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Q: speed of sound ( Answered,   4 Comments )
Question  
Subject: speed of sound
Category: Science
Asked by: kim2-ga
List Price: $5.00
Posted: 28 Nov 2004 22:35 PST
Expires: 28 Dec 2004 22:35 PST
Question ID: 435394
David is in a large stadium listening to the first words form the
astronauts who have landed on Europa. David is 80 meters from the loud
speaker. The radio station which is broadcasting the message from
Europa is 100 km from the stadium.

Assuming that Europa is at its ?average? possible position from Earth,
how long will it take the sound from the astronaut?s mouth to reach
David?s ears?
Answer  
Subject: Re: speed of sound
Answered By: redhoss-ga on 30 Nov 2004 09:14 PST
 
Hello kim2, I noticed that no one had actually answered your question
and felt that we owed you an answer. Here is some info you need to
find your answer:

Speed of light: 300,000 kilometers per second in a vacuum
http://www.school-for-champions.com/science/lightspeed.htm

Speed of sound (since you don't specify we will assume these conditions):
In air at sea level, 60% humidity and 30C, it is 344 meters per second.
http://www.sizes.com/natural/sound.htm

Distance from Earth to Jupiter:
Minimum 588.5 x 10^6 kilometers
Maximum 968.1 x 10^6 kilometers
http://www.the-planet-jupiter.com/jupiter-fact-sheet.html

Distance from Jupiter to Europa: mean distance 670,900 km
http://www.enchantedlearning.com/subjects/astronomy/planets/jupiter/moons.shtml

Using the above info plus what you give in your question we can
calculate the answer:

Time for light/radio signal to travel from Jupiter to Earth:
Maximum = 968.1 x 10^6 kilometers/300,000 kilometers per second = 3227 sec.
Minimum = 588.5 x 10^6 kilometers/300,000 kilometers per second = 1962 sec.
NOTE: This agrees with the comment made by iang-ga.

Time for light/radio signal to travel from the radio station to the
receiver at the stadium:
Time = 100 kilometers/300,000 kilometers per second = .003 sec.

Time for sound to travel from the speaker to David's ear:
Time = 80 meters/344 meters per second = .23 sec.

Now we have to decide exactly how accurate you want your answer to be.
If you want to consider that the astronauts are actually on Europa and
not Jupiter, you will introduce another variable. Since the mean
distance from Europa to Jupiter is 670,900 km we would have to include
this in our calculations.
Time = 670,900 km/300,000 kilometers per second = 2.24 sec.

To get an exact number we would have to know the exact relationship of
Europa's orbit to Earth's which is a subject I don't want to tackle.
So, we will just assume that the astronauts sent their message from
Jupiter.

Then we have two answers, one maximum and one minimum.

Maximum = 3227 + .003 + .23 = 3227.233 sec.
Minimum = 1962 + .003 + .23 = 1962.233 sec.

I hope that this is what you were looking for, Redhoss
Comments  
Subject: Re: speed of sound
From: augusta-ga on 28 Nov 2004 23:39 PST
 
Europa to loud speaker: speed of light * distance between loud speaker and Europa

+

loud speaker to ear: speed of sound * 80 meters

= length of time

right?
Subject: Re: speed of sound
From: winsplit01-ga on 29 Nov 2004 00:00 PST
 
Time = Distance / Speed

You can for all practical puposes ignore the time taken from Europa to
Stadium as the time would be negligible (Speed of light = 3 X 10^8
m/s)

Thus, in our case 

Time = 80 / 340 = 0.235 seconds   (Speed of Sound = 340 m/s)
Subject: Re: speed of sound
From: iang-ga on 29 Nov 2004 02:31 PST
 
The maximum light travel time from Europa is 54 minutes and the
minimum is 33 minutes - not exactly negligible :-)

Ian G.
Subject: Re: speed of sound
From: leperflesh-ga on 01 Dec 2004 16:59 PST
 
There is a problem with the answer: the accuracy of the answer (to
within a thousandth of a second) is totally inappropriate given the
inaccuracy of several of the variables.

The most-rounded variable is the figure used for the speed of light: a
straight 300,000 Km/sec. (the speed of light is actually 299 792 458
m/s) The figures used for distance to Jupiter are to the nearest
hundred thousand kilometers (possibly that is as accurate as we can
measure them). In science, one should never present an answer to a
higher degree of accuracy than the input variables support.

Round the answers off accordingly.

-Lep

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