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Lecture 21: Tidal Forces

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Lecture 21: Tidal Forces Lecture 21: Tidal Forces • For most of the problems we’ll treat this semester, we’ll assume that the Earth is an inertial frame of reference • But there are effects that arise precisely because that isn’t exactly true • Ocean tides are one example, and since they’re caused by gravity, we’ll look at them now • We take a simple model, where the universe consists only of the Earth and the moon, and the Earth is covered entirely by water m • R r eD Moon Earth D • The net gravitational force on the element of water m is the sum of the forces due to the Earth and the Moon: GmM GmM F = − E e − m e = m"r" m r2 r R2 R m • In principal, we could solve this equation of motion and determine exactly how this element of water moves • But what we really want to know is how someone on Earth will observe it to move – That is, we want the motion relative to the Earth • The first step is writing the equation of motion for the Earth: GM M Note: both rE and "" = − m E M E rE 2 eD rm are measured in D some inertial frame • Therefore the relative motion is given by: = − "r" "r"m "r"E GM GM GM = − E e − m e + m e r2 r R2 R D2 D • The first term is just the usual acceleration towards the center of the Earth (the one that’s equal to g near the Earth’s surface) • The other terms only exist due to the presence of the Moon • We identify these terms as the tidal acceleration: e e ’ a = −GM ∆ R − D ÷ T m «R2 D2 ◊ Magnitude of the Tidal Force • We can now find the tidal force at any point P on the Earth’s surface: y eD ξ • P Earth’s rotation e R θ To Moon x = − eD 1i = − ξ − ξ eR cos i sin j D + x D + x y cosξ = = ; sinξ = y 2 + (D + x)2 R R • Therefore, the x component of the tidal acceleration is: D + x 1 ’ a = −GM ∆ − ÷ T ,x m « R3 D2 ◊ −GM ≈ + ’ −GM ≈ + ’ = m D x − = m 1 x / D − 2 ∆ 2 1÷ 2 ∆ 3 1÷ D ≈ R ’ D ≈ R ’ ∆∆ ÷ R ÷ ∆ ∆ ÷ ÷ ««D ◊ ◊ « «D ◊ ◊ −GM ≈ + ’ = m 1 x / D − 2 ∆ 3/ 2 1÷ D 2 2 ∆»≈ y ’ ≈ D + x ’ ÿ ÷ ∆…∆ ÷ + ∆ ÷ Ÿ ÷ « «D ◊ « D ◊ ⁄ ◊ −GM ≈ + ’ = m 1 x / D − 2 ∆ 3/ 2 1÷ D 2 2 ∆» 2x ≈ x ’ ≈ y ’ ÿ ÷ ∆…1+ + ∆ ÷ + ∆ ÷ Ÿ ÷ « D «D ◊ «D ◊ ⁄ ◊ • So far we’ve kept the exact solution, but now we make use of the fact that the radius of the Earth is much less than D to get an approximate answer: −GM + ’ ≈ m 1 x / D − aT ,x 2 ∆ 3/ 2 1÷ D » 2x ÿ ∆ 1+ ÷ « D ⁄ ◊ −GM » ≈ ’ ÿ ≈ m ( + ) − 3x − 2 …1 x / D ∆1 ÷ 1Ÿ D « D ◊ ⁄ −GM » 3x x ÿ 2GM x = m 1− + −1 = m D2 … D D ⁄Ÿ D3 • Plugging through the same math for the y component yields: GM y a = − m T ,y D3 Notes on Tidal Acceleration • Note that the x component always points in the same direction as the displacement • This is why there are two “high tides” per day from the moon (one corresponds to x = +RE, and the other to x = -RE) • The Sun also produces tides on the Earth – We might expect this to be the dominant contribution, since the Sun’s mass is so much greater than the Moon’s – But, tides are due to the change in the gravitational force across the diameter of the earth – It turns out that’s the sun’s contribution to tides is about half as big as the moon’s Central-Force Motion Using Lagrangians • We can also find the equation of motion using the Lagrangian approach: 1 L = µ (r"2 + r2θ"2 ) −U (r) 2 ∂L d ∂L ∂U − = µrθ"2 − − µ"r" = 0 ∂r dt ∂r" ∂r ∂U µ ("r"− rθ"2 ) = − = F (r) ∂r • Note that this gives a relationship between the motion and the force • We can better interpret this result by making a change of variables: 1 u = r • Computing the derivatives of u, we find: du 1 dr 1 r" µ = − = − = − r" dθ r2 dθ r2 θ" l d 2u d µ ’ dt d ≈ µ ’ µ = ∆− r"÷= ∆− r"÷= − "r" dθ 2 dθ « l ◊ dθ dt « l ◊ lθ" • Using the expression for l one more time gives: d 2u µ 2 = − r2"r" dθ 2 l 2 • With this, we can now find rθ"2 and "r": 2 2 2 3 "2 ≈ l ’ l l u rθ = r ∆ ÷ = = «µr2 ◊ µ 2r3 µ 2 l 2 d 2u l 2u2 d 2u "r" = − = − µ 2r2 dθ 2 µ 2 dθ 2 • We’re now all set to substitute this back into the equation of motion: µ ("r"− r2θ") = F (r) l 2u2 d 2u l 2u3 ’ ≈ 1 ’ µ ∆− − ÷= F ∆ ÷ « µ 2 dθ 2 µ 2 ◊ «u ◊ d 2u µ 1 ≈ 1 ’ + u = − F ∆ ÷ dθ 2 l 2 u2 «u ◊ Putting this in terms of r again gives: d 2 ≈ 1 ’ 1 µr2 ∆ ÷+ = − F (r) dθ 2 «r ◊ r l 2 • If we know r(θ), this expression allows us to find the force that must be causing the motion.
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