Simple ratios dilemma; telescopes and signals from out of space

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SUMMARY

The forum discussion centers on the calculation of the power incident on a radio telescope with a diameter of 1000 feet, which is claimed to detect a signal of one picowatt spread over the Earth's surface. The original calculation incorrectly used the total surface area of the Earth instead of the appropriate cross-sectional area. The correct approach involves using the formula that incorporates the cross-sectional area, leading to the accurate power incident on the antenna being 5.6 x 10E-22 watts, rather than the initially calculated 1.43 x 10E-22 watts.

PREREQUISITES
  • Understanding of basic physics principles related to power and area.
  • Familiarity with the concept of cross-sectional area versus surface area.
  • Knowledge of the mathematical constants involved in area calculations, such as π.
  • Ability to convert units, specifically between feet and meters.
NEXT STEPS
  • Review the principles of power distribution in electromagnetic signals.
  • Study the mathematical derivation of surface area versus cross-sectional area.
  • Learn about the applications of radio telescopes in astrophysics.
  • Explore unit conversion techniques for scientific calculations.
USEFUL FOR

Students and professionals in physics, astronomy enthusiasts, and engineers working with radio telescopes or signal detection technologies will benefit from this discussion.

T7
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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 ([itex]4\pi r^2[/itex]), when you should only be using the cross-sectional area of the Earth ([itex]\pi r^2[/itex]).
 
Tom Mattson said:
You are using the total surface area of the entire Earth ([itex]4\pi r^2[/itex]), when you should only be using the cross-sectional area of the Earth ([itex]\pi r^2[/itex]).

Yep. Thanks. :)
 

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