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Calculating distance (and percent error) using parallax

  1. Sep 16, 2012 #1
    Hi all,

    This is a homework problem, but I don't think asking for help here presents any ethical issues. I'm permitted to consult my classmates and compare solutions with them as long as I do my own work, and I plan to ask my professor for help if I still feel unsure after reading your responses.

    Thanks for taking a look.

    1. The problem statement, all variables and given/known data

    In 1672, an international effort was made to measure the parallax angle of Mars at the time of opposition, when it was closest to the Earth.

    (a) Consider two observers who are separated by a baseline equal to Earth's diameter. If the difference in their measurements of Mars's angular position is 33.6", what is the distance between Earth and Mars at the time of opposition?

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

    2. Relevant equations

    none?

    3. The attempt at a solution

    I'm relatively confident of my answer for (a) -- 7.82 x 10^10 m. To find the distance between Earth and Mars, I drew an isosceles triangle with base 1.28 x 10^7 m (the diameter of Earth). The distance between Earth and Mars is equal to 6.37 x 10^6 (half the diameter of Earth) divided by the tangent of 0.00467 degrees (half the parallax angle).

    The problem is (b). I've drawn a diagram, but I'm not sure it's correct. The premise is that Observer 2 now observes Mars t seconds later than Observer 1 does. I started with the same triangle I drew for (a). Then I moved the original upper vertex (Mars 1) to the right a distance 5660t (the relative velocity of Mars times the time discrepancy t between the two observers) and called the new point Mars 2. Then I drew a line between Observer 2 and Mars 2, and I extended the original line between Observer 1 and Mars 1. (The rationale for this is that Observer 1's viewpoint hasn't changed, but Observer 2 is seeing Mars slightly later and, therefore, at a different location. In other words, Observer 1 is looking at Mars 1 and Observer 2 is looking at Mars 2. Right?) The intersection of the two lines O2-M2 and O1-M1 is the upper vertex of a new triangle, and the vertical altitude of that triangle is the new (incorrect) distance D.

    Am I right so far?

    Now I'm stuck. I need to relate t and D so that I can use the percent error to find the maximum allowed t. But I can't figure out what the relationship is. Help?
     
  2. jcsd
  3. Sep 19, 2012 #2
    Sorry for the bump, but this is due tomorrow and I'm still stuck. I talked to the professor today, and she indicated that to do (b) I should run my calculations for (a) backwards, yielding an angle measurement, and then use the angular velocity of Mars to convert the angle into a time period. I did that, using the period of Mars's orbit, but it seems wrong to ignore Earth's movement (and ignore the hint given in the problem).

    Twenty minutes after I left the professor's office, she sent me an email -- about something she'd thought about related to the problem -- that left me even more confused. I'm not sure if she was trying to restate what she'd already said or tell me that what she said this morning was not right.

    HELP! This is due tomorrow, and I would appreciate any advice. You don't have to read my long paragraphs above, just tell me briefly how you would approach (b).

    Thank you so much.
     
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