{VERSION 3 0 "APPLE_PPC_MAC" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 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 } {PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Warning " 2 7 1 {CSTYLE "" -1 -1 "" 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Author" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 16 "Kepler and Pluto" }}} {EXCHG {PARA 19 "" 0 "" {TEXT -1 33 "MTHSC 108H -- December 11, 1999 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "From the textbook, p. 695, th e orbit of Pluto has a semi-major axis of " }{XPPEDIT 18 0 "5.900*10^9 ;" "6#*&$\"%+f!\"$\"\"\"*$\"#5\"\"*F'" }{TEXT -1 92 " km and an eccent ricity of 0.2481.\nAccording to my HP48GX one astronomical unit is equ al to " }{XPPEDIT 18 0 "1.4960*10^8;" "6#*&$\"&g\\\"!\"%\"\"\"*$\"#5\" \")F'" }{TEXT -1 109 " km. Thus the book and the diagrom on\nthe hand out agree. The diagram from the handout is reproduced below.\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "5.900*10^9 / (1.4960*10^8); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+n-&Q%R!\")" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 1 "\n" }{METAFILE 264 148 148 1 ":::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::f ZjD:B_=B<@K;xY@vyyyyyA:<:FZBB:LS:B:Rb=bvLj`HoeL^[d_TJZ:B_=ZAZ:BZBB<`:<:RNl[DrWHh]Z[@ZpD ;@Z;Rb:@ZTZ:ZAR:@ZCvyyyyy=;BPXZ:B:H<@DZ;Rb:@ZAR:@ZZRLhhCS:XZ:B:H<@DZ;R b:@ZAR:@Z\\bBptsY\\[d_dj:>:?J:B::@ZDrW@f@ HZ@bBXA:@:`a>r[T^\\RXB:PkyHibrZL J>@Nr\\pll@J:R:J:Zb[TSNt[cDDHeAB:< Z]Z[:?Z:>:@ZDb?Le_brXZ:B:D\\PZ:R?we:\\\\ZBgEBNSUWc uUZ :B_ZHB<ZCc:?:;Z;b^mFah^`n>^;gD< T:Xc[d_J:N:>:@:H^msNdb\\rJB:l\\EB>KBN=EU=EWMeUYM:HZJR=tZOnittDmt<=Rc:<:XZ;R:\\_Z@ZGDb;Z[d_T: " 0 "" {MPLTEXT 1 0 19 "P := sqrt(39.44^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"+T A)oZ#!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The polar equation f or an ellipse with the orientation of the diagram is\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "assume(0 ecc*d/(1 + ecc*cos(theta));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"r GR6#%&thetaG6\"6$%)operatorG%&arrowGF(*&*&%$eccG\"\"\"%\"dG\"\"\"F/,&F 1F1*&F.F1-%$cosG6#9$F1F1!\"\"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "Now " }{XPPEDIT 18 0 "2*a = r(0)+r(Pi);" "6#/*&\"\"#\"\"\"%\"aGF &,&-%\"rG6#\"\"!F&-F*6#%#PiGF&" }{TEXT -1 19 ", so we can relate " } {XPPEDIT 18 0 "d;" "6#%\"dG" }{TEXT -1 43 " to the given mean distance to the sun. \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "assume (0%#dpG,$*&*(%#a|irG\" \"\",&F)F)%%ecc|irGF)F),&!\"\"F)F+F)F)\"\"\"F+!\"\"F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2GR6#%&thetaG6\"6$%)operatorG%&arrowGF(,$*&*(% #a|irG\"\"\",&F0F0%%ecc|irGF0F0,&!\"\"F0F2F0F0\"\"\",&F0F0*&F2F0-%$cos G6#9$F0F0!\"\"F4F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Persona lly, I prefer to write this formula as\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "r3 := theta -> a*(1-ecc^2)/(1+ecc*cos(theta));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r3GR6#%&thetaG6\"6$%)operatorG%&arr owGF(*&*&%\"aG\"\"\",&F/F/*$)%$eccG\"\"#\"\"\"!\"\"F/F5,&F/F/*&F3F/-%$ cosG6#9$F/F/!\"\"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Now le t's look at Pluto specifically." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 222 "ap := 39.44;\nep := 0.2481; \nrp := subs(a=ap, ecc=ep, r3(theta));\npd := evalf(subs(theta=0,rp)); \nad := evalf(subs(theta=Pi,rp));\nplot(rp,theta=0..2*Pi,coords=polar, title=\"The Orbit of Pluto\",\n titlefont=[TIMES,BOLD,14]);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#apG$\"%WR!\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#epG$\"%\"[#!\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %#rpG,$*&\"\"\"F',&\"\"\"F)-%$cosG6#%&thetaG$\"%\"[#!\"%!\"\"$\"+iDB,P !\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#pdG$\"++O\\lH!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#adG$\"++k]A\\!\")" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$\"10=()*f$\\lH!# 9\"\"!7$$\"1[k\"3v.K%HF*$\"1?G[WQFcS!#:7$$\"12\\_IH[()GF*$\"10\\i(e#[h vF17$$\"1$oda2]Uy#F*$\"1xj_$>6\\9\"F*7$$\"1,JVjzlOEF*$\"1,oZ9OKF:F*7$$ \"1fp`di*eW#F*$\"1$=\">=Z]&*=F*7$$\"1qZrUmjHAF*$\"1'Q9<>'QAAF*7$$\"1U. d%))ob'>F*$\"1[4sX;OUDF*7$$\"1iX!y_ F1$\"1\\L'=o+]z$F*7$$!1.f=B=B[5F*$\"1RM;q94?QF*7$$!1_%p&\\euS:F*$\"1a( QR3n;y$F*7$$!1\"ROWpKJ8#F*$\"1WcpSWH`OF*7$$!1g,xOULHEF*$\"1xt,fd*)pMF* 7$$!1eNV7#HR=$F*$\"1&3v3g;v;$F*7$$!1m$HnB'=ROF*$\"1/&QhJ=.#GF*7$$!1\\* R]yq>3%F*$\"1&ptO1exN#F*7$$!1J\"zi`z9V%F*$\"1_B._RAY=F*7$$!1_1/%H1Zq%F *$\"1nG5IX4_7F*7$$!1q\\%HNF='[F*$\"1&3Jai?in'F17$$!1H!ze\"\\ZA\\F*$\"1 'Q)GACiE:!#;7$$!1\"4G0cVF'[F*$!1?^tTP,EmF17$$!1`c3IaR6ZF*$!1'\\k;maKB \"F*7$$!1i`'*)**G;X%F*$!1Jkl$p2/\"=F*7$$!1G*)GO$344%F*$!1/cudfnYBF*7$$ !1\"y*p(f$>jOF*$!1Qil>m)))z#F*7$$!1`\"eF/$F*$!19HtWg\"*fJF*7$$!1cc/ rR*Rj#F*$!1R=E`<\"yY$F*7$$!1Jyxnd?3@F*$!1$=kCd+1m$F*7$$!1=xD$Q)fS:F*$! 1=l?cwo\"y$F*7$$!1q*>t.UB.\"F*$!1)H6EiK.#QF*7$$!19(4V4U'Q\\F1$!1(R+O@L uT@$ycNF*7$$ \"1W1_P`1$*))F1$!1RI!>cn]O$F*7$$\"1KTu[&\\jH\"F*$!1J;q5V4@JF*7$$\"1!*z Z#H\\tk\"F*$!1tnP!z#y]GF*7$$\"1yNzRgej>F*$!1(R-A#[^WDF*7$$\"1YL'fe8TB# F*$!1%e0ws3j@#F*7$$\"1EPToA%[W#F*$!1+0zB0H(*=F*7$$\"1'pysV!*>k#F*$!1#> id^Ba^\"F*7$$\"1g+fS@ZyFF*$!1mEI4:ei6F*7$$\"1c-vu.)G)GF*$!11&Q$4J0zxF1 7$$\"1eRg1_MVHF*$!1=B!*e(=M/%F17$F($!1/*4(Rt`CP!#A-%'COLOURG6&%$RGBG$ \"#5!\"\"F+F+-%&TITLEG6$Q3The~Orbit~of~Pluto6\"-%%FONTG6%%&TIMESG%%BOL DG\"#9-%%VIEWG6$%(DEFAULTGF]\\l" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "We can \+ determine the area of an ellipse by two different approaches. \nWe ca n start with the Cartesian description " }{XPPEDIT 18 0 "((x+c)/a)^2+( y/b)^2 = 1;" "6#/,&*$*&,&%\"xG\"\"\"%\"cGF)F)%\"aG!\"\"\"\"#F)*$*&%\"y GF)%\"bGF,\"\"#F)\"\"\"" }{TEXT -1 121 ", then \nset up the area integ ral for the upper half of the ellipse, and finally make a few simple c hanges of variable. \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 " A1 := 2*(b/a)*Int(sqrt(a^2-(x+c)^2),x=-(a+c)..(a-c));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A1G,$*&*&%\"bG\"\"\"-%$IntG6$*$-%%sqrtG6#,**$)% #a|irG\"\"#\"\"\"F)*$)%\"xGF5F6!\"\"*&F9F)%\"cGF)!\"#*$)F " 0 "" {MPLTEXT 1 0 146 "with(student):\nA2 := simpl ify(changevar(x=u-c,A1,u));\nA3 := simplify(changevar(u=a*sin(phi),A2, phi));\nA4 := subs(cos(phi)^2=(1+cos(2*phi))/2, A3);" }}{PARA 7 "" 1 " " {TEXT -1 29 "Warning, new definition for D" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G,$*&*&%\"bG\"\"\"-%$IntG6$*$-%%sqrtG6#,&*$)%#a|ir G\"\"#\"\"\"F)*$)%\"uGF5F6!\"\"F6/F9;,$F4F:F4F)F6F4!\"\"F5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A3G,$*(%\"bG\"\"\"%#a|irGF(-%$IntG6$*&-%% csgnG6#-%$cosG6#%$phiGF()F1\"\"#\"\"\"/F4;,$%#PiG#!\"\"F6,$F;#F(F6F(F6 " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A4G,$*(%\"bG\"\"\"%#a|irGF(-%$I ntG6$*&-%%csgnG6#-%$cosG6#%$phiGF(,&#F(\"\"#F(-F26#,$F4F7F6F(/F4;,$%#P iG#!\"\"F7,$F>F6F(F7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "For the " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 18 " in the range of " } {XPPEDIT 18 0 "-Pi/2;" "6#,$*&%#PiG\"\"\"\"\"#!\"\"F(" }{TEXT -1 5 " \+ to " }{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 15 " the function " }{XPPEDIT 18 0 "csgn(cos(phi));" "6#-%%csgnG6#-%$cosG 6#%$phiG" }{TEXT -1 169 " is simply equal to 1. Integrating this giv es us\nthe standard formula for the area of an ellipse. Notice that i t reduces to the formula for the area of a circle when\n" }{XPPEDIT 18 0 "a = b;" "6#/%\"aG%\"bG" }{TEXT -1 4 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "ellipseArea := 2*a*b*int((1+cos(2*phi))/2,phi=-Pi/2..Pi/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,ellipseAreaG*(%#a|irG\"\"\"%\"bGF'%#PiGF'" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "The second approach is to integrat e the ellipse using the area element for polar coordinates.\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "A5 := 2*Int((1/2)*r3(theta)^ 2,theta=0..Pi);\nA6 := simplify(A5);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%#A5G,$-%$IntG6$,$*&*&)%#a|irG\"\"#\"\"\"),&\"\"\"F2*$)%%ecc|irGF.F /!\"\"F.F/F/*$),&F2F2*&F5F2-%$cosG6#%&thetaGF2F2\"\"#F/!\"\"#F2F./F>; \"\"!%#PiGF." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A6G*()%#a|irG\"\"# \"\"\",(\"\"\"F+*$)%%ecc|irGF(F)!\"#*$)F.\"\"%F)F+F+-%$IntG6$*&F)F),(F +F+*&F.F+-%$cosG6#%&thetaGF+F(*&F-F))F9F(F)F+!\"\"/F<;\"\"!%#PiGF+" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "a^2*(1-2*ecc^2+ecc^4) * in t(1/(1+2*ecc*cos(theta)+ecc^2*cos(theta)^2), theta = 0 .. Pi);\nellips eArea2 := simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*()%#a|i rG\"\"#\"\"\",(\"\"\"F+*$)%%ecc|irGF(F)!\"#*$)F.\"\"%F)F+F+%#PiGF+F)*( ,&!\"\"F+F.F+\"\"\",&F+F+F.F+\"\"\"-%%sqrtG6#,&F+F+F,F6F)!\"\"F6" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%-ellipseArea2G,$*&*()%#a|irG\"\"#\" \"\",&!\"\"\"\"\"*$)%%ecc|irGF*F+F.F.%#PiGF.F+*$-%%sqrtG6#,&F.F.F/F-F+ !\"\"F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "We can also obtain th e distance from the focus to the center of the ellipse by subtracting \+ the perihelion\ndistance from " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 47 ". This in turn gives us the semi-minor axis, " }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 42 ", and shows that the area of the ellips e, " }{XPPEDIT 18 0 "Pi*a*b;" "6#*(%#PiG\"\"\"%\"aGF%%\"bGF%" }{TEXT -1 52 ",\nis the same as the area from the polar formula. \n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "c := simplify(a-r3(0));\nb : = simplify(sqrt(a^2 - c^2));\ntotalArea := ellipseArea;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%\"cG*&%%ecc|irG\"\"\"%#a|irGF'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG%#bbG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%* totalAreaG*()%#a|irG\"\"#\"\"\"-%%sqrtG6#,&\"\"\"F.*$)%%ecc|irGF(F)!\" \"F)%#PiGF." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "We now define the \+ function which returns the area of a sector from " }{XPPEDIT 18 0 "alp ha;" "6#%&alphaG" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "beta;" "6#%%betaG " }{TEXT -1 5 ".\n " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "Ar ea := unapply((1/2)*Int(r3(t)^2,t=alpha..beta),(alpha,beta));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%AreaGR6$%&alphaG%%betaG6\"6$%)opera torG%&arrowGF),$-%$IntG6$*&*&)%#a|irG\"\"#\"\"\"),&\"\"\"F9*$)%%ecc|ir GF5F6!\"\"F5F6F6*$),&F9F9*&F " 0 "" {MPLTEXT 1 0 136 "cp := subs(a=ap, ecc=ep, c);\nbp := subs(a=ap, c=cp, ecc=ep, b);\nAp := una pply(evalf(subs(a=ap, ecc=ep, Area(alpha,beta))),(alpha,beta));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#cpG$\"(k]y*!\"'" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%#bpG$\"+!e)o?Q!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ApGR6$%&alphaG%%betaG6\"6$%)operatorG%&arrowGF),$-%$IntG6$,$*&\" \"\"F3*$),&$\"\"\"\"\"!F8-%$cosG6#%\"tG$\"%\"[#!\"%\"\"#F3!\"\"$\"+[A \"*p8!\"'/F=;9$9%$\"+++++]!#5F)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 230 "We are now in a position to solve for the angle that Pluto wil l be at when it has completed 1/8 of an Pluto year\nstarting either at the perihelion or at the aphelion. Note that we need to use fsolve, \+ the numerical solve routine.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 177 "evalf(Pi*ap*bp/8);\nt0 := fsolve(Ap(0,theta)=Pi*ap*bp/8,theta );\nevalf(Ap(0,t0));\nt1 := fsolve(Ap(Pi,theta)=Pi*ap*bp/8,theta);\nev alf(Ap(Pi,t1));\nevalf(t1-Pi);\ntimeP := evalf(P/8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+eA]%# t0G$\"+by0>7!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+bA]%#t1G$\"+$)>CVO!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+bA]%&timePG$\"+,G5'4$!\") " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 566 "Starting at the perihelion, \+ Pluto sweeps out a sector with an angle of 1.21906 radians or 69.85 de grees and the\narea of the sector is 1/8 of the entire area of the ell ipse. Similarly, starting at the aphelion, Pluto sweeps out\na sector with an angle of 0.50165 radians or 28.74 degrees. Each sector has t he same area, and by Kepler's\nSecond Law, each sector will be swept o ut in the same amount of time - about 30.96 earth years.\n\nThe distan ce traveled between 0 and t0 (or Pi and t1) is simply the arclength fr om 0 to t0 (or Pi to t1).\nThe arc length integral is " }{TEXT 257 3 " not" }{TEXT -1 238 " an elementary integral. Maple will provide an an swer involving the Elliptic\nintegral functions. It is so involved as to be worthless, and a numerical approach is much simpler.\nTo see th is, simply change Int to int in the definition of " }{XPPEDIT 18 0 "s; " "6#%\"sG" }{TEXT -1 115 ". Scroll through the lengthy result\nand n otice the occurence of the special functions EllipticE and EllipticF. \+ \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "dr := diff(r3(theta ),theta);\ns := Int(sqrt(r3(theta)^2 + dr^2),theta=alpha..beta);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#drG*&**%#a|irG\"\"\",&F(F(*$)%%ecc| irG\"\"#\"\"\"!\"\"F(F,F.-%$sinG6#%&thetaGF(F.*$),&F(F(*&F,F(-%$cosGF2 F(F(\"\"#F.!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sG-%$IntG6$*$- %%sqrtG6#,&*&*&)%#a|irG\"\"#\"\"\"),&\"\"\"F5*$)%%ecc|irGF1F2!\"\"F1F2 F2*$),&F5F5*&F8F5-%$cosG6#%&thetaGF5F5\"\"#F2!\"\"F5*&**F/F2F3F2F7F2)- %$sinGF@F1F2F2*$)F<\"\"%F2FCF5F2/FA;%&alphaG%%betaG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Specializing to Pluto yields.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "sp := unapply(subs(a=ap, ecc=ep, s) , (alpha,beta));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#spGR6$%&alphaG% %betaG6\"6$%)operatorG%&arrowGF)-%$IntG6$*$-%%sqrtG6#,&*&\"\"\"F6*$),& \"\"\"F:-%$cosG6#%&thetaG$\"%\"[#!\"%\"\"#F6!\"\"$\"+[A\"*p8!\"'*&*$)- %$sinGF=\"\"#F6F6*$)F9\"\"%F6FC$\"+DWIK%)!\")F6/F>;9$9%F)F)F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Now we can answer the questions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "distance1 := evalf(sp(0,t 0));\navgSpeed1 := distance1 / timeP;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*distance1G$\"+[&f#GQ!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*a vgSpeed1G$\"+&)oZO7!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "So we \+ see that starting at the perihelion in " }{XPPEDIT 18 0 "1/8;" "6#*&\" \"\"\"\"\"\"\")!\"\"" }{TEXT -1 133 " of a Plutonian year or 30.96 ear th years, the planet will travel a distance\nof 38.2826 AU at an avera ge speed of 1.2365 AU/year.\n " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "distance2 := evalf(sp(Pi,t1));\navgSpeed2 := distance 2 / timeP;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*distance2G$\"+PfPYC! \")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*avgSpeed2G$\"+*po9!z!#5" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 211 "And we see that starting at the a phelion, Pluto only travels a distance of 24.4638 AU at an average spe ed of 0.7901 AU/year.\n\nTo convert to convert to km and km/hour we ne ed the following two conversion factors\n" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 45 "toKm := 1.4960*10^8;\ntoKmpH := toKm/(365*24);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%toKmG$\"++++'\\\"!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'toKmpHG$\"+dDw2 " 0 "" {MPLTEXT 1 0 33 "distance1*toKm;\navgSpeed1*toKmpH; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+%Gwqs&\"\"!" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#$\"+F*3;6#!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "When starting at the perihelion, Pluto travels " }{XPPEDIT 18 0 "3 .6598*10^9;" "6#*&$\"&)fO!\"%\"\"\"*$\"#5\"\"*F'" }{TEXT -1 41 " km at an average speed 13,494 km/hour .\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "distance2*toKm;\navgSpeed2*toKmpH;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#$\"+-%y(fO\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$ \"+RKQ\\8!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "By comparison, i f we simply treat the earth's orbit as circular, then the earth travel s " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 58 " AU \+ in one year at an\naverage speed of 107,301 km/hour. \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "avgEarthSpeed := evalf(2*Pi*toKmpH) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.avgEarthSpeedG$\"+h)=I2\"!\"% " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 270 "By the way, Pluto was not di scovered until 1930 by C. W. Tombaugh. (It's existence had been conje ctured by\nPercival Lowell in 1905.) Hence Kepler was stuck with Mars and its much more circular orbit -- which made\nTycho Brahe's precise measurements even more important." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "33 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }