Newton's law and non-inertial frame of reference

Loading the package

using Plots
gr()

Plots.GRBackend()

1. The demonstration of the projection

Using the difference equation to demonstrating how to solve the single particle problem

1.1 The free projection problem

Initial the parameters

x0=0.5;y0=7.;  #The initial position
Vx0=1;Vy0=2; #The initial velocity
m=2;           #The mass of the particle
g=-9.8;         #The acceleration

Fx(x,y)=0;Fy(x,y)=g.*m; #The Force in different direction

Δt=0.1;t=0:Δt:1;     #The simulation time

xt=zeros(1,length(t)+1);yt=zeros(1,length(t)+1);  #Initialization of the position vs time xt(t) and yt(t)
xt[1]=x0;yt[1]=y0;

Vxt=zeros(1,length(t)+1);Vyt=zeros(1,length(t)+1);  #Initialization of the position vs time xt(t) and yt(t)
Vxt[1]=Vx0;Vyt[1]=Vy0;

We can get the trajectory by numerical way

for i=1:length(t)
    ax=Fx(xt[i],yt[i])/m; ay=Fy(xt[i],yt[i])/m;     #The acceleration
    Vxt[i+1]=Vxt[i]+ax*Δt;Vyt[i+1]=Vyt[i]+ay*Δt;  #The velocity in next step
    xt[i+1]=xt[i]+Vxt[i]*Δt+ax*Δt^2/2;            #The numerical trajectory
    yt[i+1]=yt[i]+Vyt[i]*Δt+ay*Δt^2/2;
end

Meanwhile, we can also obtain the analytic solution

xp(t)=x0+Vx0.*t;yp(t)=y0+Vy0.*t+g.*t.^2/2;        #Numerical solution

Comparing the numerical and analytic solution

plot(xt',yt',w=2,c=:black)
plot!(xp,yp,0,1,c=:blue)

1.2 Projection with the friction

The friction has the form $\vec{F}=\kappa \vec{V}$

Initial the parameter

x0=0.5;y0=7.;   #The initial position
Vx0=1;Vy0=2;    #The initial velocity
m=2;            #The mass of the particle
g=-9.8;         #The acceleration
κ=-0.8;        #The friction coefficient

Fx(vx)=κ*vx;Fy(vy)=m*g+κ*vy;  #The Force in different direction

Δt=0.01;t=0:Δt:1;     #The simulation time
xt=zeros(1,length(t)+1);yt=zeros(1,length(t)+1);  #Initialization of the position vs time xt(t) and yt(t)
xt[1]=x0;yt[1]=y0;
Vxt=zeros(1,length(t)+1);Vyt=zeros(1,length(t)+1);  #Initialization of the position vs time xt(t) and yt(t)
Vxt[1]=Vx0;Vyt[1]=Vy0;

We can get the trajectory by numerical way

for i=1:length(t)
    ax=Fx(Vxt[i])/m; ay=Fy(Vyt[i])/m;     #The acceleration
    Vxt[i+1]=Vxt[i]+ax*Δt;Vyt[i+1]=Vyt[i]+ay*Δt;  #The velocity in next step
    xt[i+1]=xt[i]+Vxt[i]*Δt+ax*Δt^2/2;            #The numerical trajectory
    yt[i+1]=yt[i]+Vyt[i]*Δt+ay*Δt^2/2;
end

Comparing the numerical and analytic solution

plot(xt',yt',w=2,c=:black)
plot!(xp,yp,0,1,c=:blue)

2. The motion under the universal force

Numerically demonstration of the eclipsal motion under universal force

Initial the parameter

x0=1;y0=1;   #The initial position
Vx0=1;Vy0=0;    #The initial velocity
m=2;            #The mass of the particle

Fx(x,y)=-m*x/(sqrt(x.^2+y.^2))^3;Fy(x,y)=-m*y/(sqrt(x.^2+y.^2))^3;  #The Force in different direction

Δt=0.0002;t=0:Δt:5;     #The simulation time
xt=zeros(1,length(t)+1);yt=zeros(1,length(t)+1);  #Initialization of the position vs time xt(t) and yt(t)
xt[1]=x0;yt[1]=y0;
Vxt=zeros(1,length(t)+1);Vyt=zeros(1,length(t)+1);  #Initialization of the position vs time xt(t) and yt(t)
Vxt[1]=Vx0;Vyt[1]=Vy0;

We can get the trajectory by numerical way

for i=1:length(t)
    ax=Fx(xt[i],yt[i]); ay=Fy(xt[i],yt[i]);     #The acceleration
    Vxt[i+1]=Vxt[i]+ax*Δt;Vyt[i+1]=Vyt[i]+ay*Δt;  #The velocity in next step
    xt[i+1]=xt[i]+Vxt[i]*Δt+ax*Δt^2/2;            #The numerical trajectory
    yt[i+1]=yt[i]+Vyt[i]*Δt+ay*Δt^2/2;
end

Comparing the numerical and analytic solution

plot!(xt',yt',w=2,c=:black,lab="\\Delta t=0.0002")

3. The non-inertial frame of the reference

Try to demonstrate the coriolic force

θ(t)=0;r(t)=v*t;

v=0.5;

plot(θ,r,0,2π,proj=:polar,w=2,c=:red)

θ₂(t)=θ(t)-ω*t;r₂(t)=r(t);

ω=2;

plot(θ₂,r₂,0,2π,proj=:polar,w=2,c=:red)