A GPS satellite develops a fault whereby it moves into an orbit 19 800 km above the Earth’s surface and also emits a weak signal whose power is only 20 W. What will be the power per square metre of the signal received on the Earth’s surface?

Guest Nov 11, 2015

edited by
Guest
Nov 11, 2015

#3**+15 **

**A GPS satellite develops a fault whereby it moves into an orbit 19 800 km above the Earth’s surface and also emits a weak signal whose power is only 20 W. What will be the power per square metre of the signal received on the Earth’s surface? **

\(\begin{array}{rcl} I &=& \frac{P}{4\cdot \pi\cdot r^2} \qquad P = 20\ \text{watt} \qquad r = 19\ 800\ 000\ m\\ I &=& \frac{20}{4\cdot \pi\cdot 19800000^2} \\ I &=& \frac{20}{4\cdot \pi\cdot 19800000^2} \\ I &=& 4.0596608278\cdot10^{-15}\ \frac{W}{m^2}\\ \end{array}\)

heureka
Nov 11, 2015

#1**+15 **

If it emits the power equally in all directions then by the time it reaches the Earth it will be spread out over the surface of a sphere of radius 19800 km. The surface area of this is 4*pi*(1.98*10^7)^2 m^2. Divide the 20W by this area to get the power in Watts per square metre (it will be a very small number!).

Alan
Nov 11, 2015

#2**0 **

Hi Many thanks for you reply , I forgot to add:

Choose the nearest value from the list below. (W/m2 represents ‘watts per square metre’.)

Select one:

8 × 10−5 W/m2

8 × 10−8 W/m2

4 × 10−15 W/m2

2 × 10−16 W/m2

4 × 10−17 W/m2

4 × 10−9 W/m2

2 × 10−4 W/m2

Guest Nov 11, 2015

#3**+15 **

Best Answer

**A GPS satellite develops a fault whereby it moves into an orbit 19 800 km above the Earth’s surface and also emits a weak signal whose power is only 20 W. What will be the power per square metre of the signal received on the Earth’s surface? **

\(\begin{array}{rcl} I &=& \frac{P}{4\cdot \pi\cdot r^2} \qquad P = 20\ \text{watt} \qquad r = 19\ 800\ 000\ m\\ I &=& \frac{20}{4\cdot \pi\cdot 19800000^2} \\ I &=& \frac{20}{4\cdot \pi\cdot 19800000^2} \\ I &=& 4.0596608278\cdot10^{-15}\ \frac{W}{m^2}\\ \end{array}\)

heureka
Nov 11, 2015