Recent content by Digital Honeycomb

Oh! That makes sense! And then plugging that into the equation gives $$- \frac {l^2} {m^2}u^2 u'' = l^2 u^3 - 3V_o l^2 u^4 - V_o u^2$$ and that turns into $$u'' + u - \frac {V_o} {l^2} - 3V_o u^2 = 0$$ if you take ##m = 1##! Thank you so much!

Hmm, okay... so doing that I end up with the equation, $$\ddot r = \frac {l^2} {r ^3} - 3 \frac {V_o l^2} {r^4} - 2 \frac {V_0} {r^2},$$ substituting ##u = \frac {1} {r}## is simple enough for the RHS of the equation, but not for understanding exactly how to change ##\ddot r## into an equation...

Homework Statement From the homework: In General Relativity it is found that the radial equation of an object orbiting a non-rotating black hole has the form $$\dot r^2 + (1 - 2 \frac {V_o} {r} ) (\frac {l^2} {r^2} + 1) = E^2$$ where ##r## is the radial coordinate, ##l## is the angular...