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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.







 
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