tides v4

Facing Planet Problem PhL 8.13.12 This is a basic outline of how you might do this problem. This is tides on the Earth if the earth faced the moon during its rotation. Consider this picture Before doing anything, we have a lot of geometry to figure out. Assume that r12, M1, M2, R2 and θ are given quantities. Then here is how we figure out everything else shown: θ' = π-θ // determines θ' R = r12 [M1/(M1+ M2)] // determines R r1c2 = r122 + R22 - 2r12R2cosθ' // determines r1c sinβ = sinθ(R2/r1c) // determines sinβ R22 = r1c2  + r122 - 2r1cr12cosβ // determines cosβ R32 = R2+R22- 2RR2cosθ' // determines R3 sinα = (R2/R3)sinθ' // determines sinα R22 = R2+R32- 2RR3cosα // determines cosα So now we know all quantities in the picture. Next, we consider a particle at the center of object 1. We will have there this obvious force balance Fgrav1(center) = -(mM1G/r122) real Fcent(center) = mω2R fict Since a particle at the center has a' = 0, these must cancel and we get M1G/r122 = ω2R // determines ω Next, we compute the total force on the Particle at C. The magnitude of the upper vector is ω2R3 and this is a fictitious force, the angle is α as shown. The grav1 vector has magnitude mM1G/r1c2 and it is at angle β as shown. Now stare at the picture to conclude that, at point C, Fgrav1(C) = (mM1G/r1c2) [ cos(π/2-θ'- β) + sin(π/2-θ'- β) ] real Fcent(C) = (mω2R3) [ -cos(α - π/2 +θ') + sin(α - π/2 +θ') ] real Fgrav2(C) = -(mM2G/R22) fict Fsurf(C) = Fsurf real The total force F'eff acting on Particle C is the sum of these four forces F'eff = [ (mM1G/r1c2) cos(π/2-θ'- β) - (mω2R3) cos(α - π/2 +θ')] + [(mM1G/r1c2) sin(π/2-θ'- β) + (mω2R3) sin(α - π/2 +θ') – (mM2G/R22) + Fsurf ] I think this is the tidal force at point C. Now simplify by using cos(π/2-x) = sin(x) sin(π/2-x) = cos(x) Then cos(π/2-θ'- β) = cos(π/2 - [θ'+β]) = sin(θ'+β) cos(α - π/2 +θ') = cos(π/2 -α - θ') = cos(π/2-[α+θ']) = sin(α+θ') sin(π/2-θ'- β) = sin(π/2 - [θ'+β]) = cos(θ'+β) sin(α - π/2 +θ') = - sin(π/2-α - θ') = -sin(π/2- [α+θ']) = -cos(α+θ') F'eff = [ (mM1G/r1c2) sin(θ'+ β) - (mω2R3) sin(α+θ')] + [(mM1G/r1c2) cos(θ'+β) – (mω2R3) cos(α+θ') – (mM2G/R22) + Fsurf] Now we would like to know the direction of this tidal force. We finally simplify by assuming that object 1 is far away so r12 >> R2. Then our list above becomes θ' = π-θ // determines θ' R = r12 [M1/(M1+ M2)] // determines R r1c2 = r122 // determines r1c sinβ = sinθ(R2/r12) // determines sinβ 0 = r122  + r122 - 2r12r12cosβ // determines cosβ ≈ 1 so β ≈ 0 but see above R32 = R2+R22- 2RR2cosθ' // determines R3 , no simplification sinα = (R2/R3)sinθ' // determines sinα, no simplification R22 = R2+R32- 2RR3cosα // determines cosα M1G/r122 = ω2R // fact also known Then we have F'eff ≈ [ (mM1G/r122) sin(θ'+ β) - (mω2R3) sin(α+θ')] + [(mM1G/r122) cos(θ'+β) – (mω2R3) cos(α+θ') – (mM2G/R22) + Fsurf] F'eff ≈ [ (m ω2R) sin(θ'+ β) - (mω2R3) sin(α+θ')] + [(m ω2R) cos(θ'+β) – (mω2R3) cos(α+θ') – (mM2G/R22) + Fsurf] What happens at θ' = 0 ? At that point α = π and R3 = R2-R and β = 0 F'eff ≈ [ (m ω2R) sin(0) - (mω2R3) sin(π)] + [(m ω2R) cos(0) – (mω2R3) cos(π) – (mM2G/R22) + Fsurf] F'eff ≈ + [(m ω2R)+ (mω2R3) – (mM2G/R22) + Fsurf] F'eff ≈ + [(m ω2R)+ (mω2[R2-R]) – (mM2G/R22) + Fsurf] F'eff(A) ≈ + [ mω2R2 – (mM2G/R22) + Fsurf] and since A is at rest, this adds up to 0. Normally we would have Fsurf = (mM2G/R22) for something at rest on the surface, But now we have' Fsurf = (mM2G/R22) - mω2R2 so the earth presses less than normal, so something is pulling to the left, and Ftid(A) = mω2R2 to the left This is the main term I got in my version 2 document. My approx here killed off the other term which recall is 41x smaller for moon-earth. The centrifugal effect dominates. What about point B? There θ' = π, α = 0, R3 = R+R2, β= 0. So F'eff ≈ [ (m ω2R) sin(θ'+ β) - (mω2R3) sin(α+θ')] + [(m ω2R) cos(θ'+β) – (mω2R3) cos(α+θ') – (mM2G/R22) + Fsurf] F'eff ≈ [ (m ω2R) sin(π+ 0) - (mω2R3) sin(0+π)] + [(m ω2R) cos(π+0) – (mω2R3) cos(0+π) – (mM2G/R22) + Fsurf] F'eff ≈ + [- (m ω2R) + (mω2R3) – (mM2G/R22) + Fsurf] F'eff ≈ + [- (m ω2R) + (mω2[R+R2]) – (mM2G/R22) + Fsurf] F'eff(B) ≈ + [mω2R2 – (mM2G/R22) + Fsurf] Now we have Fsurf(B) = (mM2G/R22) - mω2R2 and again our particle is "lighter" by the same amount mω2R2 Now let's try a point at the top, so θ' = π/2. Then we have sinα = R2/R sinβ ≈ R2/r12 = very small R3 = [ R2 + R22]1/2 cosα = R/R3 so: F'eff ≈ [ (m ω2R) sin(θ'+ β) - (mω2R3) sin(α+θ')] + [(m ω2R) cos(θ'+β) – (mω2R3) cos(α+θ') – (mM2G/R22) + Fsurf] F'eff ≈ [ (m ω2R) sin(π/2) - (mω2R3) sin(α+π/2)] + [(m ω2R) cos(π/2) – (mω2R3) cos(α+π/2) – (mM2G/R22) + Fsurf] F'eff ≈ [ (m ω2R) 1 - (mω2R3) cos(α)] + [(m ω2R) 0 + (mω2R3)sin(α) – (mM2G/R22) + Fsurf] F'eff ≈ [ (m ω2R) - (mω2R3) cos(α)] + [(mω2R3)sin(α) – (mM2G/R22) + Fsurf] F'eff ≈ [ (m ω2R) - (mω2R3) (R/R3)] + [(mω2R3)(R2/R) – (mM2G/R22) + Fsurf] F'eff ≈ + [(mω2R3)(R2/R) – (mM2G/R22) + Fsurf] Now we find that Fsurf = (mM2G/R22) – (mω2R3)(R2/R) and again we are lighter. The amount lighter is mω2R2(R3/R) so we have an extra factor. This is a fact I have never computed before. the tidal force is outward at the top and it is larger than at A and B ! This suggests to me that the bulge will be around the belt, and we have dips at the A and B. Maybe I have made an error. OK, that is enough for now.