Timeline for Integrating radial free fall in Newtonian gravity
Current License: CC BY-SA 3.0
2 events
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| May 17, 2013 at 9:21 | comment | added | xaxa | This is not lagrangian, this is total energy ($E$) = kinetic($mv^2/2$) + potential($-GmM/r$). At $t=0$ you have kinetic energy = 0. You solve this for $v(t)^2$ algebraically, so you have $v(t)^2 = F(r)$. Then you set $v(t) = dr/dt$ and take a minus sign when extracting square root, so you have have $dr/dt = -\sqrt{F(r)}$. Then you just rearrange to get $-dr/\sqrt{F(r)} = dt$ and integrate. On the left you have a function only of $r$, on the right only of $t$. | |
| May 16, 2013 at 13:28 | history | answered | xaxa | CC BY-SA 3.0 |