In a way that's correct but a bit colorful. It is the excess of
centripetal force over gravity that causes the high tide on the far side

Cheers

Scout wrote:
I just want to make sure I'm understanding this correctly. Sorry if I'm
repeating myself here, but when you say "centripetal" aren't you implying
that the planet acts as a huge impellor in a centrifugal pump, "throwing" or
piling water up on the far side of the planet, while the planet itself acts
as a dam of sorts, preventing the water from spilling back to the low tide
areas.
Scout


"Nav" wrote in message
...

I'm going to give you the benefit of that doubt and hope you are not just
trolling. I'm sorry if you can't understand the maths. It is not
"differential gravity" -the maths are clear and unambiguous on this
point: Differentiate the gravity field equation and you just get a
monotonic function of distance from the center of system mass so that
water would only ever move in one direction, namely toward the center of
the system. It is the centripetal term that introduces the extra force
required to make a second tidal bulge. So, you need to include rotation
about the center of mass in any explanation of two tides.

If you still don't follow my argument (and accept the veracity of the
maths) then I can't help you.

Cheers

Peter S/Y Anicula wrote:

Ok, are you now ready to admit that you were mistaken, and that the
"differential gravitational explanation" does in fact explain that
there is a bulge on the side of the earth that turns away from the
moon ?

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...


Ok, then we are almost there ! The final answer on that good site
shows that it is the _imbalance_ between the radial component of

gravity


and the centripetal term. You seem to have missed the importance

(and


cause) of the term that raises the radius of the earth (r) to the

fourth


power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:



I see your point, but I keep looking at the final answer. When

all the terms


are balanced, and the minor effects ignored, what is the left is

2GmMr


cos^2/R^3, which comes from the radial component of the moon's

gravity on a


piece if the Earth m. All of the other forces, including all of

the


centrifugal forces have been balanced out. The cos^2 term is the

only thing


left that varies with latitude, which means that explains why the

bulges are at


the equator, and the pole's tides are depressed.


"Nav" wrote in message
...



Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not

interested


in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy

minima due


to the system rotation. Note the first two terms of the force

balance


equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include

the


centripetal force term (mr omega^2) across the diameter of the

earth is


simply not a correct analysis. Furtherore, if that "explanation"

is used


to show two tides it's bogus (it is clear the earth's radius does

not


cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:





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.



Jeff, That apparoach is exactly the same as the one I offered

except


that for simplicity I did not consider the earth's rotation. It

also


supports my quick and dirty proof or there being two energy minima

due


to the system rotation.