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Earth's Orbit

  1. Aug 6, 2005 #1
    How do you calculate the kinetic and potential energies of the Earth at the perihelion and aphelion?

    Given that the Earth's orbit is nearly circular

    GMm/R^2 = m(v^2)/R where M = sun's mass, m = Earth's R = orbital radius

    so

    v = √(GM/R)

    The kinetic energy, E, would be

    E = (1/2)m(v^2) = (1/2)GMm/R

    The Earth's potential energy, U, would be

    U = -GMm/R

    However, the change in E between the perihelion and aphelion would be

    (1/2)GMm/[(1/R) - (1/r)] where R and r are the respective orbital radii;

    while the change in U would be

    -GMm/[(1/R) - (1/r)] .

    Given conservation of energy, the last two terms can not both be right.

    This brings me back to the original question.
     
  2. jcsd
  3. Aug 6, 2005 #2

    Janus

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    This is not correct. What you have calculated here is the difference in E between two circular orbits of radii r and R. With a ellipitical orbit the orbital velocity at perhelion is greater than the circular orbital velocity for that radius and the orbital velocity at aphelion is less than the circular orbital velocity for that radius.
     
  4. Aug 6, 2005 #3

    Janus

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    Let's see if I can give you a push in the right direction:

    The equations:

    [tex]V = \sqrt{\frac{GM}{r}}[/tex]
    and
    [tex]E = \frac{GMm}{2r}[/tex]

    Are only true for an elipitical orbit when r = A (The semimajor axis of the orbit, or average orbital distance).

    [tex]U = -\frac{GMm}{r}[/tex]

    is true for all points of the orbit.

    So what is the total energy (E+U) at a radius A?
    is This the same as the total energy at r and R?

    Hopefully, this will give you enough to go from here.
     
  5. Aug 6, 2005 #4

    BobG

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    You're using rough estimates (treating the orbit as circular for each velocity, for example), so it won't cancel out. If you use exact values, it will.

    First off, it's easier if you use the specific energy per unit of mass to analyze the motion, then multiply the mass back in if you need an actual value for energy.
    [tex]E=m\epsilon[/tex] where [tex]\epsilon[/tex] is specific energy
    [tex]m\epsilon=m(\frac{1}{2}v^2-\frac{GM}{r})[/tex]
    [tex]\epsilon=\frac{1}{2}v^2-\frac{GM}{r}[/tex]

    If you rearrange this:

    [tex]v=\sqrt{2\epsilon+\frac{2GM}{r}[/tex]

    E is also equal to:

    [tex]\epsilon=-\frac{GM}{a}[/tex]
    where a is your average radius (semi-major axis).

    Substituting and rearranging, you get Leibniz's vis-viva equation:

    [tex]v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}[/tex]
    (If you have a circular orbit, the radius and semi-major axis are equal, which reduces to the equation you used.) If you use this equation for velocity, your calculations will show energy is conserved. (substitute the velocity equation into the specific kinetic energy equation)

    You have to calculate the specific potential energy for both perihelion and apohelion, as well. In other words, substitute the radius of perihelion for R and the radius of apohelion for R.

    If you need the actual energy of the orbit, multiply the mass back in, but it's not necessary to show energy is conserved.
     
  6. Aug 6, 2005 #5
    Thanx.

    I understand this now.
     
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