Problem Set 33: | 1 | 2 | 3 | 4 | 5 | 6 | 9 | 24 | 25 | 26 | 28 | 29 | 32 | 36 | 37 | Go up
- by Zach Peterson, 2003

1. The parallax angle of a star is 0.00017 degrees. How far away is the star?

d (parsec) = 1/p (arc-second)

0.00017 x 3600 = .612 arc-seconds

1/.612 = 1.634 parsecs

1.634 x 3.26 = 5.3 ly

2.  A star exhibits a parallax of 0.28 arc-seconds. How far away is it?

d (parsec) = 1/p (arc-second)

d = 1/.28 = 3.57 parsecs

3.57 x 3.26 = 11.64 ly

3. Using the definitions of the parsec and the light-year, show that 1 pc = 3.26 ly.

1 parsec = 3.26 ly

1 parsec = 1 A.U./tan(2.8 x 10^-4) = 3.086 x 10^13 km = (3.086 x 10^13)(9.46 x 10^12 km/ly) = 3.26 ly

4. A star is 24 parsecs away. What is its parallax angle? State (a) in seconds of arc, and (b) in degrees.

(a)    1/24 = .0416 arc seconds

(b)    .0416 x (1/3600) = 1.16 x 10^-5 degrees

5. What is the parallax angle of a star if it is 42 light-years away? How many parsecs is this?

42 ly/3.26 = 13 parsecs

1/13 = .077 arc seconds

6. A star is 25 parsecs away. How long does it take for its light to reach us?

Convert parsecs to light-years.

25 x 3.26 = 81.5 ly = 81.5 years

9.  What is the apparent brightness of the sun as seen on Jupiter? (Jupiter is 5.2 AU away)

b = L/(4pd2)

d = 7.783 x 10^6

L = 8.67 x 10^-8(4pr2)T4

L = 3.68 x 10^20

b = 3.68 x 10^20/(4p7.8 x 10^82)

b = 48.1 W/m2

24. If a galaxy is traveling away from us at 1.0 percent of the speed of light, roughly how far away is it?

v = Hd

3000 km/s = 80d

d = 3.75 x 3.26 x 10^6 = 1.2 x 10^8 ly

25. The redshift of a galaxy indicates a velocity of 2500 km/s. How far away is it?

v = Hd <- Mpc!!!
2500 = 80d

d = 31.25 Mpc

26. Estimate the speed of a galaxy (relative to us) that is near the "edge" of the universe, say 10-billion light-years away.

Convert light-years to Mpc.

10^9/3.26 x 10^6 = 1,000,000 Mpc

v = 80 x 1,000,000

v  = .8c

28. Estimate the observed wavelength for the 656-nm line in the Balmer series of hydrogen in the spectrum of a galaxy whose distance from us is (a) 106 ly, (b) 108 ly, (c) 1010 ly

The first thing to do is turn the LY into Mpc.  Since 1 pc is 3.26 LY, then
106 Ly = .306748 Mpc
108 Ly = 30.6748 Mpc and finally
1010 Ly = 3067.48 Mpc

If you use v = Hd to get the recession velocity, then you arrive at these recession veolocities
106 Ly = .306748 Mpc --> 24.54 km/s
108 Ly = 30.6748 Mpc --> 2454 km/s
1010 Ly = 3067.48 Mpc --> 245,400 km/s

Since the speed of light is 300,000 km/s the approximate formula yields

(a) Dl/l = v/c

Dl/656 nm = 24.54 km/s/300,000 km/s, Dl = .054 nm, = 656.054

(b) Dl/l = v/c

Dl/656 nm = 2454 km/s/300,000 km/s, Dl = 5.4 nm, = 661.4

(c) Well - 245,400 km/s is not small compared to 300,000 km/s (i.e. v is not << c)
So you will have to use equation 33-3 from the book (The real equation):

I get 2073 nm when I do it, they rounded a bit and got 2170 nm  I think I am right

- but since the Hubble constant is largely uncertain...

29. Estimate the speed of a galaxy, and its distance from us, if the wavelength for the hydrogen line at 434 nm is measured on Earth as being 610 nm?

use the formulae m = m0/g and g = 1/Ö1 - (v2/c2)
using the rules of perspective and simultaneity, we know that l' = 610 nm and that
l = 434 nm

610 =  434Ö1 - (v2/c2)

Do the math and you get 0.328c

Now to get the distance you use v = Hd

.328c = 80d

d = 4 x 10^9 ly

32. Calculate the wavelength at the peak of the blackbody radiation distribution at 2.7 K using the Wien law.

l(2.7) = 2.9 x 10^-3

l = .001074

36. Suppose that three main-sequence stars could undergo the three changes represented by the three arrows, A, B, and C, in the H-R diagram of Fig. 33-26. For each case, describe the changes in temperature, luminosity, and size.

<Fig 33-26 on page 1034>

A: T increases, L constant, size decreases

B: T constant, L decreases, size decreases

C: T decreases, L increases, size increases