You are here:

Physics/microwave radiation

Advertisement


Question
QUESTION: This is not a homework question please help. My question is a little wired but please answer it anyways. I'm doing the math by myself but i would like to know if I am on the right path, thank you. So transmitting 75 kilowatts of microwave energy from 20 miles above the earth surface or 105600 feet above the earth surface with a transmitting parabolic antenna with a diameter of 3 feet. with no other variables like energy loss, but if you can add that in that would be fine with me. What area in square feet would you need to receive 25 kilowatts on the ground. If you can also you may change the transmitting array area to a smaller size or bigger. The idea of the question is so i can get an idea of how to solve this on my own. Please provide an example.

ANSWER: You said transmitting from a parabolic antenna, meaning a straight beam from the transmitter.  Barring detailed antenna specs, that should be like a laser pointer, so if you want to ignore the things you say to ignore then you would need a diameter of sqrt(3) feet so you only intercepted 1/3 of the beam (area = pi*r^2).  Why wouldn't you just catch it all in a dish of 3 feet in diameter and get all 75 kW?  If you want some idea of the power spreading then I need to know a heck of a lot more about the antenna itself, and the dish, and the attenuation.  You haven't asked a problem with the simple answer you appear to be looking for.

---------- FOLLOW-UP ----------

QUESTION: Thank you for writing back, I very much appreciate it. The main purpose to my question in to find out what area I would need to receive 25 kilowatts of microwave energy from a distance of 100000 feet. The reason I pick 25 kilowatts from the 75 kilowatts at the source is because the little I know is that the power intensity spreads out as the distance increases from the source. Iím looking for a square footage area that is not greater than 500 by 500  or 250000 square feet. I can not change the power at the source and the distance is also constant. I donít know any thing about parabolic reflectors so thatís why there are no spec.

Answer
You said you were beaming it with a parabolic dish.  While you lose some in the middle of the projection process, whatever is transmitted off of the dish mostly stays in the beam unless you're going to count diffraction effects.  And a 3' diameter dish is large compared to the typical ~10cm wavelength of the microwaves.  So you should be able to capture it all in a little over 3' (for aiming issues) minus the power loss from the initial transmission and reception.  Making it 500' won't help you at all.  The losses in transmission power at the source and reception can be huge, but the size of the dish doesn't change that at all.

Physics

All Answers


Answers by Expert:


Ask Experts

Volunteer


Dr. Stephen O. Nelson

Expertise

I can answer most basic physics questions, physics questions about science fiction and everyday observations of physics, etc. I'm also usually good for science fair advice (I'm the regional science fair director). I do not answer homework problems. I will occasionally point out where a homework solution went wrong, though. I'm usually good at explaining odd observations that seem counterintuitive, energy science, nuclear physics, nuclear astrophysics, and alternative theories of physics are my specialties.

Experience

I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.

Education/Credentials
Ph. D. from Duke University in physics, research in nuclear astrophysics reactions, gamma-ray astronomy technology, and advanced nuclear reactors.

©2016 About.com. All rights reserved.