Astronomy Problems: PLEASE LOOK/HELP

[astronomystudent]

1.) The frequency of WUSC is 90.5 x 10^5 cycles per second. What is the wavelength?

I used C = wavelength (frequency)
Thus: 3x10^8 m/s = wavelength (90.5 x 10^5)
wavelength = 3.314917127 m
wavelength = 3.31 x 10^9 meters

2.) If the period for a certain satellite revolving in a polar orbit around the Earth is 2 hours, how high above the surface of the Earth does it orbit? What if this same satellite were orbiting Jupiter (mass 318 times the mass of Earth) with the same 2 hour period. How high above the center of Jupiter would it be orbiting?

I know you use Kepler's 3rd law for this and I can fill in most of the numbers for the equation by I am confused as to what is meant by "above the surface" and "above the center of Jupiter" does that have to do with semi-major axis maybe?

3.) 1.75 x 10^5 arcsec is how many degrees? (remember significant digits !)

I found this conversion: 1 arc second = 0.000277777778 degrees
I then set up a table to solve the problem:
0.000277777778 degrees/1 arc second = 1.75 x 10^5 arc seconds/x (degrees)

everything cancels out you multiply and then divide 1.75x10^5/0.000277777778
x = 6.29 x 10^8 degrees

4.) Suppose the sound from an approaching train whistle normally has a frequency of 1200 cycles per sound, but the train is approaching at 50 meters per second. How would the Doppler effect change the wavelength of the sound? (speed of sound 335 meters per second; quantitative answer here for full credit)

I know here that as the train whistle approaches the sound waves will have shorter wavelengths and higher frequencies, as it passes however, that will change and the frequencies will be lower. I am working on the math part.

 

[tony873004]

#2. I hope I don't get chastised for giving you too much information on a homework problem, but here's a web page I wrote to answer this exact question.

http://orbitsimulator.com/cmc/a2.html

You'll still have to compute altitude from SMA.

#3. Does that answer really seem reasonable? Something measured in arseconds, a very small fraction of 1 degree, equals hundreds of millions of degrees?

#4. When you understand #1, this will be easier.

 

[astronomystudent]

1.) I have since corrected it I think or maybe have a better understanding of the question.
F= 1/T
F = 1/90.5 x 10^6 cycles/second
F = 1.10 x 10^-8 Hz

Wavelength = c/f
c = 300,000,000 m/s
wavelength= 300,000,000/1.10 x 10^-8 Hz
wavelength= 2.72 x 10^16 meters
I think that is right?

2.) I used your formula, but I only have one answer. The way the question is worded it seems as if I should have two separate answers: one for "above the surface" and one for "above the surface of Jupiter"

using your equation I found the semi-major axis to be = 5.50 x 10^7 meters

the only conversion I made involving units was from hours to seconds
2 hours = 7200 seconds

3.) I have corrected the mistake I think

This time I used a different conversion:
(1/3600)(1.75x10^5) = 48.6 degrees

4.) I think maybe I understand this now if so long as I did number one correctly.
F = 1/T
F= 1/1200
F= 8.33 x 10^-4 Hz (approaching)

v= 50 meters/second
speed of sound = 335 meters/second

f(d) = f v/v-vs
8.37x10^-4 Hz ((335m/s)/(335-50))
8.37x10^-4 Hz (335/285)
Frequency = 7.09x10^-4 - new frequency

source approaching for wavelenght
wavelength = (v-vs)/f source
wavelength = (335-50)/ (8.33 x10^-4 Hz)
wavelength = 342,000 meters

Thus as the whistle approaches sound waves will have shorter wavelengths and higher frequencies.

 

[tony873004]

#2. Orbits are always measured from the center of a planet, at least for doing the math. The distance from the center of a planet and a circular orbit is called the semi-major axis (SMA). When you hear stuff like "the Space Shuttle is orbiting 300 km high", it's referring to altitude, no different that talking about how high an airplane flys.

The Earth's radius is 6378km. So standing on the surface of the Earth, you're 6378km from the center (+- a little bit since the Earth isn't perfectly spherical). So the space Shuttle orbiting 300 km high is 6378km + 300 km or 6678 km from the center. So its semi-major axis would be about 6678 km.

You got the answer correct for Jupiter. This number straight out of the calculator would represent distance from the center of Jupiter, or SMA of a circular orbit. That's what the question asks for, so you got it right.

BTW... in the calculator I linked you too, you don't need to convert seconds to hours. Simply type 2h in the period field and it will know what you mean, but it's good that you figured out the unit conversion anyway.

But question 2 also asks for how high above the surface of the Earth a satellite with a 2 hour period would orbit. Getting the SMA is the easy part since you've already demonstrated you can do it for Jupiter. Now do it for Earth. Then you need to convert that to altitude above Earth's surface.

#3. That answer seems a lot more reasonable, doesn't it? It should because it's right.

I'll look at 1 and 4 for you later.

 

[astronomystudent]

I looked at problem 2 and tried to do what you said.
2.) I did the same thing I did for "above the center of jupiter" I used the mass of the Earth - 5.97x10^24 kg and plugged numbers back into the equation I then got teh semi-major axis to be
a= 4.17 x10^5 meters

But I don't understnad what you mean by converting to altitude above the Earth's surface, or maybe I just misunderstood how to work the second part of this problem involving "above the surface"

3.) Yes, it does look much better and makes sense.

1,4.) I will wait to hear from you on these.

Thanks for helping me.

 

[tony873004]

Question 1:
Why are you doing F = 1/t? They don't give you t. They give you frequency, so you don't have to compute frequency.

Also, 3.314917127 m does not equal 3.31 x 10^9 meters.

Question 2: You didn't get the right answer for Earth. I'm surprised since you got it for Jupiter. Type the equation you used, including the numbers you plugged in.

Altitude = Semi-major axis (round orbit only) - planet radius.

The radius of the Earth is 6378 km. So if you are standing on the ground, you are 6378 km from the center of the Earth. 6378-6378=0. Your altitude is 0. If you are flying in an airplane, and your altitude is 5 kilometers, 6378 + 5 = 6383. You are 6383 kilometers from the center of the Earth. A satellite in a circular orbit whose semi-major axis is 6578 kilometers has an altitude of 200 kilometers. 6578-6378=200.

Question 4:
The question isn't asking you for what happens after the train passes. It just want to know how the wavelength of the whitle is affected while the train is approaching you.

 

[astronomystudent]

1.) c = wavelength (frequency)
3x10^8 m/s = wavelength (90.5 x 10^6)
(3x10^8 m/s)/(90.5 x 10^6)
wavelength = 3.314917127
wavelength = 3.31 meters

2.) above he center of Jupiter = 5.50 x 10^7
mass of Earth = 5.97 x10^24 kg
a = cubed root of (2hours)^2(6.67300 x 10^-11)(5.97 x10^24 kg)/(4pi^2)
a= cubed root of (51840000)(3.983781 x 10^14)/(4pi^2)
a= cubed root of (2.06519207 x10^22)/(4pi^2)
a = cubed root of (2.06519207 x 10^22)/(39.4784176)
a= cubed root of (5.231192625 x 10^20)
a= 8057498.574
a= 8.06x 10^6 m - semimajor axis

Altitude = (8.06x10^6) - (6378)
Altitude = 8051120.574
Altitude = 8.05 x 10^6 m

3.) CORRECT

4.) So for four if I just find the wavelength using the (v-vs) equation or no, I'm confused.

 

[tony873004]

#1. Correct except for a typo. The question gave you 90.5x10^5 and you used 90.5x10^6 in your computation. Change that and you'll have the right answer.

#2
You're so close. Get into the habit of putting units in your equations.

Altitude = (8.06x10^6)-6378

Put in units. Can you find your mistake?

#4. If the train were still, the sound would approach you at the speed of sound. Since its moving towards you at a given speed, what is the speed that the sound is approaching you with? Use this speed in your computation.

 

[astronomystudent]

2.) One measurement is in meters and the radius is in kilometers. I converted 6378 km = 6,378,000 meters
Altitude = (8.06x10^6)-(6,378,000)
Altitude = 1679498.574
Altitude = 1.68 x 10^6 meters - above the surface of Earth

4.) Okay maybe this is right:
c = wavelength (frequency)
335 = wavelength (1200)
335/1200 = wavelength
wavelength = 2.79 x 10^-1 meters

approaching
c = wavelength (frequency)
50 = wavelength (1200)
50/1200 = wavelength
wavelength = 4.17 x 10^-2 meters

Is that what you are saying b/c one formula shows it at rest, where the speed of sound is used and the other as it approaches at 50 meters/sec? So they are different...

 

[tony873004]

When the train is moving, the sound is not approaching you at 50 m/s. It approaching you at the speed of sound + 50m/s.

 

[durt]

No, tony, the sound is approaching at the speed of sound. Only the frequency is greater because each pulse is emitted closer to previous one because the train has moved forward during the interval between them.

 

[tony873004]

No, tony, the sound is approaching at the speed of sound. Only the frequency is greater because each pulse is emitted closer to previous one because the train has moved forward during the interval between them.

durt is right. Try it his way.

 

[astronomystudent]

4.) What? So are you saying I do one equation with the speed of sound and another with 335 +50 = wavelength (frequency)? Confused, sorry.

 

[astronomystudent]

4.) I have read what the two of you wrote again and I think maybe you are saying that I shouuld instead do: c = (wavelength)(frequency)
3 x 10^8 m/s = (wavelength)(1200 + 50) ?

 

[astronomystudent]

I still don't understand how to solve problem #4 can someone help me with that?

 

[tony873004]

Be careful. You are adding 1200 which is frequency to 50 which is velocity. You want to add 350 + 50 and divide that by the frequency.

Compare this answer to what you would get if you didn't add the 50 to the 350.

 

[SpaceTiger]

I still don't understand how to solve problem #4 can someone help me with that?

If the train is sitting still, it emits sound at a wavelength that you already calculated (0.28 meters). Let's say that, at some time, the train sends out a wavefront. How far does the train travel before it sends out the next one (remember, it emits wavefronts at a frequency of 1200 cycles per second)? What does this mean for the distance between wavefronts (that is, the wavelength)?