Jump to content

Does this work?


uncool

Recommended Posts

I'm trying to find some general things about orbits in gravitational systems in multiple dimensions. I've already found that it has to all be in one plane (easy enough), and now I'm assuming the following:

[math]F = \frac{g*m_1*m_2}{d^{#dim-1}}[/math]

[math]F_{x} = F*\frac{x}{d}[/math]

[math]F_{y} = F*\frac{y}{d}[/math]

So:

[math]\frac{d^{2}y}{y*dt^2}=\frac{d^{2}x}{x*dt^2}[/math]

[math]Let |x| = e^{f(t)}[/math]

[math]|y|=e^{g(t)}[/math]

[math]\frac{dx}{dt}=f'(t)e^{f(t)}[/math]

[math]\frac{d^{2}x}{dt^{2}}=f''(t)e^{f(t)}+f'(t)^{2}e^{f(t)}[/math]

[math]\frac{dy}{dt}=g'(t)e^{g(t)}[/math]

[math]\frac{d^{2}y}{dt^{2}}=g''(t)e^{g(t)}+g'(t)^{2}e^{g(t)}[/math]

So:

[math]f''(t)+f'(t)^{2}=g''(t)+g'(t)^{2}[/math]

[math]f''(t)-g''(t)=g'(t)^{2}-f'(t)^{2}[/math]

[math]\frac{(f'(t)-g'(t))'}{f'(t)-g'(t)}=-f'(t)-g'(t)[/math]

[math]ln|f'(t)-g'(t)|=-f(t)-g(t)+C[/math]

[math]f(t)=ln(|x|),g(t)=ln(|x|)[/math]

[math]f'(t)=\frac{x'}{x},g'(t)=\frac{y'}{y}[/math]

[math]|\frac{x'}{x}-\frac{y'}{y}|=C/|xy|[/math]

[math]|x'y-y'x|=C[/math]

Since it is constant, let us assume that x'y is always greater.

[math]x'y-y'x=C[/math]

Can anything happen from there?

-Uncool-

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.