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Gravitational Redshift Near the Earth

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that the general equation for gravitational redshift [tex]\frac{\Delta v}{v} = - \frac{GM}{c^2}(\frac{1}{r_1} - \frac{1}{r_2})[/tex] reduces to [tex]\frac{\Delta v}{v} = \frac{gH}{c^2}[/tex] near the surface of the earth.

    2. Relevant equations

    The two above, plus Newton's Law of Universal Gravitation
    [tex]F = \frac{GMm}{r^2}[/tex]

    3. The attempt at a solution

    Begin with Newton's Law and solve for GM.

    [tex]GM = \frac{F r^2}{m}[/tex]

    Expand F to mg, and cancel m.

    [tex]GM = g r^2[/tex]

    Plug this into the gravitational redshift equation.

    [tex]\frac{\Delta v}{v} = - \frac{g r^2}{c^2}(\frac{1}{r_1} - \frac{1}{r_2})[/tex]

    r_1 is the distance to the observer, which I assume to be a distant star or similar ([tex]r_1 >> r[/tex]). r_2 is the distance from the Earth's surface to the light source ([tex]r_2 = r[/tex])

    Multiplying by r,

    [tex]\frac{\Delta v}{v} = - \frac{g r}{c^2}(\frac{r}{r_1} - \frac{r}{r_2})[/tex]

    [tex]\frac{r}{r_1} = 0[/tex] and [tex]\frac{r}{r_2} = 1[/tex]

    [tex]\frac{\Delta v}{v} = - \frac{g r}{c^2}(-1)[/tex]
    [tex]\frac{\Delta v}{v} = \frac{g H}{c^2}[/tex]

    Is what I have done logical and valid? Please advise.
  2. jcsd
  3. Oct 14, 2008 #2
    I apologize for the duplicate post. When I clicked submit, it told me there was a database error. I assumed this meant the post was unsuccessfully submitted, so I tried again.

    Moderator: please remove one or the other.
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