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
)
where
M
2
= 2.89
L
2
L
1
1
Î
^(0.4˚(4.8-2.
89))=
5.807644175
Summary of Contents for EL-5230
Page 1: ...PROGRAMMABLE SCIENTIFIC CALCULATOR OPERATION MANUAL EL 5230 EL 5250 ...
Page 2: ......
Page 12: ...10 ...
Page 62: ...60 ...
Page 132: ......