earth moon and stick tides
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Part of Phil's Section 8 notes on tides and a tether. It expands the tidal force from the Moon to first order in R2/d0, removes the term that is absent in the stick case, and obtains a force proportional to (2x - d0 - y). It then builds a potential V(x,y), sets it constant to find the water surface, and plots an approximate shape, concluding that water piles up on the Moon-facing side.
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Tides if Earth and Moon don't rotate but are held apart by a stick.
Ftid = mM1G { d0/d03 – d/d3 } // the red term needs to be removed for stick situation
= mM1G { d0/d03 – [(d0 + R2 cosθ) + R2sinθ ] d0-3 [ 1 - 3(R2/d0)cosθ ]}
= (mM1G/d02) { – [(1 + (R2/d0) cosθ) + (R2/d0) sinθ ] [ 1 - 3(R2/d0)cosθ ] }
= (mM1G/d02) { – + 3(R2/d0)cosθ - (R2/d0) cosθ) - (R2/d0) sinθ) } + O((R2/d0))2
≈ (mM1G/d02) {2(R2/d0)cosθ - (R2/d0) sinθ) }
= (mM1G/d03) {2R2cosθ - R2sinθ) }
= (mM1G/d03) ( 2x - y ) - mM1G { d0/d03 }
// so here I manually subtract the red term to make it missing above
= (mM1G/d03) ( 2x - y ) - mM1G/d03 { d0 }
= (mM1G/d03) ( 2x - y -d0)
= (mM1G/d03) ( (2x - d0) - y )
= (mM1G/d03) ( (2x - d0) - y )
≈ (mM1G/d03) ( - d0 - y ) // mainly to the left
***************
Now trace the tidal shape argument:
We seek a potential V(x,y) which solves this equation (no minus sign in V=F)
V(x,y) = Fg2 + Ftid = - (GM2m/r2) + (mM1G/d03) ( -d0 - y ) (8.8.63)
where we use (8.8.24) for Ftid. The exact solution for V is the following (by inspection) :
V(x,y) = (GM2m/r) + (GM1m/d03)( -d0x - y2/2) . // (1/r) = -(1/r2), r > 0 (8.8.64)
Of interest are surfaces on which V is a constant (since we expect the water surface to be such a surface) so we let k be a constant and write
(M2/r) + (M1/d03)( -d0x - y2/2) = 1/k
(kM2/r) + (M1/kd03)( -d0x - y2/2) = 1
(kM2/r) = 1 - (kM1/d03)( -d0x - y2/2)
(kM2/r)2 = [ 1 - (kM1/d03)( -d0x - y2/2)]2
(kM2)2 = (x2+y2) [ 1 - (kM1/d03)( -d0x - y2/2)]2
So I do the same approx done in frames doc to get
(kM2)2 = (x2+y2) [ 1 - 2(kM1/d03)( -d0x - y2/2)
The d0 term in there swamps things so I guess
(kM2)2 = (x2+y2) [ 1 + 2x(kM1/d02) ]
= (x2+y2) [ 1 + 2xε' ] ε' = kM1/d02
And what does this look like? Set ε' = .05 and kM2 = 1 to get
I would have to normalize things as before, but you can see that the water piles up on the left side of the earth facing the moon, and pulls away from the far side of the Earth. So I think my conjecture is correct.