112
Example 1
What is the ratio of the sun’s luminosity to that of a star having an absolute
magnitude of 2.89?
(The sun’s absolute magnitude is 4.8.)
The former equation is equivalent to
1. Press b 0 and @ P 0.
2. Press @ Y ( 0.4 k (
4.8 - 2.89 ) ) e.
Result
5.807644175
The star is nearly six times as luminous as the sun.
Example 2
A second star has only 0.0003 times the luminosity of the sun. What is its
absolute magnitude?
The previous equation is equivalent to
1. Press 4.8 - ( l 0.0003 z 0.4 ) e.
Result
The absolute magnitude of the second star is
approximately 13.6.
Chapter 8: Application Examples
M
2
= M
1
– ————
where —— = 0.0003
L
2
L
1
L
2
L
1
log ——
0.4
4.8-(log0.0003
©0.4)=
13.60719686
—— = 10
0.4 (M
1
– M
2
)
whereM
2
= 2.89
L
2
L
1
1
Î
^(0.4˚(4.8-2.
89))=
5.807644175
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