Nav -
The two forces you are talking about are exactly the same. The centrifugal
force IS =m1m2/r^2. They are NOT two different forces. The centrifugal
force is the fictional force that is exactly opposite the real force (as viewed
from a non-accelerating system). Even if you calculate from a centrifugal
force point of view, you end up deriving the angular velocity in terms of
F=m1m2/r^2.
For anyone who wants to follow this through, here's a pretty complete version.
http://www.clupeid.demon.co.uk/tides/maths.html
The approach is to take the force the Moon's pull exerts on individual parts of
the Earth, then to add in the centrifugal force and the Earth's pull and the
Earth's daily rotation component. However, the centrifugal force is derived
from the total gravitation pull, so that component is certainly not ignored.
As is often the case, it gets messy before terms start to cancel, but the end
result is that the "differential gravity" is exactly symmetrical on the near and
far side from the Moon. It would not be fair to say the the near side component
is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the Moon.
It is when you subtract out the net pull (or add the centrifugal) that this
becomes symmetrical on both sides.
"Nav" wrote in message
...
Let me try to be clear. A gravity only argument is usually based on the
equation F=m1m2/r^2. This is a monotonic field function so that water
would only ever flow toward the point closest to the center of mass of
the system (note that the water's potential energy is proportional to
F(r).r). It cannot create two tidal bulges for that requires two
potential energy minima. Now couple that equation with F=mr omega^2 and
you get the local force on a mass of water (mw) as mw.ms/r^2 - mw r
omega^2 where ms is the system mass and r the distance to the center of
mass. Now the potential energy of water has a new local minima away from
the center of mass (plot 1/r + r^2 ; r 0). This gives the tidal bulge
on the far side of the planet. This is only a local minimum, so that if
the earth did not constantly move water to it by it's own rotation that
bulge would gradually disappear. If you like, water gets stuck at the
far side local minumum as the earth rotates.
So, as I see it, without explictly including the rotation(s) you don't
get two tides -OK?
Cheers
Jeff Morris wrote:
but the only way the system wouldn't rotate is if there was no gravity, so
I'm
not sure what your point is.
"Nav" wrote in message
...
I think if the earth moon-earth system were not rotating there would be
one tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.
Cheers
Scout wrote:
I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in
a
low high tide on the side farthest from the moon.
Scout
"Nav" wrote in message
...
Yes, so...
Cheers
Scout wrote:
If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout
"Nav" wrote in message
...
Yes, you can. Where is the center of mass of the earth moon system?
Cheers
Peter S/Y Anicula wrote:
We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.
Peter S/Y Anicula
"Nav" skrev i en meddelelse
...
Well Peter, I have to disagree there. The gravitational force acts
only
toward the center of mass of the system. This cannot by itself
produce
two bulges. To clarify this, try imagining the forces of gravity in
2D
on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the
far
side and more water as you move toward the moon... -two bulges would
not
be present.
Cheers