Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intell

Question

Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial
Intelligence) program, once said that the large radio telescope in Arecibo, Puerto Rico,
can detect a signal which lays down only a picowatt of power over entire surface of the
Earth. What is the power received by the radio telescope if the diameter of the antenna is
300 m? What would be the power of the source at the center of our galaxy which could
provide such a signal? The center of galaxy is 22000 light years away. Assume that
source is radiating uniformly in all directions and that the Earth is a disk with the radius
of 6378 km.

Explanation / Answer

A ) Here the intensity that whould be received by theArecibo antenna should be equal to intensity of that signal onearth. Then intensity received Ir = Pr/a Pr = power received a = area of cross section of antenna =?D2/4 D = 300 m Then a = ?(300m)2/4 Power on the earth Pearth = 1*10-12W R = radius of the earth = 6.37*106m Area of the earth A=4?R2= 4?(6.37*106m)2 Then Ir =Ie Pr/a =Pearth/A Power received by antenna Pr =(1*10-12W /4?(6.37*106m)2)*?(300m)2/4 = 1.4*10-22W B) r = distance of the source to the antenna =2.2*104ly*9.46 * 1015m/ly Power of the source is Ps =4?r2Ie = 4?*( 2.2*104ly*9.46 *1015m/ly)2*(1*10-12W/4?(6.37*106m )2) = 1.1 *1015W Then Ir =Ie Pr/a =Pearth/A Power received by antenna Pr =(1*10-12W /4?(6.37*106m)2)*?(300m)2/4 = 1.4*10-22W B) r = distance of the source to the antenna =2.2*104ly*9.46 * 1015m/ly Power of the source is Ps =4?r2Ie = 4?*( 2.2*104ly*9.46 *1015m/ly)2*(1*10-12W/4?(6.37*106m )2) = 1.1 *1015W

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