{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 "" -1 275 "" 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 "" 0 257 1 {CSTYLE "" -1 -1 "" 1 16 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {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$\"+_,X ')H!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Earth moves with about 30 km/s around the Sun." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 278 "Now we introduce a unit system in which the Earth 's orbit radius is of order unity, the masses are measured in multiple s of Sun's mass, and times are measured in terms of orbit times. We ca n pick three dimensionful quantities and set the scales as follows [re member: SI=MKS(A)]:" }}{PARA 0 "" 0 "" {TEXT -1 12 "M = M' * M_e" }} {PARA 0 "" 0 "" {TEXT -1 12 "T = T' * T_e" }}{PARA 0 "" 0 "" {TEXT -1 11 "X = X' * AU" }}{PARA 0 "" 0 "" {TEXT -1 78 "and with the substitut ion we express equations in terms of the new variables. " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 419 "A similar procedu re is used in modern physics to scale the atomic orbital problem; one \+ notable exception is that one uses the electron mass as a scale. In th e solar Kepler problem this is not done, because the planet's mass dro ps out of the equation of motion. In the quantum problem it does not, \+ because the force of attraction is electric, i.e., not proportional to the lighter particle's (i.e., the electron's) mass." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "What is our unit of vel ocity in these natural units for the Earth-around-the-Sun motion probl em?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "unitLength:=AU; # in troduced unit #1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+unitLengthG$\" +++++:\"\"#" }}}{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)$\"\"%F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 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attraction it moves faster (has higher kinetic energy) on accou nt of the fact that the potential energy has been lowered." }}{PARA 0 "" 0 "" {TEXT -1 218 "One can establish a relationship between the ave rage values over a period (this is called the virial theorem in classi cal mechanics, it's details depend on the force law, and it has a coun terpart in quantum mechanics)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 59 "We lift an approximate value for the peri od from the graph:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Perio d:=0.5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'PeriodG$\"\"&!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "T_avg:=evalf(Int(T_kin,0..Pe riod,5,_NCrule));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&T_avgG$\"&*yY! \"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "E_avg:=evalf(Int(E_t ot,0..Period,5,_NCrule));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_avgG $!&bn%!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "V_avg:=evalf( Int(V_pot,0..Period,5,_NCrule));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% &V_avgG$!&WN*!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V_avg/ E_avg;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+&>F2+#!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 269 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 167 "Carry out these calculations for \+ different initial conditions and verify that the relationship between \+ these average values is a feature of Newton's law of gravitation" }} {PARA 0 "" 0 "" {TEXT -1 144 "(inverse-squared force law). Ensure a co rrect choice of period by looking at the graph. Explore what happens f or slightly different power laws, " }}{PARA 0 "" 0 "" {TEXT -1 8 "such as " }{TEXT 271 1 "n" }{TEXT -1 11 "=12/10, or " }{TEXT 270 1 "n" } {TEXT -1 11 "=8/10, etc." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 257 "" 0 "" {TEXT 272 28 "2) The radial motion probl em" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "The radial problem for " }{TEXT 273 1 "r" }{TEXT -1 1 "(" }{TEXT 275 1 "t " }{TEXT -1 126 ") is derived from angular momentum conservation. It a llows us to determine the trajectory from a single differential equati on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Fi rst we calculate the angular momentum:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning , the protected names norm and trace have been redefined and unprotect ed\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "L0:=evalm(m*crosspr od([x0,y0,0],[vx0,vy0,0]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L0G- %'vectorG6#7%$\"\"!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 the calculated trajectory does observe t he conservation of angular momentum approximately. This is a test if t he numerical accuracy of the algorithm. We also see that the angular m omentum 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 s quared). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "L_mag:=norm(L0 ,frobenius);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&L_magG$\"+++++7!#9 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "The one-dimensional problem \+ of solving for the distance from the center of attraction as a functio n of time:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Vr:=-GM*m/r^n ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#VrG,$*&\"\"\"F'%\"rG!\"\"$!+++ 5v6!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Veff:=Vr+L_mag^2/ (2*m*r^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%VeffG,&*&\"\"\"F'%\"r G!\"\"$!+++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$-% 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These correspond to orbits that pass by the center of attraction once only (true comets). These \+ solutions are also relevant for charged particle scattering (Rutherfor d scattering)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "Suppose a particle is very far away from the scattering \+ center so that we can ignore the attraction. The 'free' particle is ch aracterized by:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "the asymptotic velocity vector, and associated kinetic en ergy;" }}{PARA 0 "" 0 "" {TEXT -1 118 "position vector, which together with the velocity vector and the particle mass determines the angular momentum vector." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "To illustrate the idea we use Maple to generate some nume rical solutions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart ; with(linalg):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names norm and trace have been redefined and unprotected\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "m:=1: r0:=[-100,5,0]: v0:=[1 ,0,0]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L0:=evalm(m*cross prod(r0,v0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L0G-%'vectorG6#7% \"\"!F)!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Lsq:=dotprod (L0,L0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$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-%&STYLEG6#%&POINTG-%'SYMBOLG6$%&CROSSG \"#?-%+AXESLABELSG6%Q!6\"Fa]l%(DEFAULTG-%*AXESSTYLEG6#%$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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