Question

我想写一个程序简单描述苹果通过地球的运动，或者我们称之为重力列车 ，参见（维基百科的重力训练）Gravity train by hyperphysics

我想在当时绘制移动速度和位置。

这是我制作的代码，但答案错误（速度和时间错误）。虽然我使用了下降体falling body Equations wikipedia 的方程式，加速度数值很好，但保持速度和距离的值非常大我错了什么？提前致谢最大

**我必须具备的价值时间必须等于2530.30秒（接近42.2分钟）最高速度约为每秒7 900米（28440公里/小时）。

import math
import numpy as np
import matplotlib.pyplot as plt

# Max Oceaan 2015 10 5
# Gravity train program

# Please notice that we assume ::
# -- air friction is neglected
# -- gravitational constant = 6.67408 × 10-11
# -- Earth  Mass M  5.972 × 10^24 kg


# -- the apple speed in begin is 0 (free fall)
v=0

# -- the begin time is zero
t=0


L_t=[] # list for time step
L_h=[] # list for postion (hight)
L_v=[] # list for Velocity
L_g=[] # list for acceleration

r=6371000   #Earth radius per meter
delta=100    # Length step that we walk in for_loop
for h in np.arange(0.1,2*r,delta):

# c is the correction factor for g is 
# distance of apple from center of earth / earth radius
#notice:: when the apple reach the earth center c will be negativ en -1
    c=float((r-h)/r)

# acceleration is according to law of Newten
    g= float(6.67408 * (10**-11) * 5.972 * (10**24)*(c) )/(r**2)

# time  is t=t+math.sqrt(2*(delta)/g)
# en speed is v=v+math.sqrt(2*(delta)g)
    # But must g positive and if g negative ,the laws are as follows:
# t= t + math.sqrt(2*(delta)/(abs(g)))     v=v - math.sqrt(2*(delta)*(abs(g)))
    if g>0:
        t=t+math.sqrt(2*delta/g)
        v=v+math.sqrt(2*delta*g)
    else:
        t=t + math.sqrt(2*delta/(abs(g))) 
        v=v - math.sqrt(2*delta*(abs(g)))


    L_h.append(h)
    L_t.append(t)
    L_v.append(v)
    L_g.append(g)


plt.subplot(121) 
plt.title("Velocity ")  
plt.xlabel("Time ") 
plt.ylabel("Velocity m/s") 
plt.plot(L_t, L_v, 'g-')


plt.subplot(122)
plt.title("distance ") 
plt.xlabel("Time ") 
plt.ylabel("distance  m") 
plt.plot(L_t, L_h, 'b-')

plt.tight_layout()


plt.figure(2)
plt.plot(L_t, L_g , 'b-')
plt.title("acceleration in tijd") 
plt.xlabel(" Time s") 
plt.ylabel("acceleration (a) m/ss") 


plt.show()

我试图改进它但是我在图形上得到了更糟糕的结果新代码

    # -*- coding: utf-8 -*-
import math
import numpy as np
import matplotlib.pyplot as plt

# Please notice that we assume :air friction is neglected

G = 6.67408 * (10**-11)     # gravitational constant
M = 5.972 * (10**24)        # Earth  Mass in kg
r = 6371000                 # Earth radius per meter
p = 5514                    # Mean Earth density in kg\m3

time = 0   #  the begin time is zero
a=0     #  the begin acceleration g 
v=0
t=0

L_h = [] # list for postion (hight)
L_a = [] # list for acceleration
L_v = [] # list for speed
L_t = [] # list for time

# for_loop in the path that we walk
delta=1000
for h in np.arange(1,2*r,delta):

# (M_effective) is The effective mass of earth that changes when h changes 
# (M_effective) =         new sphere volume       * density
#    M_effective = (4./3) * (math.pi) * ((r-h)**3)   *  p
    M_effective = (4./3) * (math.pi) *   p * (math.pow((r-h),3))


    a = G * M_effective / (r**2)  # (a) is the acceleration 

    if a>0:
        a=a
        t=t+math.sqrt(2*delta/a)
        v=v+math.sqrt(2*delta*a)
    else:
        a=a
        t=t + math.sqrt(2*delta/(abs(a))) 
        v=v - math.sqrt(2*delta*(abs(a)))


    L_h.append(h)
    L_t.append(t)
    L_v.append(v)
    L_a.append(a)




plt.subplot(121) 
plt.title("Velocity ")  
plt.xlabel("Time ") 
plt.ylabel("Velocity m/s") 
plt.plot(L_t, L_v, 'g-')


plt.subplot(122)
plt.title("distance ") 
plt.xlabel("Time ") 
plt.ylabel("distance  m") 
plt.plot(L_t, L_h, 'b-')

plt.tight_layout()


plt.figure(2)
plt.plot(L_t, L_a , 'b-')
plt.title("acceleration in tijd") 
plt.xlabel(" Time s") 
plt.ylabel("acceleration (a) m/ss") 


plt.show()
#Gravity train
#The travel time equals 2530.30 seconds (nearly 42.2 minutes),
#For a train that goes directly through the center of the Earth, the maximum speed is about 7 900 metres per second (28440 km/h).

在python中的简单重力列车

0 个答案: