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Chapter 8: Application Examples
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
L
2
0.4 (M
—— = 10
L
1
where M
= 2.89
2
Press b 0 and @ P 0.
1.
Press @ Y ( 0.4 k (
2.
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
log ——
M
= M
– ————
2
1
L
2
where —— = 0.0003
L
1
Press 4.8 - ( l 0.0003 z 0.4 ) e.
1.
Result
The absolute magnitude of the second star is
approximately 13.6.
112
– M
)
1
2
L
2
L
1
0.4
1
^(0.4˚(4.8-2.
Î
89))=
5.807644175
4.8-(log0.0003
©0.4)=
13.60719686
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