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?
idk bro, anyone knows????? i need help w. this as well
its 12 hours.
T=2piR/v
a=v^2/R
9.81*5.8/100=0.56898
v=sqrt(0.56898*(20000+6400(Earth's radius))*10^3)=3875.702775
T=2*pi*26400000/3875.702775=42798.97137 s
/60=713 min
/60= 11.9 h
Global Positioning System (GPS) satellites circle Earth at altitudes of approximately 20000 , where the gravitational acceleration has 5.8% of its surface value.
To determine the required percent accuracy in the distance, we need to calculate the percentage of the 2-meter uncertainty relative to the distance of 20,000 km.
1. First, convert the 2-meter uncertainty to kilometers:
2 meters = 0.002 kilometers
2. Calculate the percentage accuracy:
Percentage accuracy = (0.002 km / 20,000 km) * 100
Percentage accuracy = 0.01%
Therefore, a percent accuracy of 0.01% is required to achieve a 2-meter uncertainty relative to a satellite distance of 20,000 km.
Next, let's determine the number of significant figures needed in the distance. In scientific notation, we typically display numbers with the same number of significant figures as the precision of the measurement.
Since we desire a 2-meter uncertainty, we need to ensure that the distance is accurate to at least 2 meters. This means that the distance measurement should have at least three significant figures (to include the 2-meter uncertainty).
Hence, the distance should have at least three significant figures for accurate measurement when considering a 2-meter uncertainty.