Fall 2004
Dr. Mike Fanelli
Solutions for Assigned Problems
CHAPTER 17
PROBLEM 17- 4:
A star is determined to have a surface temperature twice that of the Sun, and a
luminosity 64 times greater. What is this star’s radius, expressed in solar units ?
ANSWER: Problem 4 is an application of the radius – luminosity – temperature
relation for stars. Given two of these values, the third is found using that relation,
described on pg 449 in the text.
In solar units, L = R2 x T4 , substituting into the expression gives
64 = R2 x 24 ,
64 = R2 x 16
Dividing each side of the expression by 16 gives, R2 = 4, therefore R = 2.
The star has twice the radius of the Sun.
PROBLEM 17- 10:
Given a star with an apparent magnitude of 10.0, and an absolute magnitude of
2.5, you are asked to find the distance to the star.
ANSWER: Stars appear fainter if located further away, just like any luminous
object. The magnitude of a star represents its brightness, either its perceived
brightness, known as its apparent magnitude, or its actual, true, brightness,
known as its absolute magnitude. More luminous stars have smaller
magnitudes, an unusual attribute of the magnitude system. The star here
appears relatively faint, at 10th magnitude. It must be further than 10 parsecs,
since absolute magnitudes are defined as the magnitude a star would have if
located at 10 parsecs from Earth. The quoted absolute magnitude is 2.5, a much
brighter value, than the observed one.
Using the expression given in MP 17-1,
D = 10 pc  10(m-M)/5
where D is the distance to be determined, expressed in parsecs, m is the
apparent magnitude of the star, and M is the absolute magnitude. This
expression says that the distance to a star can be determined from the
difference between the apparent (observed) magnitude and the absolute
magnitude.
D = 10 pc  10(10-2.5)/5 = 10 pc  101.5 = 316 pc
The star is 316 parsecs distant.