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.


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