# Waves problem

1. Nov 27, 2009

### rusny

1. The problem statement, all variables and given/known data

What is the maximum time and the minimum time it takes for communication from the space shuttle to reach the Director of NASA who is seated 20 meters away from the communication radio at Mission Control at the Johnson Space Center in Houston, Texas. You may assume that room temperature at the Center is 20 C.

2. Relevant equations

I don't know the equation I should start from

3. The attempt at a solution

If somebody could give me at least a hint how should I start, it will be great.
Thanks

2. Nov 27, 2009

### Oddbio

where is the space shuttle?

3. Nov 27, 2009

### rusny

that's the problem
that's all what is given

4. Nov 28, 2009

### Staff: Mentor

Oddbio is just trying to get you to think. If the director is communicating with the shuttle, where is the most likely place for it to be? And when they ask for the minimum and maximum delay times, what do you think they are referring to?

5. Nov 28, 2009

### rusny

so the shuttle is on the earth orbit
its speed is around 17,580 miles per hour
we know the speed of light so we can find n of the space n=c/v

is the maximum time delay is n1Sin1=n2Sin2 were n1>n2
and the minimum time is n1=n2 ?

6. Nov 28, 2009

### trixx

Correct me if I'm wrong but I don't think the shuttle's speed matters for this because it's in orbit with earth, meaning it remains above the same portion of our planet.

From there you just take the orbital distance from earth to the shuttle and find:

t=d/v using the shuttle's orbital distance to the control center
+
t=d/v using 20 meters as your distance and the speed of sound as your v

Do this twice for the min and max orbital distances and that should be your answer.

7. Nov 28, 2009

### rusny

but how is possible that shuttle had minimum and maximum distance if as you said it remains above the same portion of earth?

or did you mean the maximum and minimum distance of earth's orbit to sun?

8. Nov 28, 2009

### trixx

I suppose if it was orbiting earth further away it would have to be at a higher velocity, which can be checked with F= GMm/r^2

Wikipedia says it mostly does orbit missions, so if you're going to consider an orbiting distance and a max distance of something like 600km above the other side of earth, then your two distances would be:

190km for m in
and 600km + earth's diameter for max

I doubt your professor is asking for that wouldn't be a wrong way to look at the problem.

9. Nov 28, 2009

### Phrak

rusny, there are two delays here, sound through the air over a distance of 20 meters, and the delay due to the finite speed of light, speed of radio signals.

Secondly, a signal won't travel through the Earth, but around it.

Third, what is the orbital altitude of the space shuttle?

10. Nov 28, 2009

### trixx

11. Nov 28, 2009

### rusny

so that is what I got

t for minimum = (orbital distance/speed of light)+(distance to center/speed of sound)
=> orbital distance = 960 km = 960000 m
speed of light 3x10^8
distance to center = 20 m
speed of sound = 343 m/s
t for min = 6.15x10^-2 s

and
t for maximum = (orbital distance/speed of light)+(distance to center/speed of sound)
=>orbital distance = 960 km + half of the earth's circumference =>20998 km =20998x10^3
speed of light 3x10^8
distance to center = 20 m
speed of sound = 343 m/s
t for max = 1.28x10^-1 s

Is that right?

12. Nov 29, 2009

### Phrak

rusny. The minimum answer you get looks right, as long as you are given the orbital height is 960 km.

But who knows what the second answer should be? Are the signals able to travel around the curvature of the Earth? Are there repeating stations or satellites to relay the signal, each with their own internal electronic delays added?

Or do we need line-of-sight? In that case you want to know how far away the shuttle can be along its orbit before it goes over the horizon. It's not a well worded physics question. Maybe you need to do an internet search for a world map showing the range of Earth based receiver stations.

Last edited: Nov 29, 2009