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Eccentricity and specific mechanical energy

  1. Apr 27, 2010 #1
    Bate, Mueller, White: Fundamentals of Astrodynamics (Dover 1971), p. 29, equation 1.6-4 for the eccentricity of a satellite's orbit:

    [tex]e=\sqrt{1+\frac{2\mathcal{E}h^2}{\mu^2}}[/tex]

    [tex]e \text{, eccentricity}[/tex]

    [tex]\mathcal{E} \text{, specific mechanical energy}[/tex]

    [tex]h \text{, magnitude of specific angular momentum}[/tex]

    [tex]\mu \text{, gravitational parameter } = GM[/tex]

    (G the gravitational constant, M the mass of the planet which the satelite is orbiting.) The book describes the shape of orbits for specific mechanical energy negative (circle, ellipse), zero (parabola) and positive (hyperbola).

    My question: how do the definitions of these variables guarrantee that

    [tex]\frac{2 \mathcal{E} h^2}{\mu^2} \geq -1[/tex]

    for all physically possible orbits (given the simplifying assumptions: two bodies, the satellite much less massive than the planet, mass of planet spherically distributed, no other forces significant)?
     
  2. jcsd
  3. Apr 28, 2010 #2
    Good question! I'll take an initial stab at it.

    The short answer: There is no such thing as negative velocity, but more to the point, look to the definition of Specific Orbital Energy, which is the sum of Kinetic energy and Potential energy divided by the satellite mass.

    A cursory look at the equation cited shows that for e<1, which (by the geometry of conical sections) means you have an elliptical orbit, the specific energy must be negative (since squared values are always positive). Negative Energy? How can this be you may ask.

    We are used to thinking of Potential Energy relative to *us* but when it comes to orbits, it is defined relative to *everything else*. IOW The PE of a thing is zero if the thing is completely outside the gravity field of the planet (where g=0), and the PE is *negative* as the thing gets deeper into the gravity well of the planet.

    PE = -m*g*h

    So if the absolute value of the PE is greater than the absolute value of the KE, the total energy is negative and there is the first part of your answer, having explained how the inside of the radical, and thus in turn e, can be less than 1.0.

    The next part of the answer is to show how e cannot become an imaginary number (how the inside of the radical cannot become negative). I am not going to attempt that proof here but IINM it is relatively easy to do if you look up the "Vis-Viva equation", combine it with the equation cited and turn the algebra crank.

    This task can be simplified by considering the parabolic case where e=1.0 and thus the numerator of the second term inside the radical must be zero, meaning that the specific energy is zero, meaning that the KE is the same value as the PE, with opposite signs. This may not constitute a "proof" but hopefully will provide some understanding.
     
  4. Apr 28, 2010 #3
    Thanks, spacester. That just leaves the elliptical/circular case to consider. Combining their equation for eccentricity with the vis-viva equation, the condition we need becomes, if I've got this right,

    [tex]1+\frac{v^2h^2}{\mu} \geq \frac{1}{r}[/tex]

    If we can show that it's correct at one point on the orbit, the conservation of specific mechanical energy and specific angular momentum should ensure that it's correct generaly, since the eccentricity depends only on these variables. So if we set h=rv, as it is at the apses, the condition becomes

    [tex]\frac{v^4r^3}{\mu} \geq 0[/tex]

    and all the variables on the left are nonnegative, so the condition is fulfilled. How's that sound?
     
  5. Apr 29, 2010 #4
    NIce! You did that a lot more quickly than I would have. :-)
     
  6. Apr 29, 2010 #5
    Hmm, maybe I went too quickly... I think in my haste I made a mistake in the algebra on that first step.

    Take two:

    [tex]e=\sqrt{1+\frac{2 \mathcal{E}h^2}{\mu^2}}[/tex]

    Substitute for

    [tex]\mathcal{E}= \frac{v^2}{2} - \frac{\mu}{r}[/tex]

    and, with some rearrangement, we get the condition:

    [tex]\frac{1}{2}\left ( \frac{rv^2}{\mu} + \frac{\mu}{rv^2} \right ) \geq 1[/tex]

    The left hand side is an expression of the form

    [tex]\frac{x+1/x}{2}[/tex]

    Graphing it suggests that the condition is met for nonnegative x.

    http://www.wolframalpha.com/input/?i=%28x%2Bx^-1%29%2F2

    And this can be confirmed by checking the derivative:

    [tex]\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{x+1/x}{2} \right )=\frac{1}{2}\left ( 1-\frac{1}{x^2} \right )[/tex]

    which is zero when x=1 and -1, and we know that, in this case, x is nonnegative. And sure enough, the minimum is

    [tex]\frac{1}{2}\left ( 1+\frac{1}{1} \right )=1[/tex]
     
  7. Apr 29, 2010 #6
    Yeah, that's much better. I didn't see the algebra error at first, and while I was a little suspicious, when somebody answers their own question, it seems rude to quibble. :D

    However, math is math and what fun are forum threads if we don't find something to argue about? lol

    In that vein, I have to say that the "some rearrangement" step covers a lot of ground and I am not seeing just how you got there. Is the inequality reversed, IOW should it be less than rather than greater than?
     
  8. Apr 29, 2010 #7
    You're wise to be suspicious of my algebra ;-) But I think it checks out okay this time. Here's what I meant by rearrangement. Starting with

    [tex]e=\sqrt{1+\frac{2h^2\left ( \frac{v^2}{2} - \frac{\mu}{r} \right)}{\mu^2}}[/tex]

    the condition for e to be real becomes

    [tex]\frac{2h^2\left ( \frac{v^2}{2} - \frac{\mu}{r} \right)}{\mu^2} \geq -1[/tex]

    [tex]\frac{v^2h^2}{\mu^2}-\frac{2h^2}{r\mu} \geq -1[/tex]

    [tex]\frac{v^2h^2}{\mu^2}+1 \geq \frac{2h^2}{r\mu}[/tex]

    [tex]\frac{r \mu v^2h^2}{2h^2 \mu^2}+\frac{r \mu}{2h^2} \geq 1[/tex]

    Since, for a given orbit, specific angular momentum and specific mechanical energy are constants, we can solve for the case where h=rv at the apses, and know that the result will hold generally. So, set h=rv and do all the cancelling:

    [tex]\frac{1}{2} \left ( \frac{r v^2}{\mu}+\frac{\mu}{rv^2} \right ) \geq 1[/tex]
     
  9. May 2, 2010 #8
    OK then, we've covered the title topic rather well. In terms of problem-solving, there is another equation that people will very likely need, and that is the equation for an ellipse, given in several forms on the wikipedia article. For orbits, I like the polar equation with origin at a focus, Euler's equation:

    (I need to re-learn how to use LATEX properly . . . )

    r = a * (1-e^2) / (1 [tex]\pm[/tex] e * cos (theta))

    http://upload.wikimedia.org/math/0/b/a/0ba6cbf197ea5eb50b1ed497766b13f0.png

    r is the distance from the focus, which is also the altitude of an object on the orbital path (a spacecraft) relative to the center of the planet.

    a is the semi-major axis and is VERY handy because it is usually easy to determine, but more so because it equals the equivalent circular radius, giving you access to some simpler equations for circular orbits.

    e is eccentricity of course, and is the free parameter here.

    theta is the only other parameter, one of the two defining coordinates, and it turns out to be the angular coordinate of your location along the actual orbital path.

    So it is a nice equation for thinking about the orbital path from the perspective of a spacecraft.

    theta actually kind of hides a fourth parameter, in that you need to define where theta = 0, but that is simply taken as the point of closest approach to the primary focus, IOW the periapse.

    In addition to the geometrical equation, conservation of angular momentum often needs to be brought into the set of equations needed to solve orbital mechanics problems, as it did when you set h=rv at the apses. Energy Conservation is by the Vis-Viva equation. Let's see, what else might the student need to consider?

    The ellipse itself often needs to be defined in terms of its position and orientation in a general frame of reference. For the solar system, the plane of the ecliptic gives you a very good start, but the angular orientation can be tricky. The parameters here are inclination and 'right ascension', an angle with the zero point at 'the first point of Aries', which is the place in the sky where the Sun crosses the celestial equator at the Vernal (March) Equinox. The line referenced by the right ascension is typically the major axis of the orbit, placing the periapse at theta=0. The tricky part is when you are working with transfer orbits, where you don't know and don't want to bother with the right ascension.
     
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