaller radius.
>
> Rondo Jeffery
> Weber State University
> Ogden, UT 84408-2508
>
>
> > > > FanettT@QUEST.WITCC.CC.IA.US 11/27/01 01:33PM >>>
> 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...t=
> he change?
>
> 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

--
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