Sharp EL-5250, EL-5230 operation manual 1Î0.4˚4.8-2. 89=, 4.8-log0.0003 0.4=, Example, Result

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

L2	0.4 (M1 – M2)
—— = 10
L1

whereM2 = 2.89

1.Press b 0 and @ P 0.

2.Press @ Y ( 0.4 k (

4.8 - 2.89 ) ) e.

Result

1Î^(0.4˚(4.8-2. 89))= 5.807644175

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

L2 log —— L1

M2 = M1 – ————

0.4

L2

where —— = 0.0003 L1

1.Press 4.8 - ( l 0.0003 z 0.4 ) e.

Result

The absolute magnitude of the second star is approximately 13.6.

4.8-(log0.0003 ©0.4)= 13.60719686

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