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How can I find the length of a day on Jupiter from wavelengths?

  1. Oct 18, 2013 #1
    The question is: the wavelength of a spectral line on the left edge of Jupiter's equator is 499.98 nm, and the right edge is 500.02 for the same spectral line, what is the length of jupiter's day (in hours)?
    I know that its 10 hours but I need to show my work. I tried using many different equations such as time = dis./speed, frequency = C/wavelength, etc. but it still gave me a random number of 1.4something, I also made sure that all the units were the same, can someone please help me?
     
  2. jcsd
  3. Oct 19, 2013 #2

    Borek

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    This is about Doppler shift. Somehow I am not convinced you can solve the problem without knowing the planet radius.
     
  4. Oct 19, 2013 #3
    oh yeah sorry, i forgot to mention that more info about the planet including its radius is in my textbook, so im gonna try to use Doppler Shift formula and see if it works

    EDIT: ok so i tried but for some reason the radial velocity i got was about 24 km/s, but the actual orbit velocity of the planet is 13.1 km/s, is there any way i can convert radial velocity (24 km/s) to the actual orbit velocity (13.1 km/s)? thanks
     
    Last edited: Oct 19, 2013
  5. Oct 19, 2013 #4

    Redbelly98

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    The orbital velocity is not relevant to solving this problem. The idea is to use the relation between the radius and the speed for a spinning object.
     
  6. Oct 19, 2013 #5
    sorry, could you elaborate? because I thought that to find the time of day, you have to divide the circumference of the planet by its orbital velocity (not radial velocity)
    because time = distance(or circumference)/speed (or orbital velocity)
     
  7. Oct 19, 2013 #6
    No, the length of the day has nothing to do with orbital velocity.
    The length of the day is determined by the rotation around the planet's own axis.
    The length of the year is determined by the orbital motion.
     
  8. Oct 19, 2013 #7

    Redbelly98

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    It is correct that time = distance(i.e. circumference) / speed.

    But which speed? And the circumference of what circle?

    The orbital velocity is the speed of Jupiter along it's orbit around the Sun, taking the Sun as stationary. Using the orbital speed and the circumference of the orbit, one could calculate the time it takes Jupiter to orbit the Sun. But that is not what the question is asking for.

    The velocity you need for this problem is the speed at the surface of Jupiter, taking the center of Jupiter as stationary. Using that speed and the circumference of a circle the size of Jupiter, one could calculate the time it takes Jupiter to spin on it's axis. This is what the question is asking for.

    Hope that's clear, if not then post again.

    p.s. welcome to Physics Forums.
     
  9. Oct 20, 2013 #8
    Thank you for posting back but something that doesn't make sense is that when I plugged in Jupiter's rotation speed = 13.1 km/s into the time equation (instead of the answer I got for the radial velocity which was 24 km/s) and using Jupiter's circumference of 449,200 km for distance, the answer I get is actually right, it gives me an answer of about 10 hours, which is the length of the day on Jupiter. And also, how can you calculate the rotation speed from just the two wavelengths at each edge of Jupiter's equator? that equation (the Doppler Shift equation) will only give you the radial velocity, not the actual rotation speed of Jupiter on its axis, which is what I need to plug in into the time equation. Thank you again because this question has been driving me crazy for the past few days.
    p.s sorry I made a small mistake, I called the rotation speed "orbital velocity" in my previous posts by accident, I meant the rotation speed around Jupiter's axis, not around the sun
     
  10. Oct 20, 2013 #9

    Borek

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    Draw the Jupiter (as seen from its rotation axis) and the observer. In what direction does the surface at Jupiter's equator move at the very edge of the planet as seen by the observer?
     
  11. Oct 20, 2013 #10
    You got the wrong radial velocity. The correct radial velocity doesn't differ all that much from the orbital speed.
     
  12. Oct 20, 2013 #11

    Redbelly98

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    Ah, you're right.

    Remon, do you see where the error is? (If not, then could you show us how you calculated the 24 km/s?)
     
  13. Oct 20, 2013 #12
    No, I honestly don't see how the calculations were wrong since I checked it many times, here's exactly how I did it:
    I used the Doppler Shift equation which is: Vrad/C = (λshift - λrest)/λrest, and then I solved for Vrad
    so it equals: Vrad/(3x105 km/s) = (500.02 nm - 499.98 nm)/(499.98 nm) (I don't think it matters which one is the shift or rest wavelength since all that tells us if Jupiter is moving away or towards us and that only changes the sign, which not what the question is asking for)
    then I got: Vrad = (8x10-5)(3x105 km/s), then i got Vrad = 24 km/s (which what you guys said might be wrong)
    Then I plugged that and the circumference of Jupiter into the time equation which got me ≈ 5 hours, which I know is wrong.
    Thank you again guys, I know you're trying to help.
    p.s the radial velocity sometimes differs from the actual rotation speed because it only tells us part of what the actual speed is, and that part is only relative to us (that is, if its moving away or towards us)
     
  14. Oct 20, 2013 #13

    Borek

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    You are off exactly by the factor of two. You ignored the fact one edge gets closer while the other edge gets farther from us at the same rate.
     
  15. Oct 20, 2013 #14
    What do you mean I ignored this fact? All I did was plug in the numbers into the two equations, but the radial velocity does not help me, I need to somehow find the actual rotation speed (which is 13.1 km/s for Jupiter) from the radial velocity (which is 24 km/s), but I don't know how to do that
    Are you saying that I should divide the radial velocity by 2 to account for what you said? because that would give me 12 km/s which is close to the actual rotational speed of 13.1 km/s and give me ≈ 10.3 hours, which would be right
     
  16. Oct 20, 2013 #15
    Why would you take the wavelength from the other side as the rest wavelength?
    Both sides are moving.
     
  17. Oct 20, 2013 #16
    I didn't really consider it as the rest or shift wavelength, I just simply put it in the equation as the rest wavelength because that's what the equation calls it, for all I know, I could've just named the wavelengths: λ1 & λ2
    But I think what's wrong here is the difference between the two wavelengths, which is 0.04.
    I tried to plug in 0.02 as the difference and got a rotation speed of around 12 km/s, which gives me an answer of about 10 hours.
    So maybe what I'm doing wrong here is that I didn't consider the fact that if one edge has a λ of 500.02 nm and the opposite edge has a λ of 499.98 nm, then the center would have a wavelength of exactly 500 nm, which would mean that the λrest is actually 500 since the center is at rest and the λshift is either 500.02 nm or 499.98 nm depending on which edge you use which would give a difference of 0.02 (which is the right answer) instead of a difference of 0.04 (which is the wrong answer), which would lead to the correct number of hours, do you guys agree with me? because it seems to finally make sense.
     
  18. Oct 20, 2013 #17

    Borek

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    That's exactly what I meant - you initially ignored the fact that both wavelengths given are of moving objects.
     
  19. Oct 20, 2013 #18
    Oh ok, I think I get it now, it's just that you didn't say that the center was the rest wavelength and that it equaled 500 nm, which took me a while to figure out. The question would've been much more straight forward if it just said that the center is 500 nm and that one of the edges is 500.02 (or 499.98), but thank you all for helping.
     
  20. Oct 20, 2013 #19

    Borek

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    The question could be even more straight forward by asking "what is the length of Jupiter's day if the length of Jupiter's day is 10 hours?" :tongue2:
     
  21. Oct 20, 2013 #20
    Really? I think that would a little too hard for my tiny brain to fathom :rolleyes:
     
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