Astrophysics G: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | Go up
 - by the crazy German (Maike Scheller), class of 2004

1. Saturn is 1.427x109 km from the sun, what is that in A.U.s?

Already then, let's look at this thing... 

We know that one A.U. is 1.50x1011 m. We also know that there 1000 meters in one km...

So here is how we would convert this:

1.427x109 km (1000 m / 1 km)    because the km cancels, we know how many meters we have. 


However, we don't know our answer yet since we just did the conversion so we need to divide our answer with the value of one A.U.

Here we go:

1.427x1012 m /(1.50x1011 m) = 9.5 A.U.

(Table of contents)

2.   How many kilometers in 39.44 A.U.s?

Well, this is not much different to the previous problem just that we have a little more A.U.s and we are using the reverse way to get our answer.

We know that one A.U. is 1.50x1011 m

39.44 (1.50x1011 m) = 5.916x1012

To convert this to km we need to divide by 1000 (because there are 1000 m in a km)

5.916x1012 m / 1000 = 5.9x109 km 

(Table of contents)

3. How many arc seconds in 45o

There are (3600 arc seconds/ 1 second) so we can just use that and the 45o to find out how many seconds...

(3600 arcsec/ 1 sec) (45) = 162000 sec  

(Table of contents)

4. 152 arc seconds is how many degrees?

This is just the reverse calculation we did in number 3. 

Instead of multiplying we need to divide our given data by 3600 arc seconds to find degree...

152/3600 = .042222222 = .0422o
(Table of contents)

5. The Star Ceteus Naue has a parallax angle of .024 arc seconds, what is its distance in parsecs? 

Well, the concept is 

1 / x amount of arc seconds = parsecs

1/ .024 = 41.666666666666 = 41.7 parsecs
(Table of contents)

6. What is the parallax angle of the star Cepheus Firmea that is 450 pc away? 

Same as above just in the reverse order...

1/x = 450 pc

arc seconds = 1/450 = .0022222 = .0022 arc secs
(Table of contents)

7. The planet Imadork is 34.2 light years away from us, what is this distance in kilometers? 

We know how much meters are in one light year (9.46x1015 meters). Since we have a total of 34.2 light years we need to multiply the meters by this number so here we go:

9.46x1015 m (34.2) = 3.23532x1017

But this only gives us meters so we need to convert to km...

there are 1000 m in one km so we need to divide by 1000

3.23532x1017 m (1 km/ 1000 m) = 3.2x1014 km

(Table of contents)

8. How many light years are we from the sun?  What is this in light minutes?

Well, to solve this, we need to know what one light year is...

1 ly = 9.46x1015 m    but we also need to know how far we are from Earth (in meters) which is equal to 1 A.U. (1.50x1011 m)

This is an easy division...

1.50x1011 / 9.46x1015 m = 1.586x10-5 ly = 1.6x10-5 ly 


(Table of contents)


9. A very strong concertmaster is playing 440.00 Hz at the top of an 4.50 m tall tower on a neutron star where the “g” is 1.816 x 1014 N/kg.  We are at the bottom also playing 440.00 Hz. What is the beat frequency we hear?  Do we hear the player on the top of the tower as sharp or flat?  What frequency do we observe?

Ok... let’s think about this for a second.

  We have the h given to us which is 4.50 meters. Also, we know what “g” is ... 1.816x1014 N/kg. Since we want to know what frequency we hear when standing at the BOTTOM, we just use our given 440.00 Hz frequency.

  After looking at our given data, we can just use the formula

  Df / f = gDh / c2

  We want to find the change in frequency we hear so we solve for the change in frequency... just plug in the given data...

  Df / 440 = 1.816x1014 (4.50) / (3x108) 2

  to solve this, you get ....

  Df = 3.9952 Hz which rounds to 4.00 Hz

  Now that we found the change, we need to add this to the frequency that we are hearing which is, obviously the one coming from the neutron star or the TOP.

  440.00 Hz + 4.00 Hz = 444.00 Hz

  We observe a frequency of 444.00 Hz with a frequency difference of 4.00 Hz

  Now that we found that is this sharp or flat... just remember that at the TOP it is fast/sharp and on the BOTTOM it is slow/flat... so there is your answer... it is sharper because it is at the top coming toward the bottom so we hear a higher frequency...


(Table of contents)

10. If we are living on a neutron star, and we tune the local station “Neutrock 91.7 (MHz = x106 Hz) in at 90.2 on our FM Dial.  We know that we are at a different elevation by 35.6 m.  What is the “g” here?  Are we higher or lower than the broadcast antenna of “Neutrock”?

Well, this is basically the same concept as in number 9 just that we are finding a different variable... To be exact we want to know the value of one “g”...

  Here is the formula again :

  Df / f = gDh / c2

  Since we do not know the change in frequency yet, we need to figure that out which just takes a simply subtraction...

  Df = 91.7 – 90.2 = 1.5 MHz (so we need to multiply this answer by 106 )

  We want to know the “g” value at our elevation which means that we have to use our FM dial frequency as our frequency ... smart huh?!

The height is given as well (35.6 m) so let’s plug it all in the formula:

  1.5x106 / 90.2x106 = g (35.6) / (3x108)2

  To solve for “g” is now easy

  Do the algebra and you get

  g = 4.2x1013 N/ kg

  Now are we higher or lower?! Let’s think about this just for a second... Just by looking at our given data, we can conclude that we must be higher than the antenna...

(Table of contents)

11. A radio source near a very large black hole is exactly 1.0 AU nearer the black hole than we are.  We receive a 152.2 kHz signal from them, approximately what frequency are they transmitting if the gravitational field strength in this region is 58500 N/kg?

Ok, let's look at this problem... 

We know that the distance is 1 A.U. and we know the frequency we receive 152.2 kHz. Since we also know the value of g (58500 N/kg) we can use the following formula: 

Df / f = gDh / c2

Now, just plug in the given values and solve for the change in frequency...

Df / 152.2 = 58500(1.50x1011 ) / (3x108)2

Df = 14.8 kHz  ... but we are not done yet since we need to add this to our given frequency... 

14.8 + 152.2 = 167.0 kHz = 170 kHz

(Table of contents)

12. The Mounds galaxy is receding from us at 15,000 km/s.  What is the change in wavelength of a 415 nm spectral line?  What is the wavelength we measure?  If the Mounds were approaching us, what wavelength would we measure?

Well, we know the speed of light, the velocity and one wavelength and since we want to find the change ... there is a formula for this ....

Dl/l = v/c plug in the givens

Dl/415 = 15000 (1000) / (3x108)  .... (multiply by a 1000 to convert km to meters)

Dl = 20.75 nm

Well, what we measure is the answer + our given line which = 435.75 nm = 436 nm

Just do the opposite to find out the wavelength if Mounds is approaching us 

415 - 20.75 = 394.25 nm = 394 nm 
(Table of contents)

13. The Three Musketeers galaxy has a 525 nm line that comes in at 532 nm.  What is its recession velocity?

We can just use the same formula we used in number 12 just that we first have to figure out our change in wavelength...

Here is the formula: 

Dl/l = v/c

(532 - 525)/ 532 = v / (3x108

v = 3947368.4 m/s but we need this in km/s because that is the answer on the sheet... (divide by a 1000)

v = 3947.3 km/s = 4000 km/s 


(Table of contents)

14. The galaxy My Way is receding at 850 km/s.  At what wavelength does the 716 nm spectral line come in?

Well, same formula as above... Dl/l = v/c and we want to find the change in wavelength...

Dl / 716 = 850(1000)/ (3x108)

Dl = 2.029nm 

Now add that to the 716 and you got the wavelength we were looking for... 

716 + 2.03 = 718 nm   
(Table of contents)

15.The galaxy Androgynous is 5.18 Mpc away.  What is its recession velocity?  (Use H = 50 km/s/Mpc) 

Well, just use the simple formula 

v = Hd

v = (50)(5.18) = 259 km/s
(Table of contents)

16. The spiral galaxy Synonymous has a recession velocity of 11,520 km/s.  What distance is it from us in Mpc and light years?  At what wavelength does a 524 nm spectral line come in? (Use H = 50 km/s/Mpc)

Just use v = Hd

11,520 = (50) (d) 

d = 230.4 = 230 Mpc

However, we want this also in light years so we need to do a little conversion...

230x106 pc ( 3.26 ly / 1 pc) = 751104000 ly = 751x106 ly 

Since we know the speed and the wavelength and speed of light , we can easily solve for the change in the wavelength and then add it ti 524 nm...

Dl/l = v/c

Dl/524nm  = 11520(1000) / (3x108)

Dl = 20.1 nm

add this to the given spectral line

20.1 + 524 = 544 nm 

(Table of contents)

17. Light from the Maytag galaxy shows a redshift in the 627.0 nm spectral line to 634.1 nm.  How many light years away is it?  (Use H = 50 km/s/Mpc)

First we need to find the speed so we use this formula :

Dl/l = v/c 

(634.1 - 627.0)/(634.1) = v / (3x108)

v = 3311780.476 m / (1000) = 3311.78 km

Now, use that answer in the formula

v = Hd 

3311.78 = (50) (d) 

d = 66.23 Mpc 

Now since we want this in light years we need to convert Mpc to ly 

66.23x106 pc (3.26 ly / 1 pc) = 221x106 ly 
(Table of contents)