编辑
我仍然想要解决方案的唯一部分是让球体看起来是球形的。如果我将轴缩放相同，则球体太小但呈球形。有没有办法将未使用的部分剪掉以便放大？或者我们是否需要将轴设置得这么大并获得所需的效果？
结束编辑
通过在python 中绘制一个球体，我能够将两个球体添加到我的绘图中。
但是，无论我如何调整参数，我都无法让它们看起来像球形。此外，'g'在绘图末尾添加或任何其他颜色不会改变那里的颜色。
我也试过color='g'，但也没有用。
如何获得球形外观 ( ax autoscale) 并更改颜色？
#!/usr/bin/env python
# This porgram numerically solves the trajectory from the Earth to the moon
# with the specified flight path, true anomaly, and initial conditions.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from numpy import linspace
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 = -134.979 * 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
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[4] + omega ** 2 * u[0] + n1, # dotu[3] = that
omega ** 2 * u[1] - 2 * omega * u[3] + n2, # dotu[4] = that
0] # dotu[5] = 0
dt = np.linspace(0.0, 320000.0, 1000000.0) # 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)
# adding the moon
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)
ax.auto_scale_xyz([-50000, 400000], [0, 160000], [-130000, 130000])
# adding the earth
xe = re * np.outer(np.cos(phi), np.sin(theta)) - 4671
ye = re * np.outer(np.sin(phi), np.sin(theta))
ze = re * np.outer(np.ones(np.size(phi)), np.cos(theta))
ax.plot_surface(xe, ye, ze, 'g')
ax.auto_scale_xyz([-50000, 400000], [0, 160000], [-130000, 130000])
#
plt.show()
# The code below finds the distance between path and the moon
my_x, my_y, my_z = (384400,0,0)
delta_x = x - my_x
delta_y = y - my_y
delta_z = z - my_z
distance = np.array([np.sqrt(delta_x ** 2 + delta_y ** 2 + delta_z ** 2)])
minimum = np.amin(distance)
print(minimum)