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Total energy of elliptical orbit

  1. Oct 3, 2009 #1
    Why is the total energy of an elliptical orbit given by:
    [tex]E_{tot}=\frac{-GMm}{2a}[/tex]
    Where a=semi major axis.
    I agree for a circular orbit I can do the following:
    [tex]F_c=F_g[/tex]
    [tex]ma_c=\frac{GMm}{r^2}[/tex]
    [tex]\frac{v^2}{r}=\frac{GM}{r^2}[/tex]
    [tex]v^2=\frac{GM}{r}[/tex]
    Since the total energy also equal to the kinetic plus potential energy we have:
    [tex]E_{tot}=\frac{1}{2}mv^2-\frac{GMm}{r}=\frac{1}{2}m\frac{GM}{r}-\frac{GMm}{r}=\frac{-GMm}{2r}[/tex]
    Ok this is a similar form for circular orbit. But how can we just put a in instead of r for elliptical orbit? What is the justification?
     
  2. jcsd
  3. Oct 3, 2009 #2

    Janus

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    Consider the fact that the total energy of the orbiting object is constant for any point of its orbit. Then take the visa-vis equation for finding the orbital speed at any point of an elliptical orbit:

    [tex]\sqrt{GM \left (\frac{2}{r}-\frac{1}{a} \right )}[/tex]

    Note that when r=a, you get

    [tex] v=\sqrt{\frac{GM}{a}} [/tex]

    or

    [tex] v^2=\frac{GM}{a} [/tex]

    Thus the total energy of a elliptical orbit with a semi-major axis of 'a' is the same as a that for a circular orbit with a radius of 'a'.
     
  4. Oct 3, 2009 #3
    Well actually I was trying to derive the vis-visa equation through the total energy. So I must have an alternate way to explain it or i'll be circular logic.
     
  5. Oct 3, 2009 #4

    Janus

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  6. Oct 4, 2009 #5
  7. Feb 18, 2010 #6
    A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 5e10 m (inside the orbit of Mercury), at which point its speed is 9e4 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6e12 m from the Sun?

    please show the equations needed, if you feel generous work the problem out as well.
     
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