Cannot get RK4 to solve for position of orbiting body in Python

Question

I am trying to solve for the position of a body orbiting a much more massive body, using the idealization that the much more massive body doesn\'t move. I am trying to solve

Answer 1

You are not using rkx, rky functions anywhere! There are two return at the end of function definition you should use return [(kx1 + 2*kx2 + 2*kx3 + kx4)/6, (mx1 + 2*mx2 + 2*mx3 + mx4)/6] (as pointed out by @eapetcho). Also, your implementation of Runge-Kutta is not clear to me.

You have dv/dt so you solve for v and then update r accordingly.

for n in range(1,len(t)): #solve using RK4 functions
    vx[n] = vx[n-1] + rkx(vx[n-1],vy[n-1],t[n-1])*dt
    vy[n] = vy[n-1] + rky(vx[n-1],vy[n-1],t[n-1])*dt
    x[n] = x[n-1] + vx[n-1]*dt
    y[n] = y[n-1] + vy[n-1]*dt

Here is my version of the code

import numpy as np

#constants
G=1
M=1
h=0.1

#initiating variables
rt = np.arange(0,10,h)
vx = np.zeros(len(rt))
vy = np.zeros(len(rt))
rx = np.zeros(len(rt))
ry = np.zeros(len(rt))

#initial conditions
vx[0] = 10 #initial x velocity
vy[0] = 10 #initial y velocity
rx[0] = 10 #initial x position
ry[0] = 0 #initial y position

def fx(x,y): #x acceleration
     return -G*M*x/((x**2+y**2)**(3/2))

def fy(x,y): #y acceleration
     return -G*M*y/((x**2+y**2)**(3/2))

def rk4(xj, yj):
    k0 = h*fx(xj, yj)
    l0 = h*fx(xj, yj)

    k1 = h*fx(xj + 0.5*k0 , yj + 0.5*l0)
    l1 = h*fy(xj + 0.5*k0 , yj + 0.5*l0)

    k2 = h*fx(xj + 0.5*k1 , yj + 0.5*l1)
    l2 = h*fy(xj + 0.5*k1 , yj + 0.5*l1)

    k3 = h*fx(xj + k2, yj + l2)
    l3 = h*fy(xj + k2, yj + l2)

    xj1 = xj + (1/6)*(k0 + 2*k1 + 2*k2 + k3)
    yj1 = yj + (1/6)*(l0 + 2*l1 + 2*l2 + l3)
    return (xj1, yj1)

for t in range(1,len(rt)):
    nv = rk4(vx[t-1],vy[t-1])
    [vx[t],vy[t]] = nv
    rx[t] = rx[t-1] + vx[t-1]*h
    ry[t] = ry[t-1] + vy[t-1]*h

I suspect there are issues with fx(x,y,t) and fy(x,y,t)

This is the case, I just checked my code for fx=3 and fy=y and I got a nice trajectory.

Here is the ry vs rx plot: