"Nav" wrote in message
...
Jeff Morris wrote:
....
Before I thought you were just arguing philosophically how much we should
credit
centrifugal force, but now it appears you haven't really looked at the math
at
all. The reason why "differential gravity" is invoked is because it
represents
the differing pull of the Moon on differing parts of the Earth. Although
this
force is all obviously towards the Moon, when you subtract off the
centrifugal
force this is what is left. It is this differing pull that causes the two
tides.
Again you say that gravity causes the two tides -but I say that is not
correct. It is the DIFFERENCE between the centripetal term and the
gravity that causes the tide (How many times do I have to say this?).
You may try to malign me by saying I've not looked at the math but I
have -much closer than you have I think. Here's something for you to try
during coffee:
1) How big is Differential gravity? (1e-6 m/s/s ? It's on the web)
roughly correct
2) How big is the difference in centrifigal acceleration on each side of
the barycenter?
Centrifugal acceleration = r omega^2
The moon orbits the earth every 28 days so omega = 2 pi/ (28 * 24 * 60 *
60).
The barycenter is about at 3/4 r (1/4 r under the earth surface) so the
difference in r from one side to the other makes the imabalance.
The difference in centrifugal acceleration is therefo
1.5 * r * (2 pi/ 28 * 24 * 60 * 60)^2
r is 6.4e^6 m
Get out your calculator and work it out for yourself. It's easy and I
think you'll be surprised at the answer.
So, you you're claiming the lunar tidal forces are 65 times the accepted values.
Now, get out your calculator and run the same numbers for the Sun. The distance
to the Sun (and the E-S barycenter) is 1.5e^11 meters. The result is about 100
times less than your result for the Moon. So you're claiming that the Sun has
negligible effect on the night time tides?
Or is my equation for centrifugal acceleration wrong?
Actually, applying it in this context is your problem. Centrifugal acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.
You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making
it
simple for young children, it doesn't really explain what's going on.
But I never tried to explain it all by intertial forces, Jeff. I always
said it was the _difference_ between inertial forces and gravity. I'd
say it's you who is trying to explain it all by gravity instead!
Gravity is actually the only force at work. Any explanation must be consistent
with that. Centrifugal forces is "ficticious," it doesn't really exist. The
reality is that the Earth is in free fall towards the E-M barycenter.
What this shows is that without land masses to block tidal flow the
tides would be much bigger than they are.
Absoulute nonsense! The land masses build up the tides, they don't reduce
them. And there is no major landmass on the equator for almost half of the
Earth' circumfrence - there is plenty of room for the tides to fully devolope.
Your theory predicts island in the Pacific would be hit by 100 foot tides every
day.
That differential gravity
produces the "correct" answer really just shows how it is not the
correct sole explanation (if it had predicted higher tides that would be
expected).
Double talk - you made up a silly explanation for why we don't have 100 foot
tides and then fault the accepted explantion for not predicting the same thing.
Perhaps you would like to think about that approximation used
in the gravity term that "allows" it to cancel the centripetal term. Is
it correct to use this when we are dealing with very very small
fractions of the total acceleration??
Perhaps you'd like to explain why your approach shows that the Sun has
negligible contribution to the tides. Sorry Nav, this is looking like the
Constellation all over again.
-jeff
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