Problems in Chapter 13

Worked out for Geology/Physics 360

An Illustration of Parallax using HNSKY Sirius Now (today) in March Sirius in Sept. 2010 (6 months later)

Question: How would we find a closer star than Proxima Centauri? Problem 3. The Parallax of the red giant Betelgeuse is just barely measurable and has a value of about 0.005 arc seconds. What is its distance? Suppose the measurement is in error by + or – 0.003 arc seconds. What limits can you set on its distance.

Answer: p(Betelgeuse) = 0.005 arc seconds. Therefore its distance is d = 1/p = 200 pc.

The error in this measurement is +/–0.003", so we have to add and subtract this figure from the parallax angle to give the lower and upper limits to the distance estimate.

d(pc) = 1/0.008 = 125 pc, the larger angle corresponds to the closer distance d(pc) = 1/0.002 = 500 pc, the smaller angle corresponds to the farther distance

Problem 5. The star Rigel radiates most strongly at about 200 nm. What is its temperature? How does this compare to the Sun.

𝝀 m for Rigel = 200 nm. Use Wien’s Law to find its temperature?

T = 3 × 10 6 K nm / 𝝀 m = 3 × 10 6 K nm / 200 nm = 15,000 K.

This is much hotter than the surface of the Sun, which is about 6000 K. (2.5 times hotter). (Note that we have approximated the constant here as 3 x 10 6 , it is closer to 2.9 x 10 6

Problem 6. The bright southern star Alpha Centauri radiates most strongly at about 500 nm. What is its temperature? How does this compare to the Sun.

. Alpha Centauri has its peak radiation at 500 nm. Applying Wien’s Law, T = 3 × 10 6 K nm / 500 nm = 6000 K.

The temperature of this star is approximately the same as our Sun’s, so Alpha Centauri has a similar spectral type to the Sun (G2).

Problem 7. Arcturus is about ½ as hot as the sun but is about 100 times more luminous. What is its radius compared to the Sun.

. For Arcturus, T = T o /2, where T o is the temperature of the Sun. L = 100L luminosity, L, to its temperature, T, and radius, R. o . To find the star’s radius, use the Stefan-Boltzmann Law, which relates a star’s L = 4 𝜫 R 2 We want to find R Arc 100 L o = L Arc = 4 𝜫 R 2 Arc 𝝈 T 4 Arc 100 (4 𝜫 R 2 o 𝝈 T 4 o ) = 4 𝜫 R 2 Arc 𝝈 Therefore 𝝈 T 4 or L Arc = 4 𝜫 R 2 Arc 𝝈 100 (R 2 o T 4 o in terms of R o. ) = R 100 (R 2 o T 4 o ) = R 2 Arc 100 (R 2 o ) = R 2 Arc 2 1600 (R 2 o ) = R 2 Arc Arc T (T / 16 T 4 Arc 4 Arc 4 o T 4 Arc and L o = 4 𝜫 R 2 o 𝝈 = R 2 Arc ) / 2 4 R Arc = (1600) 1/2 R o = 40 R o Arcturus is 40 times wider than the Sun.

(T o /2) 4 T 4 o