{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 16 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 } 1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE " " -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Map le Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 16 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 16 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 14 "Kepler Problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 107 "We demonstrate the numerical solution of Newton's equation for the Kepler problem in Cartesian coordinates." }}{PARA 0 "" 0 "" {TEXT -1 192 "A unit system is chosen to make the equations dimensionless, and we assume that the Sun is infinitely heavy, i.e., we just consider the motion of a singl e body around the center of attraction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "This worksheet contains several sect ions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 147 "1) In the first section we introduce a system of coordinates that ren der the solar system dimensionless. This is achieved by choosing three scales:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "the AU (average Sun-Earth distance) as a distance measure;" }} {PARA 0 "" 0 "" {TEXT -1 37 "the solar mass M_s as a mass measure;" }} {PARA 0 "" 0 "" {TEXT -1 74 "the year (time it takes for Earth to go a round the Sun) as a time measure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 167 "We solve the Kepler problem numerically \+ for this problem, and demonstrate a few features, such as the virial t heorem (relation between kinetic and potential energies)." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 132 "2) We then explor e the problem of radial motion, including a discussion of stability of circular orbits as a function the force law;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "3) We then conclude with \+ a brief discussion of the scattering problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "We begin with sec tion 1:" }}{PARA 0 "" 0 "" {TEXT -1 54 "For the solar system we have t he following parameters:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 24 "Solar mass: 2*10^\{30\} kg" }}{PARA 0 "" 0 "" {TEXT -1 53 "Gravitational constant: 6.67*10^\{-11\} SI (N*m^2/kg^2)" }}{PARA 0 "" 0 "" {TEXT -1 24 "Earth mass: 6*10^\{24\} kg" }}{PARA 0 " " 0 "" {TEXT -1 34 "Sun-Earth distance 1.5 * 10^\{11\} m" }}{PARA 0 " " 0 "" {TEXT -1 32 "Earth's orbit time: 3.15*10^\{7\}s" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 17 "G:=6.67*10^(-11);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"GG$\"++++qm!#?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "M_s :=1.99*10^(30);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$M_sG$\"++++!*>\" #@" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "M_e:=5.97*10^(24);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$M_eG$\"++++qf\"#:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "G*M_s*M_e;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"++,;Cz\"#N" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "AU:=1.5*10^(11);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#AUG$\"+++ ++:\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "G*M_s*M_e/AU^2; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+L*[=_$\"#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Earth's attraction towards the Sun is rather lar ge when expressed in Newtons, but the acceleration in m/s^2 is not ver y large:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "G*M_s/AU^2;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+WWC**e!#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 234 "Note that this gravitational acceleration is balanced at the center of the Earth by the centrifugal force due to the rotati on about the Sun. The only net effect on an object at the surface of t he Earth is in the form of a tidal force." }}{PARA 0 "" 0 "" {TEXT -1 23 "The period is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "T_e:=evalf(365.26*24*60*60);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% $T_eG$\"++k%e:$!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "v_e: =evalf(2*Pi*AU)/T_e;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$v_eG$\"+_,X ')H!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Earth moves with about 30 km/s around the Sun." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 278 "Now we introduce a unit system in which the Earth 's orbit radius is of order unity, the masses are measured in multiple s of Sun's mass, and times are measured in terms of orbit times. We ca n pick three dimensionful quantities and set the scales as follows [re member: SI=MKS(A)]:" }}{PARA 0 "" 0 "" {TEXT -1 12 "M = M' * M_s" }} {PARA 0 "" 0 "" {TEXT -1 12 "T = T' * T_e" }}{PARA 0 "" 0 "" {TEXT -1 11 "X = X' * AU" }}{PARA 0 "" 0 "" {TEXT -1 78 "and with the substitut ion we express equations in terms of the new variables. " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 419 "A similar procedu re is used in modern physics to scale the atomic orbital problem; one \+ notable exception is that one uses the electron mass as a scale. In th e solar Kepler problem this is not done, because the planet's mass dro ps out of the equation of motion. In the quantum problem it does not, \+ because the force of attraction is electric, i.e., not proportional to the lighter particle's (i.e., the electron's) mass." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "What is our unit of vel ocity in these natural units for the Earth-around-the-Sun motion probl em?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "unitLength:=AU; # in troduced unit #1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+unitLengthG$\" +++++:\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "unitTime:=T_ e; # introduced unit #2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)unitTime G$\"++k%e:$!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "unitVel: =unitLength/unitTime; # a derived unit!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(unitVelG$\"+SI3`Z!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "The velocities V' will come out as multiples of about 4.7 km/s, \+ which means that Earth's cruising speed [about the Sun] becomes 2 Pi! " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "v_e/unitVel;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+2 `=$G'!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "What is the unit of \+ force?" }}{PARA 0 "" 0 "" {TEXT -1 73 "The natural unit should be the \+ force experienced by Earth due to gravity." }}{PARA 0 "" 0 "" {TEXT -1 78 "The unit for acceleration is clearly the velocity unit divided \+ by a time unit:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "unitAcc: =unitVel/unitTime; # a derived unit!" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%(unitAccG$\"+#*)>h]\"!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "unitMass:=M_s; # introduced unit #3 !" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)unitMassG$\"++++!*>\"#@" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 46 "unitForce:=unitMass*unitAcc; # a derived unit!" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%*unitForceG$\"+&eyr*H\"#<" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "Given this unit for the accelerat ion, it is of interest to express the acceleration that Earth experien ces from the gravitational attraction towards the Sun in this unit:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "(G*M_s/AU^2)/unitAcc;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+n\"\\o\"R!\")" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 55 "The force experienced by Earth is small in these u nits:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "(G*M_s*M_e/AU^2)/u nitForce;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+]Z0v6!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "We can find the unit for the gravitationa l constant from its dimensions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "unitG:=unitForce*(unitLength)^2/unitMass^2;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%&unitGG$\"+v$**Gq\"!#@" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 229 "Now the centrifugal force in SI units: we assume a point at which the velocity and position are at right angles, so that the m agnitude of angular momentum is given by the product of magnitudes of \+ position and velocity times mass:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "L0:=(M_e*AU*v_e);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#L0G$\"+6hOuE\"#J" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "L0/( unitMass*unitVel*unitLength); # angular momentum in our units:" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+\"fb\\)=!#9" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 37 "subs(r=AU,-diff(L0^2/(2*M_e*r^2),r));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Ogr\\N\"#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The gravitational attraction in SI units:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "-G*M_s*M_e/AU^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+L*[=_$\"#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 380 "The balance between gravitational attraction and the cen trifugal force is not perfect, it is at the level of the number of sig nificant digits carried, which measn that it implies that Earth's orbi t is not a perfect circle, and the AU averages over an ellipse with ec centricity close to 0. Earth's eccentricity is listed as 0.017. Mars h as an eccentricity of 0.092, Mercury has 0.2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "Now let us figure out the bala ncing in the unit system introduced. We calculate the centrifugal forc e from differentiating the centrifugal potential expressed as " }} {PARA 0 "" 0 "" {TEXT -1 14 "L^2/(2*m*r^2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "subs(r=AU,diff(L0^2/(2*M_e*r^2),r))*(unitMass*unit Length^3/(unitMass*unitLength*unitVel)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+GDN%=\"!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "(-G*M_s*M_e/AU^2)*(unitLength^2/(unitMass^2*unitG));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+]Z0v6!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "G/unitG;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+o\"\\o \"R!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Now, this establishes \+ what value one should use for G*M_s when working in this unit system: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 148 "We t ake a fresh start, and observe how the Earth (an object of mass m) mov es about a fixed gravitational center using a Cartesian coordinate sys tem." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "restart; GM:=39.17: m:=6*10^(24)/(2*10^(30)):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "The inverse power for the potential (here we can play if we wish):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "n:=10/10;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"nG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "V:=-GM*m/r^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&\"\" \"F'%\"rG!\"\"$!+++5v6!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 186 "Fro m the conservation of the orientation of the angular momentum vector w e know that the motion is confined to a plane, i.e., the problem is tw o-dimensional. We label the coordinates as " }{TEXT 257 1 "x" }{TEXT -1 5 " and " }{TEXT 256 1 "y" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "V:=subs(r=sqrt(x^2+y^2),V);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&\"\"\"F'*$-%%sqrtG6#,&*$)%\"xG\"\"#F'F'*$)%\" yGF0F'F'F'!\"\"$!+++5v6!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "The force is calculated as follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Fx:=-diff(V,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #FxG,$*&%\"xG\"\"\"*$),&*$)F'\"\"#F(F(*$)%\"yGF.F(F(#\"\"$F.F(!\"\"$!+ ++5v6!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Fy:=-diff(V,y); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FyG,$*&%\"yG\"\"\"*$),&*$)%\"xG \"\"#F(F(*$)F'F/F(F(#\"\"$F/F(!\"\"$!+++5v6!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Note that both components of the force depend on bot h " }{TEXT 259 1 "x" }{TEXT -1 5 " and " }{TEXT 258 1 "y" }{TEXT -1 77 ". We make this more obvious by unapplying the two expressions into functions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "FX:=unapply( Fx,x,y): FY:=unapply(Fy,x,y):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 158 "We can now evaluate the force vector at some location. Let us demonst rate that the force is central, i.e., points towards the origin of the coordinate system:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "F0:= [FX(1,2),FY(1,2)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#F0G7$,$*$-%%s qrtG6#\"\"&\"\"\"$!+++S+Z!#:,$F'$!+++!3S*F/" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 63 "The y-component of the force vector at this location gi ven by " }{TEXT 260 1 "r" }{TEXT -1 133 " = [1,2] is twice as large a s the x-component, and this is sufficient for the force vector to be p roportional to the position vector." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The Newton equation leads to a pair of coupled second-order equ ations:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "NE1:=m*diff(x(t) ,t$2)=FX(x(t),y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NE1G/,$-%%d iffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"##\"\"$\"(+++\",$*&F*\"\"\"*$),&*$)F* F1F7F7*$)-%\"yGF,F1F7F7#F3F1F7!\"\"$!+++5v6!#8" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 36 "NE2:=m*diff(y(t),t$2)=FY(x(t),y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NE2G/,$-%%diffG6$-%\"yG6#%\"tG-%\"$G6$F- \"\"##\"\"$\"(+++\",$*&F*\"\"\"*$),&*$)-%\"xGF,F1F7F7*$)F*F1F7F7#F3F1F 7!\"\"$!+++5v6!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Let us defin e some intial condition: we pick a point on the positive " }{TEXT 261 1 "x" }{TEXT -1 79 "-axis, with some velocity in the y-direction, but \+ no velocity component in the " }{TEXT 262 1 "x" }{TEXT -1 80 "-directi on, i.e., the initial position and velocity vectors are at right angle s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "x0:=1: y0:=0: vx0:=0: vy0:=6.28: # these conditions illustrate Earth's orbit" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "x0:=1: y0:=0: vx0:=0: vy0:=4.: # th ese conditions give substantial ellipticity" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "IC:=x(0)=x0,D(x)(0)=vx0,y(0)=y0,D(y)(0)=vy0;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6&/-%\"xG6#\"\"!\"\"\"/--%\"DG6# F(F)F*/-%\"yGF)F*/--F/6#F3F)$\"\"%F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 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Watch how the period of the motion changes with the four parameters that de termine the initial condition." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 264 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 81 "Change the potential function to a slightly different distance dep endence from 1/" }{TEXT 265 1 "r" }{TEXT -1 37 ". Observe what happens to the motion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "Let us visualize better what is going on by showing \+ the vectorial quantities along the trajectory." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "Np:=30; T:=0.5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NpG\"#I" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"TG$\"\"&!\"\" " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Lr:=[seq([X(T/Np*i),Y(T /Np*i)],i=1..Np)]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with( plots):" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "PL T:=plot(Lr,style=point): display(PLT,scaling=constrained);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6%7@7$$\"3Qe -(*o*eb%**!#=$\"3?qW[U'[Xl'!#>7$$\"3M)4(>RYx\"y*F*$\"3'4([$RnpNK\"F*7$ $\"3ge)zwxYs]*F*$\"3_yj.\"**[m'>F*7$$\"3*QM(foie>\"*F*$\"3Oq25#\\Tie#F 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**4N*F,7$$\"+8R#Hc&FEFcfo7$FazFc^o7$FfzFi^p7$$\"+O9)>%eFEFaho7$$\"+D]2 beFEFcfo7$$\"+==iheFE$!+;!**4N*F,7$$\"+8'o\"oeFEF`go7$$\"+2arueFEF^_p7 $$\"+-AE\")eFE$!+=!**4N*F,7$$\"+z$\\u!fFEFa\\p7$$\"+bljLfFE$!+?!**4N*F ,7$$\"+44,')fFE$!+9!**4N*F,7$F[[lFcfo7$$\"+T,1*3'FEFa\\p7$$\"+@]tRhFE$ !+'FE$!+A!**4N*F,7$$\"+Stu:iFEFi`p7$F`[l $!+:!**4N*F,7$Fe[l$!+D!**4N*F,7$F_\\l$!+I!**4N*F,7$$\"+I\"eBs'FE$!+N!* *4N*F,7$Fd\\l$!*/**4N*F[o7$$\"+E'>.$oFEFcep7$Fi\\lFcep7$F^]l$!*.**4N*F [o7$Fc]lFcin7$F]^l$!*$*)*4N*F[o7$Fg^l$!*z)*4N*F[o7$Fa_l$!*j)*4N*F[o7$F [`l$!*[)*4N*F[o7$Fe`l$!*R)*4N*F[o7$F_al$!*C)*4N*F[o7$Fial$!*6)*4N*F[o7 $Fcbl$!*.)*4N*F[o7$Fhbl$!*+)*4N*F[o7$F]cl$!*(z*4N*F[o7$Fbcl$!*)z*4N*F[ o7$Fgcl$!+&)z*4N*F,7$$\"+ee)*R%)FEF_hp7$F\\dl$!+!)z*4N*F,7$FadlFehp7$F fdl$!+')z*4N*F,7$F[elFihp7$F`el$!+(*z*4N*F,7$Feel$!+)*z*4N*F,7$Fjel$!+ #*z*4N*F,7$F_fl$!+&*z*4N*F,-Fdfl6&FfflF(FgflF(-%*AXESSTYLEG6#%$BOXG-%+ AXESLABELSG6$Q\"t6\"Q!6\"-%*THICKNESSG6#\"\"$-%%VIEWG6$;F(F_fl%(DEFAUL TG" 1 2 0 1 10 3 2 9 1 2 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "We observ e that the total energy is conserved, and that the kinetic and potenti al energies as a function of time have to be related in some way to ac hieve this." }}{PARA 0 "" 0 "" {TEXT -1 163 "When the planet is close \+ to the center of attraction it moves faster (has higher kinetic energy ) on account of the fact that the potential energy has been lowered." }}{PARA 0 "" 0 "" {TEXT -1 218 "One can establish a relationship betwe en the average values over a period (this is called the virial theorem in classical mechanics, it's details depend on the force law, and it \+ has a counterpart in quantum mechanics)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "We lift an approximate value for t he period from the graph:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Period:=0.5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'PeriodG$\"\"&! \"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "T_avg:=evalf(Int(T_ kin,0..Period,5,_NCrule))/Period;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %&T_avgG$\"+++!yN*!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "E_ avg:=evalf(Int(E_tot,0..Period,5,_NCrule))/Period;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%&E_avgG$!++++^$*!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "V_avg:=evalf(Int(V_pot,0..Period,5,_NCrule))/Period; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&V_avgG$!+++)3(=!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V_avg/E_avg;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+&>F2+#!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 269 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 167 "Carry out these calculations for different initial conditions \+ and verify that the relationship between these average values is a fea ture of Newton's law of gravitation" }}{PARA 0 "" 0 "" {TEXT -1 144 "( inverse-squared force law). Ensure a correct choice of period by looki ng at the graph. Explore what happens for slightly different power law s, " }}{PARA 0 "" 0 "" {TEXT -1 8 "such as " }{TEXT 271 1 "n" }{TEXT -1 11 "=12/10, or " }{TEXT 270 1 "n" }{TEXT -1 11 "=8/10, etc." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 257 "" 0 " " {TEXT 272 28 "2) The radial motion problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "The radial problem for " } {TEXT 273 1 "r" }{TEXT -1 1 "(" }{TEXT 274 1 "t" }{TEXT -1 126 ") is d erived from angular momentum conservation. It allows us to determine t he trajectory from a single differential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "First we calculate the an gular momentum:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(lin alg):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names nor m and trace have been redefined and unprotected\n" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 46 "L0:=evalm(m*crossprod([x0,y0,0],[vx0,vy0,0]) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L0G-%'vectorG6#7%$\"\"!F*F*$ \"+++++7!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "t0:=2;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t0G\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "evalm(m*crossprod([X(t0),Y(t0),0],[VX(t0),VY(t0),0 ]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%$!\"!\"\"!$F)F)$ \"+U+++7!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 235 "We observe that t he calculated trajectory does observe the conservation of angular mome ntum approximately. This is a test if the numerical accuracy of the al gorithm. We also see that the angular momentum vector is perpendicular to the " }{TEXT 267 1 "x" }{TEXT -1 1 "-" }{TEXT 266 1 "y" }{TEXT -1 7 " plane." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "The Euclidean norm is called the Frobenius norm in the linalg \+ package (square-root of sum of elements squared). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "L_mag:=norm(L0,frobenius);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&L_magG$\"+++++7!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "The one-dimensional problem of solving for the distance \+ from the center of attraction as a function of time:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Vr:=-GM*m/r^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#VrG,$*&\"\"\"F'%\"rG!\"\"$!+++5v6!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Veff:=Vr+L_mag^2/(2*m*r^2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%VeffG,&*&\"\"\"F'%\"rG!\"\"$!+++5v6 !#8*&$\"+++++C!#9F'*$)F(\"\"#F'F)F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Feff:=unapply(-diff(Veff,r),r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%FeffGR6#%\"rG6\"6$%)operatorG%&arrowGF(,&*&\"\"\"F.* $)9$\"\"#F.!\"\"$!+++5v6!#8*&$\"+++++[!#9F.*$)F1\"\"$F.F3F.F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "DE:=m*diff(r(t),t$2)=Feff(r( t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#DEG/,$-%%diffG6$-%\"rG6#%\" tG-%\"$G6$F-\"\"##\"\"$\"(+++\",&*&\"\"\"F7*$)F*F1F7!\"\"$!+++5v6!#8*& $\"+++++[!#9F7*$)F*F3F7F:F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "r0:=sqrt(x0^2+y0^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r0G\"\" \"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "vr0:=0; 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hq$!3))***********)**>F]]l-Fehq6&FghqF[iqF[iqFhhq-%*THICKNESSG6#\"\"$- %&TITLEG6#Q0Effective~Force6\"-%+AXESLABELSG6%Q\"rFjjuQ!6\"%(DEFAULTG- %%VIEWG6$;F[iq$\"\"&Fahq;$F3FahqF_hq" 1 2 0 1 10 3 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 " " {TEXT 268 25 "3) The scattering problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 257 "Let us return to the Kepler probl em, and consider positive-energy solutions. These correspond to orbits that pass by the center of attraction once only (true comets). These \+ solutions are also relevant for charged particle scattering (Rutherfor d scattering)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "Suppose a particle is very far away from the scattering \+ center so that we can ignore the attraction. The 'free' particle is ch aracterized by:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "the asymptotic velocity vector, and associated kinetic en ergy;" }}{PARA 0 "" 0 "" {TEXT -1 118 "position vector, which together with the velocity vector and the particle mass determines the angular momentum vector." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "To illustrate the idea we use Maple to generate some nume rical solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart ; with(linalg):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names norm and trace have been redefined and unprotected\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "m:=1: r0:=[-100,5,0]: v0:=[1 ,0,0]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L0:=evalm(m*cross prod(r0,v0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L0G-%'vectorG6#7% \"\"!F)!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Lsq:=dotprod (L0,L0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$LsqG\"#D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "n:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "V:=-1 /r^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&\"\"\"F'%\"rG!\"\"F )" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "V:=subs(r=sqrt(x^2+y^2 ),V);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&\"\"\"F'*$-%%sqrtG6 #,&*$)%\"xG\"\"#F'F'*$)%\"yGF0F'F'F'!\"\"F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "T0:=m/2*dotprod(v0,v0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#T0G#\"\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "E0:=evalf(T0+subs(x=r0[1],y=r0[2],V));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#E0G$\"+mZ7+\\!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "We observe that the particle is not entirely 'free' at a distance of 100 units. In fact, the inverse-distance-squared force la w does not fall quickly enough at large distances to really allow the \+ particle to be ever truly free at finite separations." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "F_x:=unapply(diff(-V,x),x,y);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$F_xGR6$%\"xG%\"yG6\"6$%)operatorG%& arrowGF),$*&9$\"\"\"*$),&*$)F/\"\"#F0F0*$)9%F6F0F0#\"\"$F6F0!\"\"F " 0 "" {MPLTEXT 1 0 29 "F_y:=unapply(diff(-V ,y),x,y):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "NE1:=m*diff(x( t),t$2)=F_x(x(t),y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NE1G/-%% diffG6$-%\"xG6#%\"tG-%\"$G6$F,\"\"#,$*&F)\"\"\"*$),&*$)F)F0F3F3*$)-%\" yGF+F0F3F3#\"\"$F0F3!\"\"F?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "NE2:=m*diff(y(t),t$2)=F_y(x(t),y(t)):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 54 "IC:=x(0)=r0[1],D(x)(0)=v0[1],y(0)=r0[2],D(y)(0)=v0[ 2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6&/-%\"xG6#\"\"!!$+\"/--% \"DG6#F(F)\"\"\"/-%\"yGF)\"\"&/--F/6#F4F)F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "sol:=dsolve(\{NE1,NE2,IC\},\{x(t),y(t)\},numeric,o utput=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "X: =subs(sol,x(t)): Y:=subs(sol,y(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "PL0:=plot([[0,0]],style=point,symbol=cross,color=red, symbolsize=20):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "PL1:=plo t(['X(t)','Y(t)',t=0..200],axes=boxed,color=green,thickness=3): plots[ display](PL1,PL0);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6'-%'CURVESG6%7U7$$!$+\"\"\"!$\"\"&F*7$$!+:%fRc*!\")$\"+9 0&***\\!\"*7$$!+&*\\R%=*F0$\"+1)>)**\\F37$$!+\\8Ld()F0$\"+!fh&**\\F37$ $!+U!)>F$)F0$\"+^^;**\\F37$$!+%=Y))*yF0$\"+qDh)*\\F37$$!+i&\\9](F0$\"+ ef$z*\\F37$$!+[bn*3(F0$\"+e)Rq*\\F37$$!+uvYjmF0$\"+K`'e*\\F37$$!+#eD#Q iF0$\"+MMR%*\\F37$$!+%>N.!eF0$\"+tl\\#*\\F37$$!+w==9aF0$\"+[2V!*\\F37$ $!+0x')y\\F0$\"+=;b()\\F37$$!+b!>5a%F0$\"+!R4R)\\F37$$!+wMA=TF0$\"+$\\ m%z\\F37$$!+f]TLPF0$\"+.*yV(\\F37$$!+s*3XF$F0$\"+=?`m\\F37$$!+l7L%)GF0 $\"+NEsd\\F37$$!+#G*HGCF0$\"+L2pV\\F37$$!+!o1?-#F0$\"+$=!)e#\\F37$$!+S R7s:F0$\"+%GGg*[F37$$!+!y2v8\"F0$\"+(>0w%[F37$$!+^TCEnF3$\"+:dpWZF37$$ !+5@0DXF3$\"+zAM_YF37$$!+KMbgAF3$\"+,m&3\\%F37$$\"+`(oG]#!#5$\"+\\*f$e TF37$$\"+rDOUFF3$\"+\\\\D&e$F37$$\"+>KlTvF3$\"+=d*[&>F37$$\"+#oSU9\"F0 $\"+Hfo&G%Fds7$$\"+_OX_:F0$!+B3!zA\"F37$$\"+:E_m>F0$!+01-MHF37$$\"+kV/ nBF0$!+pdh'f%F37$$\"+'F37$$\"+?`TwJF0$!+3\\_uzF37$$\"+` ]XcNF0$!+m$[_c*F37$$\"+E2%4'RF0$!+)*G-E6F07$$\"+*=SlK%F0$!+\"\\^$z7F07 $$\"+:DTDZF0$!+%\\OnW\"F07$$\"+[71+^F0$!+^8./;F07$$\"+8')4\"\\&F0$!+gz EoF07$$\"+pe>siF0$!+j3Z'4#F07$$\"+dlOcmF0$!+ 4[%zD#F07$$\"+(=&*)[qF0$!+v%fHU#F07$$\"+(RqyV(F0$!+&f.le#F07$$\"+l&f]z (F0$!+62qOFF07$$\"+ba=/#)F0$!+WYv3HF07$$\"+$o/*p&)F0$!+#3oD1$F07$$\"+# >T'f*)F0$!+y]\\EKF07$$\"+D\\\\K$*F0$!+Y:L$Q$F07$$\"+nM+L(*F0$!+r2\"=b$ F0-%'COLOURG6&%$RGBG$F*F*$\"*++++\"F0Fh[l-%*THICKNESSG6#\"\"$-F$6&7#7$ Fh[lFh[l-Fe[l6&Fg[lFi[lFh[lFh[l-%'SYMBOLG6$%&CROSSG\"#?-%&STYLEG6#%&PO INTG-%+AXESLABELSG6%Q!6\"Fa]l%(DEFAULTG-%*AXESSTYLEG6#%$BOXG-%%VIEWG6$ Fc]lFc]l" 1 2 0 1 10 0 2 9 1 2 2 1.000000 45.000000 45.000000 0 0 "Cur ve 1" "Curve 2" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "PL1:=pl ot([seq([X(4*i),Y(4*i)],i=1..50)],style=point,symbol=diamond,axes=boxe d,color=blue,thickness=3): plots[display](PL1,PL0);" }}{PARA 13 "" 1 " " {GLPLOT2D 400 300 300 {PLOTDATA 2 "6'-%'CURVESG6'7T7$$!3Oc:9;\"=**f* !#;$\"33:VL$\\e***\\!#<7$$!3%Q*zvsJm*>*F*$\"3eGLHtn#)**\\F-7$$!3gPH(o$ )>#*z)F*$\"3y@J3pDf**\\F-7$$!3'Ro&oP1d)R)F*$\"3iN^Rv7C**\\F-7$$!3_Mm3 \\bp(*zF*$\"3ebwno`v)*\\F-7$$!35KF\"\\Yrlf(F*$\"3%*znzQO6)*\\F-7$$!3W% Ht?`r^>(F*$\"315L7k,H(*\\F-7$$!3=3;o#HkMz'F*$\"3=Zo3CHD'*\\F-7$$!3#*)R Obc79R'F*$\"3#4B^+*='\\*\\F-7$$!3_/iul>(*))fF*$\"3=3V6]jO$*\\F-7$$!3Cv uG7*)3'e&F*$\"3EHG4$3,9*\\F-7$$!3yOH=&*yp#=&F*$\"3)p6$*)e2)*))\\F-7$$! 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