摘要：Calculate the precession of the perihelion of Mercury,followong the approach described in this section.
背景：
800px-Mercury_in_color_-_Prockter07-edit1.jpg
Mercuryorbitsolarsystem.gif
400px-ThePlanets_Orbits_Mercury_PolarView.svg.png
公式：
10.1.gif
数据：
10.2.gif
正文：
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import pylab as pl
import math
class solar:
def __init__(self,x0=1,y0=0,vx=0,vy=2*math.pi,dt=0.001,dbeta=0,total_time=100,alpha=0):
self.x=[x0]
self.y=[y0]
self.r=[math.sqrt(x0**2+y0**2)]
self.vx=[vx]
self.vy=[vy]
self.dt=dt
self.t=[0]
self.tt=total_time
self.db=dbeta
self.a=alpha
def run(self):
while self.t[-1]<self.tt:
self.vx.append(self.vx[-1]-4*math.pi**2*self.x[-1]*(1+self.a/(self.r[-1]**2))/self.r[-1]**(3+self.db)*self.dt)
self.vy.append(self.vy[-1]-4*math.pi**2*self.y[-1]*(1+self.a/(self.r[-1]**2))/self.r[-1]**(3+self.db)*self.dt)
self.x.append(self.x[-1]+self.vx[-1]*self.dt)
self.y.append(self.y[-1]+self.vy[-1]*self.dt)
self.r.append(math.sqrt(self.x[-1]**2+self.y[-1]**2))
self.t.append(self.t[-1]+self.dt)
def show(self):
pl.title("simulation of elliptical orbit")
pl.xlabel("x")
pl.ylabel("y")
pl.xlim(-1,1)
pl.ylim(-1,1)
pl.plot(self.x, self.y,label="$\\beta$ =%.3f"%(2+self.db))
pl.legend()
pl.grid(True)
pl.show()
def show_(self):
vs=[10.061,7.405,6.283,5.096,2.755,2.034,1.434,1.146]
rs=[0.39,0.72,1.00,1.52,5.20,9.54,19.19,30.06]
names=['mercury','venus','earth','mars','jupiter','saturn','uranus','neptune']
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
plt.title("solar system")
plt.xlabel("x")
plt.ylabel("y")
for i in range(len(vs)):
a=solar(vy=vs[i],x0=rs[i])
a.run()
ax.plot(a.x, a.y,label=names[i])
plt.legend()
plt.show()
#show the whole solar system
a=solar()
a.run()
a.show_()
#the inverse square law and the stabiliity of planetary orbits
b=solar(dbeta=0.1,vy=5)
b.run()
b.show()
#precession of the perihelon of mercury
class precession(solar):
def m_run(self):
_alphas=[0.0001,0.0004,0.003,0.0035,0.004,0.006,0.007]
omegas=[]
for i in _alphas:
a=solar(x0=0.39,vy=10.061,total_time=100,alpha=i)
a.run()
_min=1
_j=0
for j in range(len(a.r)):
if a.r[j]<_min:
_min=a.r[j]
_j=j
min_r=a.r[_j]+(a.r[_j-1]-a.r[_j])/2
temp_theta=[]
temp_t=[]
for k in range(len(a.r)):
if a.r[k]<min_r:
temp_theta.append(math.asin(a.x[k]/a.r[k])*360/2/math.pi)
temp_t.append(a.t[k])
omegas.append((temp_theta[3]-temp_theta[2])/(temp_t[3]-temp_t[2]))
print(omegas)
test=precession()
test.m_run()
_alphas=[0,0.0001,0.0004,0.003,0.0035,0.004,0.006,0.007]
_omegas=[0,2.1360957657207646,3.79859101235552, 31.586827548752012, 35.67506491602629,\
39.82020809760321,64.21130592694833, 80.79924387065746]
pl.plot(_alphas,_omegas,'.')
z=np.polyfit(_alphas,_omegas,1)
p=np.poly1d(z)
print(z)
print(p)
linspx=np.linspace(0,0.007)
linspy=z[0]*linspx+z[1]
pl.title('precession rate versus $\\alpha$')
pl.xlabel("$\\alpha$")
pl.ylabel("$d\\theta /dt$(degree/yr)")
pl.plot(linspx,linspy)
pl.grid(True)
结论：
10.3.gif
10.4.gif
10.5.gif
10.6.gif
10.7.gif
10.8.gif
10.9.gif
可见万有引力严格符合R的平方反比的关系。
致谢：
感谢秦大粤同学的帮助!
