# How Bright Is That Star (or Planet)?

Copyright © 2021–2024 by Stan Brown, BrownMath.com

Copyright © 2021–2024 by Stan Brown, BrownMath.com

In this article, I’ll set you a number of
practical real-life problems in brightness and magnitude, and walk you
through the calculations. We’ll also
develop shortcut equations relating brightness and magnitude. Finally,
we’ll relate **apparent magnitude**, the magnitude of a
star in our sky, and **absolute magnitude**, which measures a
star’s intrinsic brightness.

A free Excel workbook is available for download.

When you look at the night sky, it’s obvious that some
of those lights are brighter than others. That suggests creating a
scale of brightness, or **magnitude**, so that you can say
“This star has a magnitude of 2.1, while that star’s
magnitude is 4.8”. And in fact such a scale exists.

It all started with the ancient Greeks. (It’s amazing how many explanations start with that sentence. The ancient Greeks were hot stuff in mathematics and science, no doubt about it.) Hipparkhos (‘Ιππαρχος, about 190–120 BCE) is better known by the Latin form of his name, Hipparchus, though he lived or worked in Greek-speaking cities east of the Aegean Sea, not in Roman lands.

Hipparchus was a keen observer of the night sky, and a good mathematician. He calculated the motions of the solar system based on the Sun at the center, but abandoned the work in disgust because the orbits were not perfect circles. (Copernicus came 1700 years later.) He predicted eclipses, and he’s known as “the father of trigonometry”. But what directly concerns us in this article is his star catalog, produced around 135 BCE (and, er, “borrowed” by Ptolemy for his Almagest almost 300 years later).

In that catalog, Hipparchus didn’t just record the
positions of stars; he also recorded their brightness, or
**magnitude**. It’s not certain whether he himself invented
the system of magnitudes, but his is the first published work that we
know of that uses them.

(Before we go any further, let’s be clear that we’re
talking about the brightness of stars *as seen in our sky*.
One
star may be brighter than another not because it’s actually
brighter, but because it’s closer, or because the other
star’s light is dimmed by dust clouds between it and us. Technically
the intrinsic brightness of a star is its absolute magnitude, which
we’ll get into later, and how bright it looks
in our sky is its **apparent magnitude**, but mostly people just
say “magnitude” when they mean “apparent magnitude”.)

Hipparchus divided the visible stars into six classes. The
brightest group of stars were called “first magnitude”, the next
group “second magnitude”, and so on down to “sixth magnitude”, the faintest
group that can be seen by the naked eye under good conditions. He
chose the magnitude classes so that the first-magnitude
stars would be a certain amount brighter than the second-magnitude
stars, those would be brighter than the third-magnitude stars *by
the same amount*, and so on.

There are only 20 first-magnitude stars, more
second-magnitude stars, and so on; the largest class is
sixth-magnitude stars. (And there are *far* more stars even
dimmer, that you need a telescope or good binoculars to see, but no
one—no one that we know of, anyway—even considered
that possibility until Galileo (1564–1642) turned his
telescope on the heavens in the early 1600s and saw stars that no human had seen
before.)

The belt of Orion (at right) points to Sirius. (Hubble Space Telescope photo from Wikipedia.)

Sirius is a first-magnitude star, the brightest in the night sky, in fact; you can find it when Orion is in the sky by following the three stars of Orion’s belt. Polaris, the North Star or Pole Star, is a second-magnitude star, and you can find it by following the two “pointer stars” on the bowl of the Big Dipper. You’ve probably never heard of any sixth-magnitude stars, but the planet Uranus falls in that group; this means that a person with very good eyesight, looking at the right place in a dark sky, will just be able to see it.

The brightness of a star isn’t just one of six values. Some first-magnitude stars are brighter than others; some second-magnitude stars are brighter than others, and so on. But just with the naked eye it’s hard to pick out those fine distinctions.

Fast-forward just about 2000 years, to 1856. The English
astronomer
Norman Robert Pogson
(1826–1891) noticed that
**first-magnitude stars were about 100 times as bright as sixth-magnitude stars.**
Astronomers followed his suggestion to make that the standard.

Remember Hipparchus’ rule: as you go from magnitude
6 to 5, 5 to 4, 4 to 3, 3 to 2, and 2 to 1, you go through five
*equal steps* of increasing brightness. From 6 to 4 is not
twice as much as 6 to 5; it’s the *square* of the
increase in brightness from 6 to 5. Magnitude is a
**logarithmic scale**, because
**adding or subtracting a magnitude divides or multiplies the brightness by a fixed amount.**
(See It’s the Law Too — the Laws of Logarithms.)

Therefore, since magnitude 6 to 1, a difference of 5 magnitudes,
is a 100-fold increase in brightness, each magnitude must be
^{5}√100 = 100^{1/5} = about 2.512 times as bright as the next
dimmer magnitude. (Always remember:
**as brightness increases, magnitude decreases**, and vice
versa.)

If we’re going to assign magnitudes, can’t we do
better than just whole numbers? Sirius and
Vega
are both first-magnitude stars, but
Sirius is several times as bright as Vega. And by Pogson’s time,
astronomers *could* see and measure finer distinctions in
brightness, thanks to the photometer (a brightness-measuring
instrument) invented around 1836 by the German physicist Carl August
von Steinheil (1801–1870), This changed magnitude from a
whole-number scale to a continuous scale of real numbers.

But everything was still relative. Astronomers could say that Star A
was 0.6 magnitude brighter than Star B (or about 1.74 times as
bright), but how could they assign a specific numerical magnitude to
each star? The answer: they had to *define* one star’s
magnitude, and then all the other star magnitudes would be found based
on the difference in brightness. First they set Polaris to magnitude
2.0, but Polaris turned out to be a variable star, fluctuating in
brightness, so that was no good. There have been a few tweaks since
then, but we don’t have to worry about those.

Let’s look at how those calculations are done. We’ll start by
**converting a brightness ratio to a difference in magnitude**.

**Problem 1**: For
instance, if Polaris has a magnitude of 1.98, and Vega is about
6.026 times as bright as Polaris, what is the magnitude of
Vega?

Solution: Each change in brightness by a factor of about 2.512 is a change of magnitude by adding or subtracting 1. If we find how many times 2.512 must be multiplied by itself to get 6.026, we’ll know the difference in magnitude between Vega and Polaris. That’s not a whole number, it’s a decimal. To find it, we have to solve the equation

2.512^{x} = 6.026

To solve that, we take the logarithm of both sides. (Remember I mentioned earlier that magnitude is a logarithmic scale? This is where we start to see that. If you’re shaky on logarithms, have a look at It’s the Law Too — the Laws of Logarithms.)

*x* (log 2.512) = log 6.026

*x* = (log 6.026) / (log 2.512)

*x* = 1.95

The difference in magnitude between the two stars is 1.95. But
since the brighter star has the lower magnitude, and Vega is brighter,
Vega’s magnitude is *less* than Polaris’s 1.98 by that 1.95
difference we just computed. 1.98 − 1.95 = 0.03; the
magnitude of Vega is 0.03.

This is a little surprising at first: aren’t first-magnitude stars supposed to be the brightest? Yes, they are, but Hipparchus’ first-magnitude group, the 20 brightest stars, included a wide range of brightness. The brightest of those 20 stars is more than 2.512 times as bright as the dimmest of them, so when we assign numeric magnitudes a difference of 1 won’t be enough for the stars that Hipparchus called “first magnitude”.

In fact, there are four stars that are brighter than Vega. Since brighter stars have lower magnitudes, those four—Arcturus, Alpha Centauri, Canopus, and Sirius—all have negative magnitudes.

Now you try one. **Problem 2**: The star Sirius
is about 3.98 times as bright as Vega; what is its magnitude? (Please
try to figure it out for yourself before you look at
the solution.)

Because this article helps you,

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

So much for going from brightness ratio to magnitude
difference (though we’ll get to a shortcut later). What about
**finding the ratio of brightness when you know the difference in magnitude**?

**Problem 3**: The Hubble space
telescope can see
stars as faint as magnitude 30. How much dimmer are those stars than a
typical sixth-magnitude star? (Recall that sixth-magnitude stars are
the dimmest that normal human eyesight can see, in clear skies without light
pollution.)

Solution: The difference in magnitudes is
30 − 6 = 24. If one magnitude makes for 2.512 times as
bright, then 24 magnitudes would be 2.512 multiplied by itself 24
times. 2.512^{24} = about 4
billion. The Hubble can see objects 4 billion times as dim as a
sixth-magnitude star.

Your turn! **Problem 4**: The
magnitude of Castor, the second-brightest star in the constellation
Gemini, is 1.58. Sirius, as you calculated, has a magnitude of
−1.47. How much brighter than Castor is Sirius? (Calculate it
yourself before looking at the solution.)

(You can find Castor and its twin Pollux from Orion. Look at this page and scroll to the bottom.

Wow! We’ve gone from a simple division of stars into six classes (with some glitches in the first-magnitude class) to decimal magnitudes, magnitudes greater than 6, and negative magnitudes. We saw magnitude 30 in Problem 3, and fainter objects almost certainly exist. At the other end of the scale, there’s no definite upper limit to brightness (or lower limit to magnitude), though very bright objects will look dim to us because they’re far away and because many of them are obscured by dust and gas between us and them.

You’ve learned to convert from a difference in magnitude to a ratio of brightness, and vice versa. But it can be kind of tedious to work every problem by going back to first principles. Also, that number 2.512 isn’t the easiest to remember. Can we work out some general formulas? You betcha we can!

Let’s begin by defining the variables that will appear in our formulas:

*m*_{1}and*m*_{2}are the magnitudes of the two objects.*B*_{1}and*B*_{2}are the brightnesses of the two objects.

We’re looking for some formula that will have
*B*_{1}/*B*_{2}, the ratio of brightness, on the left side, and an
expression involving *m*_{1} − *m*_{2}, the difference in
magnitudes, on the right side.

And actually, you’ve already got this! In Problem 3 and Problem 4, you computed a brightness ratio this way:

*B*_{1}/*B*_{2} = 2.512^{(m2−m1)}

Let’s just double-check the signs; they can be tricky to get right since greater magnitude is lower brightness.

*Case 1*: star 1 is brighter. Then
*B*_{1} > *B*_{2}, and
*B*_{1}/*B*_{2} > 1. If star 1 is brighter, its
magnitude *m*_{1} is *less*, *m*_{2} − *m*_{1} is positive
(subtracting a smaller number from a bigger number), and the right
side of the equation is also greater than 1. So that’s fine.

*Case 2*: star 1 is brighter. Then
*B*_{2} > *B*_{1}, and
*B*_{1}/*B*_{2} < 1. If star 2 is brighter, its
magnitude *m*_{2} is *less*. *m*_{2} − *m*_{1} is negative
(subtracting a bigger number from a smaller number), and the right
side of the equation is also less than 1. So that’s fine.

*Case 3*: the stars are equally bright. In that case
*B*_{1} = *B*_{2} and the left side of the equation is 1. And
*m*_{1} = *m*_{2}, so *m*_{2} − *m*_{1} = 0, and
2.512^{0} = 1. That’s fine also. Yay! The signs are
correct for all possible cases.

There’s
just one more thing to do. 2.512 is not an exact number, but an
approximation. We always want to use exact numbers in an equation if
we can, to avoid or reduce rounding errors. Where did 2.512 come
from? It was ^{5}√100 or 100^{1/5}: each magnitude
was a step of 100^{1/5}, about 2.512, in brightness.

We can rewrite that in base 10 instead of base 100. That will
be a little more convenient because many calculators have a
10^{x}, antilog, or log^{−1} key.
100 = 10^{2}, and 100^{1/5} =
(10^{2})^{1/5}. Now in general,
(*c*^{a})^{b} =
*c*^{ab}. (See
It’s the Law — the Laws of Exponents if you’re a little shaky on
exponents.) Therefore 100^{1/5} =
(10^{2})^{1/5} = 10^{2/5} =
10^{0.4}.

Going back to the equation *B*_{1}/*B*_{2} = 2.512^{(m2−m1)},
replace the approximation 2.512 with the exact value
10^{0.4} to get

*B*_{1}/*B*_{2} = (10^{0.4})^{(m2−m1)}

Using that same power rule, multiply the exponents and we have:

(1) *B*_{1}/*B*_{2} = 10^{0.4(m2−m1)}

Notice that it doesn’t matter whether the two stars have
magnitudes 2.6 and 5.3 or 1 and 3.7—as long as the
*difference* in magnitudes is the same, the *ratio* of
brightness will be the same. There’s no way to get the brightness of
either star out of this equation, just the *ratio* of
brightness. (Of course astronomers have tools for getting the
brightness of an individual star, but that’s another story.)

So far, I’ve always said "star" when talking about magnitudes.
But **the planets fit into the magnitude scale too**, and so do
all the other objects in the solar system. At the right you’ll see
magnitudes of the planets and a few other objects.

Magnitudes in the Solar System | |
---|---|

Sun | −26.74 |

Moon | −12.74 (average) |

Mercury | +7.25 to −2.48 |

Venus | −2.98 to −4.92 |

Mars | +1.86 to −2.94 |

Ceres | +6.64 to +9.34 |

Jupiter | −1.66 to −2.94 |

Saturn | +1.17 to −0.55 |

Uranus | +6.03 to +5.38 |

Neptune | +8.00 to +7.67 |

Pluto | +16.3 to +13.65 |

Source: Wikipedia articles on the named bodies. |

**The Sun**’s magnitude doesn’t vary much,
because the Sun itself is pretty stable, and the Earth orbits the Sun at
very nearly a constant distance. But all the other objects change in
brightness with changes in their distance from us, their distance from
the Sun, and their phases.

**The Moon**’s magnitude obviously varies with the Moon’s
phase. But its distance from us matters too, and that’s on a
different time cycle from the Moon’s phases. Thus one full Moon
doesn’t have the same magnitude as another. A so-called **“supermoon”** is
simply a full Moon that occurs when the moon is near its minimum
distance from Earth in its orbit (“perigee”). Such a full Moon can be
about 30% brighter than a full Moon when the Moon is at a point in its
orbit far from us (“apogee”).

Timeanddate.com has a very nice animation showing how the planets’ distances, sizes, and magnitudes vary over time.

As you can see from the table, **Venus and Jupiter** are
always brighter than any star in the night sky. (The brightest star is
Sirius, at magnitude −1.47.) In fact, Venus is so bright that,
if you know where to look, you may see it in blue sky after
sunrise or before sunset. Jupiter is a brilliant white “star” brighter
than any of the actual stars. With a pair of binoculars, you can see
Jupiter as a tiny disc rather than a point like the other planets, and
its four largest moons as well—what Galileo saw with his telescope.

The dwarf planet **Ceres** is listed in the table because
you should sometimes be able to see it with binoculars. **Uranus** is a
faint naked-eye object under favorable conditions, but always visible
in binoculars if you’re not in a city.

Naturally, whether a given planet is visible on a given night depends on where it is in its orbit relative to us. To learn which planets will be visible tonight and where to look, Earthsky is a good source. And the good old Farmers’ Almanac lets you customize your location and tells you when the planets will rise and set.

Let’s use the planets to try finding the ratio of brightness from the difference in magnitudes.

**Problem 5**: How much brighter is
Mars at its brightest than at its dimmest?

Solution: Use formula (1).
*B*_{2}/*B*_{1} = 10^{0.4(m2−m1)} =
10^{0.4(1.86−(−2.94))} =
10^{0.4(4.8)} = about 83. Mars at its brightest is 83
times as bright as at its dimmest.

Incidentally, Mars has such a wide variation because its distance from us can be as little as 35 million miles (56 million km) or as much as 248 million miles (400 million km), and its maximum distance from the Sun (“aphelion”) is 120% of its minimum distance (“perihelion”).

**Problem 6**: When I was inspired
to write this article, on
13 October 2020, Mars was at magnitude −2.6, brighter than it
will be again till the year 2035 and outshining even Jupiter, at
magnitude −2.3. How much brighter was Mars than Jupiter
on that night? (Solution here.)

**Problem 7**: But Jupiter is
usually brighter than Mars. At their mean brightnesses, Jupiter’s
magnitude is −2.20, and Mars’s is +0.71. How much brighter is
Jupiter than Mars, on average? (Solution here.)

What if you have a brightness ratio between two objects, and you want the difference in their magnitudes? In other words, what’s the reverse of formula (1)? Start with that formula:

*B*_{1}/*B*_{2} = 10^{0.4(m2−m1)}

You want to solve that formula for *m*_{2} − *m*_{1}, but
the magnitude difference is locked up in the exponent, so you
unlock it by taking the logarithm of both sides:

log(*B*_{1}/*B*_{2}) = 0.4(*m*_{2} − *m*_{1})

Multiply both sides by 2.5 to get rid of the 0.4, and swap left and right sides:

(2)*m*_{2} − *m*_{1} = 2.5 log(*B*_{1}/*B*_{2})

What if you have a brightness ratio, and the magnitude of one
object, and you want the magnitude of the other? Nothing easier: just
add *m*_{1} to both sides and you get

(3)*m*_{2} = *m*_{1} + 2.5 log(*B*_{1}/*B*_{2})

**Problem 8**:
Mira is a red
giant star in the
constellation Cetus
(“the Whale”). It’s a variable star, which grows dim and then bright
again over a period of about 11 months. At its dimmest, it’s only a
10th-magnitude star, but it grows 1600 times as bright during its
cycle. What is its magnitude at its brightest? (Solution
here.)

**Problem 9**:
According to Isaac Asimov,
“Nova Aquilae
appeared during World War I, just as the Germans’ last great offensive
on the Western Front was beginning to run out of steam. Five months
later, Germany surrendered and Nova Aquilae was called ‘the star of
victory’ by the Allied soldiers at the front.”

The nova brightened very quickly to 52,000 times the star’s normal brightness. How many magnitudes does that represent, and was it an increase or decrease in magnitude? (Solution here.)

Binary and multiple stars (two or more stars bound to each
other by gravity) are fairly common. The closer ones can be
distinguished by binoculars. The closest of all is Alpha Centauri, one
of the brightest in our skies, but *the* brightest is our old
friend Sirius.

Sirius A and Sirius B orbit their common center of gravity with a period of about 50 years. Sirius A is about six times the size of our Sun, and has a magnitude of −1.45. Sirius B is a white dwarf, with a mass slightly greater than our Sun’s but a volume less than that of Earth; it has a magnitude of 8.68.

Because this article helps you,

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

If we know the individual magnitudes of two companion stars, can we find
the magnitude of the pair? If we know the combined magnitude of the
pair, and the individual magnitude of one companion, can we find the
magnitude of the other? The answer is *yes* to both
questions.

Of course we can’t just add magnitudes: that would be
like *multiplying* brightnesses. Instead, we have to compare the
brightness of the combination to the brightness of one star. From
that we can find out how much lower is the magnitude of the combination than
the magnitude of that same star.

Well, the brightness of the pair is just

*B*_{C} = *B*_{2} + *B*_{1}

and the brightness *ratio* of the combination to star 2
alone is

*B*_{C}/*B*_{2} = (*B*_{2} + *B*_{1})/*B*_{2}

*B*_{C}/*B*_{2} = 1 + *B*_{1}/*B*_{2}

But formula (1) tells us that the ratio of
brightness between the two stars of Sirius (or any binary star) is
*B*_{1}/*B*_{2} = 10^{0.4(m2−m1)}, so we
can substitute that in:

*B*_{C}/*B*_{2} = 1 + 10^{0.4(m2−m1)}

Given the ratio of brightness and the magnitude of star 2, what is the magnitude of the combination? formula (3) tells us that

*m*_{C} = *m*_{2} + 2.5 log(*B*_{2}/*B*_{C})

We don’t know *B*_{2}/*B*_{C}, but we do know *B*_{C}/*B*_{2}
from a few lines above, and *B*_{2}/*B*_{C} = 1/(*B*_{C}/*B*_{2}).
log(1/*x*) = −log(*x*), so we can make the
substitution:

*m*_{C} = *m*_{2} − 2.5 log(*B*_{C}/*B*_{2})

And substituting in the value of *B*_{C}/*B*_{2} that we figured
above, we get

(4a)*m*_{C} = *m*_{2} − 2.5 log(1 + 10^{0.4(m2−m1)})

When I looked at
Wikipedia’s formula
after deriving formula (4a), I was struck by how different the two
formulas look.
Nevertheless, they’re equivalent, and here’s how you can
see that.
Multiply inside the logarithm by
10^{0.4m2}·10^{−0.4m2}.
That’s equal to
10^{(0.4m2−0.4m2)} =
10^{0} = 1, and multiplying by 1 doesn’t change
anything.

*m*_{C} =
*m*_{2} − 2.5 log(10^{0.4m2}·10^{−0.4m2}·[1 + 10^{0.4m2}·10^{−0.4m1}])

Next, use log(*a**b*) =
log(*a*) + log(*b*) to pull that first factor into a
separate logarithm, and distribute the −2.5 coefficient:

*m*_{C} =
*m*_{2} − 2.5 [log(10^{0.4m2}) + log(10^{−0.4m2}·[1 + 10^{0.4m2}·10^{−0.4m1}]) ]

*m*_{C} =
*m*_{2} − 2.5 log(10^{0.4m2}) − 2.5 log(10^{−0.4m2}·[1 + 10^{0.4m2}·10^{−0.4m1}])

Now we’re getting somewhere! (*Finally!* I hear you
mutter.) log(10^{0.4m2}) is just 0.4*m*_{2}, which
simplifies the first logarithm, and then 2.5×0.4 = 1:

*m*_{C} =
*m*_{2} − 2.5×0.4*m*_{2} − 2.5 log(10^{−0.4m2}·[1 + 10^{0.4m2}·10^{−0.4m1}])

*m*_{C} =
*m*_{2} − *m*_{2} − 2.5 log(10^{−0.4m2}·[1 + 10^{0.4m2}·10^{−0.4m1}])

*m*_{C} =
−2.5 log(10^{−0.4m2}·[1 + 10^{0.4m2}·10^{−0.4m1}])

Multiplying out the 10^{−0.4m2} in the second
logarithm will simplify that one too. (Remember that
10^{−0.4m2}·10^{0.4m2} =
10^{0} = 1.)

*m*_{C} =
−2.5 log(10^{−0.4m2} + 10^{−0.4m2}·10^{0.4m2}·10^{−0.4m1})

*m*_{C} =
−2.5 log(10^{−0.4m2} + 10^{−0.4m1})

And swapping the *m*_{1} and *m*_{2} terms to put *m*_{1} first, we
have Wikipedia’s formula:

(4b)*m*_{C} =
−2.5 log(10^{−0.4m1} + 10^{−0.4m2})

Formulas (4a) and (4b) may look different, but they’re completely equivalent. You can use whichever one you prefer.

**Problem 10**: Sirius A and Sirius B look like a single star to
the naked eye. Their individual magnitudes (from telescopes) are
−1.45 and +8.68. What’s their combined magnitude?

Solution: You could use formula (4a) or formula (4b); I’ll use the first one.

*m*_{C} = *m*_{2} − 2.5 log(1 + 10^{0.4(m2−m1)})

*m*_{C} = 8.68 −
2.5 log(1 + 10^{0.4(8.68−(−1.45)})

*m*_{C} = 8.68 − 2.5 log(1 + 10^{0.4(10.13)})

*m*_{C} = −1.45

Hmm—that’s the same magnitude as Sirius A alone! Is something wrong with the calculation? No, it’s just that Sirius B makes a negligible contribution to the brightness of the pair. Their magnitudes differ by more than 10, which means that Sirius B is less than 1/10,000 as bright as Sirius A. (Sirius B, a white dwarf, is about 2½ times as hot as Sirius A, but its surface area is less than 1/40,000 of Sirius A’s, and therefore it puts out far less energy.)

By the way, sharp-eyed readers will have noticed that in other places I gave the magnitude of Sirius as −1.47, and here I said −1.45. Different sources tend to give slightly different magnitudes for objects. Also, astronomers recalibrated the zero point of the magnitude scale, which tweaked every magnitude by a couple of notches in the second decimal place.

Well, that was a bit of a letdown. Let me give you an example
where the companion *does* contribute measurably to the total
brightness.

**Problem 11**: Alpha Centauri is the nearest star to us—apart from
the Sun, of course. It is actually a triple star. Alpha Centauri A and
Alpha Centauri B look like a single star to the naked eye. (Alpha
Centauri C is separated from them by more than the width of four full
Moons, and has a magnitude of +11.13, so we can ignore it.) If Alpha
Centauri A and B have magnitudes +0.01 and +1.33, what is the
magnitude of the single visual object Alpha Centauri AB? Solution
here.)

Everything we’ve done so far has related to *how
bright a star or planet looks when viewed from Earth’s
surface*, which is called its **apparent magnitude**. The simple
“magnitude” almost always means apparent magnitude.

But apparent magnitude doesn’t just depend on how bright
a star actually is, how much light it’s putting out.
(The amount of light and energy that a star emits, inside or outside
the visible part of the spectrum, is its **luminosity**.)
It also depends on how far away the star is. A bright star that’s very
far away will look dimmer to us than a dimmer star that’s closer.

So how can we compare the amount of light two stars put
out—their **luminosities**? Astronomers do this by
defining **absolute magnitude**. This is simply the apparent
magnitude that a star *would have*—how bright it would
look to us on Earth—if it were located 10 parsecs (32.616
light-years) from us. Astronomers use the symbol *m* for apparent
magnitude, and *M* for absolute magnitude.

(Apparent magnitude also depends on whether there’s anything between the star and us that would absorb some of the light. Space has lots of dust clouds, but to keep things simple I’ll ignore them.)

There is a formula for absolute magnitude in terms of apparent magnitude and distance. As a preliminary, let’s see how a star’s apparent magnitude varies with distance. Then we can plug “10 parsecs” into that formula, and have the formula for absolute magnitude.

Light follows the inverse-square law: to somebody twice as far away the light source will look 2² = 4 times as faint; to somebody 3 times away it will look 3² = 9 times as faint, and so on.

Why should that be? Well, picture the Earth, 93 million miles
(150 million km) from the sun. How much of the sun’s total
output hits the Earth? The Earth’s radius is 4000 miles (6400
km). If you are in a rocket ship floating far enough away that
you can see the whole Earth, it would look like a circle
π·4000² = just over 50 million square
miles (81 million km²) in area. (The surface area of the
Earth is four times this; we’re just looking at its
cross-sectional area.)

Now picture a hollow sphere, centered on the sun, and with
radius equal to the Earth’s distance from the sun. The area of
that sphere is 4π·(93×10^{6})² =
1.1×10^{17} sq. mi. (1.7×10^{17}
km²). The fraction
of the sun’s output that hits the Earth equals
the Earth’s cross-sectional area divided by the area of that
imaginary sphere, 50×10^{6}/1.1×10^{17} =
4.62×10^{−10}.

What if the Earth were twice as far from the sun? The
Earth’s cross section is still the same, but the area of that
imaginary sphere is now
4π·(186×10^{6})² =
4.3×10^{17} sq. mi., and the fraction of the sun’s
light that hits the Earth is
50×10^{6}/4.3×10^{17} =
1.16×10^{−10}, one-fourth as much. The size of that
imaginary sphere grows as the square of the distance, so the fraction
of sunlight that hits the Earth (or an observer’s eyeballs)
shrinks as the square of the distance.

For any given shiny object, if we write *B*_{1} for its
brightness when it’s at a distance *d*_{1} from us, and *B*_{2} for its
brightness when distance *d*_{2} away from us, then the inverse-square
relationship looks like this:

(5)*B*_{2}/*B*_{1} = (*d*_{1}/*d*_{2})²

How does the magnitude of that object at distance *d*_{2}
relate to its magnitude at distance *d*_{1}? formula (3)
relates magnitude difference to brightness ratio:

*m*_{2} = *m*_{1} + 2.5 log(*B*_{1}/*B*_{2})

in formula (5) we have *B*_{2}/*B*_{1}, not *B*_{1}/*B*_{2}. However,
*B*_{1}/*B*_{2} = 1/(*B*_{2}/*B*_{1}), and
log( 1/(*B*_{2}/*B*_{1}) ) =
−log(*B*_{2}/*B*_{1}), so that formula is the same as

*m*_{2} = *m*_{1} − 2.5 log(*B*_{2}/*B*_{1})

Now replace *B*_{2}/*B*_{1} with the right-hand side of formula (5):

*m*_{2} = *m*_{1} − 2.5 log( (*d*_{1}/*d*_{2})² )

log(*x*²) = 2 log(*x*), so that

*m*_{2} = *m*_{1} − 2.5 × 2 log(*d*_{2}/*d*_{1})

And multiplying the two constants together, we have:

(6)*m*_{2} = *m*_{1} − 5 log(*d*_{2}/*d*_{1})

If a star’s actual magnitude is *m*_{1} and its actual distance
is *d*_{1}, formula (6) tells you what its magnitude *m*_{2} would be at
a distance of *d*_{2}.

Now make some substitutions into formula (6):
*m*_{1} → *m* is its apparent magnitude at its actual
distance *d*_{1} → *d*, and *m*_{2} → *M* is its
absolute magnitude, what its magnitude would be at the reference
distance *d*_{2} → 10 parsecs or 32.616 light-years:

(7a)*M* = *m* − 5 log(*d*_{parsecs}/10)

or:

(7b)*M* = *m* − 5 log(*d*_{light-years}/32.616)

Just a quick check for plausibility: if a star is closer than 10 parsecs or 32.616 light-years, the fraction inside parentheses will be less than 1, its log will be negative, and absolute magnitude will be greater than apparent magnitude. Greater magnitude means a fainter object, so we’re saying that the star would be fainter if moved to 10 parsecs or 32.6 light-years. So everything’s consistent.

**Problem 12**: What’s the absolute magnitude of our Sun?

Solution: The sun’s apparent magnitude is −26.74, and
its distance from Earth is 9.30 × 10^{7} miles
(1.50 × 10^{8} km). We’ll have to convert that distance
to light-years or parsecs. A light-year is
5.88 × 10^{12} miles (9.46 × 10^{12}
km), so the sun’s distance is
1.50 × 10^{8}/9.46 × 10^{12}) = 0.0000158
light-year. Using formula (7b),

*M* = −26.74 − 5 log(0.0000158/32.616)

*M* = 4.83

If we were looking at it from 10 parsecs or 32.616 light-years away, the Sun would be an undistinguished fifth-magnitude star.

**Problem 13**: What’s the absolute magnitude of Alpha Centauri
A, lying 4.37 light-years or 1.34 parsecs from us, with an apparent
magnitude of +0.01? How much more or less energy does it put
out than our Sun? (Solution here.)

It may happen that you know the absolute magnitude of an object and want to find its distance. For example, there’s a class of variable stars—stars that grow dimmer and then brighter again—called Cepheid variables. There’s a known relationship between a Cepheid’s intrinsic brightness and the length of its brighter-dimmer cycle. By comparing its absolute magnitude with its apparent magnitude in our sky, astronomers can tell how far the star is from us.

And the math’s not hard at all. We just have to solve formula (7a) or formula (7b) for distance. Starting from the formula in parsecs,

*M* = *m* − 5 log(*d*_{parsecs}/10)

5 log(*d*_{parsecs}/10) = *m* − *M*

log(*d*_{parsecs}/10) = (*m* − *M*)/5 = 0.2(*m* − *M*)

*d*_{parsecs}/10 = 10^{0.2(m − M)}

*d*_{parsecs} = 10 × 10^{0.2(m − M)}

And since 10 × 10^{x} =
10^{1+x},

(8a)*d*_{parsecs} = 10^{1 + 0.2(m − M)}

We can play the same game with formula (7b), except there are no powers of 10 to combine:

*M* = *m* − 5 log(*d*_{light-years}/32.616)

5 log(*d*_{light-years}/32.616) = *m* − *M*

log(*d*_{light-years}/32.616) = (*m* − *M*)/5 = 0.2(*m* − *M*)

*d*_{light-years}/32.616 = 10^{0.2(m − M)}

(8b)*d*_{light-years} = 3.2616*d*_{parsecs} or 32.616 × 10^{0.2(m − M)}

**Problem 14**: Sirius A has an apparent magnitude of
−1.47. If its absolute magnitude is +1.42, how far away from
us is it? (Solution here.)

**Problem 15**: A
Cepheid variable star
in the Lesser Magellanic Cloud (LMC) has a mean apparent magnitude of
15.56. Its brightness varies with a period of 4.76 days, and following
the relationship discovered by Henrietta Swan Leavitt in 1912 the star
must have a mean absolute magnitude of −3.57. How far is that
star—and therefore the LMC—from us?

Solution: Using formula (8b),

*d*_{light-years} = 32.616 × 10^{0.2(m − M)}

*d*_{light-years} = 32.616 × 10^{0.2(15.56 − (−3.57))}

*d*_{light-years} = 32.616 × 10^{0.2(19.13)}

*d*_{light-years} = 218,500

The Lesser Magellanic Cloud is about 218,500 light-years from us. (Dividing by 3.2616, that is about 67,000 parsecs.)

Quoting from the article linked above, “In practice astronomers would try and observe as many Cepheids as possible in another galaxy in order to determine a more accurate distance. As the number of stars observed go up the uncertainties involved in calculations for individual stars can be statistically reduced.”

Try this free Excel workbook. It’s set up to do all the types of calculations discussed in this article.

The red numbers in each section are the inputs to the calculations, and you’ll change them to match your problem. All the worksheets are protected so that you can’t accidentally click into the wrong cell and mess up something. But there’s no password, so you can unprotect any sheet if you want to “look under the hood”. You can also press Alt+F11 to see the program code.

In preparing this article, I consulted the following sources:

- “Absolute Magnitude” in Wikipedia, accessed 2021-05-22.
- “Apparent Magnitude” in Wikipedia, accessed 2020-10-13.
- “Castor (star)” in Wikipedia, accessed 2020-10-12.
- “Cepheid Variable Stars & Distance Determination on the Australia Telescope National Facility site, accessed 2021-05-28.
- “Galileo Galilei” in Wikipedia, accessed 2020-10-12.
- “Hipparchus” in Wikipedia, accessed 2020-10-12.
- “N. R. Pogson” in Wikipedia, accessed 2020-10-12.
- “Uranus Information”, at TheSkyLive.com, accessed 2020-10-12.
- Asimov, Isaac. “Brightening Stars” in The Relativity of Wrong (Doubleday, 1988). Originally published July 1987 in The Magazine of Fantasy and Science Fiction.
- Asimov, Isaac. “The Comet That Wasn’t” in Quasar, Quasar, Burning Bright (Doubleday, 1978). Originally published November 1976 in The Magazine of Fantasy and Science Fiction.
- Asimov, Isaac. “How Little?” in The Sun Shines Bright (Doubleday, 1981). Originally published September 1979 in The Magazine of Fantasy and Science Fiction.
- Asimov, Isaac. The essay “Quasar, Quasar, Burning Bright” in the book Quasar, Quasar, Burning Bright (Doubleday, 1978). Originally published October 1976 in The Magazine of Fantasy and Science Fiction.
- Miles, Richard. “A light history of photometry: from Hipparchus to the Hubble Space Telescope.” Journal of the British Astronomical Association, vol.117 (2007), no.4, pp. 172-186. Accessed 2021-11-22.
- Ratcliff, Martin, and Alister Ling. “Mars Reaches Brilliant Opposition.” Astronomy, October 2020, pp.36 and 42.

**2021-11-15/22**: Verified links and updated the changed ones.**29 May 2021**: A few tweaks to the workbook:- In section B, ask for Object #2’s magnitude and then display Object #1’s magnitude after the magnitude difference.
- In section E, display “parsec” or “parsecs” as needed rather than the lazy “parsec(s)”.
- In section F, ask for apparent magnitude then absolute magnitude instead of the other way around.

**28 May 2021**: New article, with nifty Excel workbook.

**Solution to problem 2**: If one star
is 3.98 times as bright as another, you need to know how many times
2.512 must be multiplied by itself to yield 3.98:

2.512^{x} = 3.98

*x* (log 2.512) = log 3.98

*x* - (log 3.98) / (log 2.512) = about 1.50

Sirius is brighter than Vega so its magnitude is lower. 0.03 − 1.50 = −1.47. Sirius is one of four nighttime stars with negative magnitudes.

(back)

**Solution to problem 4**: The
difference in magnitudes is 1.58 − (−1.47) =
1.58 + 1.47 = 3.05. One magnitude of difference is
a brightness ratio of 2.512, so 3.05 magnitudes is a brightness ratio
of 2.512^{3.05} = about 16.6. Sirius is 16.6 times as
bright as Castor, so Hipparchus’ class of first-magnitude stars
actually extends through a brightness ratio of 16.6.

(back)

**Solution to problem 6**: Let Mars be
“star” 1 and Jupiter be “star” 2. Then
*m*_{1} = −2.6, and *m*_{2} = −2.3. Using
formula (1),

*B*_{1}/*B*_{2} =
10^{0.4(−2.3 − (−2.6))} =
10^{0.4(−2.3 + 2.6)} =
10^{0.4×0.3} = about 1.32.

Mars was 32% brighter than Jupiter, or about 1.32 times as bright.

That may not seem like a big difference, but for Mars to be
brighter than Jupiter *at all* is unusual. See
Problem 7.

(back)

**Solution to problem 7**: You want to
know how much brighter Jupiter is, so make it “star” 1.
*m*_{1} = −2.20 (Jupiter) and *m*_{2} = +0.71 (Mars).
Again using formula (1),

*B*_{1}/*B*_{2} =
10^{0.4(0.71 − (−2.20))} =
10^{0.4×2.91} = 14.6. At their average brightnesses, Jupiter is
14.6 times as bright as Mars.

Jupiter is so bright that its *minimum*
magnitude, −1.66, is still brighter than Sirius
(−1.47) and much brighter than Mars’s average (+0.71).

(back)

**Solution to problem 8**: Use formula (3):

*m*_{2} = *m*_{1} + 2.5 log(*B*_{1}/*B*_{2})

*m*_{2} (to be found) is Mira’s magnitude at its brightest.
*m*_{1} = 10 is its magnitude at its dimmest, and
*B*_{1}/*B*_{2} = 1/1600 is the ratio of brightness.

*m*_{2} = 10 + log(1/1600)

Since the brightness ration 1/1600 is less than 1, the logarithm
is negative, so *m*_{2} < *m*_{1}, which makes sense since
*m*_{2} is the magnitude of Mira when it’s brighter. Punching the
numbers into a calculator, you’ll get *m*_{2} = 1.9897, which is
2.0 after rounding.

(back)

**Solution to problem 9**: Use formula (2):

*m*_{2} − *m*_{1} = 2.5 log(*B*_{1}/*B*_{2})

Here *B*_{1}/*B*_{2} is 52,000, and 2.5 log 52,000 is
about 11.8 magnitudes. Since lower magnitudes indicate brighter
objects, Nova Aquilae dropped 11.8 magnitudes.

(Its peak brightness was −0.5, about as bright as Saturn at its brightest; this was the brightest nova in over 300 years. If it dropped 11.8 magnitudes, its normal magnitude was about 11.3, only visible in a fairly good telescope.)

(back)

**Solution to problem 11**: You could use either formula (4a) or formula (4b),
but I’ll use the second one for variety.

*m*_{C} =
−2.5 log(10^{−0.4m1} + 10^{−0.4m2})

*m*_{C} =
−2.5 log(10^{−0.4(0.01)} + 10^{−0.4(1.33)})

*m*_{C} = −0.27

Alpha Centauri AB is the third brightest star in the sky—again, not counting the Sun—but unfortunately it’s in the southern skies and never visible north of the southern edge of the continental United States. If you vacation in Hawaii, though, be sure to look for it!

(back)

**Solution to problem 13**: You could use formula (7a) or formula (7b), but I’ll
use the first one for variety:

*M* = *m* − 5 log(*d*_{parsecs}/10)

*M* = 0.01 − 5 log(1.34/10)

*M* = 4.37

If they were at the same distance from us, Alpha Centauri A would shine about half a magnitude brighter than our Sun. We can be more precise by using formula (1):

*B*_{1}/*B*_{2} = 10^{0.4(m2−m1)}

*B*_{1}/*B*_{2} = 10^{0.4(4.83−4.37)}

*B*_{1}/*B*_{2} = 1.53

Alpha Centauri A is 1.53 times as bright as our Sun, or 53% brighter.

(back)

**Solution to problem 14**: Using formula (8a),

*d*_{parsecs} = 10^{1 + 0.2(m − M)}

*d*_{parsecs} = 10^{1 + 0.2(−1.47 − 1.42)}

*d*_{parsecs} = 2.64

If you use formula (8b) instead, you get 8.62 light-years.

(back)

Because this article helps you,

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

Updates and new info: https://BrownMath.com/bsci/