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tide potential calculation
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Handwritten-style derivation typed in a Word file by Phil, dated 1.18.17, of the tidal force and potential on a body of radius R2 due to a distant mass M1. He expands d^-2 and d^-3 to first order in R2/d0, gets a polar-coordinate force proportional to (-sinθ + 2cosθ), and compares with Butakov's deformation formula. A later check of the integrated potential gives the wrong force, so he notes the calculation must be redone and that Taylor's approach finds the potential by inspection.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
Tide potential Calculation PhL 1.18.17
Start with
Ftid = mM1G ( 0/d02 – /d2)
Omit the leading constant since it won't matter
Q = ( d0/d03 – d/d3)
and here is our picture
In Cartesian coordinates we can write
dx = d0 + R2 cosθ dy = R2sinθ
d = (d0 + R2 cosθ) + R2sinθ
d0 = d0
Here I spend a lot of time trying to get the potential by doing line integration of the force.
The potential is
-V = !Syntax Error, IQ ds ds = R2dθ
then
-V = !Syntax Error, I( d0/d03 – d/d3) R2dθ
= !Syntax Error, I( d0/d03 – [(d0 + R2 cosθ) + R2sinθ ]/d3) R2dθ
= !Syntax Error, I{[ 1/d02 - (d0 + R2 cosθ)/d3 ] + [R2sinθ/d3] } R2dθ
= R2 !Syntax Error, I{[ 1/d02 - (d0 + R2 cosθ)/d3 ] + [R2sinθ/d3] }dθ
= R2 !Syntax Error, I{[ 1/d02 - (d0 + R2 cosθ)/d3 ](-sinθ) + [R2sinθ/d3] cosθ}dθ // (E.5.5)
= R2 !Syntax Error, I { -sinθ/d02 + sinθd0/d3 + R2sinθcosθ/d3 + R2sinθcosθ/d3} dθ
= R2 !Syntax Error, I { -sinθ/d02 + sinθd0/d3 + 2R2sinθcosθ/d3} dθ
= R2 !Syntax Error, I { -sinθ/d02 + sinθd0/d3 + R2sin2θ/d3} dθ
Now write
d2 = d02+R22 - 2d0R2cos(π-θ) = d02+ R22 + 2d0R2cosθ = d02[ 1 + (R2/d0)2 + 2 (R2/d0)cosθ ]
≈ d02[ 1 + 2 (R2/d0)cosθ ]
d ≈ d0 [ 1 + 2(R2/d0)cosθ]1/2
d-2 ≈ d0-2 [ 1 + 2(R2/d0)cosθ]-1 ≈ d0-2 [ 1 -2(R2/d0)cosθ ]
d-3 ≈ d0-3 [ 1 + 2(R2/d0)cosθ]-3/2 ≈ d0-3 [ 1 - 3(R2/d0)cosθ ]
Then have
= R2 !Syntax Error, I { -sinθ/d02 + sinθd0/d3 + R2sin2θ/d3} dθ
= R2!Syntax Error, I {-sinθ/d02 + sinθd0 * d0-3 [ 1 - 3(R2/d0)cosθ ] + R2sin2θ * d0-3 [ 1 - 3(R2/d0)cosθ ] }dθ
= (R2/d03)!Syntax Error, I {-sinθd0 + sinθd0 [ 1 - 3(R2/d0)cosθ ] + R2sin2θ[ 1 - 3(R2/d0)cosθ ] }dθ
= (R2/d03)!Syntax Error, I { sinθd0 [- 3(R2/d0)cosθ ] + R2sin2θ[ 1 - 3(R2/d0)cosθ ] }dθ
= (R2/d03)!Syntax Error, I { -3R2sinθcosθ + R2sin2θ - 3R2(R2/d0)sin2θcosθ }dθ
= (R22/d03)!Syntax Error, I { -3sinθcosθ + sin2θ - 3(R2/d0)sin2θcosθ }dθ
= (R22/d03)!Syntax Error, I { - (3/2) sin2θ + sin2θ - 3(R2/d0)sin2θcosθ }dθ
= (R22/d03)!Syntax Error, I { - (1/2) sin2θ - 3(R2/d0)sin2θcosθ }dθ
≈ (R22/d03)!Syntax Error, I { - (1/2) sin2θ }dθ
= (-1/2) (R22/d03)!Syntax Error, I sin2θdθ
= (-1/2) (R22/d03)(1/2)[ 1 - cos2θ]
Putting back the constant factor, here is my result for the potential function which goes with the tidal force
V(θ) = + (1/4) mM1G (R22/d03)[ 1 - cos2θ]
Butakov gives
R(θ) = R2 + a cos(2θ) , (8.7.27)
a = (3/4)R2 (M1/M2)(R2/r12)3 . (8.7.28)
Let's go back to the start,
Ftid = mM1G ( d0/d03 – d/d3)
d = (d0 + R2 cosθ) + R2sinθ
d0 = d0
Ftid = mM1G ( d0/d03 – [ (d0 + R2 cosθ) + R2sinθ ]/d3)
now use
d-3 ≈ d0-3 [ 1 + 2(R2/d0)cosθ]-3/2 ≈ d0-3 [ 1 - 3(R2/d0)cosθ ]
so then
Ftid = mM1G ( d0/d03 – [ (d0 + R2 cosθ) + R2sinθ ] ) d0-3 [ 1 - 3(R2/d0)cosθ ] )
= (mM1G/d03) ( d0– [ (d0 + R2 cosθ) + R2sinθ ] [ 1 - 3(R2/d0)cosθ ] )
= (mM1G/d02) ( – [ (1 + (R2/d0) cosθ) + (R2/d0)sinθ ] [ 1 - 3(R2/d0)cosθ ] )
= (mM1G/d02) ( – - (R2/d0) cosθ) - (R2/d0)sinθ + 3(R2/d0)cosθ )
= (mM1G/d02) ( - (R2/d0) cosθ) - (R2/d0)sinθ + 3(R2/d0)cosθ )
= (mM1G/d02)(R2/d0) ( - cosθ - sinθ + 3cosθ )
= (mM1G/d02)(R2/d0) ( - sinθ + 2cosθ )
Now check this at the compass points
Ftid(0) = (mM1G/d02)(R2/d0) 2 point B to the right
Ftid(π) = - (mM1G/d02)(R2/d0) 2 point A to the left
Ftid(π/2) = - (mM1G/d02)(R2/d0) down at top
Ftid(-π/2) = +(mM1G/d02)(R2/d0) up at bottom
where I toss terms of order (R2/d0)2 . Compass agrees. I like the general formula:
Ftid(θ) ≈ (mM1G/d02)(R2/d0) ( - sinθ + 2cosθ )
Well, I guess this is my first statement of Ftid in polar coordinates.
Now let's try again doing the integral
-V = !Syntax Error, IFtid(θ) ds ds = R2dθ
So then
-V = (mM1G/d02)(R2/d0)!Syntax Error, I( - sinθ + 2cosθ ) R2dθ
= R2 (mM1G/d02)(R2/d0)!Syntax Error, I( - sinθ + 2cosθ ) dθ
= R2 (mM1G/d02)(R2/d0)!Syntax Error, I( - sinθcosθ + 2cosθ[-sinθ] ) dθ
= R2 (mM1G/d02)(R2/d0)!Syntax Error, I( - 3sinθcosθ ) dθ
= R2 (mM1G/d02)(R2/d0)!Syntax Error, I( - (3/2) 2sinθcosθ ) dθ
= R2 (mM1G/d02)(R2/d0)!Syntax Error, I( - (3/2) sin2θ ) dθ
= - (3/2) R2 (mM1G/d02)(R2/d0)!Syntax Error, Isin2θ dθ
= - (3/2)R2 (mM1G/d02)(R2/d0)(1/2)(1-cos2θ)
= - (3/2)R22 (mM1G/d03)(1/2)(1-cos2θ)
= - (3/4)R22 (mM1G/d03)(1-cos2θ)
And therefore
V = (3/4)r2 (mM1G/d03)(1-cos2θ) r = R2
Ftid(θ) = -V = - (1/R2)∂θV - ∂rV
= - (1/R2) (3/4)r2 (mM1G/d03)2sin2θ - (3/4)2r (mM1G/d03)(1-cos2θ)
= (3/4)(mM1G/d03) [ - R22sin2θ - 2R2(1-cos2θ) ]
= (3/4)(mM1G/d03) [ - ( -sinθ + cosθ ) R22sin2θ - (cosθ + sinθ )2R2(1-cos2θ) ]
= - (3/4)(mM1G/d03) [ ( -sinθ + cosθ ) R22sin2θ + (cosθ + sinθ )2R2(1-cos2θ) ]
= - (3/4)(mM1G/d03) R22 [ ( -sinθ + cosθ ) sin2θ + (cosθ + sinθ )(1-cos2θ) ]
= - (3/4)(mM1G/d03) R22 [ ( -sinθ sin2θ + cosθ(1-cos2θ) + ( cosθ sin2θ + sinθ (1-cos2θ) ]
Now:
-sinθ sin2θ + cosθ(1-cos2θ) = -sinθ 2sinθcosθ + cosθ 2 sin2θ
= -2sin2θcosθ + 2cosθ sin2θ = 0
cosθ sin2θ + sinθ (1-cos2θ) = cosθ2sinθcosθ + sinθ 2sin2θ
= 2sinθcos2θ + 2sinθ sin2θ = 2sinθ
so the result is then
Ftid(θ) = - (3/4)(mM1G/d03) R22 [ (0) + ( 2sinθ) ]
which is the wrong answer. The correct answer is
Ftid = (mM1G/d02)(R2/d0) ( - sinθ + 2cosθ )
So things are just hugely messed up in this entire doc, I guess I need to start over.
I later realized that in hybrid notation I could find the potential by inspection! I got this hint from reading how Taylor did it.