Astrophysics Tutorial 4 - Stellar Distances and Magnitudes


Astronomical distances

The Parsec

Apparent Magnitude

An Equation

Absolute Magnitude


Astronomical distances

Space is very big.  The table gives the time for light to travel various distances.  Light travels at 3.0 × 108 m s-1  = 300 000 km s-1.




Sun to Earth

500 s

Sun to Jupiter

40 min

Sun to Pluto

6 hours

Sun to Proxima Centauri (nearest star)

4 years

Across the Milky Way

100 000 years

Galaxy to Galaxy

1 000 000 years



Kilometres are far too small to be useful.  Measuring space distances is called astrometry.


The Parsec

Astronomers use the idea of parallax to estimate the distances of stars.  You will know about parallax as a source of error in reading instruments.  If your eye is not in the right place, your reading will be offset (either too high or too low) as your eye is not in line with the pointer.  On a train, you will see a tree in a nearby field moving relatively quickly compared to the hills in the distance.  This is due to parallax.




Stars appear offset against the background of other stars, depending on the season:


Autumn                            Spring


In the autumn, the following pattern is observed.  The background of distant stars remains the same, but the nearby star appears to have moved to the right.  In the spring, the background pattern is observed as before.  However the nearby star appears to have moved to the left.


In the winter you might not see the star as it will be behind the Sun.


In Summer, the nearby star will be halfway between the two positions.


The closer the star, the more rapidly it will appear to move, and the variations in position will be greater.  


To work out the parallax angle, we can draw a diagram like this:



Let the distance from the Sun to the star be d.  Let the radius of the Earth’s orbit be R.  The angle can be worked out using the tangent function:



Since q sin qtan q for small angles in radians, we can write:




The distance d is very much bigger than R, so the angle is going to be very small. 


Although we have stressed the radian as the units to measure angles in, astronomers often use degreesAngles of less than 1 degree are often measured in minutes, where 1 minute is 1/60 degree.  An even smaller angle is used called an arc second, where 1 arc second = 1/60 minute = 1/3600 degree.  This is the basis of the parsec (pc) which is defined as:


the distance to a star that subtends an angle of one arc second to the line from the Earth to the Sun


The diagram shows the idea:


On this diagram, the distance, R, is the distance from the Earth to the Sun in astronomical units


1 astronomical unit = 1.496 × 1011 m


Now 1 degree = 2p ÷ 360 = 0.0175 rad


1 arc second = 0.0175 ÷ 3600 = 4.85 × 10-6 rad.


Since d = R/q,

d = 1.496 × 1011 m ÷ 4.85 × 10-6 rad = 3.09 × 1016 m


1 light year = 3.00 × 108 m s-1 × 365 dy × 86400 s dy-1 = 9.46 x 1015 m


Therefore d = 3.09 × 1016 m ÷ 9.46 x 1015 m ly-1 = 3.26 light years.



If we measure the angle of parallax, we can measure the distance from the Sun by the equation:



The figure 1 refers to the distance from the Sun in astronomical units.  The term d is the distance in parsecs, and q is the angle in arc-seconds.  The smaller the angle in arc seconds, the further the star.



Before using this equation, make sure that the angle is in parsec, NOT radians.


The table shows the units used in astronomy and astrophysics. 



What it is



Astronomical Unit

Distance between Earth and Sun


1 AU = 1.496 x 1011 m

Light Year

Distance travelled by light in 1 year


1 ly = 9.46 x 1015 m


An object at 1 pc subtends an angle of 1 arc second for a distance of 1 AU


1pc = 3.086 x 1016 m

1 pc = 3.26 ly


Parallax methods work well up to about 100 pc (which is quite a long way).  Beyond that, the measurement of the parallax angle is not at all easy.


Question 1

What is 10 parsec in kilometres?  A supersonic plane is travelling at 3000 km h-1.  How long would it take to travel 10 parsec? 




Apparent Magnitude

The Greek astronomer Hipparchus (190 BC - 120 BC)  classified stars according to their apparent brightness to the naked eye, about two thousand years ago. (In some texts you will see his name written as Hipparchos.)  Its scale was 1 to 6.  The scale is still used today and is called the apparent magnitude scale.  The apparent magnitude is given the code m.  Magnitude 1 stars are about 100 times brighter than magnitude 6 stars.  A change in 1 magnitude is a change of 2.512 (1001/5 = 2.512).  The scale is logarithmic because each step corresponds to multiplying by a constant factor.


The apparent magnitude is the brightness of a star as it appears to the observer.


To work out the apparent magnitude we need to:

Worked example

The Sun has a magnitude of -26.7.  The moon has a magnitude of - 12.7.  What is the ratio of their brightnesses?


Difference in magnitudes = dimmest - brightest = -12.7 - -26.7 = +14

Ratio of brightnesses = 2.51214 = 398 000


Therefore the Sun is about 400 000 time brighter than the Moon.  Notice that bright objects have negative magnitudes.  The scale looks like this:




Question 2

How much brighter is Sirius (m = -1.46) than Betelgeuse (m = +0.50)?


Question 3

What do you think is a problem with the apparent magnitude scale?



An Equation

Consider two stars that have apparent magnitudes mA and mB.   They have intensities IA and IB respectively. The difference in magnitude is:



Each difference in magnitude of 5 gives an intensity increase of 100 times.

This can be rewritten as:

Take logarithms to get rid of the powers:


This rearranges to:


Maths Note

The term 'lg' stands for log10.  Do not use natural logs here.



Absolute Magnitude

In Nuclear Physics Tutorial 4, we saw that the intensity of gamma radiation reduced as an inverse square law.  This is true for any radiation, including visible light.  The relationship is:



[I – intensity; I0 intensity at the source; k – constant; x – the distance from the source]


When doing intensity calculations, it is more common that we do not know what I0 is.  Instead we have a count I1 at point 1 and a count I2 at point 2 at distances x1 and x2 respectively. 

So we can write:





We can combine these in a rearranged form to give us:


I1(x1)2 = kI0 = I2(x2)2


So we can write:


If we rearrange further we can get a ratio:




The apparent magnitude represents the intensity, i.e. the power per square metre.



Consider two stars A and B that have intensities IA and IB respectively.  Suppose the stars had apparent magnitudes mA and mB.   The difference in magnitude is:


 Dm = mA - mB


Now, every difference of 5 magnitudes gives an increase in intensity of 100 times.  We can give a general rule as:



We can rewrite this as:




Each increase in magnitude does NOT give an increase in intensity of 20 times.


We can get rid of the power by taking logarithms.  This time we use log10 (lg) not ln.  lg (100) = 2.  If one number is divided by a second number, the log of the second number is subtracted. 


Therefore we can write the equation as:

lg(IA) - lg(IB) = 2 × 0.2 Dm = 0.4 Dm


Rearranging in terms of Dm:



we can write:



where d is the distance of the star.  The log of a number squared is twice the log of the number, so:





When we use absolute magnitude, we "place" all stars at an arbitrary distance of 10 parsec.  We give the absolute magnitude the code M.  So we are comparing the brightnesses of all stars as if they were at 10 pc.


So let's think of another star at a distance d which has an apparent magnitude of m.  We can work out the absolute magnitude by moving the star to 10 parsec (in our minds, of course):


We can rewrite this as:



The absolute and apparent brightnesses of stars are related by this formula:


[ m - apparent magnitude; M - absolute magnitude; d - distance (pc)]




Make sure that you use log10 not loge.



Worked Example

The apparent magnitude of Sirius = -1.46.  It is at a distance of 4 ly from Earth.  What is its absolute magnitude?


Convert ly to pc: 4 ÷ 3.26 = 1.23 pc

-1.46 - M = 5 log 1.23/10

-M = (5 x -0.910) + 1.46

-M = -4.55 + 1.46 = - 3.09

M = +3.09


If the star is less than 10 pc away, then the absolute magnitude tends to move in the positive sense, which means that they become dimmer.  Stands to reason.  If you move a star further away, the dimmer it gets.  Stars over 10 pc will have absolute magnitudes that are brighter than their apparent magnitude.


Question 4 

The apparent magnitude of the Sun is -26.7  What is its absolute magnitude?  How does it compare with the absolute magnitude for the star Alpha Centauri which is +4.38?



The answer to this question should tell you that the Sun viewed from 10 pc would glow with a brightness similar to many other stars.


Question 5 

Bellatrix and Elinath are two stars with the same apparent magnitude. The distance from the

Earth to Bellatrix is 470 light years and its absolute magnitude is -4.2.

(i) Calculate the distance to Bellatrix in parsecs.

(ii) Calculate the apparent magnitude of Bellatrix.

(iii) Elinath has an absolute magnitude of -3.2. State, giving a reason, which of the two stars is closer to the Earth. 

(AQA Past Question)