Simple ratios dilemma; telescopes and signals from out of space

AI Thread Summary
The discussion revolves around calculating the power incident on a radio telescope with a diameter of 1000 feet, which is claimed to detect a signal of one picowatt across the Earth's surface. The initial calculation incorrectly uses the total surface area of the Earth instead of the appropriate cross-sectional area. The correct approach should apply the cross-sectional area formula, leading to a revised power calculation. The participant acknowledges the mistake after receiving clarification. The conversation highlights the importance of using the correct geometric parameters in such calculations.
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Hi,

Here's the question:

a) There is a radio telescope that is 1000 feet in diameter. It is claimed that it can detect a signal that lays down on the entire surface of the Earth a power of only one picowatt. What power would be incident on the antenna in this case?


It seems to me that this can be solved simply by considering the ratio of the surface area of the Earth to the power of the signal, and applying that to the surface of the telescope.

As far as I can see, given radius of Earth = 6.37 x 10E6 m and 1 foot = 0.3048m,

{ [ 1 x 10E-12 ] / [ 4 x pi x (6.37x10E6)^2 ] } x [ pi x (1/2 * 1000 * 0.3048)^2 ]

would give the answer. It actually gives 1.43 x 10E-22 when what is required is 5.6 x 10E-22 (4 times larger).

Can anyone spot what I'm doing wrong?

Cheers
 
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You are using the total surface area of the entire Earth (4\pi r^2), when you should only be using the cross-sectional area of the Earth (\pi r^2).
 
Tom Mattson said:
You are using the total surface area of the entire Earth (4\pi r^2), when you should only be using the cross-sectional area of the Earth (\pi r^2).

Yep. Thanks. :)
 
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