moments of inertia of special oblate spheroid

Inertial moments of an oblate spheroid (approx) PhL 3.19.17 Texas reference: http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node109.html http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node108.html Here is the picture Use spherical coordinates (r,θ,φ), Assume unit mass density ρ. Then dm = dV = ρr2drdφsinθdθ. Note that z = rcosθ The moment of inertia of dm is given by dI = dm(rsinθ)2 = [ρr2drdφsinθdθ](rsinθ)2 The boundary of the oblate spheroid is assumed to be this, where ε << 1 R(θ) = a[1 - (2/3)ε P2(cosθ)] Here then is the moment of inertial about the vertical axis I3 = ρ !Syntax Error, Idφ !Syntax Error, Idθ sin3θ !Syntax Error, Ir4 = 2πρ !Syntax Error, Idθ sin3θ !Syntax Error, Ir4 = (2πρ/5) !Syntax Error, Idθ sin3θ R5(θ) Now R(θ) = a[1 - (2/3)ε P2(cosθ)] R5(θ) ≈ a5 [1 - (10/3)ε P2(cosθ)] Then I3 ≈ (2πρ/5) a5 !Syntax Error, Idθsin3θ [1 - (10/3)ε P2(cosθ)] The first integral is this We know that P2(cosθ) = (1/2)(3cos2θ-1) Then we want !Syntax Error, Idθsin3θ P2 = (1/2) !Syntax Error, Idθsin3θ (3cos2θ-1) Maple says Therefore I3 ≈ (2πρ/5) a5 !Syntax Error, Idθsin3θ [1 - (10/3)ε P2(cosθ)] = (2πρ/5) a5 [ (4/3) - (10/3)ε (-2/3) ] = (2πρ/5) a5 [ (4/3) + (20/9)ε ] = (2πρ/5) (4/3) a5 [ 1 + (5/3)ε ] = ρ (8π/15) a5 [ 1 + (5/3)ε ] Now approximate that M = (4/3)πa3ρ ρ = (3/4) M/(πa3) Then have I3 ≈ (3/4) M/(πa3) * (8π/15) a5 [ 1 + (5/3)ε ] = (3/4) M (8/15) a2 [ 1 + (5/3)ε ] = (2/5) M a2 [ 1 + (5/3)ε ] I now go check the list of inertias on wiki, So that is good and we the correction for oblateness to first order in ε. But now I back up because I see this result in that list If we have spheroid, set A = B and you then get I3 = (m/5) (A2+C2) I1 = (m/5) (2A2) Thus we have I3 - I1 = (m/5) (A2+C2) - (m/5) (2A2) = (m/5) ( C2-A2) In our model we have R(θ) = a[1 - (2/3)ε P2(cosθ)] Therefore C = a[1 - (2/3)ε P2(0)] = a[1 - (2/3)ε (-1/2)] = a [1 +ε/3] A = a[1 - (2/3)ε P2(1)] a[1 - (2/3)ε 1] = a[1 - 2ε/3] C+A ≈ 2a C-A = εa since P2(cosθ) = (1/2)(3cos2θ-1). Then I3 - I1 = (m/5)(C+A)(C-A) ≈ (m/5)(2a)(εa) = (2/5)mεa2 THAT took a long time for me to get!! ****************************** On scratch I have showed that the torque on an oblate Earth is given by N = 3(2/5) GMa2ε/r3cosθsinθ = 3{(2/5)Mεa2 } G/r3cosθsinθ = 3{I3 - I1} G/r3cosθsinθ Compare this to Williams result I think I am on to it finally. Note that cosθsinθ = cos(π/2-δ)sin(π/2-δ) = sinδ cosδ So I have lost a mass somewhere. *********************** Here is the calculation of the torque on the Earth ************* STEP 1 Away from sources of charge, the electrostatic potential must satisfy the Laplace equation. The same is true for the gravitational potential away from mass sources. In general any such potential can be expressed as a sum over the "atomic forms" (harmonics) of a given coordinate system. For spherical coordinates the atomic forms can be taken as [rn , r-n-1] [ Pnm(z), Qnm(z) ] [eimφ, e-imφ] z = cosθ where n and m are non-negative integers. These are appropriate for regions of space which have the full range of angle φ. In such a region, for the potential to be single valued in azimuth one must have m = integer so exp(im(φ+2π)) = exp(imφ). This in turn forces n to be an integer, and m ≤ n. If θ = 0 is part of the range of interest, one cannot have Qnm(z) terms since they blow up at z = 0. In this case one has [rn , r-n-1] [ Pnm(z)] [eimφ, e-imφ] z = cosθ which gives the familiar spherical harmonics expansion of a potential. If the potential is azimuthally symmetric, one must have m = 0 reducing the atomic form set to [rn , r-n-1] Pn(z) If a region of interest extends to r = ∞, only the r-n-1 powers are allowed. We then arrive at this expansion for the gravitational potential outside an azimuthally symmetric mass distribution. V(r,θ) = Σn=0∞ Vn r-m-1 Pn(cosθ) If the potential is also symmetric under z reflections, only even n contribute in the sum, and this is the situation for the potential of an oblate spheroid with z as the symmetry axis. If the spheroid is only very slightly oblate, we keep just the first two terms to get V(r,θ) = V0/r + V2 r-3 P2(cosθ) STEP 2 Consider an ellipse of the form x2/A2 + z2/B2 = 1. If we measure polar angle θ down from the z axis, then x = rsinθ and z = rcosθ and the ellipse equation becomes r2 = For an "oblate" (wide) ellipsoid we have A > B. If the ellipse is only slightly oblate, write A = B+δ In this case we get r2 = ≈ ≈ ≈ A2 [ 1-2(δ/B)cos2θ ] ≈ A2 [ 1-2(δ/A)cos2θ ] Then r ≈ A [ 1-(δ/A)cos2θ ] The eccentricity of the ellipse is ε2 = (A2-B2)/A2 = (A+B)(A-B)/A2 ≈ 2Aδ/A2 = 2δ/A Then have r = A [ 1- (1/2)(2δ/A)cos2θ ] = [ 1- (1/2)ε2cos2θ ] STOP. My Texas document claims and you see the ε is not squared, and is called ellipticity which other sides say is the usual eccentricity Now write P2 = (1/2) (3cos2θ - 1) ≡ 2P2 = 3cos2θ -1 3cos2θ = 2P2 + 1 so cos2θ = (1/3)(2P2+ 1) Then have r = [ 1- (1/2)ε2cos2θ ] = [ 1- (1/6)ε2(2P2+ 1) ] How am I ever doing to get this result: Try to do it: r = 1 - (1/3)ε2P2 - (1/6)ε2 = (1- (1/6)ε2) - (1/3)ε2P2 = (1- (1/6)ε2) [ 1 - (1/3)ε2P2 / (1- (1/6)ε2) ] = d [ 1 - (1/3)ε2P2 /d ] d = (1- (1/6)ε2) ≈ d [ 1 - (1/3)ε2P2 ] So the fraction is wrong and the power of ε is wrong! What is the "mean radius" of this oblate spheroid? Go back to r = [ 1- (1/2)ε2cos2θ ] Then <r> = [ 1- (1/2)ε2<cos2θ> ] = [ 1- (1/2)ε2(1/3)] if I did it right = [ 1- (1/6)ε2] if I did it right = a of Texas author But this is exactly my d from above! Conclusion: ether the Texas author has goofed up, or I have goofed up royally. I now need another source on the Texas claim which is the STARTING point of his presentation: I found a doc which says this: This author is using the "flatness" factor f. If I go back to r ≈ A [ 1-(δ/A)cos2θ ] then f = (A-B)/A = δ/A which compare to ε2 = (A2-B2)/A2 = (A+B)(A-B)/A2 ≈ 2Aδ/A2 = 2δ/A Thus ε2 = 2f and then my result is r = d [ 1 - (1/3)ε2P2 ] = d [ 1 - (1/3)2fP2 ] = d[ 1 - (2/3) f P2 ] so the Texas offer is using symbol f to be "flatness", not eccentricity. That would explain it! In fact, later Texas does say this. confirming that his ε is really the flattening factor This same new author says ******************************************* Another approach: in curvilinear /sphericals I have V(r,θ,φ) = Σn=0∞ Σm=-nn anm(r) Pnm(cosθ) eimφ anm(r) = (2n+1) f(n,-m) (1/4π) ∫dΩ V(r,θ,φ) e-imφ Pnm(z) If V goes not depend on φ, only m = 0 can contribute in the projection so an0(r) = (2n+1) f(n,0) (1/4π) ∫dΩ V(r,θ)Pn(z) V(r,θ) = Σn=0∞ an0(r) Pn(cosθ) In this case just write as an(r) = (2n+1)f(n,0) (1/4π) ∫dΩ V(r,θ)Pn(z) V(r,θ) = Σn=0∞ an(r) Pn(cosθ) Note that Pnm(z) = f(n,m) Pn-m(z) f(n,m) = Γ(n+m+1)/(n-m+1) f(n,0) = Γ(n+1)/(n+1) = 1 So here we are vn(r) = (2n+1) (1/4π) ∫dΩ V(r,θ)Pn(z) V(r,θ) = Σn=0∞ vn(r) Pn(cosθ) ≈ V0/r + V2 r-3 P2(cosθ) Now above I said that V(r,θ) = V0/r + V2 r-3 P2(cosθ) Now are things connected here? I guess it is this v0(r) = V0/r v2(r) = V2/r3 Then for example v0(r) = (1/4π) ∫dΩ V(r,θ)P0(z) = (1/4π) ∫dΩ V(r,θ) But this is not helping because I want a different thing. Start over: V(r) = ∫dV' ρ(r') / |r-r'| Jackson scale electrostatics 1/R = Σn=0∞ Σm=0∞ εm f(n,-m) (r<)n(r>)-n-1 Pnm(z) Pnm(z') cos[m(φ-φ')] (**) If mass density ρ(r') does not depend on φ', only the m=0 terms survive so in effect 1/R = Σn=0∞ ε0 f(n,0) (r<)n(r>)-n-1 Pn(z) Pn(z') = Σn=0∞ (r<)n(r>)-n-1 Pn(z) Pn(z') Then put this in to get V(r) = ∫dV' ρ(r') Σn=0∞ (r<)n(r>)-n-1 Pn(z) Pn(z') Now if r is outside the entire mass distribution, r> = r and you have V(r) = ∫dV' ρ(r') Σn=0∞r'nr-n-1 Pn(z) Pn(z') = Σn=0∞ r-n-1 Pn(z) ∫dV' r'nρ(r') Pn(z') = Σn=0∞ r-n-1 Pn(z) { 2π ∫r'2+ndr' ∫sinθ'dθ' ρ(r',cosθ') Pn(z') } = Σn=0∞ r-n-1 Pn(z) { 2π∫r'2+ndr' ∫dz' ρ(r',z') Pn(z') } Add factor -G on the right to get V(r) = Σn=0∞ r-n-1 Pn(z) { -2πG∫r'2+ndr' ∫dz' ρ(r',z') Pn(z') } = Σn=0∞ r-n-1 Pn(z) Vn where Vn = -2πG∫r'2+ndr' ∫dz' ρ(r',z') Pn(z') So I am now agreeing with Texas who says So I could do the rest of his page and conclude as he does that So I think I can derive this given his oblate form where I have shown that ε is really the flatness. I would next like to show now the projection integrals relate to moments! Vn = -2πG∫r'2+ndr' ∫dz' ρ(r',z') Pn(z') V0 = -2πG∫r'2dr' ∫dz' ρ(r',z') = -GM M = total mass V2 = -2πG∫r'4dr' ∫dz' ρ(r',z') P2(z') = ??? Remember that z' = cosθ, it is not a Cartesian z'. I think the inertial radius is R = r' sinθ' I33 ≡ ∫dxdydz ρ(x)[ r2 - z2] I11 ≡ ∫dxdydz ρ(x)[ r2 - x2] I33 - I11 = ∫dxdydz ρ(x) [x2-z2] In sphericals maybe have x = rsinθcosφ and z = rcosθ. Then I33 - I11 = ∫dφ ∫sinθdθ ∫r2dr ρ(x) [ r2sin2θcos2φ - r2cos2θ ] Now go from the Cartesian z to the spherical z and write this as I33 - I11 = ∫dφ ∫dz ∫r4dr ρ(x) [ (1-z2)cos2φ - z2 ] Then use ∫dφ cos2φ = π to get I33 - I11 = ∫dz ∫r4dr ρ(x) [ π(1-z2)cos2φ - 2πz2 ] = π ∫dz ∫r4dr ρ(x) [ (1-z2)- 2z2 ] = π ∫dz ∫r4dr ρ(x) [ (1-3z2) ] But P2 = (1/2) (3cos2θ - 1 so P2(z) = (1/2)(3z2-1) so (1-3z2) = -2P2(z) and then I get I33 - I11 = -2π ∫dz ∫r4dr ρ(x)P2(z) Compare this to result above V2 = -2πG∫r'4dr' ∫dz' ρ(r',z') P2(z') Therefore I think V0 = -GM V2 = G (I33 - I11) Then I get V(r,θ) = V0/r + V2 r-3 P2(cosθ) = -GM/r + G (I33 - I11) P2(cosθ)/ r3 This agrees with so I guess I have obtained this thing two ways now. The C and A are inertia moments. Then all you have to do is compute the torque on this as I have already done on scratch.