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Relativistic frequency shift of satellite signal

  1. Jun 24, 2014 #1
    Relativistic frequency shift of satellite signal

    1. The problem statement, all variables and given/known data
    An Earth satellite used in the Global Positioning System moves in a circular orbit with period 11 h 58 min. (a) Determine the radius of its orbit. (b) Determine its speed. (c) The satellite contains an oscillator producing the principal nonmilitary GPS signal. Its frequency is 1575.42 MHz in the reference frame of the satellite. When it is received on the Earth's surface, what is the fractional change in this frequency due to time dilation, as described by special relativity? (d) The gravitational blueshift of the frequency according to general relativity is a separate effect. The magnitude of that fractional change is given by

    ##\frac{\delta f}{f} = \frac{\delta U_g}{mc^2}##

    where ##\delta U_g/m## is the change in gravitational potential energy per unit mass between the two points at which the signal is observed. Calculate this fractional change in frequency. (e) What is the overall fractional change in frequency?

    ##c = 2.998\times 10^8 m/s\\
    G = 6.671\times 10^{-11}\frac{Nm^2}{kg^2}\\
    r_{Earth} = 6.3781\times 10^6 m\\
    m_{Earth} = 5.97219\times 10^{24} kg\\##

    2. Relevant equations
    ##F_g = \frac{Gm_{sat}m_{Earth}}{r^2}\\
    F_c = \frac{m_{sat}v^2}{r}\\
    rate = \frac{distance}{time}\\
    f_{obs} = \frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}f_{source}\\
    t' = \gamma t = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}}\\
    T = \frac{1}{f}\\
    fractional\ change = \frac{f_{new}-f_{original}}{f_{new}}\\
    U = \frac{Gm_{sat}m_{Earth}}{r}##

    3. The attempt at a solution
    (a) ##F_{sat} = 0 = F_c - F_g = \frac{m_{sat}v^2}{r} - \frac{Gm_{sat}m_{Earth}}{r^2}\\
    \frac{m_{sat}v^2}{r} = \frac{Gm_{sat}m_{Earth}}{r^2}\\
    v^2 = \frac{Gm_{Earth}}{r}\\
    rate_{sat} = \frac{distance_{sat}}{time_{sat}} = \frac{2\pi r}{T}\\
    \frac{Gm_{Earth}}{r} = v^2 = (\frac{2\pi r}{T})^2 = \frac{4\pi^2r^2}{T^2}\\
    r^3 = \frac{Gm_{Earth}T^2}{4\pi^2}\\
    r = \sqrt[3]{\frac{Gm_{Earth}T^2}{4\pi^2}}\\
    r = \sqrt[3]{\frac{(6.671\times 10^-11\frac{Nm^2}{kg^2})(5.97219\times 10^{24} kg) (43080s)^2}{4\pi^2}} = 2.656 \times 10^7 m##

    (b) ##v = \frac{2\pi r}{T} = \frac{2\pi(2.656\times 10^7m)}{43080s} = 3874 m/s##

    (c) ##fractional\ change = \frac{f_{new}-f_{original}}{f_{original}}\\
    f_{new} = \frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}f_{original}\\
    fractional\ change = \frac{\frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}f_{original}-f_{original}}{f_{original}}
    =\frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}-1\\
    =\frac{\sqrt{1 + \frac{3874m/s}{2.998\times 10^8m/s}}}{\sqrt{1 - \frac{3874m/s}{2.998\times 10^8m/s}}}-1 = -1.292\times 10^{-5}##

    (d) ##\frac{\Delta f}{f} = \frac{\Delta U_g}{mc^2}
    = \frac{U_{satellite} - U_{surface}}{mc^2}\\
    = \frac{\frac{Gm_{satellite}m_{Earth}}{r_{orbit}} - \frac{Gm_{satellite}m_{Earth}}{r_{surface}}}{m_{satellite}c^2}\\
    = \frac{Gm_{Earth}}{c^2}\left({\frac{1}{r_{orbit}} - \frac{1}{r_{surface}}}\right)\\
    = \frac{\left(6.671\times 10^-11\frac{Nm^2}{kg^2}\right)(5.97219\times 10^ {24}kg)}{(2.998\times 10^8m/s)^2}\left(\frac{1}{6.3781m} - \frac{1}{2.656\times 10^7m}\right)\\
    = 5.281\times 10^{-10}##

    (e) ##fractional\ change_{total} = fractional\ change_{special} + fractional\ change_{general} = -1.292\times 10^{-5} + 5.281\times 10^{-10} = -1.292\times 10^{-5}##

    The problem I'm having is with part (c) (and by extension, then, part (e)). This question comes from a book (Modern Physics, 3rd edition, Serway, Chapter 1, #39), and its answers are in the back of the book. My answers for (a) and (b) match with the back within calculator error, but its answer for (c) is -8.34 x 10-11. I've looked online for help on this problem, and found the following solution at http://web.unbc.ca/~hussein/Phys_205_Fall_2004/Selected_Ch01.pdf: [Broken]

    ##T = \gamma T'\\
    f = \frac{1}{T}\\
    fractional change = \frac{f-f'}{f}' = \frac{\frac{1}{T} - \frac{1}{T'}} {
    {\frac{1}{T'}}}
    =\frac{\frac{1}{\gamma T'} - \frac{1}{T'}} {{\frac{1}{T'}}}
    =\frac{1}{\gamma} - 1 = \sqrt{1-\frac{v^2}{c^2}}-1\\
    =\sqrt{1-\frac{(3874m/s)^2}{(2.998\times 10^8m/s)^2}} - 1 = -8.35\times 10^{-11}##

    My issues, then, are the following:
    My attempt at part (c) seems correct (if for no other reason than that it only requires use of one equation given in the book, as opposed to having to derive one), but gives an answer far different than that in the back of the book. Is the back incorrect?
    The pdf answer is far closer to the given one, but it doesn't make sense why that should be the case: this is a relativistic frequency shift question, so the relativistic frequency shift equation (as I used) should provide the answer. Is the pdf answer correct?
    If the relativistic frequency shift equation isn't applicable here, when *would* it be applicable? It seems the pdf derivation could be used everywhere there's a relativistic frequency shift, and since it's a different equation, it implies to me that one equation is generally correct for all of these types of problems, and one isn't. Is that the case? And if so, does that mean that the relativistic frequency shift equation is simply incorrect, or generally inapplicable?

    Thanks in advance for any assistance!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 24, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Your equation requires that the satellite moves directly towards the observer (or directly away). That would make the satellites crash - not good for the system.

    The question asks for the frequency shift due to time dilation only. This is equivalent to the frequency shift with a perfectly orthogonal motion (satellite motion relative to viewing direction).


    There is a more general formula for arbitrary angles between line of sight and relative motion, but that is not needed here.
     
  4. Jun 25, 2014 #3
    That answered my question. Thank you!
     
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