Python \u3067\u6955\u5186\u8ecc\u9053\u4e0a\u306e\u6642\u523b\u3054\u3068\u306e\u4f4d\u7f6e\u3092\u6570\u5024\u7684\u306b\u6c42\u3081\u3066\u30a2\u30cb\u30e1\u3092\u3064\u304f\u308b<\/a><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\u3053\u308c\u3092\u5fdc\u7528\u3057\u3066\uff0c\u8fd1\u70b9\u79fb\u52d5\u304c\u3042\u308b\u5834\u5408\u306e\u30a2\u30cb\u30e1\u30fc\u30b7\u30e7\u30f3\u3092\u4f5c\u6210\u3059\u308b\u3002<\/p>\n
\u5468\u671f<\/h3>\n
${\\color{blue}{\\Phi}} \\equiv \\gamma \\phi$ \u3068\u304a\u304f\u3068\uff0c<\/p>\n
$$r({\\color{blue}{\\Phi}}) =\u00a0 \\frac{a(1-e^2)}{1 + e \\cos({\\color{blue}{\\Phi}}) }$$<\/p>\n
${\\color{blue}{\\Phi}} = 0$ \u306e\u3068\u304d\u306b $r$ \u304c\u6700\u5c0f\u5024 $r_{\\rm min} = a(1-e)$\u3002<\/p>\n
\u6b21\u306b\u518d\u3073 $r=r_{\\rm min}$ \u306b\u306a\u308b\u307e\u3067\u306e\u56fa\u6709\u6642\u9593\u9593\u9694\u3092 $P_{\\tau}$ \u3068\u3059\u308b\u3068\uff0c<\/p>\n
\\begin{eqnarray}
\nr^2({\\color{blue}{\\Phi}}) \\frac{d{\\color{blue}{\\Phi}} }{d\\tau} &=& \\gamma \\ell \\\\
\n\\int_0^{2\\pi} r^2( {\\color{blue}{\\Phi}} ) \\,d {\\color{blue}{\\Phi}} &=& \\gamma \\ell \\int_0^{P_{\\tau}} d\\tau \\\\
\n2 \\pi a^2 \\sqrt{1-e^2} &=& \\gamma \\ell P_{\\tau} \\\\
\n\\therefore \\ \\ P_{\\tau} &=& \\frac{2 \\pi a^2 \\sqrt{1-e^2}}{\\gamma \\ell}
\n\\end{eqnarray}<\/p>\n
\u7121\u6b21\u5143\u5316<\/h3>\n
$P_{\\tau}$\u3092\u4f7f\u3063\u3066\u7121\u6b21\u5143\u5316\u3055\u308c\u305f\u56fa\u6709\u6642\u9593 $T$ \u3092<\/p>\n
$$T \\equiv \\frac{\\tau}{P_{\\tau}}$$<\/p>\n
\u3068\u3059\u308b\u3068\uff0c<\/p>\n
$$\\frac{d{\\color{blue}{\\Phi}} }{dT} = 2 \\pi \\frac{(1+e \\cos\\phi)^2}{(1-e^2)^{3\/2}} $$<\/p>\n
\u3053\u308c\u3092\u6570\u5024\u7684\u306b\u89e3\u304d\uff0c${\\color{blue}{\\Phi}} (T)$ \u304c\u308f\u304b\u3063\u305f\u3089\uff0c$a$ \u3067\u898f\u683c\u5316\u3057\u305f\u5ea7\u6a19<\/p>\n
\\begin{eqnarray}
\nR(T) &\\equiv& \\frac{r}{a} = \\frac{1-e^2}{1+e\\cos{\\color{blue}{\\Phi}}(T)} \\\\
\nX(T) &\\equiv& \\frac{r}{a} \\cos\\phi \\\\
\n&=& \\frac{1-e^2}{1+e\\cos{\\color{blue}{\\Phi}}(T)}\\cos\\frac{{\\color{blue}{\\Phi}}(T)}{1 – \\Delta}\u00a0 \\\\
\n&\\simeq&\\frac{1-e^2}{1+e\\cos{\\color{blue}{\\Phi}}(T)}\\cos\\left(\\left(1+\\Delta\\right) {\\color{blue}{\\Phi}}(T)\\right)\u00a0 \\\\
\nY(T) &\\equiv& \\frac{r}{a} \\sin\\phi \\\\
\n&=& \\frac{1-e^2}{1+e\\cos{\\color{blue}{\\Phi}}(T)}\\sin\\frac{{\\color{blue}{\\Phi}}(T)}{1 – \\Delta} \\\\
\n&\\simeq&\\frac{1-e^2}{1+e\\cos{\\color{blue}{\\Phi}}(T)}\\sin\\left(\\left(1+\\Delta\\right) {\\color{blue}{\\Phi}}(T)\\right)
\n\\end{eqnarray}<\/p>\n
\u3067\u30b0\u30e9\u30d5\u306b\u3059\u308b\u3002\uff08\u53b3\u5bc6\u306b\u8a00\u3048\u3070\uff0c\u30b7\u30e5\u30d0\u30eb\u30c4\u30b7\u30eb\u30c8\u6642\u7a7a\u306a\u306e\u3067\u5e73\u5766\u3067\u306f\u306a\u3044\u304c\uff0c\u8fd1\u70b9\u79fb\u52d5\u3092\u8996\u899a\u7684\u306b\u8868\u73fe\u3059\u308b\u306e\u306b\u306f $x = r \\cos\\phi, \\ y = r \\sin\\phi$ \u3068\u3057\u305f\u30c7\u30ab\u30eb\u30c8\u5ea7\u6a19\u3067\u63cf\u3044\u3066\u3088\u308d\u3057\u3044\u304b\u3068\u3002\uff09<\/p>\n
Python \u3067\u30a2\u30cb\u30e1\u30fc\u30b7\u30e7\u30f3\u3092\u4f5c\u6210\u3059\u308b<\/h3>\n\n
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\u7121\u6b21\u5143\u5316\u3055\u308c\u305f\u5f0f<\/h3>\n
scipy.integrate.solve_ivp()<\/code> \u306f $\\displaystyle \\frac{dy}{dt} = f(t, y)$ \u306e\u5f62\u306e\u5f0f\u3092\u89e3\u304f\u306e\u3067\uff0c\u5909\u6570\u3092\u63c3\u3048\u308b\u3002<\/p>\n$ {\\color{blue}{\\Phi}} = \\gamma\\phi \\rightarrow y, \\ T \\rightarrow t$<\/p>\n
\\begin{eqnarray}
\n\\frac{dy}{dt} &=& f(t, y) = 2 \\pi \\frac{(1+e \\cos y)^2}{(1-e^2)^{3\/2}} \\tag{1}\\\\
\nR(y) &\\equiv& \\frac{r}{a} = \\frac{1-e^2}{1+e\\cos y} \\tag{2}\\\\
\nX(y) &\\equiv& \\frac{x}{a} = R(y) \\cos \\left( \\left(1+\\Delta \\right) {y}\\right) \\tag{3}\\\\
\nY(y) &\\equiv& \\frac{y}{a} = R(y) \\sin \\left( \\left(1+\\Delta \\right) {y}\\right) \\tag{4}
\n\\end{eqnarray}<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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\u5fc5\u8981\u306a\u30e2\u30b8\u30e5\u30fc\u30eb\u306e import<\/h3>\n<\/div>\n<\/div>\n<\/div>\n