Shadow Posted March 21, 2009 Posted March 21, 2009 Hey all, I just need someone to check the following: [math]F_a= G \frac{m_a \cdot m_b}{r^2}[/math] [math]a_a(t)= \frac{F_a}{m_a}[/math] [math]v_a(t)= \int G \cdot \frac{m_b}{r^2} \,dt = G \cdot \frac{m_b}{r^2} \cdot t + v_0[/math] [math]s_a(t)=\int G \cdot \frac{m_b}{r^2} \cdot t + v_0 \,dt = G \cdot \frac{m_b}{2 \cdot r^2} \cdot t^2 + v_0 \cdot t + s_0[/math] Cheers, Gabe
the tree Posted March 21, 2009 Posted March 21, 2009 Seems all right to me. Object b comes across as a fixed point of reference, I think.
timo Posted March 21, 2009 Posted March 21, 2009 Hey all, I just need someone to check the following: What exactly do you want checked? - If you want to know whether you made a calculation mistake somewhere: No, I see none. - If you want to know whether what you wrote makes sense: Strictly speaking you did not even say what the letters stand for, what can be assumed as given, what you want to arrive at, etc. Of course it is relatively obvious in your case. But "obvious" is a nasty bugger that stabs you in the back when you expect it the least. E.g.: Are you aware that you assumed r being independent of time and of the implications that has? - If you want to know if you wrote it down in a correct manner: It is obvious what you mean, so basically "yes". Purists would possibly prefer rewriting statements like [math]v_a(t)= \int G \cdot \frac{m_b}{r^2} \,dt = G \cdot \frac{m_b}{r^2} \cdot t + v_0[/math] as [math]v_a(t)= v_0 + \int_0^t G \cdot \frac{m_b}{r^2} \,dt' = G \cdot \frac{m_b}{r^2} \cdot t + v_0[/math].
Shadow Posted March 21, 2009 Author Posted March 21, 2009 I meant a calculation mistake. And if I understand correctly, the r issue should have no effect since this will be used in a program where r (and therefor F) is constantly recalculated in a loop. Cheers and thanks, Gabe
Recommended Posts
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 accountSign in
Already have an account? Sign in here.
Sign In Now