As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces. Here's a site that uses that approach:

http://co-ops.nos.noaa.gov/restles3.html


"Nav" wrote in message
...
Jeff, you really cannot explain two tides a day unless you also include
the centripetal forces of the earth moon pair -this is the key that is
seems repeatedly lost.

Cheers


Jeff Morris wrote:

"DSK" wrote in message
...

Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:

On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.


Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple
way:
http://burro.astr.cwru.edu/Academics...ity/tides.html