TIDAL FORCES AND THE EXPANDING ORBIT OF MOON

Transcription

1 TIDAL FORCES AND THE EXPANDING ORBIT OF MOON CHRISTOPHER YANG DR. ROBIN REHAGEN FALL 206 LAS POSITAS COLLEGE PHYSICS 8A HONORS PROJECT

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3 Table of Contents Project Abstract 4 Chapter : Integrating Gravitation on dm Masses. Differential Gravitational Force on a Point Mass.2 Calculating the Total Gravitational Force on The Earth 5 5 Chapter 2: Fluid Tidal Bulge and Gravitational Potential 2. Gravitational Potential Introduction 2.2 Combined Effect of the Earth and Moon 2.3 Modeling Tidal Bulges and Subtracting a Fictitious Potential 2.4 Computer Equipotential Line and Interpretations Chapter 3: Coordinate Setup for Chapters 4 and 5 3. Coordinate setup of Earth-Moon System 3.2 Validity of Rotational Dynamic Equations 20 2 Chapter 4: Dipole Approximation and Torque on Earth 4. Newton s Third Law of Torques and Torque on Earth 4.2 Center of Mass of the Tidal Bulges 4.3 The Mass of Each Bulge 4.4 Finding τ $%&'() +, - from τ.%/0( +, ) with the Dipole Approximation 4.5 Angular Acceleration and Torque Relation to Changing Angular Velocity Chapter 5: Angular Momentum and the Orbit of the Moon 5. Earth and Moon Spherical Component Spin Angular Momenta Projections 5.2 Momenta Without the Need to be Projected 5.3 Conservation of Angular Momentum Appendix A Quantities Used in Calculations 43 3

4 Project Abstract This research project calculates the decrease in angular velocity of Earth to determine the rate of increase of the Earth-Moon distance; the research finds that the decrease in angular velocity results from the tidal friction caused by uneven gravitational forces from the Moon on Earth. The project uses gravitational equipotential lines to model the tidal bulges of Earth and calculates the torque on Earth by approximating the tidal bulges as two point-masses; it concludes with the use of the conservation of angular momentum to calculate the effect of the torque on the distance between the Earth and Moon. The results of the report show a tidal bulgeight of ~0.5 m on two locations along the equatorial line of the Earth, a decreasing angular velocity of the Earth by a current rate of ~5 0 (: rad/s, and an increase in Earth- Moon distance of approximately (A cm/year. The rate of increase in Earth-Moon distance is about 0 A times smaller than current estimates (spaceanswers.com); an assumption of the Earth as completely covered by water may be an explanation. Despitaving such inaccuracy, the project still shows how the force between the Earth and Moon decreases over time, predicting that the Moon could more easily leave its course in the future. 4

5 Chapter : Integrating Gravitation On dm Masses. Differential Gravitational Force on a Point Mass For an object with mass dm on planet Earth, the differential force it experiences from the Moon is expressed in the following equation, df F(G = GM G R LM(G dm where G is the gravitational constant, M G is the mass of the moon, and R LM(G is the distance from the dm mass object to the center of mass of the moon (CM MOOP ). The purpose of this chapter is to find the total gravitational force the moon exerts on Earth by summing all gravitational forces exerted on objects of mass dm..2 Calculating the Total Gravitational Force on The Earth Considering the Earth as a perfect spherical mass, it is possible to divide the Earth into vertical discs of simpler parameters. The coordinate system is designed in Figure.. The discs of thickness dx, with center a distance of R LRST away from CM MOOP, can again be divided into rings of height dx, thickness dr, and variable inner radius r, as described in Figure.2. Figure. The coordinate system, a differential disk, and the variables on a diagram. All points on the same ring (Figure.2) experience the same x-component of force from the Moon and magnitude of force from the Moon because, three-dimensionally, the R LM(G and θ for each point on the ring is constant. Thus, it is reasonable to consider the whole ring as one object and use Newton s Law of Gravitation to find the x-component of the force it experiences from the moon. Any ring with radius r, height dx, and thickness dr will have the differential volume of dv = π r + dr πr dx = π r + 2rdr + dr r dx 5

6 where dr is negligibly small. Thus, wave dv = π 2rdr dx = 2πr dr dx Figure.2 An exaggerated differential ring with vertical thickness dr, horizontal height dx. Assuming Earth has constant density ρ, a ring with inner radius r would have differential mass dm = ρ dv = ρ 2πr dr dx Now, we can choose the ring as our differential mass dm object, and insert the above equation into the equation for differential masses (df F(G ) _RPF(` = GM G R LM(G From the diagram in Figure.2, we can see that ρ 2πr dr dx cos θ Substituting this expression, we get cos θ = R LRST R LM(G (df F(G ) _RPF(` = GM G R LM(G ρ 2πr dr dx R LRST R LM(G (df F(G ) _RPF(` = It is also necessary to solve R LM(G for r: GM G R LM(G A R LRST ρ 2πr dr dx R LM(G = R LRST + r R LM(G = R LRST + r 6

7 Replacing R LM(G, we get (df F(G ) _RPF(` = GM G R LRST + r A R LRST ρ 2πr dr dx To find the total x-component of force the disc experiences, the use of an integral is necessary. For a disc with radius R (Figure ), it is true that: (F F(G ) LRST(` = f e GM G R LRST + r A R LRST ρ 2πr dr dx Note that R is a variable: for different discs, R changes. dx is a constant for this integral (but is not one for the next integral, which will be introduced later). To find the total x-component of force on the whole Earth, another integral must be applied. Assuming Earth has a constant radius R g, the variable x ranges from R g to R g. (F F(G ) LRST(` = ( f e GM G R LRST + r A R LRST ρ 2πr dr dx First, we integrate the inner integral with respect to dr, using the inverse Chain Rule. = 2πρGM G R LRST r ( = 2πρGM G i ( = 2πρGM G i ( R LRST R LRST f e f e R LRST + r A 2r R LRST + r A R LRST + r (j k i = 2πρGM G R LRST R LRST + r (j k ( = 2πρGM G R LRST R LRST + R (j k R LRST (j k dx ( = 2πρGM G R LRST R LRST + R (j k + R LRST (j k dx ( Now, in the second integral, note that R LRST becomes a variable. dr dr e f e f dx dx dx dx 7

8 Figure.3 The relationship between R E, R, and x. To integrate with respect to x, we replace R with an expression containing only variable x, as shown in Figure.3. R g = R + x R = R g x Replacing R, we get (F F(G ) lol(` = 2πρGM G R LRST R LRST + R g x (j k + R (j k LRST dx ( R LRST can be replaced if we know the distance from the center of mass of the Earth to the center of mass of the Moon,. Replacing R LRST, we get R LRST = x (F F(G ) lol(` = 2πρGM G ( x x + R g x (j k + x (j k dx = 2πρGM G x x + R g x (j k + x x (i dx ( 8

9 = 2πρGM G x x + R g x (j k + dx ( = 2πρGM G ( x x + R g x + dx x = 2πρGM G dx + dx ( D (e g(g 2x + R h g = 2πρGM G x ( 2x + R g dx ( 2x + R g dx + 2R g = 2πρGM G x ( 2x + R g dx 2 ( 2 2 x + R g dx + 2R g = 2πρGM G x ( 2x + R g dx 2 (D g(g 2 x + R g ) i + 2Rg 2 ( x = 2πρGM G dx + D g(g 2 x + R g ( D g(g 2x + R g ( + 2R g

10 = 2πρGM G x ( 2x + R g dx + 2 R g + R g + 2 R g + R g + 2R g x = 2πρGM G dx + R g + R g ( D g(g 2x + R g + 2R g x = 2πρGM G dx + R g + R g + 2R g ( D g(g 2x + R g x = 2πρGM G dx + R g + R g + 2R g ( D g(g 2x + R g x = 2πρGM G dx 2R g + 2R g ( D g(g 2x + R g = 2πρGM G x ( 2x + R g dx Using integrating by parts, the following can be determined u = x du = dx dv = 2 x + R g dx v = x + R g dx v = 2 D g(g 2 x + R g 2 v = 2 x + R g 0

11 r s u dv = uv r s s v du r ( x 2 x + R g = x 2 x + R g ( D 2 x + R g dx g(g ( = x 2 x + R g ( + 2D 2 D g(g 2 x + R g dx g(g ( = R g R g + R g + R g + 2D 2 D g(g 2 x + R g dx g(g ( = R D g R g + R g + R g g(g e h + 2D 2 D g(g 2 x + R g dx g(g = 2R g + ( 2D 2 D g(g 2 x + R g dx g(g ( = 2R g 2D 2 D g(g 2 x + R g dx g(g ( = 2R g 2 2 x + R g 3 2 A (

12 = 2R g 2 x + R g 3 A = 2R g = 2R g = 2R g 3 2 x + R g 3 2 R g + R g = 2R g = 2R g A ( A ( + 2 R g + R g R g A + R g A A 3 R g + 3R g R g A A + 3 R g + 3R g + R g A A 3 R g + 3R g R g A A 3 R g 3R g R g A A = 2R g = 2R g = 2R g + 3 R g R g A 3 R g R g A 3 R g + R g A + 3 R g + R g A 3 6 R g + 2R g A = 2R g + 2 R g + 2R A g 3D = 2R g + 2R g + 2R A g g(g 3 = 2R g A 3 Plugging this into the original equation, we get (F F(G ) lol(` = 2πρGM G because ρ v πr A g A is the mass of the Earth M g, 2R g A 3 (F F(G ) lol(` = GM GM g 2

13 Because the total x-component of the gravitational force the Moon exerts on Earth exactly matched Newton s Law of Gravitation, we reasoned that the total y- and z-components of the gravitational force from Moon were zero. To visualize the cancellations of force vector components, we removed the x-components of force vectors on various differential masses, to create the grey, dashed components in Figure.4. We observed that pairs of opposite-facing force vectors canceled their y- and zcomponents. Since this was true for all point masses on a disc, we knew the net forces in the yand z- directions were zero. Figure.4 The combination of y- and z-components are shown in grey; as vectors, they cancel with another, opposite-direction but same magnitude, grey vector. The red vectors represent the combination of all x- y- and z-components. 3

14 Chapter 2: Gravitational Potential Within the Earth-Moon System 2. Gravitational Potential Introduction To quantify the dimensions of liquid tidal bulges on Earth, we used equipotential lines to predict the positions of outer molecules in the liquid formation. We used the gravitational potential (V) from a non-inertial frame of reference as a model for the Earth-Moon system. The gravitational potential is defined as the gravitational potential energy per unit mass ( kg) of a mass m that shares potential energy with another mass M. We use the conventional system by defining all potentials as zero when m is infinitely far from M. With mass m and the gravitational potential energy U it shares, gravitational potential is V = U m We conclude that the gravitational potential energy shared by the particle m is the gravitational work M does on m to move m from an infinitely far distance from the CM of M to the current position of the CM of m. The current position r must be measured from the CM of M, where r = 0 represents a position at that CM. The work, W, is positive. If the particle m does not enter the interior of M, the following is valid V = W m = _ m F Gdr = } m _ GMm r dr = } _ GM r dr } lim s } _ GM dr = lim r GM _ = lim s s } r s s } GM r + GM b = GM r Figure 2. Coordinate system used in this chapter. 4

15 By defining an x, y, and z coordinate system, as shown in Figure 2., we wrote V = GM x + y + z We found the vector form of gravitational acceleration a of any massʹ acceleration by reasoning that, since V x = GMx x + y + z A ( V y = GMy x + y + z ( A and V z = GMz x + y + z ( A V x + V y + V z = GM x x + y + z A + y x + y + z A z x + y + z A = GM x + y + z x + y + z A = GM x + y + z = GM r = a the components of a might have been a` = V x or V x a = V y or V y a = V z or V z To determine which set of equations were valid, we observed that: As x increases from 0 As y increases from 0 As z increases from 0 V is positive Œ is positive Œ is positive x a x is negative a is negative a is negative As x increases to 0 As y increases to 0 As z increases to 0 V is negative Œ is negative Œ is negative x a x is positive a is positive a is positive Table 2. We chose the correct a x, a y, and a z expressions by observing the sign of the partial derivatives of V and the components of a with respect to variable changes in value. 5

16 Thus, a` = V x a = V y a = V z and thus a = V x ı + V y ȷ + V z k or simply a = V x V V ı + ȷ + y z k = V 2.2 Combined Effect of the Earth and Moon Both the Earth and Moon produce gravitational fields that overlap. To find the superimposition of gravitational potentials mathematically, we found the sum of the gravitational potentials on the Earth and Moon. Using the coordinate system shown in Figure 2., we first found the gravitational potential V /% from the mass M g Earth, centered about the origin, V /% = GM g x + y + z To define the gravitational potential V ++, from the mass M G Moon, we chose a position of the CM of Moon at x =, y = 0, z = 0, such that V ++, = GM G x + y + z The total V value was thus V + = GM g x + y + z GM G x + y + z 2.3 Modeling Tidal Bulges and Adding a Fictitious Potential In space, water bodies reorganize to position their outermost molecules at points of equal gravitational potential. We assumed the same situation for the Earth, and used this concept to find the dimensions of the tidal bulges. 6

17 Throughout the process of calculating the bulge dimensions, we ignored the effect of tidal friction which caused an angular offset on the bulges. We knew that the tidal bulges, and the gravitational potential, would be symmetrical about the x-axis, so we efficiently worked in two dimensions, x and y, using the expression at z = 0 for V + : V + = GM g x + y GM G x + y Since our goal was to interpret gravitational potential in the reference frame of Earth, we knew wad to work in a non-inertial perspective, since the Earth was constantly accelerating toward the center of mass of the Earth-Moon system. The V + expression was completely based on an inertial frame of reference. To enter the non-inertial reference frame, wad to consider a fictitious potential. Figure 2.2 shows the CMs of the Earth and Moon with black s, and the CM of the Earth-Moon system with a green. The figure also replaces the blue-colored, inertial axes with red colored ones, to indicate that the axes move with the center of Earth (and are thus non-inertial). Throughout the orbit of the Moon, the Earth constantly rotates about the center of mass of the Earth-Moon system: The gravitational force of the Moon causes a centripetal acceleration on the center of Earth, which points toward the CM of the Earth-Moon system. The fictitious centrifugal force acts along the negative x direction in Figure 2.2. It is a force vector opposite in direction, but same in magnitude, as the force the Moon exerts gravitationally on the Earth. The force the Moon exerts on the Earth in the x-direction is F` = GM gm G which is the centripetal force that causes the Earth to orbit around the Earth-Moon center of mass it is the force that causes the reference frame of the Earth to be non-inertial. Figure 2.2 Center of masses and the orbit of Earth about the Earth-Moon CM. 7

18 To balance out the centripetal force experienced by the Earth, we introduced a fictitious centrifugal force in the opposite direction. The centrifugal force points in the x direction, and its vector component is F $&'( = GM gm G The acceleration caused by this fictitious force on the Earth is a $&'( = GM G To find the corresponding fictitious potential, we noted that a $&'( was a constant acceleration: it was the acceleration of the reference frame. We thus reasoned that since dv $&' x dx = a $&'( the corresponding V $&' x must be V $&' x = GM G x š We calculated the gravitational potential on the non-inertial frame of Earth, V +, by adding V $&' š V + x, y = GM g x + y GM G x D g(g + y + GM G x 2.4 Computer Equipotential Line and Interpretations To find the amount at which water bulged on the equatorial sides of Earth, we used Mathematica as a tool. We used numerical values listed in A, and used the following procedures to find the tidal bulge maximum height: š. We found the gravitational potential V + y = R g, the radius of the Earth. 0, R g at the North and South poles, when 2. We defined d, the amount at which water bulged out at the equatorial position closest to the Moon, and d œ, the amount at which water bulged out at the equatorial position farthest form the Moon. Thus, based on the coordinate system in Figure 2., at the equatorial position closest to the Moon, the gravitational potential was š V + R g + d, 0, and at the equatorial position farthest from the Moon, the š gravitational potential was V + R g d, 0. 8

19 š š 3. We set V + R g + d, 0 = V + 0, R g GM g R g + d GM G R g + d D + GM G š R g(g D g + d = V + 0, R g g(g š š to solve for d and we set V + R g d, 0 = V + 0, R g GM g R g d GM G R g d D + GM G š R g(g D g d = V + 0, R g g(g Our results were the following: d = m d œ = m With an average bulging height, d /0. = 0.53 m As we interpreted our results, we noted that d /0. was derived from an assumption of the Earth being completely covered by water: land forms contribute to rising water levels. Our results of d and d œ were considerably close, with percent difference of 2.2 %, suggesting that our model was quite reasonable.

20 Chapter 3: Coordinate Setup for Chapters 4 and 5 3. Coordinate Setup of Earth-Moon System For the next two chapters, we use a unified method of interpreting the Earth-Moon system. We ignore the effect of the Sun on the Earth-Moon system. Per data from ccar.colorado.edu, the following is the natural Earth-Moon system setup: Figure 3. The natural inclinations of the Earth-Moon system from a side view. In Figure 3., a blue line connects the center of masses of the Earth and Moon; orange dashed lines represent axes normal to the blue line; and solid black lines represent axes of rotation that pass-through a center of mass. We set up a coordinate system with the CM to CM axis oriented horizontally. The resulting diagram is Figure 3.2, with the orange normal lines vertical, and the blue linorizontal. The spin direction of the Earth and Moon are added as well. Figure 3.2 The rotated view of the Earth-Moon system. 20

21 In Figure 3.2, a third, central, orange dashed-line is added. The axis passes through the center of mass of the Earth-Moon system, and stands parallel to the other axes. In the next chapters, we use the orange, dashed lines, labeled as z-axes, in Figure 3.2 as our axes of rotation, because each axis is a symmetry axis that passes through a center of mass, and has no component of Earth-Moon system torque in its direction (ignoring the effect of the Sun). In Chapter 5, we project all angular momenta along this direction, and use some of the conditions detailed in Section 3.2. We considered the tidal bulges, and noted that, due to friction between spherical Earth and the tidal bulges (the spherical Earth spins much faster than the tidal bulges), the tidal bulges slightly tilt by an angle of θ, into the page in Figure 3.2 (detailed in Chapter 4). 3.2 Validity of Rotational Dynamic Equations The equations we use in Chapters 4 and 5 include L = Iω, L = Iω, τ = and τ = Iα. The validity of the equations is summarized as follows,. L is the component of angular momentum along the direction of vector ω. The component equation L = Iω is always true, whether the axis about which ω lies is a symmetry axis or not. This is also valid with direction changes of ω. This, however, does not guarantee there aren t other components of L along different axes other than ω. 2. L = Iω is only valid relative to an origin along a symmetry axis. It is valid even if the axis changes direction. 3. τ = is only valid through an inertial (non-accelerating) axis or a non-inertial axis that passes through the center of mass of an object being considered. 4. τ = Iα is only valid through a direction-constant inertial axis or a direction-constant moving axis that passes through the center of mass of the object being considered. (Resources Used in this Section and Further Information: physics.arizona.edu; University Physics, Young and Freedman; Physics for Scientists & Engineers, Giancoli) 2

22 Chapter 4: Dipole Approximation and Torque on Earth 4. Newton s Third Law of Rotational Motion and Torque on Earth To analyze the effects of the tidal bulges of Earth on the spin rotation of spherical Earth, we approximated the bulges of Earth as two point masses. The foundation of our calculations is the rotational analog to Newton s Third Law: for an applied torque, there exists an equal and opposite reaction torque. By experimental observations, it has been found that the axis passing through the CM (center of mass) of Earth and both of its tidal bulges have an angle of inclination of approximately θ = 3 (lhup.edu) above the axis that passes through both the CM of the Earth and Moon, if viewed from the top of the Earth-Moon system. The offset is shown in Figure 4.. A possible explanation to this phenomenon is a collection of complex resistance forces between the surface of Earth and the water bodies that surround it, causing the slight angular offset from the Axis CM of Figure 4.. Such an offset causes an uneven distribution of gravitational forces on the water bulge component of the Earth, which in turn causes a torque on the water bulge, τ.%/0( +, ). In this experiment, we interpreted the Earth as composed of two major components: its water bulge component and its spherical component. We reasoned that both components exerted a friction torque on each other an action-reaction friction torque (opposite in direction yet equal in magnitude) that allowed the water bulge to stop rotating (at rotational equilibrium) when positioned at the angle θ. Throughout our calculations, we used τ.%/0( +, ) to denote the torque created by the uneven distribution of gravitational forces on the water body, caused by the θ = 3 inclination, τ $%&'(- +, ) as the torque created by the friction from spherical Earth on the water bulges, and τ $%&'() +, - as the torque created by the friction from the water bulges on the spherical Earth. Figure 4. The top view of the Earth-Moon system with exaggerated angle of inclination θ. The water bulges experience rotational equilibrium. 22

23 We reasoned from the observation of the rotational equilibrium of the water bulges that Στ = 0 = τ.%/0( +, ) + τ $%&'(- +, ) Based on the coordinate system defined by the right-hand rule, we reasoned that Στ = 0 = τ.%/0( +, ) + τ $%&'(- +, ) τ $%&'(- +, ) = τ.%/0( +, ) The second of the action-reaction pair for τ $%&'(- +, ) is τ $%&'() +, -, such that their magnitudes satisfy the following τ $%&'() +, - = τ $%&'(- +, ) = τ.%/0( +, ) This relationship allowed us to understand the torque τ $%&'() +, - that was creating the slowing of the rotation of spherical Earth a cause of the elongations of day lengths on Earth. 4.2 Center of Mass of a Tidal Bulge From the last section, we knew that τ $%&'() +, - could be quantified by first finding the component torque τ.%/0( +, ) along the z-direction defined in Figure 3.2. To find τ.%/0( +, ), we simplified the two water bulges into a dipole, with two point masses rotating about the CM of the Earth during Moon s orbit. We used each point mass as a representation of the center of mass of a bulge. Figure 4.2 The coordinate system we used to find the CM of a bulge. The dark blue component contained both water and solid. 23

24 We found the center of mass of one water bulge with the coordinate setup in Figure 4.2, defining the x axis along the Axis Inclination of Figure 4.. We only found the center of mass in 2D because we knew that the bulges were symmetrical between the +z and z axes. We were only looking for the CM of the bulge located on the +x axis. Finding the equations that bounded the bulge in 2D, the first of equation for the determination was x a + y =, x 0 b x a = y b, x 0 x = a y b, x 0 x = a y b = a b b y where a and b were defined quantities in Figure 4.2. a is one-half the end-of-bulge to end-ofbulge distance we found with Mathematica in Section 3.6, while b is one-half the vertical distance of the ellipse found in that same section. The second equation that bounded the bulges in 2D was the circle formula The area between the two curves was, x + y = b, x 0 x = b y, x 0 x = b y A = s (s a b b y b y dy s = b y a b dy (s s = a b b y dy (s 24

25 Figure 4.3 The triangle used for trigonometric substitution. Forming a right triangle in Figure 4.3 and performing trigonometric substitution, sin δ = y b When y = b y = b sin δ dy = b cos δ dδ cos δ = b y b b y = b cos δ b = b sin δ δ = π 2 When y = b, δ =, or simply. Altering the integral, A A = a b b cos δ cos δ dδ ( = b a b ( + cos 2δ dδ 2 25

26 = b 2 a b δ + 2 sin 2δ ( = b 2 a b π sin π π 2 + sin π 2 = bπ 2 a b the area traced out by one bulge in 2D. To find the center of mass position along the y axis, y TM = A s (s y a b b y b y dy = 2A s (s 2y b y a b dy = 2A s a b 2y b y dy (s = 2A a b s 2y b y (s dy = 3A a b b y A s (s = 3Ab a b b y A s (s = 3Ab a b 0 0 = 0 Which was reasonable because the area was evenly distributed above and below the x axis. We found the x-coordinate by x TM = 2A s (s a b b y b y dy = 2A s (s = 2A a b b y b y s (s b y a b dy dy 26

27 = 2A a s b b y (s dy = 2A a b b y ya s 3 (s = 2A = 2A a b ba 3 ba b A + 3 ba a b ba 3 ba + b A 3 ba = A a b ba 3 ba = Ab a b 2 3 ba = 2 3A b a b Plugging in the expression for A, x TM = 4 3bπ a b b a b 4 = 3π a b a b a + b = 4 3π a + b The result made sense because v had a numerical value not too far from 0.5: the x A TM was close to the average between a and b, or was close to the point exactly in between the points x = a and x = b. 4.3 The Mass of Each Bulge We found the mass of each water bulge by first finding their volumes. We formed the formula V = 4 3 πab 4 3 πba = 4 3 πb a b The three-dimensional density ρ of water allowed us to write an expression for the mass of the ocean bulges, m s(lol = 4 3 πρ b a b 27

28 Each bulgad one-half the mass, thus for each bulge, m s = 2 3 πρ b a b 4.4 Finding τ $%&'() +, - from τ.%/0( +, ) with the Dipole Approximation Figure 4.4 (a top view) shows the approximation of the two bulges as two point-masses, rotating about the CM of the Earth. To find the total torque component τ F_r±(G OP ², we split the problem into finding two separate torques: first, the torque component caused by the gravitational force on the mass farthest away from the Moon τ ($/%, and second, the torque component caused by the gravitational force on the mass closest to the Moon τ (,³/%. Figure 4.5(a) on the next page sets up variables and constants to find the torque component caused by gravitational force on the mass nearest from the Moon τ (,³/%. θ is the constant angle of inclination between the axis passing through the bulges and the axis passing through CMs of the Earth and Moon. We reasoned that, l = L cos θ and by the Pythagorean Theorem, h = L sin θ l + h = d,³/% where d,³/% represents the distance from the CM of the Moon to the nearest point mass. Combining expressions, L cos θ + L sin θ = d,³/% Figure 4.4 The dipole approximation for tidal bulges. Each dimensionless point mass takes the position of the center of mass of a water bulge; each have a mass equal to that of a bulge. 28

29 (a) (b) Figure 4.5 (a) The variables used to find F near. (b) The variables used to find F far. (Diagrams not to scale; (a) and (b) do not have the same scale factor.) The gravitational force magnitude from the Moon on the point mass is F,³/% = Gm sm G = d $/% Gm s M G L cos θ + L sin θ Thus, the torque τ (,³/% was τ (,³/% = Gm s M G L cos θ + L sin θ 4 3π a + b sin γ We noted that the angle γ = tan (i h l = tan (i L sin θ L cos θ Using Figure 4.6(a), we reasoned that the angle between the two vectors, when tail-to-tail, was θ + tan (i L sin θ L cos θ Thus, τ (,³/% = Gm s M G L cos θ + L sin θ 4 3π a + b sin θ + L sin θ tan(i L cos θ 2

30 (a) (b) Figure 4.6 (a) F near and r near placed tail-to-tail. (b) F far and r far placed tail-to-tail. Figure 4.5(b) on the previous page sets up variables and constants to find the torque component caused by gravitational force on the mass farthest away from the Moon τ ($/%. θ is the constant angle of inclination between the axis passing through the bulges and the axis passing through the CMs of the Earth and Moon. We reasoned that, and by the Pythagorean Theorem, l = L cos θ h = L sin θ l + D g(g + h = d $/% where d $/% represents the distance from the CM of the Moon to the farthest point mass. Combining expressions, L cos θ + D g(g + L sin θ = d $/% The gravitational force magnitude from the Moon on the point mass is F $/% = Gm sm G = d $/% Gm s M G L cos θ + + L sin θ Thus the torque τ ($/% was τ ($/% = Gm s M G L cos θ + + L sin θ 4 3π a + b sin β 30

31 We noted that the angle φ = tan (i h + l = tan (i L sin θ + L cos θ Using Figure 4.6(a), we reasoned that the angle between the two vectors, when tail-to-tail, was tan (i L sin θ + L cos θ θ = 80 + tan(i L sin θ + L cos θ θ Thus, τ ($/% = Gm s M G L cos θ + + L sin θ 4 3π a + b sin L sin θ tan(i + L cos θ θ Combining the two torques to find the net torque, we found that τ.%/0( +, ) Gm s M G = L cos θ + L sin θ + tan (i L sin θ θ L cos θ Gm s M G L cos θ + D g(g + L sin θ 4 3π a + b sin θ 4 3π a + b sin L sin θ tan(i + L cos θ = 4Gm sm G 3π a + b sin θ + tan (i L sin θ L cos θ L cos θ + L sin θ sin tan (i L sin θ D + g(g + L cos θ θ L cos θ + D g(g + L sin θ Plugging in L = v A a + b, we obtained the equation on the next page. 3

32 = 4Gm sm G 3π a + b 4 a + b sin θ sin θ + tan (i 3π 4 a + b cos θ 3π 4 3π a + b cos θ + 4 3π a + b sin θ + 4 a + b sin θ sin tan (i 3π + 4 a + b cos θ 3π 4 3π a + b cos θ + θ + 4 3π a + b sin θ We concluded that the τ $%&'() +, - had the same sign as τ.%/0( +, ), and so τ $%&'() +, - = 4Gm sm G 3π a + b 4 a + b sin θ sin θ + tan (i 3π 4 a + b cos θ 3π 4 3π a + b cos θ + 4 3π a + b sin θ + 4 a + b sin θ sin tan (i 3π + 4 a + b cos θ 3π 4 3π a + b cos θ + θ + 4 3π a + b sin θ Which is the expression for the component torque that causes solid Earth to slow in angular speed: the component torque that causes days to become longer. Using current values of, the torque on spherical Earth was calculated to be τ $%&'() +, - = ia N m (current value) which is a negative number: what we expected. This is a current value because, as will be described in Chapter 5, increases over time. 4.5 Angular Acceleration and Torque Relation to Changing Angular Velocity The first step was to determine whether the torque was constant. After interpreting all variables in the final equation of Section 4.4, we concluded that the torque was not constant, and thus the resulting angular acceleration would not be constant, due to changing values. However, since the change in torque would be small over a short period ( year), we assumed the torque to be constant and used constant-acceleration kinematics equations. 32

33 Since the axis z never changed direction and passed through the center of mass of the two bulges and the spherical Earth, we knew that the vector relation τ = Iα would be valid. We noted that τ $%&'() +, - was the one and only torque component in the z direction on the spherical Earth component. Thus, Στ = τ $%&'() +, - = 2 5 m Sa α Where m S represents the mass of the solid Earth. (Please note that a in this equation represents a distance, not an acceleration). It followed that, α = 5 τ $%&'() +, - 2m S a = (: rad/s (current value) Since α, τ $%&'() +, - and the moment of inertia were assumed to be constant, we wrote ω t = ω ÊË&,(-,'Ì%%³,,Í + 5 τ $%&'() +, - 2m S a t (for small t) for the angular velocity component of spherical Earth, a time t after the current rotational angular velocity component, ω ÊË&,(-,'Ì%%³,,Í. 33

34 Chapter 5: Angular Momentum and the Orbit of the Moon 5. Earth and Moon Spherical Component Spin Angular Momenta Projections To evaluate the effects that the decreasing angular velocity of spherical Earth created on the orbital radius of the Moon, we used the conservation of angular momentum as a tool. In preparation to accurately use rotational dynamic equations, the z axes we chose for this chapter, introduced in Chapter 3, have fixed directions and pass through center of masses. To find the total angular momentum of the Earth system, we again interpreted the Earth as a combination of two separate components: Its water bulge and its spherical component. The spherical component contained both water and solid. We first determined the spin angular velocities of the spherical component of Earth and Moon about their rotation axes, and found the corresponding angular momentum of spin, L ÊË&,. Then, we projected the angular momenta along the z-direction. We ignored the effect of the Sun in all calculations. Considering the spin angular momenta of the Earth and Moon about their natural spin axes both slightly tilted from what we defined as the z-axis in Chapter 3 we knew that the equation L ÊË&, = Iω (angular momentum about spin axis) would hold, since the spin axes were symmetry axes. Even though the spin axes changed direction throughout the Earth-Moon orbit about the Earth-Moon CM, by the rules introduced in Chapter 3, L ÊË&, = Iω would still hold. For this reason, we allowed ourselves to project the L ÊË&, vector along the z-axis in the same way we would project the ω vector along the direction, because L ÊË&, had the same direction as ω. SPIN OF SPHERICAL EARTH To find the magnitude of spin angular momentum of spherical Earth, we first determined the magnitude of angular velocity of spherical Earth about its spin axis, ω ÊË&,(- = 2π T ÊË&,(- The subscript spin SE represents of spherical Earth, about spin axis. The corresponding magnitude of spin angular momentum about the spin axis is L ÊË&,(- = I ÊË&,(- 2π T ÊË&,(- = 2 5 M gb 2π T ÊË&,(- b was defined in Figure

35 Setting φ i = 28.5, the angle of inclination of the spin of Earth (described in Chapter 3), the spin angular momentum projection on the z-axis was L ÊË&,(-,Í = 2 5 M gb 2π T ÊË&,(- cos φ i The component is positive. We knew that another way to express the angular momentum component was L ÊË&,(-,Í = 2 5 M gb ω ÊË&,(- cos φ i or alternatively L ÊË&,(-,Í = 2 5 M gb ω ÊË&,(- (Í and we knew that wad a function of ω ÊË&,(- (Í with respect to time from Chapter 4, for any time t after this current moment. Plugging in that expression (which approximated constant angular acceleration for small t), L ÊË&,(-,Í = 2 5 M gb ω ÊË&,(-,'Ì%%³,,Í + α t ω ÊË&,(-,'Ì%%³,,Í is the current value of the projection of ω ÊË&,(- along the z-direction. Thus, we wrote ω ÊË&,(-,'Ì%%³,,Í = ω ÊË&,(-,'Ì%%³, cos φ i and plugged it in, L ÊË&,(-,Í = 2 5 M gb ω ÊË&,(-,'Ì%%³, cos φ i + α t SPIN OF SPHERICAL MOON To find the magnitude of spin angular momentum of spherical Moon, we first determined the magnitude of angular velocity of spherical Moon about its spin axis, ω ÊË&,(- = 2π T ÊË&,(- The subscript spin SM represents of spherical Moon, about spin axis. 35

36 The corresponding magnitude of spin angular momentum about the spin axis is L ÊË&,(- = I ÊË&,(- 2π T ÊË&,(- = 2 5 M GR G 2π T ÊË&,(- R G is the radius of the Moon, and M G is the mass of the Moon. Setting φ = 6.688, the angle of inclination of spin of Moon (described in Chapter 3), the spin angular momentum projection on the z-axis was L ÊË&,(-,Í = 2 5 M GR G 2π T ÊË&,(- cos φ The component is positive. T ÊË&,(-, the spin period of the Moon, is equal to T +% & (-, the orbit period of the Moon about the Earth-Moon center of mass because the Moon is tidally locked to the Earth. Using this relation, L ÊË&,(-,Í = 2 5 M GR G 2π T +% & (- cos φ We wanted to find a relation between T +% & (- and the distance between the Moon and Earth. We defined the variable d G as the orbital radius of the Moon, or the distance between the Moon and the center of mass of the Earth-Moon system. The cause of the orbit of Moon was the gravitational force between the Moon and Earth. The corresponding centripetal acceleration was GM g = 4π d G Solving for T +% & (-, T +% & (- T +% & (- = 2π d G GM g We find an expression for d G in terms of : x TM = d G = M G 0 + M g M g + M G d = M g d G = M g M g + M G 36

37 Plugging in the expression, T +% & (- = 2π M g M g + M G GM g = 2π L ÊË&,(-,Í = 2 5 M GR G cos φ D g(g 5.2 Momenta Without the Need to be Projected Some components of the total Earth-Moon angular momentum had angular velocities ω already in the direction of the z-axis. Though some were not symmetrical about a z-axis, the relation L Ø = Iω permitted calculations of component angular momenta in such situations. SPIN OF THE BULGES OF EARTH To find the z-component angular spin momentum of the bulges of Earth, we first considered the magnitude of angular velocity of its spin, ω ÊË&,(Ù = 2π T +% & (- The subscript spin BE represents of Earth bulges, about spin axis. The period of spin of the bulges of Earth (combined) were T +% & (-, because the bulges had to maintain a θ inclination with the Moon. The corresponding z-component of spin angular momentum about the spin axis is L ÊË&,(-,Í = I ÊË&,( I ÊË&,(- 2π = I T ÊË&,( 2 +% & (- 5 M 2π gr g T +% & (- where I ÊË&,( is the total moment of inertia of ellipsoid Earth. Plugging in a corresponding expression for I ÊË&,(, L ÊË&,(-,Í = M g 5 a + b 2 5 M 2π gr g T +% & (- both a and b were defined in Figure 4.2. The component is positive. Using the relation T +% & (- = 2π 37

38 from the previous section, we got L ÊË&,(-,Í = M g 5 a + b 2 5 M gr g ORBIT OF MOON. To find the z-component of the angular momentum of the Moon due to its orbit about the Earth-Moon CM, we first considered the magnitude of angular velocity of its orbit, ω +% & (- = 2π T +% & (- The subscript orbit SM represents of spherical Moon, about orbital axis. We noted that the axis passing through the Earth-Moon center of mass, the orbital axis, about which we calculated the orbital angular momentum of Moon, was not a symmetry axis for the Moon. However, noting that L +% & (-,, the component angular momentum of the orbit of spherical Moon along the direction of ω +% & (-, had the relation L +% & (-, = Iω +% & (- and that we concluded that L +% & (-, = L +% & (-,Í L +% & (-,Í = I +% & (- 2π T +% & (- We approximated the spherical Moon as a point mass revolving about the Earth-Moon CM. The moment of inertia was thus I +% & (- = M G d G Thus, L +% & (-,Í = M G d G 2π T +% & (- The component is positive. Using the relation T +% & (- = 2π 38

39 from the previous section, we got We also used the relation L ÊË&,(-,Í = M G d G D g(g and plugged it in d G = M g M g + M G L ÊË&,(-,Í = M G M g M g + M G ORBIT OF EARTH To find the z-component of the angular momentum of Earth due to its orbit about the Earth- Moon CM, we first considered the magnitude of angular velocity of its orbit, ω +% & ( = 2π T +% & ( = 2π T +% & (- The subscript orbit E represents of Earth, about orbital axis. T +% & ( is equal to T +% & (- due to the maintaining of the position of the center of mass of the Earth-Moon system. We noted that the axis passing through the Earth-Moon center of mass, the orbital axis, about which we calculated the orbital angular momentum of the Earth, was not a symmetry axis for the Earth. However, noting that L +% & (,, the component angular momentum of the orbit of Earth along the direction of ω +% & (, had the relation L +% & (, = Iω +% & ( and that we concluded that L +% & (, = L +% & (,Í L +% & (,Í = I +% & ( 2π T +% & (- We approximated the Earth as a point mass revolving about the Earth-Moon CM. The moment of inertia was thus 3

40 Thus, I +% & (- = M g d g L +% & (,Í = M g d g 2π T +% & (- The component is positive. To find an expression of d g in terms of, we used the center of mass equation, defining x = 0 at the CM of Earth, and +x in the direction of the Moon x TM = d g = M g 0 + M G M g + M G And thus, Using the relation L +% & (,Í = M g d g = M G M g + M G M G M g + M G 2π T +% & (- from the previous section, we got T +% & (- = 2π L +% & (,Í = M g M G M g + M G TOTAL ANGULAR MOMENTUM. The total momentum in the z-direction is L +, = 2 5 M gb ω ÊË&,(-,'Ì%%³, cos φ i + α t M GR G + M g 5 a + b 2 5 M gr g + M G M g M g + M G + M g M G M g + M G cos φ 40

41 Or, alternatively L +, = 2 5 M gb 2π cos φ T i + 2 ÊË&,(- 5 M GR G + M g 5 a + b 2 5 M gr g + M G M g M g + M G + M g M G M g + M G 5.3 Conservation of Angular Momentum cos φ Our goal was to find the distance by which the Moon moved away from Earth one year after the current moment. To perform this task, we knew that Στ = 0, Στ = 0, Στ = 0, and Στ` = 0 on the Earth-Moon system (we ignored the effect of the Sun). Consequently, the sum of momenta in the z-direction of the Earth-Moon system would be constant. We first found the current angular momentum in the z-direction using the equation (without t variable) L +,,'Ì%%³, = 2 5 M gb 2π cos φ T i + 2 ÊË&,(- 5 M GR G,Û + M g 5 a + b 2 5 M gr g + M G M g,û M g + M G + M g M G,Û M g + M G,Û,Û,Û,Û,Û cos φ,û by using the current value of,,û. 4

42 We found that, L +,,'Ì%%³, = Av kg m /s We equated it to the angular momentum equation after year, L +,,'Ì%%³, = 2 5 M gb ω ÊË&,(-,'Ì%%³, cos φ i + α t Ü M GR G + M g 5 a + b 2 5 M gr g + M G M g M g + M G + M g M G M g + M G cos φ Where t Ü represents the seconds in a year. To find the distance e by which the Moon was leaving the Earth in a year, we plugged in e +,Û for and solved for e. Our result was e = cm Compared to the current approximate of 3.8 cm per year (spaceanswers.com), this was a considerably small number. We noted, however, that solid mountain formations, or lands of higher altitude, of heights much larger than our tidal bulge 0.5 m height approximation, would have also significantly contributed to the tidal effects on Moon. 42

43 Appendix A: Quantities Used in Calculations Constant Value Description Source G (ii m A /s kg Gravitational Constant universetoday.com M M kg Mass of Moon google.com M E v kg Mass of Earth nssdc.gsfc.nasa.gov R E (b) m Polar Radius of Earth nssdc.gsfc.nasa.gov θ 3 Angle Offset: Tidal Bulges lhup.edu ρ W 000 kg/m A Density of Water saylor.org D E(M ä m CM-CM Dist. Earth Moon spaceplace.nasa.gov φ (Chapter 3) ccar.colorado.edu φ (Chapter 3) ccar.colorado.edu T y s Seconds in a Year rapidtables.com T spin s Period of the Earth Spin reference.com 43