# A Satellite Phone Call

1. Feb 28, 2014

### blue_lilly

1. The problem statement, all variables and given/known data
A communications satellite is in a synchronous orbit that is 3.33×107 m directly above the equator. The satellite is located midway between Quito, Equador, and Belem, Brazil, two cities almost on the equator that are separated by a distance of 3.30×106 m. Calculate the time it takes for a telephone call to go by way of satellite between these cities. Ignore the curvature of the earth.

2. Relevant equations

Time= Speed / distance
Speed of Sound: 340.29 m/s
a^2 + b^2 = c^2

3. The attempt at a solution

So, I started by drawing what was going on and it turned into a equilateral triangle which I made into two right triangles.

I divided the distance between Quito and Belem in half (3.30 E 6)/2 = 1,650,000 m and I solved for the distance between Quito and the Satellite by using Pythagorean theorem: c^2=b^2+a^2
c^2 = (1,650,000)^2 + (3.33 E 7)^2​
c^2 = (2.72 E 12) + (1.108 E 15)​
Sqrt(c^2) = Sqrt((2.72 E 12) + (1.108 E 15))​
c = 5.49 E 13 m​
C is the distance from Quito to the Satellite and so then, the distance from the Satellite and Belem would also be 5.49 E 13 m. Which would mean the total distance that the call would have to travel is 1.0989 E 14 m.

The formula for Speed is Speed=Distance/Time which can be rewritten as Time=Distance/Speed. The speed of sound constant is 340.29 m/s.
t= (1.0989 E 14) / (340.29)​
t= 3.2293 E 11 seconds​

This answer is incorrect and I am not really sure where I went wrong.

Any help would be greatly appreciated C:

2. Feb 28, 2014

### SteamKing

Staff Emeritus
You don't use sound to communicate with satellites in orbit.

Also, look carefully at the time to make the call: 3.23E11 sec.

That's over 10,000 years, quite a long time to wait for a response.

Last edited: Feb 28, 2014
3. Feb 28, 2014

### voko

Cannot be correct. 1.108 E 15 is much greater than 2.72 E 12, so the result should not be much different from 3.33 E 7.