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Second Kepler law for hyperbolas

  1. Jul 23, 2013 #1
    A body can describe 3 types of orbits around another (considering only the gravitational force), a elipse, a hyperbole or a parabola.
    Does the second kepler law (area law) work for hyperboles or parabolas too?
     
  2. jcsd
  3. Jul 23, 2013 #2

    WannabeNewton

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    Yes. Kepler's 2nd law holds for any central force motion. Here is one proof:

    Consider a particle moving under a central force ##F = f(r)\hat{r}##. We see immediately that ##\frac{\mathrm{d} L}{\mathrm{d} t} = r\times F = 0## hence ##L = r\times p## is a constant of motion. Now consider the position of the particle at an instant ##t## and another instant ##t + \Delta t## so that the position of the particle is given in polar coordinates by ##(r,\theta)## and ##(r + \Delta r, \theta + \Delta \theta)## respectively. For small ##\Delta \theta##, the area swept out during this interval will be given by the triangle swept out of side length ##r + \Delta r## and height ##r\Delta \theta## so ##\Delta A = \frac{1}{2}r^{2}\Delta \theta + \frac{1}{2}r\Delta r \Delta \theta ## hence ##\frac{\Delta A}{\Delta t} = \frac{1}{2}r^{2}\frac{\Delta \theta}{\Delta t} + \frac{1}{2}r\frac{\Delta r \Delta \theta }{\Delta t}## so ##\frac{dA}{dt} = \lim_{\Delta t\rightarrow 0 }\frac{\Delta A}{\Delta t} = \frac{1}{2}r^{2}\frac{d\theta}{d t} = \text{const.} ## as desired.
     
  4. Jul 23, 2013 #3

    mathman

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    Spelling error. You are talking about hyperbolas. Hyperbole is a completely different thing.
     
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