Magnitude Calculations

Demonstrate an understanding of, and perform simple calculations involving, apparent magnitude (m), absolute magnitude (M) and distance (d in pc), using this formula: M = m + 5 - 5 log d involving powers of 10 only (students are not required to calculate d using this equation, only M and m)

There are different types of calculation you will be asked to make. Working out differences in apparent magnification, finding absolute and apparent magnification and also distance.


To work out Absolute Magnitude:
M = m + 5 - 5 log d

To work out Apparent Magnitude:
m = M-5+5 log d

To work out Distance using Magnitude:
10(m-M+5)/5


Example 1
Spica has an apparent magnitude of 0.98 and an absolute magnitude of - 3.55. Which is brighter when viewed from a distance of 10 parsecs? Highlight the below text to see the solution.

10 parsecs is the brightness measured at absolute magnitude. A smaller (or even negative) number is brighter. Spica's absolute magnitude is therefore brighter.

Example 2
Star A is apparent magnitude 2.3. It is 2.5 times brighter than B. What is B's apparent magnitude? Highlight below.

2.5 brighter = 1 magnitude
B = 3.3

Example 3
Two stars, A and B have different apparent magnitudes: A= 1.8, B = 4.8
a) How many degrees of apparent magnitude is A brighter than B?
b) How much brighter is A than B Highlight below..

a) 4.8 minus 1.8 = 3
b) 2.53 = 2.5 x 2.5 x 2.5 = 16

Example 4
The star, Rigel is 238 parsecs from Earth. Its apparent magnitude is 0.15. What is Rigel's absolute magnitude? Highlight below..

M = m + 5 - 5 log d
M = 0.15 + 5 -5 log 238
M = -6.73

Example 5
The star, Regelus has an absolute magnitude of 0.54. It is 23.8 parsecs from Earth. What is its apparent magnitude from Earth? Highlight below..

m = M-5+5 log d
m = 0.54 -5 + 5 log 23.8
m = 2.42

Example 6
Deneb has an apparent magnitude of 1.25 and an absolute magnitude of -8.75. How far away from Earth is Deneb in parsecs? Highlight below..

10 (m-M+5)/5
(m-M+5)/5
1.25 - -8.75 + 5 = 15 (Subtracting a negative number produces a positive)
15/5 = 3
103 = 1000 parsecs

 

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