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Calculate frequency change due to gravity

  1. Sep 24, 2011 #1
    I'm trying to solve the frequency change due to gravity of a radio signal transmitted at 2x10^9 Hz from geosynchronous orbit (H = 35780 km). This (would be) easy if I could use the value g = 9.8m/s^2, the catch is "do not assume constant g".

    The equation assuming constant g is:

    (f_rec - f_trans)/f_rec = gH/c^2 (Eq 1)

    (Note this is the form derived from the original energy conservation relation: hf = hf' + mgH, use m=E/c^2=hf/c^2).

    My initial stab at it was to substitute an expression using the universal gravitational constant:

    (f_rec - f_trans)/f_rec = (GM)*(1/r_earth - 1/(r_earth + H))/(c^2) (Eq 2)

    where M is the mass of the earth and G is the universal gravitational constant. This results in a frequency shift of ~+1 Hz (so small I needed to use a num2str command in MATLAB to see it given the magnitude of f_trans).

    I then thought perhaps I was being cute with assuming H and an inverse distance would cancel w/o deleterious effects, so I went back to the unsimplified form:

    (f_rec - f_trans)/f_rec = (GMH)*(1/r_earth - 1/(r_earth + H))^2/(c^2) (Eq 3)

    This yielded a frequency shift of ~+5 Hz (again so small I needed to use a num2str command in MATLAB to see it given the magnitude of f_trans).

    My issue is that I'm not sure I'm accounting for non-constant gravity correctly; if I am then I've solved the problem, if not...

    Please forgive if this is an easy problem or there is a "well-known" treatment for non-constant gravity I am just not finding, I am a mathematician with only a high school physics background taking a masters level physics class.

    Thanks for the help.
  2. jcsd
  3. Sep 25, 2011 #2

    rude man

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    See link below.

    According to that, the red shift should be
    Δf/f = 1/{sqrt(1 - rs/R1) - 1/sqrt(1 - rs/R2)}

    where rs is the Schwarzschild radius for Earth = 2GM/c^2, M = mass of Earth, or about 9 mm; R1 is the distance from Earth center to the satellite, and R2 is the radius of Earth.
    Since Δf/f < 0 this amounts to a blue shift, as you also found. So Δf/f is actually > 0.

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