Magnitude Calculations

13.3 - Be able to use the distance modulus formula to determine the absolute (M) or apparent magnitude (m) of a star, given the distance to the star (d): M = m + 5 − 5log d where d is the distance in parsec

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.


Question 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? 

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

 

Question 2

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

2.5 brighter = 1 magnitude 
B = 3.3

 

Question 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

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

 

Question 4

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

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

 

Question 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?

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

 

Question 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?

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

 

 

Summary

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