A GPS satellite develops a fault whereby it moves into an orbit 19 500 km above the Earth’s surface and also emits a weak signal whose power is only 23 W. What will be the power per square metre of the signal received on the Earth’s surface? Choose the nearest value from the list below. (W/m2 represents ‘watts per square metre’.)

Select one:

2 x 10−5 W/m2

4 x 10−4 W/m2

4 x 10−16 W/m2

2 x 10−15 W/m2

5 x 10−17 W/m2

5 x 10−8 W/m2

5 x 10−15 W/m2


I've read over and over how to try and figure this question out and I just don't understand it. Can anybody please help me / explain this to me please?

 Nov 28, 2017

Go online to this link and read about the attenuation of the electromagnetic signals from space or other sources: https://en.wikipedia.org/wiki/Free-space_path_loss



I = W/d^2 =23/19,500^2 =~6 x 10^-8 - That is the closest I get !!.

 Nov 28, 2017
edited by Guest  Nov 28, 2017


This post has an explanation of the physics by Sir Alan and an example by Heureka.


Here is a reproduction of  Heureka’s math, using the data from this question.


\(\begin{array}{rcl} I &=& \frac{P}{4\cdot \pi\cdot r^2} \qquad P = 23\ \text{watt} \qquad r = 19\ 500\ 000\ m\\ I &=& \frac{23}{4\cdot \pi\cdot 19500000^2} \\ \\ I &=& 4.8\cdot10^{-15}\ \frac{W}{m^2}\\ \end{array}\)


Closest answer from list:  5 x 10−15 W/m2



It’s usually not necessary to reinvent the wheel, but you must be careful not to inflate the tire on the wheel with hot air from a blarney bag. 

 Nov 28, 2017
edited by GingerAle  Nov 28, 2017

See here:  http://web2.0calc.com/questions/calculate_1

 Nov 28, 2017

It’s time to . . . re-tire, Mr. BB. At least change the air . . .indecision

GingerAle  Nov 28, 2017

LOL Time to re-tire or at least change the air [error] hahahaha!

Guest Nov 29, 2017

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