Re: TIDES, was Asteroid Problem

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

"Paul O. Johnson" wrote:

> Ludwik, I don't understand your second sentence: "In this case . .. "
> Furthermore, I don't understand your argument about the asteroid
> one earth-diameter away. What direction wrt the earth-moon line
> are you assuming the asteroid is located?

1) One word was lost in the second sentence; see the corrected
    version below.
2) Clearly, I am not the only one who needs some tutoring on
    tides. JohnD made a useful contributions but not enough.
3) On my picture the point-like asteroid was located on the
    horizontal line passing through the center of the earth.

> Our moon is about 30 earth-diameters away. In this case
> HORIZONTAL components of forces exerted on any little
> part of our planet are very small in comparison with
> perpendicular components (on my picture where the
> moon-earth line is horizontal). Thus lunar tidal forces
> (gradient of F) are essentially horizontal (on my picture).
> But this would not be so for an asteroid located only one
> earth-diameter away from the center of our planet. Would
> the vertical force components reduce the tidal effect?

Stephen Muray referred to the the 1/r^3 tidal effect; I was
under the impression that this is true only when the tidal
source (asteroid) is far away from us (with respect to the
earth diameter). I hope to learn more about this (and
about bulge being a little larger than another) under the
current tread. I no longer think that an m=10^20 asteroid,
located two or three diameters away can produce tides as
strong as those caused by moon or sun. Recall that this
was the issue several days ago. That is why I changed
the subject line above.
Ludwik Kowalski