Required Precision for GPS Distance Measurement

[DSG3002]

Homework Statement

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?

Homework Equations

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.

 

[queenofbabes]

And 2m is how many percent of 20,000,000m?

 

[Chrisas]

See the section titles "Relative Uncertainty" at this link...

http://www.batesville.k12.in.us/Physics/APPhyNet/Measurement/UncertaintyDictionary.html

After looking at the link, post what you think the answer should be.

 

[DSG3002]

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.)

 

[Chrisas]

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?

 

[DSG3002]

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?

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.

 

[Chrisas]

I would say 8 is correct. No problem, hope it's right :)

 

[diazona]

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?

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.