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Measuring the distance to Mars with parallax.

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    a) Two observers are separated by a baseline equal to earth's diameter. If the differences in their measurements of Mars' position at opposition (i think this means when it is closest to earth) is 33.6" (arcsec). What is the distance between earth and Mars at opposition?

    b)If the distance to Mars is to be measured within 10%, how closely must the clocks used by the two observers be synchronized? (Ignore the rotation of the earth. THe orbital velocities of earth and Mars are 29.79 km/s and 24.13km/s respectively.)


    2. Relevant equations
    [tex]tan\frac{\theta}{2}=\frac{R_{Earth}}{d}[/tex]



    3. The attempt at a solution
    a)I think I know this part, pls check.
    [tex]tan\frac{\theta}{2}=\frac{R_{Earth}}{d}[/tex]
    since theta is small:
    [tex]\frac{\theta}{2}\approx\frac{R_{Earth}}{d}[/tex]
    [tex]d=\frac{2 R_{Earth}}{\theta}[/tex]
    Converting theta back into radians I get 7.82x10^10 m.

    b) This I'm having trouble with. Well I want 10% error; this means that my error should be 7.82x10^9 m. Observers on earth will see that Mars is moving at 5.66 km/s (the difference in the orbital velocity between Earth and Mars). And this is how far I got.
     
  2. jcsd
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