2. Sect. 5.5: A Simple Model of the Ocean Tides Tides are caused by the gravitational attraction of the ocean to both the Moon & the Sun. This is a Complicated Problem! –The physics of ocean tides was first discussed by Newton! –Instead of the full, complicated problem, we’ll look at a simple model! –This model gets the Qualitative Physics correct. A Major Complication!! The surface of the Earth is not an inertial system! (& Newton’s 2 nd Law is valid in inertial systems only!) Motion in non-inertial systems in general is discussed in Ch. 10! Related Complications: –Earth & Moon rotate about their center of mass + also orbit the Sun. –The Earth is rotating + the Moon is orbiting the Earth. –Assume: The water nearest the Moon is is ~ “pulled away” from the Earth & the water farthest away is ~ “pushed towards” the Earth. Before discussing even the simple model in Sect. 5.5, we’ll first discuss the general Qualitative Physics of the Tides.

3. Qualitative Physics The Moon & the Sun pull Earth’s oceans toward them. –Because the gravitational force falls off rapidly with distance (  r -2 ), the pull of the Moon or the Sun is stronger on the side of Earth that is closer to, or faces, the Moon or the Sun. This is illustrated in the figures with billiard balls attracted to a planet. Effect of the Moon on the Ocean Tides

4. Effect of the Moon’s Gravitational Attraction on the Earth’s Oceans Viewed from a perspective outside the Moon-Earth system: Viewed from a perspective at the center of the Earth:

5. This Effect is Superimposed on the Earth’s Rotation

6. A qualitative depiction of the net effect of the Moon’s Gravity + the Earth’s Rotation on the Tides. Is it obvious that the “bulges” are HIGHLY exaggerated?

7. Spring Tides & Neap Tides The Sun also contributes to Earth’s tides. –When the Sun & Moon line up to produce higher tides, this is called Spring Tides. –Neap Tides occur when the gravitational forces due to the Moon & the Sun partially cancel each other. Question: During what phases of the Moon do Spring Tides & Neap Tides occur?

8. Spring Tide Neap Tide

9. Tidal Friction This “sloshing” back & forth of Earth’s oceans twice a day, due to the tides, actually slows down the rotation speed of the Earth! Due to this effect, the day is lengthening by about 0.002 seconds per century! Discussion Why do you think the Moon keeps the same face toward Earth all the time? –Is it just coincidence? –Actually, we can see 59% of the Moon’s surface from the Earth. –This can be explained using Kepler’s laws!!!

10. Comment It is known that tidal forces also affect galaxies

11. First: Consider the effect of the Moon only. Later: Add in the effect of the Sun. Make some VERY simple (!) model assumptions: –The Earth’s surface is completely covered with water(!) –The Earth rotates, but neglect this at first; add it in later. Combine Newton’s 2 nd Law with Newton’s Universal Law of Gravitation to model tides. –To apply Newton’s 2 nd Law, we need an inertial reference frame! –Set up an inertial reference frame in the “background” space, as in the figure on the next page. –Warning! The notation gets a bit “thick” from here on! Sect. 5.5: A Simple (Crude) Model of the Tides

12. Inertial reference frame: x, y, z M m = Moon mass M E = Earth mass D = Distance from the Moon’s center to the Earth’s center. Calculate: The gravitational effect of the Earth + the Moon on a small (test) mass m on the Earth’s surface. R = Position vector of m from the Moon. r = Position vector of m from the Earth’s center. The radius of the spherical Earth = r. r m = Position vector of m from the inertial system origin. r E = Position vector from the inertial system origin to Earth’s center

13. Newton’s 2 nd Law (in the inertial frame) for mass m: m(d 2 r m /dt 2 ) = (F total ) m = (F G ) m-E + (F G ) m-M (1) (F G ) m-E = Gravitational force between the mass m & Earth. (F G ) m-M = Gravitational force between the mass m & Moon. (F total ) m = - G(mM E r -2 )e r - G(mM m R -2 )e R (2) Newton’s 2 nd Law (in the inertial frame) for the Earth M E (attraction to the Moon): M E (d 2 r E /dt 2 ) = (F total ) E (3) (F total ) E = Gravitational force between the Moon & the Earth. (F total ) E = - G(M E M m D -2 )e D (4) All F’s & r’s are vectors. All e’s are unit vectors! (arrows, hats left off!)

14. Goal: Find the vector acceleration (d 2 r/dt 2 ) as measured in the non-inertial system with origin at the Earth’s center. From the figure: r = r m - r E  (d 2 r/dt 2 ) = (d 2 r m /dt 2 ) – (d 2 r E /dt 2 ) (5) Combining Eqs. (1) - (5) & doing some algebra gives: (d 2 r/dt 2 ) = -(GM E r -2 )e r -(GM m )[(e R R -2 ) - (e D D -2 )] (6) 1 st term  acceleration of m due to Earth’s gravitational force 2 nd 2 terms  acceleration from the Tidal Force (responsible for tides) = difference between the gravitational acceleration due to the Moon, felt at Earth’s center & that due to the Moon, felt on Earth’s surface. These are what we are interested in!

15. The effect of the tidal force at various points on the Earth: The figure shows a polar view, with polar axis = z-axis & looks at the force at different points on the Earth. The tidal force on a mass m on the Earth’s surface (using only the 2 nd 2 terms from Newton’s 2 nd Law, Eq. (6)): F T = - GmM m [(e R R -2 ) - (e D D -2 )] (7)  --------------------- D ---------------------- 

16. The Tidal Force on a mass m on the Earth’s surface: F T = -GmM m [(e R R -2 ) - (e D D -2 )] (7) Look at various points on the Earth’s surface: At point a : The farthest point on Earth from the Moon: At a, e R, e D are (approximately) in the same direction, along the x-axis. R > D  The 2 nd term dominates & F T is along the + x-axis, as in the diagram.  ------------ D -------------------   ------------------ R -------------------- 

17. The Tidal Force on a mass m on the Earth’s surface: F T = -GmM m [(e R R -2 ) - (e D D -2 )] (7) At point b : The nearest point on Earth from the Moon: At b, e R, e D are in (approximately) in the same direction, along the x-axis. R < D  The 1 st term dominates & F T is along the + x-axis, as in the diagram. Also, r << D  F T has approximately the same magnitude as at a, but it is in the opposite direction.  ------------- D -------------------   ------------ R -------------- 

18. The magnitude of the tidal force on the x-axis: F Tx  - GmM m [(R -2 ) - (D -2 )] On the x-axis at a: R = D +r  F Tx  -GmM m [(D+r) -2 - D -2 )] = -G(mM m D -2 )[(1+{rD -1 }) -2 - 1] Note that:|rD -1 |  0.02 <<1  Expand (1+{rD -1 }) -2 in a Taylor’s series for small |rD -1 |  F Tx  -G(mM m D -2 ) [1- 2(r/D)+ 3(r/D) 2 -....-1] or F Tx  + 2GmM m rD -3 (8) On the x-axis at b: By the reasoning we just had, the force there is (approx.) the same magnitude as in (8), but opposite in direction.  ------------ R --------------   ----------- D ----------- 

19. The Tidal Force on a mass m on the Earth’s surface: F T = -GmM m [(e R R -2 ) - (e D D -2 )] (7) At point c: e R is approximately parallel to e D  x-components of the 2 terms in F T approximately cancel because R  D & the x-components of e R & e D are approximately the same. The y component of (7) at c, using e Ry  (rD -1 ) j is  F Ty  -G(mM m )rD -3 (Along the + y-axis, towards Earth’s center) At point d: F Ty has approximately the same magnitude as at d, but it is along the negative y-axis, away from Earth’s center.  ------------- D -------------------   -------------- R ------------------- 

20. Summary of the (approximate) directions of the tidal forces at points a, b, c, & d

21. The Tidal Force on a mass m on the Earth’s surface: F T = -GmM m [(e R R -2 ) - (e D D -2 )] (7) At an arbitrary point e: The x & y components of (7) are obtained by the replacements r  x in F Tx (for points a & b) & r  y in F Ty (for points c & d):  F Tx  2G(mM m )(xD -3 ) F Ty  -G(mM m )(yD -3 )

22. The tidal force at an arbitrary point e for arbitrary angle θ, using x = r cosθ, y = r sinθ  F Tx  2G(mM m )(xD -3 ) = 2G(mM m )(r cosθ)D -3 F Ty  -G(mM m )(yD -3 ) = -G(mM m )(r sinθ)D -3

23. Schematic Representation of Tidal Forces Is it obvious that the “bulges” are HIGHLY exaggerated?

24. This simple model of tidal forces predicts that the water along the y-axis is shallower than along the x-axis. Of course, the figure is greatly exaggerated! The Earth rotates about its axis once per day  2 high tides per day.

25. This simple model of tidal forces has neglected the effect of the Sun’s gravitational attraction! A simple calculation shows: –The Sun’s gravitational attraction is approximately 175 times stronger than the Moon’s at Earth’s surface. –However, a calculation similar to our calculation with the Moon shows that the Tidal Force due to the Sun is approximately 0.46 that due to the Moon. (Prob. 5.18) –The Sun gives a stronger total force, but a weaker tidal force. Also, the gradient of the Sun’s tidal force over the Earth’s surface is smaller than for the moon because of the larger distance to the Sun.

26. Example 5.5 Calculate the maximum height change in the tides caused by the Moon. Use the same simple model along with a solution proposed by Newton himself! Imagine 2 wells; 1 in the direction of the high tide (x-axis) & 1 in the direction of the low tide (y-axis).

27. h  height change wanted. ΔPE for mass m = mgh. (Move m from c to a, in the figure). Work: W = ΔPE mgh = W = I 1 + I 2 I 1 = ∫ (F Ty )dy (r < y < 0) F Ty = -G(mM m )(yD -3 ) I 2 = ∫ (F Tx )dx (0 < x < r) F Tx = 2G(mM m )(xD -3 )  mgh = (3/2)G(mM m )(r 2 D -3 ) h = (3/2)(G/g)M m (r 2 D -3 ) Putting in numbers: h  0.54m r = Earth radius

28. Real Tides Real tides are affected by both the Moon & the Sun! The highest tides (“Spring Tides”) occur when the Earth, Moon & Sun are lined up (at new & full Moons). The lowest tides (“Neap Tides”) occur in the 1 st & 3 rd quarters of the Moon (when the Sun & Moon are at right angles to each other, partially canceling out their effects). The maximum tide (every 2 weeks) should have a height (including the effect of the Sun = 0.46 the effect of the Moon) of 1.46 h  0.83m

29. Actual tides are higher! Because of neglected effects, our simple model is much too simple! For example: 1. Earth is not covered with water! 2. Land masses play role. 3. Local effects can increase high tide levels to several meters! 4. Tidal friction between the water & the Earth causes energy loss by Earth. 5. Earth land masses aren’t rigid & are distorted by tidal forces. 6. The Earth rotates. 7. The Moon orbits the Earth. All of these together mean that there are not exactly 2 tides per day. (Time between them: 12 hr. 26 min.) ALSO: 8. The plane of the Moon orbit around the Earth is not  to the rotation axis  One high tide per day is higher than the other! 9. Tidal friction results in the Earth “dragging” the ocean with it as it rotates.  This results in high tides not being along the Earth-Moon axis, but several degrees apart, as in the figure on the next page.

30. Cartoon of “Actual” Tides Is it obvious that the “bulges” are HIGHLY exaggerated?