1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Specific Orbital Energy confusion

  1. Apr 9, 2010 #1
    If specific orbital energy for an elliptical orbit is -[tex]\frac{\mu}{2a}[/tex] , shouldn't that mean that the further from the Earth an orbit is the lower its specific orbital energy? The Wikipedia page on specific orbital energy (admittedly not the best source) indicates the opposite, and if it were true that SOE decreased with semi-major axis wouldn't it be the case that it would be easier (in terms of energy needed) to reach the Moon than low earth orbit?

    I'm guessing my confusion comes from the negative sign... what does negative specific orbital energy even mean, anyway?

    Thanks for reading.
     
  2. jcsd
  3. Apr 9, 2010 #2

    Filip Larsen

    User Avatar
    Gold Member

    As you probably know, the (constant) total mechanical specific energy for body in a keplerian orbit is the sum of the kinetic and potential specific energy. Since potential specific energy is [itex]C-\frac{\mu}{r}[/itex], with [itex]C[/itex] being an arbitrary constant of integration, tradition has it to select a constant of zero (so the zero level of the potential is at infinity). The potential specific energy is then written simply as [itex]-\frac{\mu}{r}[/itex] and while this means it is always a negative value, it also means that the total specific energy is negative for bound elliptical orbits, zero for parabolic orbits and positive for hyperbolic orbits. In calculations you are of course free to choose a different reference level or constant of integration for the potential and total energy if that is a benefit for you.
     
    Last edited: Apr 9, 2010
  4. Apr 9, 2010 #3
    So what you're saying is that -[tex]\frac{\mu}{2a}[/tex] is the energy with the 0 reference point set at an infinite distance from the Earth, i.e. that it's the energy it would take to reach that orbit from infinity? Or was that just an explanation of the negative sign in the equation?
     
  5. Apr 9, 2010 #4

    Filip Larsen

    User Avatar
    Gold Member

    The first, but I can see my answers wasn't that helpful and that I should probably have answered your direct questions instead, so let me try that now

    Try make (or find) a sketch of the specific mechanical energy as a function of the semi-major axis a and compare energy values for different a values. As you can see, the further from earth (the bigger a is) the bigger the mechanical becomes. You could also try calculate the energy difference as

    [tex] \epsilon(a_2) - \epsilon(a_1) = (-\frac{\mu}{2a_2}) - (- \frac{\mu}{2a_1}) = \frac{\mu}{2}\left(\frac{1}{a_1}-\frac{1}{a_2}\right) = \frac{\mu}{2}\frac{a_2-a_1}{a_1 a_2} [/tex]

    As you can see, if [itex]a_2 > a_1[/itex] the change in mechanical energy from going from [itex]a_1[/itex] to [itex]a_2[/itex] is positive.
     
  6. Apr 9, 2010 #5
    OK, I think I see now. So my problem is that I was interpreting it as an absolute value (kinda), thinking that if [tex]|-\frac{\mu}{2a}|[/tex] was smaller, then it took less energy to reach that orbit. But because it becomes a smaller negative number, it is actually taking more energy, as shown by the energy difference you calculated. And to get the specific energy to reach an orbit from Earth's surface, I'd have to do the integration for the PE with different bounds so that r=0 means PE=0, and add the KE for the orbit's velocity (and also account for the non-zero initial launch speed, unless launching from the north or south pole).

    Is the above correct?

    Thanks a lot for the help!
     
  7. Apr 9, 2010 #6

    Filip Larsen

    User Avatar
    Gold Member

    Yes, that is correct (though a more precise wording would be "because it becomes a negative number with a smaller absolute value" or "because it becomes a bigger number")
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook