在python中绘制轨道轨迹 [英] plotting orbital trajectories in python

问题描述

如何在python中设置三体问题?如何定义求解ODE的函数?

How can I setup the three body problem in python? How to I define the function to solve the ODEs?

这三个方程是
x'' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * x，
y'' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * y和
z'' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * z.

The three equations are
x'' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * x,
y'' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * y, and
z'' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * z.

写成我们有6个一阶

x' = x2，

y' = y2，

z' = z2，

x2' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * x，

y2' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * y和

z2' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * z

我还想在地线"或地球轨道"和火星"的路径中添加我们可以假定为圆形的路径. 地球距离太阳149.6 * 10 ** 6公里，距离火星227.9 * 10 ** 6公里.

I also want to add in the path Plot o Earth's orbit and Mars which we can assume to be circular. Earth is 149.6 * 10 ** 6 km from the sun and Mars 227.9 * 10 ** 6 km.

#!/usr/bin/env python                                                             
#  This program solves the 3 Body Problem numerically and plots the trajectories      

import pylab
import numpy as np
import scipy.integrate as integrate
import matplotlib.pyplot as plt
from numpy import linspace

mu = 132712000000  #gravitational parameter
r0 = [-149.6 * 10 ** 6, 0.0, 0.0]
v0 = [29.0, -5.0, 0.0]
dt = np.linspace(0.0, 86400 * 700, 5000)  # time is seconds

推荐答案

如所示，您可以将其编写为由六个一阶ode组成的系统:

As you've shown, you can write this as a system of six first-order ode's:

x' = x2
y' = y2
z' = z2
x2' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * x
y2' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * y
z2' = -mu / np.sqrt(x ** 2 + y ** 2 + z ** 2) * z

您可以将其另存为矢量:

You can save this as a vector:

u = (x, y, z, x2, y2, z2)

并因此创建一个返回其派生函数:

and thus create a function that returns its derivative:

def deriv(u, t):
    n = -mu / np.sqrt(u[0]**2 + u[1]**2 + u[2]**2)
    return [u[3],      # u[0]' = u[3]
            u[4],      # u[1]' = u[4]
            u[5],      # u[2]' = u[5]
            u[0] * n,  # u[3]' = u[0] * n
            u[1] * n,  # u[4]' = u[1] * n
            u[2] * n]  # u[5]' = u[2] * n

给定初始状态u0 = (x0, y0, z0, x20, y20, z20)和时间t的变量，可以将其按如下方式馈入scipy.integrate.odeint:

Given an initial state u0 = (x0, y0, z0, x20, y20, z20), and a variable for the times t, this can be fed into scipy.integrate.odeint as such:

u = odeint(deriv, u0, t)

，其中u将是上面的列表.或者，您可以从头开始解压缩u，并忽略x2，y2和z2的值(必须先用.T移调输出)

where u will be the list as above. Or you can unpack u from the start, and ignore the values for x2, y2, and z2 (you must transpose the output first with .T)

x, y, z, _, _, _ = odeint(deriv, u0, t).T

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