If I understand correctly, I think you're saying that if the e-m system stop
rotating, the water on the far side of the planet would eventually "fall" or
stabilize at fixed levels, with low tide being farthest from the moon ;
i.e., net forces would eventually equal zero? I guess I can see that.
Does inertia provide the rest of the answer (i.e., the water on the far side
tends to stay at rest (temporarily) while the earth accelerates toward the
moon?)
Scout


"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