Simple sound intensity difficulty

[Vanessa23]

According to the text, at a distance of 50 m from a jet engine, the sound intensity is 10 W/m2. This corresponds to a loudness (or an intensity level) of 130 dB. [You should convince yourself that this is true.] If you find the sound too loud and you want to reduce the intensity by a factor of 10, to what new distance (approximately) from the jet engine should you move to achieve this reduction?


I=P/Area

Intensity level = 10db*log(I/Io)


to find the power of the jet engine:
p=10*((50^2)*4pi)
=314159

a factor of 10 reduction would mean 13dB
13=10log(I/10^-12)
1.3=log(I)-log(10^-12)
-10.7=log(I)
I=10^-10.7

A=P/I
A= 314159/ 10^-10.7
A=1.57x10^16

1.57x10^16 = 4pi r^2
r=397289

so does that mean that you really can't make the intensity decrease by a factor of 10?

 

[catkin]

I don't think dB need to come into the calculation; the question asks for a reduction in intensity, not in "intensity level" or loudness.

The requisite new level is a factor of 10 less than the first level of 10 W/m^-2, that is 1 W/m^-2.

Assuming the jet engine is on a plane surface and there are no reflections then the noise power is radiating through the surface of a hemisphere ...