Re: Black Holes and tidal forces

[Rondo JEFFERY <RJEFFERY@WEBER.EDU>]

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