Image talk:Tidal-forces-calculated.png
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[edit] Precisions?
Could someone clarify the orientation of the diagram? We're told the primary is off to th eleft, but are we looking from above the orbital plane, or from behind the moon on its orbit?
Urhixidur 03:38, 2004 Dec 3 (UTC)
- What makes you think there would be any difference? The moon doesn't even need to be in an orbit to experience tidal forces, it could be just falling towards the primary. Theresa Knott (The snott rake) 05:24, 3 Dec 2004 (UTC)
The assessment that the tidal force at the poles(or on the great circle thru the poles and perpindicular to the sun-moon line is 1/2 of that along the sun-moon line does not appear to be correct. It should be a factor r/R(where r is the radius of the moon, and R is the distance from the center of the sun to the center of the moon) based upon a second order expansion - not 1/2.
[edit] Inconsistency
"generated by a mass off the bottom" "the mass generating the gravitational field is indeed off to the left, not the right"
Occultations 12:43, 20 December 2005 (UTC)
[edit] Orientation (new version)
Some while ago, BD rotated the pic (so the mass was at the bottom); but left the left/right bit in (as Occultations noticed, above; sorry I didn't pick up on that).
Now I've uploaded a higher rez version, and (to match the schematic on tidal forces) have put the mass off to the right.
Hope thats not too confusing...
William M. Connolley 22:08, 19 February 2006 (UTC)