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Homework Help: The ionosphere

  1. Feb 23, 2006 #1
    An ionosonde on a satellite orbiting at 1000 km probes the topside of the F2 layer. If the ionosonde transmits radio pulses downwards at 5.6 MHz and 7.1 MHz and receives radio echoes with delay times of 2 msec and 2.667 msec respectively, determine (a) the F2 electron density at an altitude of 500 km, and (b) the exospheric temperature. You may ignore the effects of any magnetic fields and assume that the temperature is constant above the F2 peak which lies well below the 400 km level.

    i dont understand how a radio echo can be used to determine anything...
    i think that to find the answer for part b i have to find the scale height of the electrons because h = RT/Mg

    i suspect it is similar to something that was answered in this thread
  2. jcsd
  3. Feb 24, 2006 #2


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    At the very least, the relative delay time can be used to find the separation between the critical layers for the two frequencies.
  4. Feb 25, 2006 #3
    does the delay time to the cutoff frequency in some way?
  5. Feb 25, 2006 #4


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    I think you left out a verb in your question so I am not quite sure what you're asking.

    Perhaps this will help. If the two waves have different frequencies then they will reflect from different places in the ionosphere. Each will reflect from its corresponding critical surface or critical density. Since they are different, one wave will have to travel farther to reach its reflection point and so it will take longer for its reflected signal to be returned to the detector.

    The simplest approximation you can make is that the two waves travel at the speed of light down to and back from their respective reflection points. This isn't quite correct since the group velocity of a wave tends to zero as it approaches the reflection point. It is for you to decide how much error you can accept for the respective travel times.

    In either case, the fact that one wave takes somewhat longer to return gives you some information about the spatial separation of the two reflection points.
  6. Feb 26, 2006 #5
    ok i can find the separation between the layers.
    i m not too sure on how that leads to finding the elctron density, however
  7. Feb 26, 2006 #6


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    Here's a simple example. Suppose the electron density varies linearly (your density model is more complicated so you'll have to work harder but I'm just demonstrating the principle.)

    [tex]n = n_0 \frac {x}{L}[/tex]

    The reference density [itex]n_0[/itex] and scale length L are unknown.

    However, you do know the travel times to and from each critical surface so

    [tex]x_i = c t_i / 2[/tex]

    where [itex]t_i[/itex] is the delay time for each wave (i = 1, 2) and [itex]x_i[/itex] is the location of each reflection point.

    Now we have

    [tex]n_1 = n_0 \frac {c t_1}{2L}[/tex]

    [tex]n_2 = n_0 \frac {ct_2}{2L}[/tex]

    with two unknows [itex]n_0[/itex] and L. You can easily solve these equation and once those two quantities are determined you can find the electron density at any location!

    As I pointed out earlier, you may want to improve on your calculation by integrating [itex]dx/v_g[/itex] to more accurately reflect the variable speed of each wave as it approaches the critical density. My guess is that the difference will be relatively small since the scale length will turn out to be fairly large but you should check anyway.
  8. Feb 26, 2006 #7
    [tex] n_{e}(h,\chi) = \sqrt{\frac{q_{max}}{keff}} \exp[0.5(1-z-\sec \chi \exp(-z))] [/tex]

    since h1 = c (2msec) /2
    [tex] n_{e}(h_{1},\chi) = \sqrt{\frac{q_{max}}{keff}} \exp[0.5(1-z_{1}-\sec \chi \exp(-z_{1}))] [/tex]

    the second one is h2 = c (2.667msec)/2
    [tex] n_{e}(h_{2},\chi) = \sqrt{\frac{q_{max}}{keff}} \exp[0.5(1-z_{2}-\sec \chi \exp(-z_{2}))] [/tex]

    but the thing is the satelite is shooting from the top of the atmosphere so this value of h is 1000 - h above the surface of earth

    is this teh way to go?
    but arent there a lot more unknons here?
    Last edited: Feb 26, 2006
  9. Feb 26, 2006 #8
    and we can find the Ne values for the heights using the relation between the density of electrosn and the cirtical frequency. Right?
  10. Feb 26, 2006 #9


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    I don't know how you defined z but it looks like it is a normalized height so it will likely contain a scale factor (what I called L) and the factor

    [tex]\sqrt {\frac {q_{max}}{k_{eff}}}[/tex]

    must be a reference density so you appear to have only two unknowns.

    And, yes, you will need to do a transformation from "ground based" to "space based" coordinates.
  11. Feb 26, 2006 #10
    s iz defined like this
    [tex] z = \frac{h - h_{max}}{H} [/tex]
    and [tex] H = \frac{RT}{Mg} [/tex]
    H is the sacle height where T is the temperature also unknown
    Keff is the reaction coefficnet of the electron contributing reaction
    q max is the maximum ion concentration
    and chi is the angle of attack of the signal
    we're assuming here that the angle of attack is the same, but unknown

    so then
    [tex] n_{e}(h_{1},\chi) = \sqrt{\frac{q_{max}}{keff}} \exp[0.5(1-\left(\frac{h_{1}-h_{max}}{H}\right)-\sec \chi \exp\left(\frac{h_{max}-h}{H}\right))] [/tex]
    seems to have more than 2 unknowns... or are they related to each other somehow?
  12. Feb 26, 2006 #11


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    You have a relation between the scale height and temperature so one of the unknowns is eliminated. Also, if you are getting a return signal at the source then you also know the "angle of attack" (you must have normal incidence in order for the signal to return to the source).

    That leaves the square rooted quantity (it is the reference density which you will treat as a single quantity), H and hmax as your unknowns. You do need more information to solve for three unknowns. You may have additional knowledge at your disposal that I am unaware of. Is there anything else in your model that might be used to eliminate one of the remaining unknowns?
  13. Feb 26, 2006 #12
    i can find Ne (h,chi) and h1 and h2 can be found out using what was discussed later

    so that only leves the refernce quantity and h max and H which is related to the temperature
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