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Debye Potential and Jupiter encounter

  1. Feb 20, 2004 #1
    1)
    It is desired to use an encounter with Jupiter to send a spacecraft on a trajectory which barely escapes the solar system. The space craft is launched from Earth onto a trajectory whose aphelion is at Jupiter’ s orbit. Determine how close to the surface of Jupiter the spacecraft must pass during the encounter. I seem to feel that energy conservation would be required here but am not entirely sure as to what the approach should be exactly.

    2)
    Given the Debye potential V( r ) = (A/r) exp(-r/lambda) where r is spherical radius and A and lambda are constants, find the range of radii for which circular orbits are stable in this potential.

    For on this question, I am genuinely lost and am open to all options and suggestions !
     
  2. jcsd
  3. Feb 21, 2004 #2
    Anything here/...come on guys please!
     
  4. Feb 21, 2004 #3
    Possible Debye solution

    Ok,,,for the Debye potential, here' s what I think......
    Can someone please check it.

    E is the total energy

    E = (1/2)m(vsubr)^2 + V(r) + L^2/2mr^2 ,call this the energy equation

    V(r) is given and vsubr = 0 for circular orbits....

    Then dE/dr = 0 so upon differentiating E for r I would then get an expression for L ^ 2 ....


    Plug that expression into the energy equation to get an expression for the Energy that does not include L. Since for stability, E < 0 is required, I would then simply solve for that expression for E < 0 and get some r range...


    I get that r cannot equal 0 and r < lambda - 2 so r must be greater than 2. Can someone please check this?
     
  5. Feb 22, 2004 #4
    I'm probably getting in over my head here, but since no-one else seems to be just yet...

    What's your definition of "stable?" Do you mean a circular orbit such that the orbiting object, if perturbed, would maintain orbit? Or do you mean a non-degenerate, bounded orbit?

    Requiring E < 0 simply provides that the orbit is elliptical and not hyperbolic (E > 0) or parabolic (E = 0), i.e. that it is bounded.

    cookiemonster
     
  6. Feb 22, 2004 #5
    I think that it must be the one for the perturbed bit! When we discussed the topic in class, it was titled Stability of circular orbits....there was no other stuff mentioned anywhere!
    If there is another way of approaching the question, I am open to suggestions!In class, he used a general potential equation...something like -alpha^(-n) or something and the way that things turned out, only E<0 returned stable circular orbits.....I used the same analysis but just used the given potential instead of the one we used in class.
     
  7. Feb 22, 2004 #6
    I think I can help with the stability of orbits, then. As for the Jupiter slingshot, keep in mind that even though the spacecraft is never coming back, it's still in an orbit. Particularly an orbit with E>0, and since it's in orbit, the total energy will remain constant. The only two components of that energy are kinetic energy and potential energy, however potential energy comes from two places: the sun's gravitational field and Jupiter's.

    [tex]
    m\stackrel{..}{r} = F = -\frac{dU(r)}{dr}
    [/tex]
    [tex]
    m\stackrel{..}{r} = \frac{ml^2}{r^3} + f(r)
    [/tex]

    That's just F = ma, using the effective potential to define force and f(r) is the central force function. Note that [itex]m\stackrel{..}{r} = 0[/itex] and r = a constant (I'll use r = a) for circular orbits. Then,

    [tex]
    -\frac{ml^2}{a^3} = f(a)
    [/tex]

    Now let's change the coordinates to one more convenient for expressing perturbations. Particularly, we're going to redefine the coordinates as the difference between the expected circular radius and the actual radius.

    [tex]
    x = r - a
    [/tex]
    Re-express F = ma using the new notation:
    [tex]
    m\stackrel{..}{r} = F = -\frac{dU(r)}{dr}
    [/tex]
    [tex]
    m\stackrel{..}{x} = ml^2(x+a)^{-3} + f(x+a)
    [/tex]

    Expand the (x+a)^(-3) as a power series,

    [tex]
    m\stackrel{..}{x} = ml^2a^{-3}(a - 3\frac{x}{a} + \cdots) + [f(a) + f'(a)x + \cdots]
    [/tex]

    Truncate to only the first power of x and substitute in the expression we got for f(a) to yield

    [tex]
    m\stackrel{..}{x} + [\frac{-3}{a}f(a) - f'(a)]x = 0
    [/tex]
    [tex]
    m\stackrel{..}{x} = -[\frac{-3}{a}f(a) - f'(a)]x
    [/tex]

    Note that this is the equation for a simple harmonic oscillator iff the coefficient of x is positive. If the coefficient is not, then the solution to x is not stable and neither is the orbit. Therefore, require the coefficient to be greater than 0.

    [tex]
    \frac{-3}{a}f(a) - f'(a) > 0
    [/tex]

    Think you can solve it from there?

    cookiemonster
     
    Last edited: Feb 22, 2004
  8. Feb 22, 2004 #7
    Did you get that from Analytical Mechanics by Fowles...the exact thing is in there...wow.. a lot of people on here are using that one..the same book that I am using!
     
  9. Feb 22, 2004 #8
    Thanks a lot with the Debye problem as it cleared so many things up. Very useful stuff really man.

    Now can you help me a bit more with the Jupiter problem. I understand what you say about the energy being constant but am unsure as to what comes next, know what I mean? Anything useful would be great!
     
  10. Feb 22, 2004 #9
    Symbol for symbol with just a little bit of elaboration, mark.

    How about starting by modeling the energy at a few locations, student?

    cookiemonster
     
  11. Feb 23, 2004 #10
    Ok here goes...

    At location 1) Energy at Earth...KE+PE, where KE = 0 and PE = - GMm/R where R is the radius of the Earth and M is the mass of the Earth. m is the mass of the spacecraft.

    At location 2) Now, for the encounter with Jupiter, I am not sure what goes on here! The spacecraft just enters the gravity field in which case it has PE and KE here...Ke is the usual (1/2)mv^2 but the expression for PE is puzzling me.

    At location 3) Now to excape, E = 0 so at the point where it has just barely excaped, according to me , KE = -PE...should they both(KE and PE) be zero since it is not moving at a speed, just drifting and does not experience any potential....


    Unfortunately, that is all I can think of Any feedback would be great
     
  12. Feb 23, 2004 #11
    Does that make sense to you? I am still not convinced that it is infact the correct analysis so any help would be incredibly useful.
     
  13. Feb 23, 2004 #12
    Really, anything would help me..please !
     
  14. Feb 23, 2004 #13
    Ok this Jupiter question is really driving me nuts...please some one help me !
     
  15. Feb 23, 2004 #14
    Heh, I was hoping somebody else would hop in here because I've never actaully done one of these problems. But since that doesn't seem to be the case, this is how I'd approach it:

    Just barely escaping tells me: [itex]E_\textrm{total} = 0[/itex], beacuse the potential energy is 0 and the kinetic energy is 0.

    This suggests that, at any given time, [itex]T + U = 0[/itex].

    Invoke this at the Earth location: [itex]U_\textrm{Earth} + U_\textrm{Sun} + T = 0[/itex]

    I guess we have to assume that the spacecraft is launched from the average radius of the earth, so that will give us the potential on the Earth. The potential due to the Sun is pretty constant, so treat is as such. We should be able to use this to get the angular momentum, which will stay constant throughout the orbits.

    Invoke this at the Jupiter location: [itex]U_\textrm{Jupiter} + U_\textrm{Sun} + T = 0[/itex].

    The Sun's potential at Jupiter is pretty much constant, so let's treat it as one. Jupiter's potential energy is a function of the approach distance, which you're solving for. You should be able to find the kinetic energy as a function of approach distance and angular momentum, which means you will have an equation in which the only variable is approach distance.

    I hope this helps. I don't know if it does because I haven't worked it out and I've never worked a problem like this (stupid freshman, eh?). If it's not working out, then I'm probably wrong and leading you in the wrong direction. If I am, I apologize, and hopefully somebody will cut in and mention it.

    cookiemonster
     
  16. Feb 24, 2004 #15
    Yeah that does make sense to me but I have a few questions...

    "The potential due to the Sun is pretty constant, so treat is as such. We should be able to use this to get the angular momentum, which will stay constant throughout the orbits."

    I am not sure that I understand this fully. The Potential due to the Sun do you mean at the Earth' s location? That would be -GMsunMearth/Earth-Sun distance right? Now I am confused as to how I would use the Angular momentum thing the way that you say so.

    "Invoke this at the Jupiter location: .
    Jupiter's potential energy is a function of the approach distance, which you're solving for."

    That would be -GMsunMJupiter/Jupiter-Sun distance right?

    "You should be able to find the kinetic energy as a function of approach distance and angular momentum, which means you will have an equation in which the only variable is approach distance. "

    Yeah I need some more clarification here as well please.

    The rest of it does make sense.
     
  17. Feb 24, 2004 #16
    Heh, sorry. I should have been much more clear. Re-reading my post, it is quite ambiguous.

    All of the potentials are resultant from forces acting upon the spacecraft, originating from their subscript. The separation you'll be looking for will always be between the spacecraft and the body in the subscript. For instance, [itex]U_\textrm{Jupiter}[/itex] is the potential energy of the spacecraft in Jupiter's gravity well. As such, the radius you're looking for here would be the distance between the center of Jupiter and the spacecraft. [itex]U_\textrm{Sun}[/itex] would be the potential energy of the spacecraft in the Sun's gravity well and, as such, you would want the separation between the spacecraft and the Sun.

    Another note: [itex]U_\textrm{Sun}[/itex] in the 3rd and 4th equations are not the same. The equation to get them will be the same, but the separation will be much greater (and hence the [itex]U_\textrm{Sun}[/itex] will be smaller) in the 4th equation. It wouldn't surprise me if both [itex]U_\textrm{Sun}[/itex] terms are so insignificant compared to the [itex]U_\textrm{Earth}[/itex] and [itex]U_\textrm{Jupiter}[/itex] terms that they can be ignored. I put them there just to be safe.

    Now let's rewrite the 4th equation as functions of their independent variable.

    [tex]U_\textrm{Jupiter}(a) + U_\textrm{Sun} + T(a) = 0[/tex]
    where "a" is the minimum approach to Jupiter (how close it gets).

    It can be shown (hopefully, I haven't done it, although I will if I find time) that [itex]U_\textrm{Jupiter}[/itex] depends only on "a". Additionally, [itex]T[/itex] can be shown to depend only on "a". [itex]U_\textrm{Sun}[/itex] is a constant. Therefore, the only variable in the equation is "a". So rewrite each term in that form and solve for "a" and you should arrive at an answer.

    cookiemonster
     
  18. Feb 24, 2004 #17
    It would be great if you could find the time to do it the way you did the Debye potential one...that was a very clear explanation...hopefully you can do the same here ...please .
     
  19. Feb 24, 2004 #18
    Well, I went and worked on the Jupiter problem a bit. The conclusion? I think I was running in circles before...

    I'm not going to copy the derivations for the stuff I use. It'll probably be in your book if you look.

    "Barely escapes" implies that Total Energy as r->infinity is 0 because "barely" means escapes with no velocity (and hence no kinetic energy) and "escapes" means potential = 0, thus:

    [tex]U + T = E[/tex]
    [tex]0 + 0 = E[/tex]

    Now we've established somewhere back there that E = 0 is the condition for a parabolic orbit. A parabola is a conic section with eccentricity of 1. So, let's write the general conic section and apply eccentricity = 0:

    In general,
    [tex]r = \frac{\alpha}{1+\epsilon\cos\theta}[/tex]
    where [itex]\alpha = \frac{L^2}{km}[/itex], m is mass, L is angular momentum, and k is GMm.
    Apply [itex]\epsilon = 0[/itex] to get:
    [tex]r = \frac{\alpha}{1+\cos\theta}[/tex]
    Which is minimized when [itex]\cos\theta = 1[/itex]. Thus,
    [tex]r_\textrm{min} = \frac{\alpha}{2} = \frac{L^2}{2km}[/tex]

    The angular momentum (as far as I can tell) is not uniquely determined by the problem, as the particular orbit used is not described, so that means that this is as simple as we can make it.

    Maybe somebody will show up with a definite answer, but this is the best this senior can do.

    cookiemonster
     
    Last edited: Feb 24, 2004
  20. Feb 24, 2004 #19
    I understand....finally...it is starting to make a lot of sense...now for one final question..is what you found how close to the surface of Jupiter the spacecraft must pass during the encounter ? i.e. is rmin given by how close to Jupiter if must pass?
     
  21. Feb 24, 2004 #20
    Yes, what I solved should be the distance of closest approach. I forgot to note that I treated the potential due to the sun as negligible and didn't include it.

    I was re-reading the problem, particularly the part about "The space craft is launched from Earth onto a trajectory whose aphelion is at Jupiter’ s orbit."

    The "aphelion" is the maximum distance from the orbited body. Considering that, the question doesn't make sense. The aphelion, if the spacecraft is to escape, must be infinity. However, the problem states that the farthest the spacecraft ever gets is Jupiter, so it clearly cannot escape.

    Odd, huh.

    cookiemonster
     
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