Recent content by Binaryburst

Say I have a special type of black hole, an infinetly small one. I am standing still in deep space and i let a point mass to fall. What is it's equation of motion? I'm interested in the equation itself not just an explanation. It should be sth like this x=f(t).

Thank you! At last :D

@ mfb I am out of ideas. I have a problem that I have been struggling with for half a year and still haven't solved it. I posted it once but didn't get an exact complete solution. https://www.physicsforums.com/showthread.php?t=668751&goto=newpost It has to be done without conservation of...

This is what wolfram alpha says: http://m.wolframalpha.com/input/?i=x%27%27%28t%29%3D-x%2F%28x%5E2%2B1%29&x=10&y=2

The solution to this equation should be: x(t) = sin(t) I simply don't know how to get there ... I have integrated but I get something like integ of 1/sqr(-ln(1+x^2)) dx?! :(

If have this equation: \frac {d^2x}{dt^2}=-\frac{x}{1+x^2} How do I solve it?

I did a simulation with vx and it matched the sinusoidal, it seems. Any help from the experts?

In the consevation of Energy formula you got the speed with respect to the height only which is x^2 so we actually got vx :~> when you replace y with x^2 and vy when you leave y unchanged. PS: I'm not an expert either. :D

It is right. That's what I got too... That's just v*cos(theta). It results that v is equal to f(x).

Omg! What a humongous mistake in my formulation of the conservation of energy!

Actually i got the total speed dependent only on the x-axis so saying that v.total is dx/dt is correct because it's no longer the slope of the parabola but the slope of f(x).

I am super puzzelled as well. I can't figure out what I did.

Hmmm.. That's interesting I actually got the vx. Sorry for the blunder. I was too excited :) correcting the mistake.

I had v=f(x). Rewritten it as follows: v/f(x)=1 ; 1/f(x)*dx/dt=1. ; Integrate with respect to t Int( 1/f(x)* dx/dt * dt ) = int( 1 dt )

I'm thinking how could i get it without using the conservation of energy and using forces.