# Homework Help: Homework question about radio waves

1. Oct 6, 2012

### lordvipomme

"At what minimal height must be situated the point of reflection of a radio wave emitted in Montreal and received in Paris? The distance which separates both cities is 5 400 km and the radius of the Earth is 6 400 km."

I don't even know where to start, we didn't learn this. I would assume the point is figuring the answer out by ourselves but I've been staring at the problem and all I could do was draw the situation. Please help!

EDIT: Does it go something like this? http://imgur.com/5d7Mm
EDIT2: I don't think it's an antenna, more like a satellite... I found... 6 400 km and that leaves me as confused as ever. Is it correct?

Last edited: Oct 6, 2012
2. Oct 6, 2012

### voko

Explain how you got your result.

3. Oct 6, 2012

### lordvipomme

Yes.
This is the drawing: http://imgur.com/yfNVc
This is what I came up with: http://imgur.com/x61XV (I forgot something, The intersection of the lines is O)
The more I look at it the sillier it seems.

First, I know that θi = θr
From that, I concluded that angle BAO = angle DAO
Thus, segment AO bisects angle BAD in half, cutting segment BD in half as well. 5400/2=2700.

Second, I was messing around a bit and was wondering what I would get if I made the situation look like a lozenge. I got what I got and started calculing using the Pythagorean theorem. I got that segment BC (because I knew OC and BO's lenght) equals 6946,22.

Third, since in a lozenge all sides are the same lenght, AD = 6946,22 as well.
I now had triangle AOD of which I knew the hypotenuse and the base. I used the Pythagorean theorem and got AO's length, 6 400.

Seems extremely silly to me that the answer to the problem was in the problem itself and that it was this easy. Could someone correct me? Tell me where I went wrong and why what I did isn't correct?

Thank you!

PS: If I'm not clear in any steps, please do tell me and I'll edit my post.

4. Oct 6, 2012

### lordvipomme

I got help on another website. I was told about the arclength formula and that I was to figure out the angle spanned by the 5400 km arc between Montreal and Paris. I'll try working it out from there.

5. Oct 6, 2012

### Simon Bridge

Welcome to PF;
http://imgur.com/5d7Mm
won't come up :(

I'm not sure that quite illustrates what is being described in the problem ... the transmitter is at point M(ontreal) and the reciever is at point P(aris). P and M are on the surface of the Earth ... the distance to walk from one to the other along the Earth's surface is S=5400km.

The radio beam leaves M and reflects of some spot in the sky (off the ionosphere, or whatever) so that it can be received at P.

This should give you a bunch of triangles - the main problem is to determine the geometry when the reflection point is at it's lowest.

You can figure it out quickly by trial and error:
... draw M and P on a circle radius R representing the Earth - if the center of the Earth is point O, then you can draw lines OP and OM easily enough.

... you can also draw another radial line that the reflection point has to be on.
... you can guess a reflection point and draw the incident and reflected rays.

What is the closest the reflection point can get to the Earth's surface and still reflect between points P and M?

edit: while I was typing you got back to us with:
The fact that the distance is an arc-length is important, yep.

6. Oct 6, 2012

### voko

You should realize that the absolute minimum of the reflector's altitude is reached where the tangent (to the surface of the Earth) lines from the two cities meet. It cannot be lower, because then they would have to pierce the ground, which radiowaves cannot.

7. Oct 6, 2012

### lordvipomme

Yes, thinking that M and P had a straight line between them was a silly error.

Thanks a bunch, everyone! I'll try to work it out!

8. Oct 6, 2012

### Simon Bridge

@lordvipomme: you have received a pretty big hint there - the rest is geometry. Let us know how you get on.