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Homework Help: Accuracy and Uncertainty

  1. Aug 10, 2009 #1
    1. The problem statement, all variables and given/known data

    GPS can be used to determine positions with great accuracy. The system works by determining the distance between the observer and each of the several satellites orbiting Earth. If one of the satellites is at a distance of 20,000 km from you, what percent accuracy in the distance is required if we desire a 2-meter uncertainty? How many significant figures do we need to have in the distance?

    2. Relevant equations

    3. The attempt at a solution

    I'm new to Physics and I'm just taking it this year, so all I could figure is converting 20,000 km to 20,000,000 meters. That's all I know to do, though.
    Last edited: Aug 10, 2009
  2. jcsd
  3. Aug 10, 2009 #2
    And 2m is how many percent of 20,000,000m?
  4. Aug 10, 2009 #3
  5. Aug 10, 2009 #4
    So that makes the percent accuracy 1 x 10^5 % (or 0.00001%)?

    What about significant figures in the distance? I don't understand what the question means by that... (This is from a worksheet I have.)
  6. Aug 10, 2009 #5
    Watch out for your typo...that should be 10^(-5)%. But other than that, I think that's right.

    See this for significant digits rules/examples
    http://www.batesville.k12.in.us/Phy...t/Significant_Digits.html#counting sig digits

    With an uncertainty of 2 meters, the range measurement could be any where from
    20,000,002 meters to
    19,999,998 meters.

    How many digits are required to write those numbers?
  7. Aug 10, 2009 #6
    So, 8?

    Thanks for your help. I figured this would be overlooked since it looks like cake compared to the other questions I've seen asked here.
  8. Aug 10, 2009 #7
    I would say 8 is correct. No problem, hope it's right :)
  9. Aug 10, 2009 #8


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    Homework Helper

    Technically it's asking how many digits are needed to write the distance to the satellite so that the precision represented by the last digit is less than the desired uncertainty. In this case, if you write 8 digits, the last of those digits represents 1-meter accuracy (remember the place value system?). If you only had 7 significant digits, the last of those digits would represent 10-meter accuracy, so the distance 20,000,000 with seven significant digits (the last zero being an insignficant digit) would have an inherent uncertainty of 10 meters. That's too large to be able to tell where you are to 2-meter precision.

    But bottom line, I agree with 8 digits being the correct answer here.
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