How Long Does a Call Travel via Satellite Between Quito and Belem?

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The discussion centers on calculating the time it takes for a telephone call to travel via satellite between Quito, Ecuador, and Belem, Brazil. A communications satellite is positioned 3.33×10^7 m above the equator, with the two cities 3.30×10^6 m apart. The user attempted to apply the Pythagorean theorem to find the distance from each city to the satellite, resulting in a total distance of approximately 1.0989×10^14 m. However, the calculated time of 3.23×10^11 seconds is deemed incorrect, as it suggests an impractically long duration for a satellite call. The discussion highlights the need to use the speed of light for satellite communications rather than the speed of sound.
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Homework Statement


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.

Homework Equations



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


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:
 
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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:
blue_lilly said:

Homework Statement


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.

Homework Equations



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


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​

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.
 
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