How Many Angles Maintain 2.00 W/m² Intensity in GPS Satellite Tests?

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SUMMARY

The forum discussion centers on calculating angles that maintain an intensity of 2.00 W/m² during GPS satellite tests using 1575.42 MHz UHF transmitters. The wavelength (λ) was determined to be 0.19 m using the formula λ = c/f, where c is the speed of light. The calculations revealed that there are four angles in the range 0°<θ<90° where the intensity is also 2.00 W/m²: 2.11°, 4.22°, 6.33°, and 8.44°. A more precise value for λ is necessary for accurate intensity calculations at specific angles.

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This discussion is beneficial for physics students, engineers working with GPS technology, and professionals involved in satellite communications and signal processing.

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Homework Statement


The GPS (Global Positioning System) satellites are approximately 5.18 m across and transmit two low-power signals, one of which is at 1575.42 MHz (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42 MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 W/m2.

(a) At how many other angles in the range 0∘<θ<90∘ is the intensity also 2.00 W/m2?
dsin(theta) = m*lamda
rearrange to find m, but lamda unknown so need to find that first using lamda=c/f
lamda=(3*10^8)/(1575.42*10^6) = 0.19m
m = 5.18(1)/0.19 = 27
CORRECT

(b) Find the four smallest( positive) angles in the range 0∘<θ<90∘ for which the intensity is 2.00 W/m2.
dsin(theta) = m*lamda
rearrange to find theta for m = 1,2,3,4
m=1 --> 2.11 degrees
m=2 --> 4.22 degrees
m=3 --> 6.33 degrees
m=4 --> 8.44 degrees
CORRECT

(c) What is the intensity at a point on the circle at an angle of 4.90 degrees from the centerline?
photo of attempt attached.
INCORRECTNot sure what to do at this point, any help appreciated.
 

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I get the same answer as you.
 
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haruspex said:
I get the same answer as you.

yup turns out it was just the system being super precise, the value needed to be calculated with a more precise lamda value, not 0.19. But otherwise, it is correct.
 

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