# Thread: satellite, find time it takes a call between two cities.

1. ## satellite, find time it takes a call between two cities.

A communications satellite is in an orbit that is 4.10 x 10^7 m directly above the equator. Consider the moment when the satellite is located midway between Quito, Equador, and Belem, Brazil; two cities almost on the equator that are separated by a distance of 3.40 x 10^6 m.

Find the time it takes for a telephone call to go by the way of satellite between these cities. Ignore the curvature of the earth.

in seconds.

heres what I believe it looks like: http://img518.imageshack.us/img518/6439/46101437mc6.png

also, i know that electromagnetic waves propagate through a vacuum at a speed given by c = 3.00 x 10^8

i wasn't sure why the equation would not be one of the given distances d/c.

Anyone who can explain this one please.

2. Originally Posted by rcmango
A communications satellite is in an orbit that is 4.10 x 10^7 m directly above the equator. Consider the moment when the satellite is located midway between Quito, Equador, and Belem, Brazil; two cities almost on the equator that are separated by a distance of 3.40 x 10^6 m.

Find the time it takes for a telephone call to go by the way of satellite between these cities. Ignore the curvature of the earth.

in seconds.

heres what I believe it looks like: http://img518.imageshack.us/img518/6439/46101437mc6.png

also, i know that electromagnetic waves propagate through a vacuum at a speed given by c = 3.00 x 10^8

i wasn't sure why the equation would not be one of the given distances d/c.

Anyone who can explain this one please.
This is nearly a trick question: You have to "harmonize" all given distances.

I take as a distance unit $10^6\ m$

Let a denote the distance of the satellite above the equator: $a = 41 \cdot 10^6 \ m$

Let g denote the distance on the ground between Quito and Belem: $g = 3.4 \cdot 10^6\ m$

The communication signal is running from Quito via satellite to Belem. Use Pythagorean Theorem to calculate the distance s of the signal:

$s = 2 \cdot \sqrt{\left(\frac12 \cdot 3.4 \cdot 10^6 \right)^2 + \left(41 \cdot 10^6\right)^2}$

And now calculate the time needed by the signal to cover the distance s.

3. okay i get 82070457.53 for s, the distance.

now i tried using the formula i had to divide s by c = 3.00 x 10^8

to get a final answer of about .274 seconds.

Thankyou!