vpython中的椭圆轨道

2024-05-20 11:12:35 发布

您现在位置:Python中文网/ 问答频道 /正文

我有以下代码。这个代码是模拟环绕其他物体的轨道物体,例如太阳系。当你运行它的时候,物体会以圆形轨道运行。在

import math
from vpython import *
lamp = local_light(pos=vector(0,0,0), color=color.yellow)
# Data in units according to the International System of Units
G = 6.67 * math.pow(10,-11)

# Mass of the Earth
ME = 5.973 * math.pow(10,24)
# Mass of the Moon
MM = 7.347 * math.pow(10,22)
# Mass of the Mars
MMa = 6.39 * math.pow(10,23)
# Mass of the Sun
MS = 1.989 * math.pow(10,30)
# Radius Earth-Moon
REM = 384400000
# Radius Sun-Earth
RSE = 149600000000
RMS = 227900000000
# Force Earth-Moon
FEM = G*(ME*MM)/math.pow(REM,2)
# Force Earth-Sun
FES = G*(MS*ME)/math.pow(RSE,2)
# Force Mars-Sun
FEMa = G*(MMa*MS)/math.pow(RMS,2)

# Angular velocity of the Moon with respect to the Earth (rad/s)
wM = math.sqrt(FEM/(MM * REM))
# Velocity v of the Moon (m/s)
vM = wM * REM
print("Angular velocity of the Moon with respect to the Earth: ",wM," rad/s")
print("Velocity v of the Moon: ",vM/1000," km/s")

# Angular velocity of the Earth with respect to the Sun(rad/s)
wE = math.sqrt(FES/(ME * RSE))
# Angular velocity of the Mars with respect to the Sun(rad/s)
wMa = math.sqrt(FEMa/(MMa * RMS))

# Velocity v of the Earth (m/s)
vE = wE * RSE
# Velocity v of the Earth (m/s)
vMa = wMa * RMS
print("Angular velocity of the Earth with respect to the Sun: ",wE," rad/s")
print("Velocity v of the Earth: ",vE/1000," km/s")


# Initial angular position
theta0 = 0

# Position at each time
def positionMoon(t):                                     
    theta = theta0 + wM * t
    return theta

def positionMars(t):                                     
    theta = theta0 + wMa * t
    return theta

def positionEarth(t):
    theta = theta0 + wE * t
    return theta


def fromDaysToS(d):
    s = d*24*60*60
    return s

def fromStoDays(s):
    d = s/60/60/24
    return d

def fromDaysToh(d):
    h = d * 24
    return h

# Graphical parameters
print("\nSimulation Earth-Moon-Sun motion\n")
days = 365
seconds = fromDaysToS(days)
print("Days: ",days)
print("Seconds: ",seconds)

v = vector(384,0,0)
E = sphere(pos = vector(1500,0,0), color = color.blue, radius = 60, make_trail=True)
Ma = sphere(pos = vector(2300,0,0), color = color.orange, radius = 30, make_trail=True)
M = sphere(pos = E.pos + v, color = color.white,radius = 10, make_trail=True)
S = sphere(pos = vector(0,0,0), color = color.yellow, radius=700)

t = 0
thetaTerra1 = 0
dt = 5000
dthetaE = positionEarth(t+dt)- positionEarth(t)
dthetaM = positionMoon(t+dt) - positionMoon(t)
dthetaMa = positionMars(t+dt) - positionMars(t)
print("delta t:",dt,"seconds. Days:",fromStoDays(dt),"hours:",fromDaysToh(fromStoDays(dt)),sep=" ")
print("Variation angular position of the Earth:",dthetaE,"rad/s that's to say",degrees(dthetaE),"degrees",sep=" ")
print("Variation angular position of the Moon:",dthetaM,"rad/s that's to say",degrees(dthetaM),"degrees",sep=" ")

while t < seconds:
    rate(500)
    thetaEarth = positionEarth(t+dt)- positionEarth(t)
    thetaMoon = positionMoon(t+dt) - positionMoon(t)
    thetaMars = positionMars(t+dt) - positionMars(t)
    # Rotation only around z axis (0,0,1)
    E.pos = rotate(E.pos,angle=thetaEarth,axis=vector(0,1,0))
    Ma.pos = rotate(Ma.pos,angle=thetaMars,axis=vector(0,1,0))
    v = rotate(v,angle=thetaMoon,axis=vector(0,1,0))
    M.pos = E.pos + v
t += dt

我想知道如何将轨道轨道轨道改为椭圆轨道? 我试过好几种方法,但都找不到解决办法。在

谢谢。 谢谢你


Tags: ofthetoposdtmathcolorsun
1条回答
网友
1楼 · 发布于 2024-05-20 11:12:35

这看起来更像是一个物理问题,而不是编程问题。问题是当计算速度和线性积分位置时,假设每个轨道都是圆的(例如v*dt)。这不是计算轨道物体轨迹的方法。在

为了简单起见,我们假设所有的质量都是点质量,所以没有任何奇怪的重力梯度或姿态动力学需要解释。在

从那里,你可以参考麻省理工学院的网页。(http://web.mit.edu/12.004/TheLastHandout/PastHandouts/Chap03.Orbital.Dynamics.pdf)在轨动力学。在第七页,有一个方程,它把你的中心体的径向位置,作为许多轨道参数的函数。好像你有所有的参数除了轨道的偏心率。如果你有详细的短命数据或近点/近点信息,你可以在线查找或计算它。在

从这个方程中,你会看到分母中的φ-phi_0项。这就是俗称的真正的卫星异常。代替时间,你将迭代这个真正的异常参数从0到360来找到你的径向距离,从真正的异常,倾角,到上升点的直角,和近点的参数,你可以找到一个特定的真实异常的三维笛卡尔坐标。在

从真正的异常中走出来就不那么琐碎了。你需要找出偏心异常,然后在每个偏心异常步骤找到平均异常。你现在有平均异常作为时间的函数。可以在“节点”之间进行线性插值,在“节点”处用v*dt计算位置。你可以用vis-viva方程来计算速度,dt就是计算出的时间步长之间的差。在

在每一个时间步,你可以在python程序中更新卫星的位置,它会正确地绘制出你的轨迹。在

关于真正异常的更多信息,wikipedia有一个很好的描述:https://en.wikipedia.org/wiki/True_anomaly

关于轨道元素的更多信息(需要从径向位置转换为笛卡尔坐标):https://en.wikipedia.org/wiki/Orbital_elements

相关问题 更多 >