{"id":25606,"date":"2026-08-25T11:40:00","date_gmt":"2026-08-25T16:40:00","guid":{"rendered":"https:\/\/blog.espol.edu.ec\/algoritmos101\/?p=25606"},"modified":"2026-08-28T21:24:58","modified_gmt":"2026-08-29T02:24:58","slug":"s2eva2026paoi_t2-edo-problema-dos-cuerpos","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2026paoi_t2-edo-problema-dos-cuerpos\/","title":{"rendered":"s2Eva2026PAOI_T2 EDO Problema de dos cuerpos"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2026PAOI_T2 EDO Problema de dos cuerpos<\/a><\/p>\n\n\n\n

literal a<\/h2>\n\n\n\n

Simplificar las expresiones:<\/p>\n\n\n\n

\n

G = 4.982 x 10-19<\/sup><\/p>\n\n\n\n

m1<\/sub> = 5.9722 x 1024<\/sup><\/p>\n\n\n\n

m2<\/sub> = 7.3490 x 1022<\/sup> <\/p>\n<\/div>\n\n\n\n \\frac{d^2 x(t)}{dt^2}=\\frac{G m_1}{\\left( \\sqrt{x(t)^2+y(t)^2}\\right)^3}\\left(0-x(t)\\right)<\/span>\n\n\n\n \\frac{d^2 y(t)}{dt^2}=\\frac{G m_1}{\\left( \\sqrt{x(t)^2+y(t)^2}\\right)^3}\\left(0-y(t)\\right)<\/span>\n\n\n\n

Se empieza reemplazando las constantes para simplificar la ecuaci\u00f3n..<\/p>\n\n\n\n \\frac{d^2 x}{dt^2}=\\frac{4.982 ( 10^{-19}) 5.9722 (10^{24})}{\\left( \\sqrt{x^2+y^2}\\right)^3}\\left(-x\\right)<\/span>\n\n\n\n \\frac{d^2 y}{dt^2}=\\frac{4.982 (10^{-19}) 5.9722 x (10^{24})}{\\left( \\sqrt{x^2+y^2}\\right)^3}\\left(-y\\right)<\/span>\n\n\n\n

quedando de la siguiente forma:<\/p>\n\n\n\n

\n\n\n\n \\frac{d^2 x}{dt^2}=-2.9754(10^6)\\frac{x}{\\left( \\sqrt{x^2+y^2}\\right)^3} <\/span>\n\n\n\n \\frac{d^2 y}{dt^2}=-2.9754(10^6) \\frac{y}{\\left( \\sqrt{x^2+y^2}\\right)^3}<\/span>\n\n\n\n

Condiciones iniciales cuando t0<\/sub> =0<\/p>\n\n\n\n

\n

x0<\/sub> = 384.4<\/p>\n\n\n\n

y0<\/sub> = 0<\/p>\n\n\n\n

vx0<\/sub> = 0<\/p>\n\n\n\n

vy0<\/sub> = 88.128<\/p>\n<\/div>\n\n\n\n

Genera la tabla a desarrollar:<\/p>\n\n\n\n

ti<\/sub><\/th>xi<\/sub><\/th>yi<\/sub><\/th>vxi<\/sub><\/th>vyi<\/sub><\/th><\/tr><\/thead>
0<\/td>384.4<\/td>0<\/td>0<\/td>88.128<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

literal b<\/h2>\n\n\n\n

El sistema de ecuaciones es de 2da derivada, por lo que se aplica el m\u00e9todo de Runge-Kutta de 2do orden a cada expresi\u00f3n.<\/p>\n\n\n\n

\n
\n f_x(t,x,y,vx,vy) = v_x <\/span>\n\n\n\n f_y(t,x,y,vx,vy)= v_y<\/span>\n<\/div>\n\n\n\n
\n g_x(t,x,y,vx,vy)=-2.9754(10^6)\\frac{x}{\\left( \\sqrt{x^2+y^2}\\right)^3} <\/span>\n\n\n\n g_y(t,x,y,vx,vy) = -2.9754(10^6) \\frac{y}{\\left( \\sqrt{x^2+y^2}\\right)^3} <\/span>\n<\/div>\n<\/div>\n\n\n\n

Los pasos del m\u00e9todo se adaptan al ejercicio:<\/p>\n\n\n\n

\n
K1_{x} = h (vx) <\/span>\n\n\n K1_{y} = h (vy) <\/span>\n\n\n K1_{vx} = h \\left( -2.9754(10^6)\\frac{x}{\\left( \\sqrt{x^2+y^2}\\right)^3} \\right)<\/span>\n\n\n K1_{vy} = h \\left( -2.9754(10^6) \\frac{y}{\\left( \\sqrt{x^2+y^2}\\right)^3} \\right)<\/span>\n<\/div>\n\n\n\n
K2_{x} = h (vx+K1_{vx}) <\/span>\n\n\n K2_{y} = h (vy+K1_{vy}) <\/span>\n\n\n K2_{vx} = h \\left( -2.9754(10^6)\\frac{x+K1_x}{\\left( \\sqrt{(x+K1_x)^2+(y+K1_y)^2}\\right)^3} \\right)<\/span>\n\n\n K2_{vy} = h \\left( -2.9754(10^6) \\frac{y+K1_y}{\\left( \\sqrt{(x+K1_x)^2+(y+K1_y)^2}\\right)^3} \\right)<\/span>\n<\/div>\n<\/div>\n\n\n\n
\n
x_{i+1}=x_i+\\frac{K1_{x}+K2_{x}}{2}<\/span>\n\n\n y_{i+1}=y_i+\\frac{K1_{y}+K2_{y}}{2}<\/span>\n<\/div>\n\n\n\n
vx_{i+1}=vx_i+\\frac{K1_{vx}+K2_{vx}}{2}<\/span>\n\n\n vy_{i+1}=vy_i+\\frac{K1_{vy}+K2_{vy}}{2}<\/span>\n<\/div>\n<\/div>\n\n\n t_{i+1}=t_i+h<\/span>\n\n\n\n

literal c<\/h2>\n\n\n\n
ti<\/sub><\/th>xi<\/sub><\/th>yi<\/sub><\/th>vxi<\/sub><\/th>vyi<\/sub><\/th><\/tr><\/thead>
0<\/td>384.4<\/td>0<\/td>0<\/td>88.128<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

itera =0<\/p>\n\n\n\n

\n
K1_{x} = 0.1 (0) = 0 <\/span>\n\n\n K1_{y} = 0.1 (88.128) =8.8128 <\/span>\n\n\n K1_{vx} = 0.1 \\left( -2.9754(10^6)\\frac{384.4}{\\left( \\sqrt{384.4^2+0^2}\\right)^3} \\right) =-2.0136 <\/span>\n\n\n K1_{vy} = 0.1 \\left( -2.9754(10^6) \\frac{0}{\\left( \\sqrt{384.4^2+0^2}\\right)^3} \\right) = 0<\/span>\n<\/div>\n\n\n\n
K2_{x} = 0.1 (-2.0136) = -0.2014<\/span>\n\n\n K2_{y} = 0.1 (88.128+0) = 8.8128 <\/span>\n\n\n K2_{vx} = 0.1 \\left( -2.9754(10^6)\\frac{384.4+0}{\\left( \\sqrt{(384.4+0)^2+(0+8.8128)^2}\\right)^3} \\right) = -2.0121 <\/span>\n\n\n K2_{vy} = 0.1 \\left( -2.9754(10^6) \\frac{0+8.8128}{\\left( \\sqrt{(384.4+0)^2+(0+8.8128)^2}\\right)^3} \\right) =-0.0461<\/span>\n<\/div>\n<\/div>\n\n\n\n

actualiza las variables de la iteraci\u00f3n:<\/p>\n\n\n\n

\n
x_{1}=384.4+\\frac{0-0.2014}{2} =384.30<\/span>\n\n\n y_{1}=0+\\frac{8.8128+8.8128}{2} = 8.8128<\/span>\n<\/div>\n\n\n\n
vx_{1}=0+\\frac{-2.0136-2.0121}{2} = -2.0128 <\/span>\n\n\n vy_{1}=88.128+\\frac{0+-0.0461}{2} = 88.105<\/span>\n<\/div>\n<\/div>\n\n\n t_{1}=0+0.1 = 0.1<\/span>\n\n\n\n
ti<\/sub><\/th>xi<\/sub><\/th>yi<\/sub><\/th>vxi<\/sub><\/th>vyi<\/sub><\/th><\/tr><\/thead>
0<\/td>384.4<\/td>0<\/td>0<\/td>88.128<\/td><\/tr>
0.1<\/td>384.3<\/td>8.8128<\/td>-2.0128<\/td>88.105<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

itera=1<\/p>\n\n\n\n

\n
K1_{x} = 0.1 (-2.0128) = -0.2013 <\/span>\n\n\n K1_{y} = 0.1 (88.105) = 8.8105 <\/span>\n\n\n K1_{vx} = 0.1 \\left( -2.9754(10^6)\\frac{384.3}{\\left( \\sqrt{384.3^2+8.8128^2}\\right)^3} \\right) =-2.0131<\/span>\n\n\n K1_{vy} = 0.1 \\left( -2.9754(10^6) \\frac{8.8128}{\\left( \\sqrt{384.3^2+8.8128^2}\\right)^3} \\right) = -0.0462 <\/span>\n<\/div>\n\n\n\n
K2_{x} = 0.1 (-2.0128-2.0131) = -0.4026 <\/span>\n\n\n K2_{y} = 0.1 (88.105-0.0462) = 8.8059<\/span>\n\n\n K2_{vx} = 0.1 \\left( -2.9754(10^6)\\frac{384.3-0.2013}{\\left( \\sqrt{(384.3-0.2013)^2+(8.8128+8.8105)^2}\\right)^3} \\right) = -2.0105<\/span>\n\n\n K2_{vy} = 0.1 \\left( -2.9754(10^6) \\frac{8.8128+8.8105}{\\left( \\sqrt{(384.3-0.2013)^2+(8.8128+8.8105)^2}\\right)^3} \\right) = -0.0922<\/span>\n<\/div>\n<\/div>\n\n\n\n
\n
x_{2}=384.3+\\frac{-0.2013-0.4026}{2} = 384.0<\/span>\n\n\n y_{2}=8.8128+\\frac{8.8105+8.8059}{2}= 17.621<\/span>\n<\/div>\n\n\n\n
vx_{2}=-2.0128+\\frac{-2.0131-2.0105}{2} = -4.0246<\/span>\n\n\n vy_{2}=88.105+\\frac{-0.0462-0.0922}{2}= 88.036<\/span>\n<\/div>\n<\/div>\n\n\n t_{2} = 0.1+0.1 = 0.2<\/span>\n\n\n\n
ti<\/sub><\/th>xi<\/sub><\/th>yi<\/sub><\/th>vxi<\/sub><\/th>vyi<\/sub><\/th><\/tr><\/thead>
0<\/td>384.4<\/td>0<\/td>0<\/td>88.128<\/td><\/tr>
0.1<\/td>384.3<\/td>8.8128<\/td>-2.0128<\/td>88.105<\/td><\/tr>
0.2<\/td>384.0<\/td>17.621<\/td>-4.0246<\/td>88.036<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

literal d<\/h2>\n\n\n\n

Algoritmo en Python<\/p>\n\n\n

\n# 2Eva2026PAOI_T2 EDO Problema de dos cuerpos\n# Trayectoria dos cuerpos en espacio\nimport numpy as np\n\nG  = 6.6740e-11*((60*60*24)**2\/(1000000**3)) # constante gravitacion\nm1 = 5.9722e24  # masa tierra\nm2 = 7.3490e22  # masa luna Cuerpo2\n\nr = lambda x,y: np.sqrt(x**2 + y**2)\n# ecuacion cuerpo 2\nfx = lambda t,x,y,vx,vy: vx\ngx = lambda t,x,y,vx,vy: G*m1*(0-x)\/(r(x,y))**3\nfy = lambda t,x,y,vx,vy: vy\ngy = lambda t,x,y,vx,vy: G*m1*(0-y)\/(r(x,y))**3\n\n# condiciones iniciales\nx0 = 3.84400e8\/1000000  # coordenadas Cuerpo2\ny0 = 0\nvx0 = 0 # velocidad Cuerpo2\nvy0 = 1.02e3*(60*60*24)\/1000000\nt0 = 0\nh  = 0.1\nmuestras = 250+1\n \n# Algoritmo como funci\u00f3n\ndef rungekutta2_fg(fx,fy,gx,gy,t0,x0,y0,vx0,vy0,h,muestras):\n    ''' solucion a EDO d2y\/dx2 con Runge-Kutta 2do Orden,\n    adaptado al ejercicio\n    '''\n    tamano = muestras + 1\n    tabla = np.zeros(shape=(tamano,5+8),dtype=float)\n    tabla[0,:5] = [t0,x0,y0,vx0,vy0]\n     \n    ti = t0 # valores iniciales\n    xi = x0\n    yi = y0\n    vxi = vx0\n    vyi = vy0\n\n    for i in range(1,tamano,1):\n        \n        K1x = h * fx(ti,xi,yi,vxi,vyi)\n        K1y = h * fy(ti,xi,yi,vxi,vyi)\n        K1vx = h * gx(ti,xi,yi,vxi,vyi)\n        K1vy = h * gy(ti,xi,yi,vxi,vyi)\n\n        K2x = h * fx(ti+h,xi+K1x,yi+K1y,vxi+K1vx,vyi+K1vy)\n        K2y = h * fy(ti+h,xi+K1x,yi+K1y,vxi+K1vx,vyi+K1vy)\n        \n        K2vx = h * gx(ti+h,xi+K1x,yi+K1y,vxi+K1vx,vyi+K1vy)\n        K2vy = h * gy(ti+h,xi+K1x,yi+K1y,vxi+K1vx,vyi+K1vy)\n\n        xi = xi + (K1x+K2x)\/2\n        yi = yi + (K1y+K2y)\/2\n\n        vxi = vxi + (K1vx+K2vx)\/2\n        vyi = vyi + (K1vy+K2vy)\/2\n        \n        ti = ti + h\n\n        tabla[i] = [ti,xi,yi,vxi,vyi,\n                    K1x,  K1y,  K1vx,  K1vy,\n                    K2x,  K2y,  K2vx,  K2vy ]\n    return(tabla)\n \n# PROCEDIMIENTO\ntabla = rungekutta2_fg(fx,fy,gx,gy,\n                       t0,x0,y0,vx0,vy0,h,muestras)\nn = len(tabla)\n# SALIDA\nprint('Trayectoria de dos cuerpos')\nnp.set_printoptions(precision=4)\nprint('EDO f,g con Runge-Kutta 2 Orden')\nprint('i ','[ ti, xi,  yi,  vxi,  vyi',']')\nprint('   [ K1x,  K1y,  K1vx,  K1vy ]')\nprint('   [ K2x,  K2y,  K2vx,  K2vy ]')\ntamano = 5\nfor i in range(0,tamano,1):  \n    txt = ' '\n    if i>=10:\n        txt = '  '\n    print(str(i),tabla[i,:5])\n    print(txt,tabla[i,5:9])\n    print(txt,tabla[i,9:])\n\n\n# GRAFICA\nimport matplotlib.pyplot as plt\nti = tabla[:,0]\nxi = tabla[:,1]\nyi = tabla[:,2]\n\nplt.plot(xi,yi,color='blue', label='Trayectoria')\n\nplt.plot(0,0,'o',color='blue',label='Cuerpo1')\nplt.plot(xi[0],yi[0],'*',color='red',label='Cuerpo2[0]')\nplt.plot(xi[-1],yi[-1],'o',color='red',label='Cuerpo2[n]')\n\n# entorno de gr\u00e1fica\nplt.axhline(0,color='gray',linestyle='dashed')\nplt.axvline(0,color='gray',linestyle='dashed')\nplt.xlabel('x [1000Km]')\nplt.ylabel('y [1000Km]')\nplt.title('Trayectoria 2 Cuerpos. h='+str(h)+', muestras='+str(muestras)+', vy0:'+str(vy0))\nplt.legend(loc='lower left')\nplt.tight_layout()\nplt.show()\n<\/pre><\/div>\n\n\n
resultados.txt<\/p>\n\n\n\n
Trayectoria de dos cuerpos\nEDO f,g con Runge-Kutta 2 Orden\ni  [ ti, xi,  yi,  vxi,  vyi ]\n   [ K1x,  K1y,  K1vx,  K1vy ]\n   [ K2x,  K2y,  K2vx,  K2vy ]\n0 [  0.    384.4     0.      0.     88.128]\n  [0. 0. 0. 0.]\n  [0. 0. 0. 0.]\n1 [ 1.0000e-01  3.8430e+02  8.8128e+00 -2.0128e+00  8.8105e+01]\n  [ 0.      8.8128 -2.0136  0.    ]\n  [-0.2014  8.8128 -2.0121 -0.0461]\n2 [ 2.0000e-01  3.8400e+02  1.7621e+01 -4.0246e+00  8.8036e+01]\n  [-0.2013  8.8105 -2.0131 -0.0462]\n  [-0.4026  8.8059 -2.0105 -0.0922]\n3 [ 3.0000e-01  3.8349e+02  2.6420e+01 -6.0343e+00  8.7920e+01]\n  [-0.4025  8.8036 -2.0115 -0.0923]\n  [-0.6036  8.7943 -2.0078 -0.1383]\n4 [  0.4    382.7905  35.2051  -8.0407  87.7591]\n  [-0.6034  8.792  -2.0088 -0.1384]\n  [-0.8043  8.7782 -2.0041 -0.1843]<\/code><\/pre>\n\n\n\n
\"EDO<\/figure>\n\n\n\n
literal e<\/h2>\n\n\n\n
El resultado con el algoritmo se aproxima a la orbita de un cuerpo mas peque\u00f1o orbitando sobre uno mucho mas grande. Semejante a lo sugerido en el enunciado sobre el movimiento de la luna alrededor de la tierra. La simplificaci\u00f3n no considera el movimiento de la tierra que se observa por ejemplo en las mareas, que es un modelo mas complejo.<\/p>\n\n\n\n
literal f<\/h2>\n\n\n\n
Solo para comprobar que el algoritmo presentado considera la condici\u00f3n de iniciar con una velocidad tangencial menor, hace que la \"luna\" pierda su \u00f3rbita y comience a alejarse de la tierra.<\/p>\n\n\n\n
El resultado gr\u00e1fico con el algoritmo es:<\/p>\n\n\n\n
\"EDO<\/figure>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2026PAOI_T2 EDO Problema de dos cuerpos literal a Simplificar las expresiones: G = 4.982 x 10-19 m1 = 5.9722 x 1024 m2 = 7.3490 x 1022 Se empieza reemplazando las constantes para simplificar la ecuaci\u00f3n.. quedando de la siguiente forma: Condiciones iniciales cuando t0 =0 x0 = 384.4 y0 = 0 vx0 = 0 […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[49],"tags":[58,54],"class_list":["post-25606","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/25606","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=25606"}],"version-history":[{"count":7,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/25606\/revisions"}],"predecessor-version":[{"id":25648,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/25606\/revisions\/25648"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=25606"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=25606"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=25606"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}