Astr 2100 Descriptive Astronomy
Fall 2017  Assignment 11  Answers
Assignment
Find the distances to the following

Star A, with a parallax angle of 0.1 arc seconds d
= 1/0.1 = 10 pc

Star B, with a parallax angle of 0.012 arc seconds d
= 1/0.012 = 83.3 pc

The parallax angle for the star Arcturus

The parallax angle is 0.0883 arc seconds

d = 1/0.0883 = 12 pc

Star C with a surface temperature of 4,500 K and a relative magnitude of
9

From the HR diagram, for a temperature of 4,500 K
the absoluet magnitude is about 5.2

m  M = 5 log(d)  5

9  5.2 = 5 log(d)  5

log(d) = 1.76

d = 10^{1.76} = 58 pc

Star D with a surface temperature of 20,000 K and a relative magnitude
of 7.6

From the HR diagram, for a temperature of 4,500 K
the absoluet magnitude is about 25

m  M = 5 log(d)  5

9  (2.5) = 5 log(d)  5

log(d) = 3.02

d = 10^{3.01} = 1050 pc

Star E with a light curve whose maximum is at 450 nm and a relative magnitude
of 5.9

Using Wien's Law T = 2,898,000 / 450 = 6,440 K

From the HR diagram, for a temperature of 6,440 K
the absoluet magnitude is about 3

m  M = 5 log(d)  5

5.9  3 = 5 log(d)  5

log(d) = 1.58

d = 10^{1.58} = 38 pc

Star F with a light curve whose maximum is at 800 nm and a relative magnitude
of 8

Using Wien's Law T = 2,898,000 / 800 = 3,625 K

From the HR diagram, for a temperature of 3,625 K
the absoluet magnitude is about 7

m  M = 5 log(d)  5

8  7 = 5 log(d)  5

log(d) = 1.2

d = 10^{1.2} = 16 pc

A Cepheid variable G whose period is 8 days and whose relative magnitude
is 11

With a period of 8 days the absolute magnitude is
about 2.6

m  M = 5 log(d)  5

11  (2.6) = 5 log(d)  5

log(d) = 3.72

d = 10^{3.72} = 5,250 pc

A Cepheid variable H whose period is 40 days and whose relative magnitude
is 15

With a period of 40 days the absolute magnitude is
about 5.1

m  M = 5 log(d)  5

15  (5.1) = 5 log(d)  5

log(d) = 5.02

d = 10^{5.02} = 105,000 pc

A Type I supernova with a relative magnitude of 8.9 (Hint, first search
for the absolute magnitude of a Type I supernova)

All supernovae have an absolute magnitude of 19.3

m  M = 5 log(d)  5

8.9  (19.3) = 5 log(d)  5

log(d) = 6.64

d = 10^{6.64} = 4.4 x 10^{6} pc =
4.4 Mpc
Back