在python中繪製一個球體軌道軌跡

Question

如何將一個半徑爲1737的球體放在(384400,0,0)的位置？

這個球體將是我的軌跡中的月球。

與代碼的其他一切都很好，我只是不知道如何在該半徑的位置添加一個球體。

import numpy as np 
from scipy.integrate import odeint 
import matplotlib.pyplot as plt 
from mpl_toolkits.mplot3d import Axes3D 

me = 5.974 * 10 ** (24) # mass of the earth          
mm = 7.348 * 10 ** (22) # mass of the moon          
G = 6.67259 * 10 ** (-20) # gravitational parameter        
re = 6378.0 # radius of the earth in km           
rm = 1737.0 # radius of the moon in km           
r12 = 384400.0 # distance between the CoM of the earth and moon     
M = me + mm 

pi1 = me/M 
pi2 = mm/M 
mue = 398600.0 # gravitational parameter of earth km^3/sec^2      
mum = G * mm # grav param of the moon           
mu = mue + mum 
omega = np.sqrt(mu/r12 ** 3) 
nu = -129.21 * np.pi/180 # true anomaly angle in radian      

x = 327156.0 - 4671 
# x location where the moon's SOI effects the spacecraft with the offset of the 
# Earth not being at (0,0) in the Earth-Moon system        
y = 33050.0 # y location              

vbo = 10.85 # velocity at burnout            

gamma = 0 * np.pi/180 # angle in radians of the flight path     

vx = vbo * (np.sin(gamma) * np.cos(nu) - np.cos(gamma) * np.sin(nu)) 
# velocity of the bo in the x direction           
vy = vbo * (np.sin(gamma) * np.sin(nu) + np.cos(gamma) * np.cos(nu)) 
# velocity of the bo in the y direction           

xrel = (re + 300.0) * np.cos(nu) - pi2 * r12 
# spacecraft x location relative to the earth   
yrel = (re + 300.0) * np.sin(nu) 

# r0 = [xrel, yrel, 0]               
# v0 = [vx, vy, 0]                
u0 = [xrel, yrel, 0, vx, vy, 0] 


def deriv(u, dt): 
    n1 = -((mue * (u[0] + pi2 * r12)/np.sqrt((u[0] + pi2 * r12) ** 2 
               + u[1] ** 2) ** 3) 
     - (mum * (u[0] - pi1 * r12)/np.sqrt((u[0] - pi1 * r12) ** 2 
               + u[1] ** 2) ** 3)) 
    n2 = -((mue * u[1]/np.sqrt((u[0] + pi2 * r12) ** 2 + u[1] ** 2) ** 3) 
     - (mum * u[1]/np.sqrt((u[0] - pi1 * r12) ** 2 + u[1] ** 2) ** 3)) 
    return [u[3], # dotu[0] = u[3]            
      u[4], # dotu[1] = u[4]            
      u[5], # dotu[2] = u[5]            
      2 * omega * u[5] + omega ** 2 * u[0] + n1, # dotu[3] = that   
      omega ** 2 * u[1] - 2 * omega * u[4] + n2, # dotu[4] = that   
      0] # dotu[5] = 0              


dt = np.arange(0.0, 320000.0, 1) # 200000 secs to run the simulation    
u = odeint(deriv, u0, dt) 
x, y, z, x2, y2, z2 = u.T 

fig = plt.figure() 
ax = fig.add_subplot(111, projection='3d') 
ax.plot(x, y, z) 
plt.show() 

Answer 1

您可以添加以下代碼繪製領域，plt.show()前：

phi = np.linspace(0, 2 * np.pi, 100) 
theta = np.linspace(0, np.pi, 100) 
xm = rm * np.outer(np.cos(phi), np.sin(theta)) + r12 
ym = rm * np.outer(np.sin(phi), np.sin(theta)) 
zm = rm * np.outer(np.ones(np.size(phi)), np.cos(theta)) 
ax.plot_surface(xm, ym, zm) 

然而，你的月亮看起來都伸了出來，因爲規模不等於所有軸。爲了改變軸的刻度，可以plt.show()之前添加類似

ax.auto_scale_xyz([-50000, 400000], [0, 160000], [-130000, 130000]) 

。結果仍然不是完全正確，但我留給你去玩，以獲得更好的結果 - 我只是選擇了一些數字，使它看起來更好。

Answer 2

的球所有的代碼是在這裏：

http://matplotlib.org/examples/mplot3d/surface3d_demo2.html

從那裏，你只需要在任何你想要的方向移動，改變10，你想用的任何半徑。