Orbital Dynamics of the Moon

Transcription

1 Orbital Dynamics of the Moon Matthew Rappeline KapCC Stem Program Kapiolani Community College Faculty Advisor: Dr. Herve Collin Spring 2012

2 Contents I. Introduction 3 A. Newton s Model of Motion and Gravity B. Lagrangian Mechanics II. Purpose of the Project 6 III. Methods 7 A. Simplifying the Model Shared Aspects Velocity and Acceleration for the Moon s Orbit B. Earth s COM as the Center of Rotation Newtonian Analysis of the Moon s Orbit Lagrangian of the Moon s Orbit C. Earth/Moon System COM as the Center of Rotation Reduced Mass using Newtonian Analysis Reduced Mass Lagrangian F. Extracting Experimental Data from APOLLO Experimental Distances Experimental Angles IV. Results Eccentricity of the Moon s Orbit A. Plotting the Earth COM as the Center of Rotation B. Plotting the Earth/Moon COM as the Center of Rotation

3 1. Plotting the Equations of Motion Comparing Theoretical and Experimental Data V. Conclusion 33 A. Equations of Motion B. Accuracy of the Obtained Equations VI. Appendix 37 A. Data Extraction Program B. Octave Program C. Elliptical Conic Section - Reduced Mass D. Deriving the Lagrange Equations of Motion VII. Bibliography 45 A. Citations B. References

4 I. Introduction Gravity, a central force, governs celestial phenomena from orbital motion to black holes. The orbits of the Earth and Moon, stars around a black hole at the center of the Milky Way and perhaps all galaxies around a central black hole are created by mass grip on spacetime, gravity. Understanding orbits and other two-body system dynamics is essential to a greater understanding of gravity and the universe abroad. Binary Star Systems, modeled as a reduced mass, are essential to the search for more tangible evidence of black holes. Investigating these types of astral anomalies is the core of the National Aeronautics and Space Administration (NASA). Sending, both manned and unmanned, spacecraft into the depths of space to observe these phenomena is one of NASA s primary methods of data collection. Several spacecraft and satellites are in low-earth orbit (leo), geosynchronous orbit (geo), the Moon s orbit, as with GRAIL-A and B, and even Mars orbit and beyond learning more about the universe. Highly controlled by gravity these orbits can be calculated using two body system dynamics. Even our search to understand the apparent expansion of the universe, and dark energy involves the relative motion between many systems of two bodies. These will be at the forefront of NASA s attention and mission objectives as it continues to seek the unknown. A. Newton s Model of Motion and Gravity In 1687 Isaac Newton gave the world Philosophiae Naturalis Principia Mathematica[1] in which he mapped out his concepts for absolute space and time, laws of motion and law of gravity. Newton s Three Laws of motion can be stated as thus: 1) F = 0 2) F = d dv (m v) = m = m a dt dt 3) F a/b = F b/a. 3

5 The first law states that if there are no external forces acting on a mass, then there will be no resulting acceleration on that mass. In words a mass already at rest will remain at rest and a mass in motion will remain in the same motion, unaffected or acclerated by an external force. Thus the second law states that if there are external forces acting on a mass there will be a resulting acceleration on that mass. The magnitude and direction of the acceleration will depend on the sum, in magnitude and direction, of the external forces acting on the mass. The third law is commonly stated as for every action there is an equal and opposite reaction. It can be said that forces are interactions between two bodies. Newton s Law of Gravity can be stated as thus: F g = G m 1m 2 r 2 ê. [1] Through this new model of gravity and the laws of motion Newton would give the work of Kepler mathematical form and support. The dynamics laid out by Newton would permeate all aspects of scientific inquiry and is the basis for what we now call classical mechanics. Studies on the implications and extensions of Newton s laws would continue for centuries and his laws would be refined through the process. B. Lagrangian Mechanics Joseph-Louis Lagrange ( ) refined classical mechanics by including this principle of least action, with Total Energy as the quantity to be conserved. His Lagrangian mechanics, detailed in Mechanique Analytique (1788) [2] simplified analysis especially pertaining to systems with non-conservative forces. This adaptation made it much easier to get the general equations of motion for a system. The Lagrangian for a system can be defined as L = T U where T is the kinetic energy and U is the potential energy. The general form of the Euler-Lagrange equation in generalized coordinates is: (See Appendix D for more information.) d L L = 0. dt q q William Rowan Hamilton ( ) expanded on this work by generalizing dynamics. In two papers to the Royal Society in 1834 and 1835, fittingly titled On a General Method in Dynamics, Hamilton lays out his principle of Varying Action. [3] It can be stated as thus: 4

6 Of all possible paths along which a dynamical system may move from one point to another within a specified time interval (consistent with any constraints), the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energies.[4] This principle broadened the scope of classical mechanics and still finds influence in the field of quantum mechanics. We can relate Hamilton s principle of least action to the Lagrangian via δs = δ t 2 t 1 L = 0 For the Lagrangian, essentially this means that the particle will follow the path that wastes the least energy or no energy. Gravity is a conservative force, so indeed no energy is wasted in the moon s orbit (i.e. total energy and angular momentum are conserved) so it follows the path of least action. Newton, Lagrange and Hamilton gave us incredibly powerful, descriptive tools which we can use to analyze the dynamics of almost any system. In this project I will apply these mathematical tools to the Earth-Moon system. 5

7 II. Purpose of the Project The purpose of this project is to obtain the equations of motion for the Earth/Moon system using Newtonian Analysis and Lagrangian Mechanics, first with a geocentric moon and second with a barycentric system. The theoretical data these equations generate will be compared with experimental data from the Apache Point Lunar Laser Ranging Observatory (APOLLO) and NASA to comment on the accuracy of the model. The mathematical and graphical analysis will be posted to a website that will include automatically updating scripts. 6

8 III. Methods A. Simplifying the Model 1. Shared Aspects The models share a few assumptions that simplify the calculations. The models also have some common aspects that should be detailed before proceeding. 1) All calculations will be made in polar coordinates. 2) I will be defining the initial conditions of the system as follows θ = 0, t = 0, r o = km, v o = km/s. [5] These correspond to the Moon being at the apogee in its orbit, with the coordinate system centered at the Moon s centroid in the r and θ directions. 3) I will assume that the Earth and Moon are homogenous. The mass of the Earth m e = kg. The mass of the moon m m = kg. 4) Based on the fact that they are homogenous, the r distances will be measured from the Earth s center of mass, or centroid, to the Moon s centroid. For all values of θ, r > 0.The radius of the Earth R e = 6, 371km This is the volumetric mean radius as obtained from the NASA database. [5] The radius of the moon r m is the volumetric mean radius of the moon defined as 1, 737.1km, again obtained from the NASA database. 11 m3 5) In Newton s Law of Gravity, G is the universal gravitational constant and equals kgs 2 The acceleration due to gravity from the earth g is equal to 9.81 m s 2. 6) A different convention will be used to denote derivatives. First derivatives are marked by: Second derivatives will be marked as: dr dt = ṙ and dθ dt = θ d 2 r dt 2 = r and d2 θ dt 2 = θ 7

9 2. Velocity and Acceleration for the Moon s Orbit To model the dynamics of the moon s orbit, equations for the position, velocity and acceleration must be obtained. ê θ dê r dt dê θ dt 7 moon θ r 7 θ earth Diagram2 ê r Based on figure 2, the moon s position vector can be modeled as r = rê r [3.1]. Where ê r is a unit vector parallel to the direction of motion of r and likewise ê θ is the unit vector parallel to the direction of motion for θ. These two unit vectors must be mutually orthogonal at all points in time (ê r ê θ ). Starting from 3.1, differentiating with respect to time yields velocity: It should be noted here that dê θ dt v = ṙ = ṙê r + r dêr dt.[3.2] = θê r and that dêr dt = θê θ. Mechanically we can see this in the dynamics. As θ changes with time i.e. θ it moves in the positive direction. As this happens the tendency of ê r is to move in the positive θ direction and the tendency of ê θ is to move in the negative r direction. Mathematically this can be expressed by looking at the components of r. From figure 2 ê r can be defined as < cosθ, sinθ >. For the dot product to be zero, thereby satisfying the condition that ê θ ê r, ê θ must be defined as < sinθ, cosθ >. Taking the derivatives with respect to time: 8

10 dê r dt = d dt < cosθ, sinθ >=< sinθ θ, cosθ θ >= θê θ dê θ dt = d dt < sinθ, cosθ >=< cosθ θ, sinθ θ >= θ < cosθ, sinθ >= θê r From equation 3.2 above, we can now simplify to obtain velocity: v = ṙ = ṙê r + r dêr dt v = ṙê r + r θê θ. [3.3] Differentiating again with respect to time yields the acceleration: a = dv = rê dt r + ṙ dêr dt + [(ṙ θ + r θ)ê θ ] + r θ dê θ dt a = rê r + ṙ θê θ + ṙ θê θ + r θê θ + r θ( θê r ) a = rê r r( θ) 2 ê r + 2ṙ θê θ + r θê θ a = [ r r( θ) 2 ]ê r + [r θ + 2ṙ θ]ê θ [3.4] The first term [ r r( θ) 2 ] [3.5] represents the acceleration in the r direction and the second term [r θ+2ṙ θ] [3.6] represents the acceleration in the θ direction. These values will be referred to several times in the following sections. B. Earth s COM as the Center of Rotation 1. Newtonian Analysis of the Moon s Orbit e r e o Moon Fg Earth Diagram 3: Free Body Diagram for the Moon A free-body diagram shows all external forces acting on a particular body. Looking at the free-body diagram for the moon in figure 3, gravity F g from the Earth is the only force acting on the moon in this 9

11 model. This force acts in the negative r-direction and thus from Newton s second law, the moon will have a resulting acceleration in the r direction, a r. Thus the sum of the forces in the r direction can be expressed as Fr = m m a r. As gravity is the only force- it can be written that: Fr = m m a r = F g = Gmemm r 2 Substituting the acceleration in the r direction, from equation 3.5, into the above equation and simplifying: m m [ r r( θ) 2 ] = Gmemm r 2 r r( θ) 2 = Gme r 2 r = Gme r 2 [4.1] As the change in θ is constant any derivative will be zero, thus r( θ) 2 = is the equation for the moon in the r direction. Referring to the free-body diagram in figure 2, there are no forces acting in the θ direction. Mathematically this can be expressed as Fθ = 0. Referring to equation 3.6, the accleration in the θ direction is: m m (r θ + 2ṙ θ) = 0. r θ + 2ṙ θ = 0. [4.2] 4.2 is the equation of motion for the moon in the θ direction. As θ moves in the positive direction r changes. Therefore r must be a function of θ, r(θ). Starting with the prodect rule let u = r 2 and v = θ. From this it follows that du dt this into the form of the product rule yields: = 2rṙ and dv dt = θ. Substituting r 2 θ + θ(2rṙ) r 2 θ + 2rṙ θ [4.3] 10

12 Multiplying equation 4.3 by 1 produces the acceleration in the θ direction, as defined in equation 3.6. r Thus equation 4.2 can be written in the form, 1 d r dt (r2 θ) = 0 [4.4]. Examining equation 4.4 further, d dt (r2 θ) must equal zero since 1 r means that (r 2 θ) [4.5] must be constant with respect to time. cannot because r cannot be zero. This This can be arraranged as r(r θ) = r(rω). Recognizing rω as v θ = r θ, the term can be re-written as rv θ. The fact that this product must be constant with time means that the angular momentum of the moon is conserved in its orbit. This is Kepler s Second Law. Since gravity is a central force and it is the only force acting on the moon, in this model, the moon s angular momentum must be conserved. Physically it follows that the above is constant. We can see that the position vector r multiplied by the velocity v θ or r θ must be constant. As r decreases or the Moon draws closer to the Earth v θ must increase to keep the product constant and vice versa. At t = 0, r = r o, v r = ṙ = 0, then v = v θ = v o given the initial conditions. Equation 4.5 can then be rewritten as r 2 θ = rvθ = r o v o. Solving for θ yields: θ = rovo r 2. [4.6] Substituting equation 4.6 into equation 4.1 and simplifying brings a new equation of motion in the r direction: a r = r r( θ) 2 = Gme r 2 r r[ rovo ] 2 = Gme r 2 r 2 r r2 o v2 o r 3 = Gme r 2. [4.7] Keeping r(θ) in mind, change of variable can be used to simplify equation 4.7. u can be defined as u = 1 r. To replace r in equation 4.7 the second derivative of r = 1 u must be taken. Taking the first derivative of r with respect to time: ṙ = 1 u 2 du dt = 1 u 2 du dθ θ. Equation 4.6 can be rewritten with the definition of u as r o v o u 2. It can now be said that 11

13 ṙ = r o v o du dθ. [4.8] Differentiating equation 4.8 with respect to time: r = d ( r du dt ov o ) = r d dθ ov o ( du) = r d dt dθ ov o θ ( du) = r d dθ dθ ov 0 θ 2 u dθ 2 Again substituting equation 4.6 nets the equation: r = r 2 ov 2 ou 2 d2 u dθ 2. [4.9] Plugging equation 4.9 into 4.7 produces: [( rov 2 ou 2 2 d2 u) r2 ov 2 dθ 2 o ] = Gme r 3 r 2 Remembering that u = 1 r, u2 = 1 r 2, r = 1 u, r2 = 1 u 2, I multiplied the above equation by 1 u 2 the negative signs to obtain: and cancelled Multiplying this equation by 1 r 2 ov 2 o brings: [r 2 ov 2 o d2 u dθ 2 + r2 o v2 o r ] = Gm e. d 2 u dθ 2 d 2 u dθ r = Gme r 2 ov 2 o + u = Gme. [4.10] rov 2 o 2 This linear differential equation finds u as a function of θ. Equation 4.10 can be left in terms of u to obtain the specific solution of the equation. The general solution to equation 4.10 takes the form: u = Asinθ + Bcosθ + Gme r 2 o v2 o [4.11] Equation 4.11 introduces A and B. These can be found by plugging the initial conditions, θ = 0, du dθ = 0, u = 1, into the first two derivatives. Differentiating with respect to θ: r du dθ = Acosθ + Bsinθ [4.12] Differentiating again with respect to θ: d 2 u = Asinθ Bcosθ [4.13] dθ 2 Looking at equation 4.12, with the initial conditions, gives 12

14 0 = Acos(0) + Bsin(0). Cos(0) = 1, Sin(0) = 0 so the previous equation simplifies to A = 0. Examining equation 4.11 at θ = 0, remembering that A = 0, gives: u = Bcosθ + Gme. ro 2v2 o With the initial conditions this can be expressed as: 1 r o = B(1) + Gme r 2 o v2 o or B = 1 r o Gme. ro 2v2 o Plugging these results for A and B into equation 4.11 grants, Expanding the first term, u = [ 1 r o Gme r 2 o v2 o ]cosθ + Gme. ro 2v2 o u = 1 r o cosθ Gme r 2 ov 2 o cosθ + Gme. rov 2 o 2 The specific solution of u can be found by rearranging terms: u = 1 r o cosθ + Gme (1 cosθ) [4.14] rov 2 o 2 Recalling that u = 1, we can rewrite equation 4.14 as: r 1 = 1 r r o cosθ + Gme (1 cosθ) [4.15] r 2 o v2 o To condense equation 4.15, the eccentricity e can be modeled as e = rov2 o Gm e 1. [4.16] Manipulating equation 4.16 to solve for Gme : rov 2 o e = rov2 o Gm e 1 = Gme 1+e r ovo 2. [4.17] Substituting equation 4.17 into equation 4.15 brings: 1 r = 1 r o [(1 1 1+e )cosθ e ] 13

15 In the term (1 1 1+e )cosθ, 1 can be defined as. This simplifies the above equation to: 1+e 1+e 1 = 1 r r o [( e cosθ) + 1 ] 1+e 1+e Noticing the common denominator (1+e) present on the left hand side of the equation and multiplying through by r o yields: r or = 1+(ecosθ) 1+e. Taking the inverse leaves: r r o = 1+e 1+(ecosθ) [4.18] Moving r o back to the left hand side of the equation brings forth the final equation for r in terms of θ: r(θ) = ro(1+e) 1+(ecosθ). [4.19] Newtonian analysis has yielded equation 4.19 for the position of the moon as a function of θ. Note that at θ = 0, cos(θ) = 1 so the denominator reduces to 1 + e. Equation 4.19 then becomes r(0) = r o which checks out. 2. Lagrangian of the Moon s Orbit From Appendix C the Euler-Lagrange equations of motion take the general form d L L dt ẋ x = 0. [5.1] The Lagrangian of any system can be defined simply as L = T U [5.2]. Here T is the kinetic energy of the system and U is the potential energy of the system. V theta Vr e theta e r h Diagram 4: Lagrangian Diagram 14

16 The kinetic energy of the moon can be defined as, 1 2 m mv 2 t In this term, v 2 t refers to the total velocity of the moon which has a v r component and a v θ component. From equation 3.3 above, v r = ṙ and v θ = r θ. The magnitude of v t equals: (vr ) 2 + (v θ ) 2. Since we are looking for v 2 t, squaring both sides gives v 2 t = (v r ) 2 + (v θ ) 2. Thus in our model: v 2 t = (ṙ) 2 + (r θ) 2 = (ṙ) 2 + (r 2 ( θ) 2 ). This makes the kinetic energy, T = 1 2 m m[(ṙ) 2 + (r 2 ( θ) 2 )]. The potential energy of the moon can be defined as, m m gh = Gmemm r. The potential can be expressed as F g over a particular distance r or (F g ) (r). This can be re-written as ( Gmemm ) (r). Cancelling r leaves r 2 Gmemm r. Plugging these two results into equation 5.2 yields the the Lagrangian of the moon s orbit In polar coordinates, equation 5.1 takes the form L = [ 1 2 m m((ṙ) 2 + (r 2 ( θ) 2 )] + Gmemm r. [5.3] d dt L L θ = 0 [5.4] and d L L = 0 [5.5]. θ dt ṙ r Applying 5.4 to 5.3 for the θ direction: L = 0 [5.6] θ L θ = m mr 2 θ [5.7] d L dt θ = m m d dt (r2 θ) [5.8] 15

17 Recalling equations 4.3 and 4.4, equation 5.8 can be rewritten as: Applying 5.5 to 5.3 for the r direction: d L dt θ = m m[r 2 θ + 2rṙ θ] [5.9] L r = m mr( θ) 2 Gmemm r 2 [5.10] L = m ṙ mṙ [5.11] d L = m dt ṙ m r. [5.12] Now the above results can be substitued into the Lagrange equations of motion for this system. Substituting 5.6 and 5.8 into equation 5.4 and simplifying: m m [r 2 θ + 2rṙ θ] (0) = 0 r θ + 2ṙ θ = 0 [5.13] Note that 5.13 is the same as equation 4.2 for a θ. Substituting equation 5.10 and 5.12 into equation 5.5 yields: [m m r [m m r( θ) 2 Gmemm r 2 ] = 0 r r( θ) 2 = Gme r 2. r = Gme r 2. [5.14] Again note that 5.14 is the same as equation 4.1 for a r. C. Earth/Moon System COM as the Center of Rotation a. Reduced Mass Values and Assumptions In the reduced mass sections the total mass can be defined as, M = m e + m M = kg kg = kg. The ratio of the moon s mass to the total mass is, m m M = kg kg = The ratio of the earth s mass to the total mass is, 16

18 The reduced mass can be defined as µ = m 1m 2 m 1 +m 2. m e M = kg kg = In the reduced mass section the r e and r m will be measured to the respective centroids from the center of mass (COM) of the Earth-Moon system COM e/m. Also in these sections r is used as the distance between the moon and the earth. This actually equals r e + r m + d. d is the distance between the Earth and the Moon and this accounts for the change in r. d varies between 357,000 km and 407,000 km, with the average at 383,000 km [5]. b. Earth-Moon System COM y r e d r m C.O.M. x Diagram 1: Earth-Moon COM I started by centering the coordinate system at the far equatorial edge of the Earth to find the COM of the Earth-Moon system, as seen in figure 1. (Refer to Appendix F for more on center of mass.) With the coordinate system here, the earth s centroid is r e away from the origin and the moon s is (2r e r m ) = 397, 479.1km away. Then the COM in the x direction can be found, x = mere+mmrm M x = ( kg)(6371km)+( kg)(1737.1km) kg x = kgkm kgkm kg x = 11, km. Now subtracting x from 2r e leaves, 2r e x = 12742km 11, km = km. Therefore in figure 1 the COM of the Earth-Moon system is km below the Earth s surface. This will be used as the COM in the reduced mass model. 17

19 1. Reduced Mass using Newtonian Analysis The preceding two sections have dealt with the moon s orbit around the Earth, assuming the Earth to be fixed, i.e. its does not translate nor rotate. This is obviously an assumption and it neglects the effects of the moon s gravity on the Earth. In reality, although the moon s gravitational pull on the Earth is small it is not negligible. Indeed the interaction between the gravitational fields, the moon s and the earth s, is what produces the phenomenon we understand as gravity (prior to General Relativity). I want to now include the effects of the Moon s gravitational pull on the Earth. Reduced Mass is used to model the orbital dynamics of a two-body system united by a central force as a one-body system. The orbit occurs around the center of mass (com) of the two bodies. Therefore this concept can be used in two ways: 1) to model the two-body system as a one-body system, with individual position vectors based off the com, or 2) to model the dynamics of the two-body system around the com. As gravity is the only force in this model, from Newton s Third Law it can be said that the magnitude of the two forces exerted on each mass are equal, F e/m = F m/e = F (6.1) From Newton s Second Law it can be said that, Fmoon = m m r m = F e/m (6.2) Fearth = m e r e = F m/e (6.3) Diagram 5: Reduced Mass In diagram 5 COM represents the COM of the Earth-Moon system as calculated in section II. r e is the position vector for the earth which represents the distance from the Earth/Moon COM to the COM of the earth. r m is the position vector for the moon which represents the distance from the Earth/Moon COM to the COM of the moon. 18

20 In terms of the model, the com can be expressed as a weighted average to find the individual position vectors. r com = (mere)+(mmrm) m e+m m.(6.4) In this section, the term M will be defined as m e + m m. The total distance r between the Earth and the Moon is measured from the com of the Earth to the com of the moon. This can be written as, r = r m r e. (6.5) With the coordinate system centered at COM, r e will always be in the negative r direction. This means that the term r e will always be negative so equation 6.5 will be a sum of terms. Given equation 6.5, equation 6.4 can be solved for r e and r m to obtain the equations for the individual position vectors. Solving equation 6.4 for r e, Multiplying by 1, and solving produces, M r com = (mere)+(mmrm) M (M)r com = (m e r e ) + (m m r m ) (M)r com = (m e r e ) + (m m (r + r e )) (M)r com = (m e r e ) + (m m r) + (m m r e ) (M)r com = r e (m e + m m ) + (m m r) r com = r e + mmr M r e = r com mm M r (6.6) This is the equation for the Earth s position in the orbit. Now solving equation 6.4 for r m, r com = (mere)+(mmrm) M (M)r com = (m e r e ) + (m m r m ) (M)r com = (m e (r m r)) + (m m r m ) (M)r com = (m e r m ) (m e r) + (m m r m ) (M)r com = r m (m e + m m ) (m e r) 19

21 Again multiplying by 1, and solving leaves, M r com = r m me M r r m = r com + me M r (6.7) This is the equation for the moon s position in the orbit. In this model COM e/m is fixed, i.e. does not move with respect to time, so any time derivatives of r com will be zero. Also, with the coordinate system centered at COM, there is no position vector to the COM so r com = 0. Then equation 6.6 can be written as, and equation 6.7 can be rewritten as, r e = mm r, (6.8) M r m = me M r (6.9). Now the acceleration needed for equations 6.2 and 6.3 can be obtained. Taking the first and second time derivatives of equation 6.8 produces: v e = r e = mm ṙ (6.10) M a e = r e = mm r (6.11) M Equation 6.11 is the acceleration of the Earth in the r direction. There is no acceleration in the θ direction. Taking the first and second time derivatives for equation 6.9 yields: v m = r m = me ṙ (6.12) M a m = r m = me r (6.13) Equation 6.13 is the acceleration of the Moon in the r direction. Again there is no acceleration in the θ direction. M The reduced mass shows the equivalence of the forces. Note here that equation 6.2 and 6.3 reduce to identical equations. Substituting 6.13 into 6.2, Fmoon = m m ( me M r) = F µ r = F (6.14) Likewise substituting 6.11 into 6.3, 20

22 Fearth = m e ( mm r) = F M µ r = F (6.15) Note that equations 6.14 and 6.15 are identical. Here F is the magnitude of the force, which satisfies Newton s Third Law. This states that the force, experienced by either mass, is identical when measured from the COM of the Earth-Moon system COM. The force due to gravity can then be modeled as, Setting equation 6.2 equal to equation 6.16 Gmemm ˆr(6.16) r 3 m m r m = Gmemm ˆr r 3 r m = Gme (6.17) r 2 Equation 6.17 can be recognized as the gravitational acceleration due to the Earth. Likewise setting equation 6.3 equal to 6.16 m e r e = Gmemm ˆr r 3 r e = Gmm r 2 (6.18) Equation 6.17/6.18 can be recognized as gravitational acceleration for an orbiting body around the Earth/Moon respectively. As orbits occur in ellipses, the equations of motion for m e and m m take on the form of equation 4.19, r(θ) = ro(1+e) 1+(ecosθ) The model with Earth s COM at the center of the system defines r o at the perigee of the orbit. The radius at the perigee r p of an elliptical orbit can be modeled as, r o = r p = ǎ(1 e) (6.19) Likewise the radius at the apogee r a can be defined as, Substituting 6.19 into equation 4.19 r max = r a = ǎ(1 + e) r(θ) = (ǎ(1 e))(1+e) 1+(ecosθ) r = ǎ(1 e2 ) (1+ecos(θ)). (6.20) 21

23 Here ǎ represents the semi-major axis of the respective orbits. This notation is used to differentiate it from a the acceleration for the Earth and Moon. It is the convention to use equations of the form of 6.20 with the principle of reduced mass. This is due to the way that the r value is defined and that the respective position vectors (r e and r m ) represent the ǎ value for their respective orbits. Therefore the orbits of the Earth and Moon, with the Earth/Moon COM as the center of the system, are represented graphically according to equation The distance that the r values represent, in both models, is the same, but the way in which it is defined differs. With the Earth s COM as the center of the system, the r value is the radial distance to the Moon from the origin. As the moon changes position in its orbit, this value of r likewise changes. Therefore this r value is dynamic. The r value of the reduced mass section is dynamic. However, these dynamics are based on initial conditions defined by a fixed value of r. 2. Reduced Mass Lagrangian In this section the coordinate system will remain centered at the COM of the Earth-Moon system. Given this, and equation 7.13, the kinetic energy of the Moon can be modeled as, The potential energy for the Moon is, T m = 1 2 m mv 2 m = 1 2 m m[ me M ṙ]2. (7.1) U m = m m gr m. (7.2) Therefore the Lagrangian of the Moon can be written as, L m = T m U m = 1 2 m m[ me M ṙ]2 + m m gr m L m = 1 2 m m( me M )2 (ṙ) 2 + m m g( me M r) L m = 1 2 m m( me M )2 (ṙ) 2 + gµr. (7.3) The kinetic and potential energy of the Earth can be defined as, T e = 1 2 m ev 2 e = 1 2 m e[ mm M ṙ]2. (7.4) U e = m e gr e (7.5) Thus, the Lagrangian for the Earth is, 22

24 L e = T e U e = 1m 2 e[ mm M ṙ]2 m e gr e L e = 1 2 m e( mm M )2 (ṙ) 2 m e g( mm M r) L e = 1 2 m e( mm M )2 (ṙ) 2 gµr (7.6) The position and movement of the Earth and Moon do not depend on θ, but only on their position in the r direction relative to the com. Therefore there is only one equation of motion, in the r direction for the Earth and Moon. From equation 7.3 the necessary values for the Moon s equation can be obtained, L m ṙ L m r = gµ (7.7) = mmm2 e M 2 ṙ (7.8) d dt ( Lm ṙ ) = mmm2 e M 2 r. (7.9) Applying equations 7.7 and 7.9 to equation 5.5 the Lagrangian for the Moon can be obtained, Recall that µ = memm M. d ( L) L = 0 dt ṙ r m mm 2 e M 2 r gµ = 0 µ me r gµ = 0 M m e r = g (7.10) M From equation 7.6 the necessary values can be obtained for the Earth s equation of motion, L e ṙ L e r = gµ (77.11) = mem2 m M 2 ṙ (7.12) d dt ( Le ṙ ) = mem2 m M 2 r. (7.13) Applying 7.11 and 7.13 to equation 5.5 the Lagrangian for the Earth is, d dt ( Le ṙ ) Le r = 0 mem2 m M 2 r + gµ = 0 µ mm r + gµ = 0 M m m M r = g (7.14) 23

25 F. Extracting Experimental Data from APOLLO The experimental data was taken from the Apache Point Observatory Lunar Laser Operations (APOLLO) project located in New Mexico. (For more information on the project please visit [ tmurphy/apollo/].) This site conducted weekly Lunar Laser Ranging (LLR) experiments. This involves shooting very intense, short bursts of photons to one of the five retroreflector sites on the moon. A small fraction of these initial photons reflect make it back to Earth and are collected by the APOLLO detector. The data collected by APOLLO are recorded via a series of normal points. A small sample of normal points used as the experimental distances is included below. [8] The normal points can be deciphered according to the table below. Note that the two-way time of flight is the pertinent data required to calculate the experimental distances to the moon. Data after these fourteen digits represent weather and atmospheric condtions which are not considered in this experiment so I do not detail them here. Digits Interpretation 1-2 Station Identifier 3-6 Year 7-8 Month 9-10 Day Hour Minute Seconds (10 7) Two-way Time of Flight (10 13 seconds) To extract all this experimental data from APOLLO, I wrote a program in PERL (see Appendix A). The program extracts all necessary data from the APOLLO site, converts the data to the units of the theoretical data and plots this data in Matlab. 24

26 1. Experimental Distances To compare the experimental data with the theoretical data from the Newtonian and Lagrangian sections, I first converted the fourteen digit value of two-way time of flight to seconds. (Note: The first line of the sample normal points will be used as the example conversion.) / = seconds This represents the one way time of flight. Multiplying this by the speed of light (in km) produces the experimental distance from this data point seconds ( km/s 2 ) = 351, km As a correction in the data I also added 8, km to each experimental distance to keep the model consistent. The theoretical distances are measured from the Earth s centroid to the Moon s centroid and thus include both r e and r m as defined above. The experimental distances are measured from APOLLO, on the surface of the Earth, to a retroreflector on the surface of the moon. Therefore both r e and r m must be added to the experimental distances to have the same axes as the theoretical data. r e + r m = 6, , = 8, km This will help draw more accurate conclusions but is still a possible source of error in this experiment. 2. Experimental Angles The experimental angle was more tedious to collect than the experimental distances. To get to an angle approximation I started with the time of firing for the laser. First, I extracted the four digit year value and two digit month data into one data file and the two digit day of the month data into a second data file. To normalize the time data to one lunar period, days, the two time data files were converted to a sequential day from 2006 to 2010, the range from which the data was taken. To accomplish this, first I converted the year data into a sequential day value from representing the days from January 01, December 31,2010. Second, I converted the month and day data to a day value from Adding these two values gives a sequential value to each of the data points for these time values. The time values need to be converted to an approximate angle against which to plot the experimental distances. A lunar period is about days. Dividing this period by 2π, 25

27 π = yields a value of rad. This value multiplied by the sequential day value gives an approximate day angle for each data point from the time data. The item-by-item matrix manipulation was done in Octave for ease. The code in its entirety is contained in Appendix B. 26

28 IV. Results 1. Eccentricity of the Moon s Orbit Starting with equation 4.18, r r o = e+1. (ecosθ)+1 the apogee (max) can be modeled at 180. Equation 4.18 then becomes, Solving for e yields, r max r o = 1+e 1 e r max r o (1 e) = 1 + e r max r o r max r o r max r o rmax r o (e) = 1 + e 1 = rmax r o (e) + (e) 1 = (e)( rmax r o + 1) e = rmax ro 1 rmax (4) +1 ro Using NASA data r max = and r o = Plugging these into equation 4 in yields the eccentricity, e = e = (5) A. Plotting the Earth COM as the Center of Rotation Remembering the initial conditions, equation 4.19 becomes, r(θ) = ( ) 1+( cosθ) r(θ) = ( cosθ). (8.1) 27

29 This equation is represented graphically below via Octave. Earth s COM as the Center of the System B. Plotting the Earth/Moon COM as the Center of Rotation 1. Plotting the Equations of Motion As defined above the total distance r can be defined as 386,108.1 km. Plugging this value into equation 6.9 yields, r e = ( )(386, 108.1km) = km Likewise substituting into equation 6.10 gives, r m = ( )(386, 108.1km) = 381, km These are the initial positions for the Earth and the Moon for the dynamics of the Earth/Moon System. This corresponds to the Moon being at the perigee in its orbit which aligns with the initial conditions of 28

30 the model with the Earth as the center of the system. Therefore the dynamics are based on these initial values of the position vectors. These are also the values of the radii of the semimajor axes ǎ for the respective, elliptical orbits of the Earth and Moon. So, ǎ e = km and aˇ m = 381, km. The eccentricity of the moon s orbit and the earth s orbit will be the same. e = e m = e e From Appendix C the equations of motions take the general form, The equation for the moon can be written as, r = a(1 e2 ) (1+ecos(θ)). (8.2) r m = The equation for the earth can be written as, am(1 (em)2 ˇ (1+e mcos(θ)) r m = (381, )(1 ( )2 ) (cosθ) r m = (cosθ). (8.3) r e = ǎe(1 (ee)2 (1+e ecos(θ)) r e = ( )(1 ( )2 ) (cosθ) r e = (cosθ). (8.4) These are the equations for the Earth and Moon with the coordinate system centered at COM e/m. They are represented graphically below via QT Octave, with the Moon s Orbit in blue and the Earth s in green. Reduced Mass COM as the center of the system 29

31 2. Comparing Theoretical and Experimental Data a. Data Comparison: Earth as the center of the system Figure 1: Equation 8.1 Theoretical Data vs. APOLLO Data The theoretical data values in the table above are generated by substituting the values for the approximate angles into equation

32 r(θ) = ( cos( ) = To compile the experimental data, one data point for each approximate angle was selected to produce a distribution for one full revolution (360 ). The percent error in the theoretical data is calculated according to following equation, with the final result in percent. The first row from figure 5 is used as an example: theoretical distance experimental distance theoretical distance x = percent. The absolute value of the difference in distance is taken to increase the accuracy of the percent error. If the absolute value were not taken, data points where the experimental data is greater than the theoretical data would decrease the overall percent error. This would represent a greater accuracy in the analysis than the experiment actually displays. The average percent error in the theoretical data for equation 8.1 is percent when compared to the experimental data. The standard deviation is percent which is rather high given the low value for percent error. This is due mostly to the fact that there are a few outliers in the data set and that the sample size for the data is relatively small. b. Data Comparison: Earth/Moon COM as the center of the system The theoretical data values in the table below are generated by plugging the approximate angles into equation 8.3. r m = cos( ) = km The average percent error for the theoretical data from equation 8.3, is percent. This means that on average equation 8.3 is closer to the experimental data than equation 8.1. Due to the fact that the reduced mass model more closely resembles the actual dynamics of the orbit, it would be expected that this model displays less error. c Final Graphical Representation Below is the final graphical compilation. In light blue is equation 8.1, equation 8.3 is in red, equation 8.4 is in light green. The blue crosses represent the experimental data points. 31

33 Figure 2: Equation 8.3 Theoretical Data vs. APOLLO data Figure 3: Equations 8.1, 8.3, 8.4 and Experimental Data vs. Angle 32

34 V. Conclusion A. Equations of Motion As predicted, Newtonian analysis and Lagrangian Mechanics produced the same equation of motion for the Moon and the Earth, whether the Earth was the center of the system or the Earth/Moon COM was the center of the system. In the first case, where the Earth is the center of the system, Newtonian analysis and Lagrangian Mechanics agreed in both the r and θ directions. In the r direction equation 4.1 and 5.14, respectively for the two methods, are In the θ direction equation 4.2 and 5.13 are r = Gme r 2. r θ + 2ṙ θ = 0. In the second case, where the Earth/Moon COM is the center of the system, both methods of analysis brought equations of motion for the Earth and Moon that simplify to the reduced mass equation. µ r = F Using the Lagrangian required a bit of reconfiguration to obtain the reduced mass equation above. Starting with equation 7.10 and 7.14, Multiplying both sides by m m, (7.10) me M ( M m e r m = g r m ) = g m m r m = m m g m m r m = F (7.15) 33

35 Likewise for the Earth, (7.14) mm M ( M m m r e ) = g r e = g m e r e = m e g m e r e = F (7.16) Laws. Notice that 7.15 and 7.16 mirror equations 5.2 and 5.3. This supports Newton s Second and Third Also note that substituting 6.13 back into 7.15 yields: m m r m dt 2 = F m m ( me M r) = F µ r = F (7.17) Substituting equation 6.11 into 7.16: m e r e = F m e ( mm r) = F M µ r = F (7.18) Equations 7.17 and 7.18 are identical to equations 6.15 and This means that the force on either mass is equal when measured from the Earth-Moon COM which is what the principle of reduced mass would predict. B. Accuracy of the Obtained Equations The reduced mass model better reflects the actual dynamics of the orbit of the Moon and the Earth and should therefore be more accurate than the first model. The Earth is not fixed and indeed its orbit is affected by the sun, moon and even other planets. The moon s orbit is affected by the same masses. Therefore the reduced mass model, which removes an assumption from the geocentric model and considers the effect of the moon s gravity on the Earth, should exhibit a lower error. THEORETICAL AND EXPERIMENTAL ERROR 34

36 The data selection process may have influenced the average percent error for better or worse. Taking the absolute value of the difference between the theoretical and experimental really eliminates most if not all of the positive error (lowering of the average percent error) in favor of this experiment. That being said the theoretical data did display a very high level of similarity compared to the experimental data from Apache Point. This shines well on the theoretical data as the experimental error in the APOLLO LLR data is 1 cm in the range of km. The low average percent error for the theoretical data shows the accuracy of the results from this research. The continued ability of these models to predict such astronomical events with precision will find them continued relevance. 35

37 36

38 VI. APPENDIX A. Data Extraction Program 37

39 38

40 39

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42 41

43 B. Octave Program C. Elliptical Conic Section - Reduced Mass An ellipse centered at the origin, with major radius a and minor radius b, takes the form: x 2 + y2 = 1 (1). a 2 b 2 Recalling Appendix A equation 1 can be written in terms of polar coordinates, r 2 cos 2 (θ) a 2 + r2 sin 2 (θ) b 2 = 1 (2). Foci Theta c r b a In the above diagram, a and b are the semimajor and semiminor axes respectively and c is the distance from the origin to the foci. r is the distance from one of the foci to the curve and theta is the angle between r and the zero-axis. 42

44 (3). The eccentricity e equals c. The distance from a foci to one of the major verteces is a c or a(1 e) a Combining equation 3 with the identity b 2 = a 2 c 2, it can then be written that b 2 = a 2 (1 e 2 ). (4) In the reduced mass sections, the center of rotation is at the com, one of the foci of each orbit. This center of rotation is a distance c from the origin. Then it can be written that, x = rcos(θ) + c (5) and y = rsin(θ) (6). Equation 2 then becomes: (rcos(θ)+c) 2 a 2 + r2 sin 2 (θ) b 2 = 1 (7) Moving a 2 and b 2 to the right hand side and expanding the first term yields, (b 2 )(r 2 cos 2 (θ) + c 2 + 2crcos(θ)) + (a 2 )r 2 sin 2 (θ) = a 2 b 2 b 2 r 2 cos 2 (θ) + b 2 c 2 + 2b 2 crcos(θ) + (a 2 )r 2 sin 2 (θ) = a 2 b 2 (8) Remembering the identity sin 2 (θ) = 1 cos 2 (θ), equation 8 can be rewritten as, b 2 r 2 cos 2 (θ) + b 2 c 2 + 2b 2 crcos(θ) + a 2 r 2 (1 cos 2 (θ) = a 2 b 2 b 2 r 2 cos 2 (θ) + b 2 c 2 + 2b 2 crcos(θ) + a 2 r 2 a 2 r 2 cos 2 (θ) = a 2 b 2 Now substituting b 2 = a 2 (1 e 2 ) and c = ea, a 2 (1 e 2 )e 2 a 2 + 2eaa 2 (1 e 2 )rcos(θ) + a 2 (1 e 2 )r 2 cos 2 (θ) + a 2 r 2 a 2 r 2 cos 2 (θ) = a 2 a 2 (1 e 2 ) Now multiplying by 1 a 2 and putting the quantity to be solved for r on the right hand-side, (1 e 2 )a 2 e 2 + 2aer(1 e 2 )cos(θ) + (1 e 2 )r 2 cos 2 (θ) a 2 (1 e 2 ) r 2 cos 2 (θ) = r 2 Simplifying and rearranging terms, [a 2 e 2 a 2 e 4 ] + 2aer(1 e 2 )cos(θ) + [r 2 cos 2 (θ) e 2 r 2 cos 2 (θ)] [a 2 a 2 e 2 ] r 2 cos 2 (θ) = r 2 a 2 e 4 + 2a 2 e 2 a 2 + 2aer(1 e 2 )cos(θ) e 2 r 2 cos 2 (θ) = r 2 a 2 (1 2e 2 + e 4 ) + 2aer(1 e 2 )cos(θ) e 2 r 2 cos 2 (θ) = r 2 a 2 (1 e 2 ) 2 + 2aer(1 e 2 )cos(θ) e 2 r 2 cos 2 (θ) = r 2 [ercos(θ) a(1 e 2 )] 2 = r 2 Taking the square root now leaves, 43

45 r = ±[ercos(θ) a(1 e 2 )]. If e = 0 the equation becomes r = ±( a). Because r must be positive and mathematically because the directrix of the ellipse lies to the right, the negative must be taken which yields, Finally the equation can be solved for r: r = a(1 e 2 ) ercos(θ) (9). r + ercos(θ) = a(1 e 2 ) r(1 + ecos(θ)) = a(1 e 2 ) r = a(1 e2 ) (1+ecos(θ)). (10) This is the general form of the equation of motion for a body in the reduced mass sections. D. Deriving the Lagrange Equations of Motion The Lagrangian is defined as L = T U. (1) T is the kinetic energy and can be defined as 1 2 mv2 = 1 2 m(ẋ)2. This only depends on velocity, the derivative of position (ẋ), so we can write T (ẋ). U is the potential energy and can be defined as mgh = mgx, where g is the acceleration due to gravity on Earth, and h is the distance above or below the origin of the coordinate system of reference. h is also the position x of the body in question so we can write U(x). Here it should be noted that the partial derivative of U with respect to x, U x = mg. This is the weight of the object or the potential force created by that object in the field. In a conservative field, like a gravitational field, the potential of the object mg equals the sum of the forces on that object (to hold it at the current position). Since gravity F g is the only force acting on the moon it can be equated to the potential, so we can write F = Fg = U x (2). Taking the partial derivative of T with respect to ẋ, T = 1 m2ẋ = mẋ (3). ẋ 2 44

46 The last term can be recognized as p momentum. The derivative with respect to time of equation 3 d ( T dt ẋ ) = mẍ = dp dt. (4) Note that because U and T do not depend on ẋ and x respectively, U ẋ = 0 (5) and T x = 0 (6) From Newton s second law dp dt = F. With equations 2 and 4 it can now be written that d ( T ) = U. (7) dt ẋ x From equations 5 and 6 and using common denominator equation 7 can be rewritten as d dt (T U) ( ) = ẋ (T U) x. (8) Moving both terms to one side and recalling the identity from equation 1 the final Lagrange equations of motion can be obtained. d ( L) L = 0 (9) dt ẋ x 45

47 Bibliography A. Citations [1] Isaac Newton, Synopsis of Life and Work - Wikipedia Newton [2] Joseph-Louis Lagrange, Synopsis of Life and Work - Wikipedia Louis Lagrange [3] William Rowan Hamilton, Synopsis of Life and Work - Wikipedia Rowan Hamilton [4] Marion, Jerry and Stephen Thornton. Classical Dynamics of Particles and Systems. Saunders College Publishing 4th. edition, Orlando, [5] Parameter Fact Sheet on the Moon and its Orbit - NASA [6] History of the Principle of Least Action - Wikipedia of least action [7] Home website for the APOLLO Project tmurphy/apollo/ [8] Normal points from APOLLO LLR from tmurphy/apollo/norm pts.html B. References [9] Fitzpatrick,Richard.Newtonian Dynamics. University of Texas at Austin, Online. [10] Bedford,Anthony and Wallace Fowler. Engineering Mechanics:Dynamics. Pearson Prentice Hall 2nd edition, New Jersey, [11] Lagrange s Method - MIT Lectures S. Widnall Dynamics Fall 2009 Version Lecture 20 ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecturenotes/mit16 07F09 Lec20.pdf 46

48 [12] Ellipses and Mathematical Proofs - Wolfram s World of Physics [13] Kepler s Laws of Planetary Motion - Wikipedia laws of planetary motion [14] Lagrange Points Hyperphysics - Georgia State University [15] Lagrangian Mechanics - Berkman Institute, University of Illinois PDF/chp1.pdf [16] Overview of Orbital Mechanics - Rocket and Space Technology [17] Shape of a Newtonian Orbit with Lagrangian Applied shapes 1.html [18] Gravitational Interactions of the Earth and Moon: Barycentric Motion [19] Moon s Effect of the Earth s Orbit and Motion [20] Least Action Principle - Wolfram s World of Physics Action Principle.html [21] Binary Stars - Physics Net [22] Ch. 2 Lagrange s and Hamilton s Equations - Rutgers University 47