How Bright Is That Star (or Planet)?
Copyright © 2021 by Stan Brown
Copyright © 2021 by Stan Brown
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 = 1001/5 = 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
2.512x = 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.)
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. 2.51224 = 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:
We’re looking for some formula that will have B1/B2, the ratio of brightness, on the left side, and an expression involving m1 − m2, 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:
B1/B2 = 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 B1 > B2, and B1/B2 > 1. If star 1 is brighter, its magnitude m1 is less, m2 − m1 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 B2 > B1, and B1/B2 < 1. If star 2 is brighter, its magnitude m2 is less. m2 − m1 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 B1 = B2 and the left side of the equation is 1. And m1 = m2, so m2 − m1 = 0, and 2.5120 = 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 1001/5: each magnitude was a step of 1001/5, 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 10x, antilog, or log−1 key. 100 = 102, and 1001/5 = (102)1/5. Now in general, (ca)b = cab. (See It’s the Law — the Laws of Exponents if you’re a little shaky on exponents.) Therefore 1001/5 = (102)1/5 = 102/5 = 100.4.
Going back to the equation B1/B2 = 2.512(m2−m1), replace the approximation 2.512 with the exact value 100.4 to get
B1/B2 = (100.4)(m2−m1)
Using that same power rule, multiply the exponents and we have:
(1) B1/B2 = 100.4(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.
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). B2/B1 = 100.4(m2−m1) = 100.4(1.86−(−2.94)) = 100.4(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:
B1/B2 = 100.4(m2−m1)
You want to solve that formula for m2 − m1, but the magnitude difference is locked up in the exponent, so you unlock it by taking the logarithm of both sides:
log(B1/B2) = 0.4(m2 − m1)
Multiply both sides by 2.5 to get rid of the 0.4, and swap left and right sides:
(2)m2 − m1 = 2.5 log(B1/B2)
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 m1 to both sides and you get
(3)m2 = m1 + 2.5 log(B1/B2)
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.
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
BC = B2 + B1
and the brightness ratio of the combination to star 2 alone is
BC/B2 = (B2 + B1)/B2
BC/B2 = 1 + B1/B2
But formula (1) tells us that the ratio of brightness between the two stars of Sirius (or any binary star) is B1/B2 = 100.4(m2−m1), so we can substitute that in:
BC/B2 = 1 + 100.4(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
mC = m2 + 2.5 log(B2/BC)
We don’t know B2/BC, but we do know BC/B2 from a few lines above, and B2/BC = 1/(BC/B2). log(1/x) = −log(x), so we can make the substitution:
mC = m2 − 2.5 log(BC/B2)
And substituting in the value of BC/B2 that we figured above, we get
(4a)mC = m2 − 2.5 log(1 + 100.4(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 100.4m2·10−0.4m2. That’s equal to 10(0.4m2−0.4m2) = 100 = 1, and multiplying by 1 doesn’t change anything.
mC = m2 − 2.5 log(100.4m2·10−0.4m2·[1 + 100.4m2·10−0.4m1])
Next, use log(ab) = log(a) + log(b) to pull that first factor into a separate logarithm, and distribute the −2.5 coefficient:
mC = m2 − 2.5 [log(100.4m2) + log(10−0.4m2·[1 + 100.4m2·10−0.4m1]) ]
mC = m2 − 2.5 log(100.4m2) − 2.5 log(10−0.4m2·[1 + 100.4m2·10−0.4m1])
Now we’re getting somewhere! (Finally! I hear you mutter.) log(100.4m2) is just 0.4m2, which simplifies the first logarithm, and then 2.5×0.4 = 1:
mC = m2 − 2.5×0.4m2 − 2.5 log(10−0.4m2·[1 + 100.4m2·10−0.4m1])
mC = m2 − m2 − 2.5 log(10−0.4m2·[1 + 100.4m2·10−0.4m1])
mC = −2.5 log(10−0.4m2·[1 + 100.4m2·10−0.4m1])
Multiplying out the 10−0.4m2 in the second logarithm will simplify that one too. (Remember that 10−0.4m2·100.4m2 = 100 = 1.)
mC = −2.5 log(10−0.4m2 + 10−0.4m2·100.4m2·10−0.4m1)
mC = −2.5 log(10−0.4m2 + 10−0.4m1)
And swapping the m1 and m2 terms to put m1 first, we have Wikipedia’s formula:
(4b)mC = −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.
mC = m2 − 2.5 log(1 + 100.4(m2−m1))
mC = 8.68 − 2.5 log(1 + 100.4(8.68−(−1.45))
mC = 8.68 − 2.5 log(1 + 100.4(10.13))
mC = −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.
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·(93×106)² = 1.1×1017 sq. mi. (1.7×1017 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, 50×106/1.1×1017 = 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·(186×106)² = 4.3×1017 sq. mi., and the fraction of the sun’s light that hits the Earth is 50×106/4.3×1017 = 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 B1 for its brightness when it’s at a distance d1 from us, and B2 for its brightness when distance d2 away from us, then the inverse-square relationship looks like this:
(5)B2/B1 = (d1/d2)²
How does the magnitude of that object at distance d2 relate to its magnitude at distance d1? formula (3) relates magnitude difference to brightness ratio:
m2 = m1 + 2.5 log(B1/B2)
in formula (5) we have B2/B1, not B1/B2. However, B1/B2 = 1/(B2/B1), and log( 1/(B2/B1) ) = −log(B2/B1), so that formula is the same as
m2 = m1 − 2.5 log(B2/B1)
Now replace B2/B1 with the right-hand side of formula (5):
m2 = m1 − 2.5 log( (d1/d2)² )
log(x²) = 2 log(x), so that
m2 = m1 − 2.5 × 2 log(d2/d1)
And multiplying the two constants together, we have:
(6)m2 = m1 − 5 log(d2/d1)
If a star’s actual magnitude is m1 and its actual distance is d1, formula (6) tells you what its magnitude m2 would be at a distance of d2.
Now make some substitutions into formula (6): m1 → m is its apparent magnitude at its actual distance d1 → d, and m2 → M is its absolute magnitude, what its magnitude would be at the reference distance d2 → 10 parsecs or 32.616 light-years:
(7a)M = m − 5 log(dparsecs/10)
(7b)M = m − 5 log(dlight-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 × 107 miles (1.50 × 108 km). We’ll have to convert that distance to light-years or parsecs. A light-year is 5.88 × 1012 miles (9.46 × 1012 km), so the sun’s distance is 1.50 × 108/9.46 × 1012) = 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(dparsecs/10)
5 log(dparsecs/10) = m − M
log(dparsecs/10) = (m − M)/5 = 0.2(m − M)
dparsecs/10 = 100.2(m − M)
dparsecs = 10 × 100.2(m − M)
And since 10 × 10x = 101+x,
(8a)dparsecs = 101 + 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(dlight-years/32.616)
5 log(dlight-years/32.616) = m − M
log(dlight-years/32.616) = (m − M)/5 = 0.2(m − M)
dlight-years/32.616 = 100.2(m − M)
(8b)dlight-years = 3.2616dparsecs or 32.616 × 100.2(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),
dlight-years = 32.616 × 100.2(m − M)
dlight-years = 32.616 × 100.2(15.56 − (−3.57))
dlight-years = 32.616 × 100.2(19.13)
dlight-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:
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:
2.512x = 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.
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 2.5123.05 = 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.
Solution to problem 6: Let Mars be “star” 1 and Jupiter be “star” 2. Then m1 = −2.6, and m2 = −2.3. Using formula (1),
B1/B2 = 100.4(−2.3 − (−2.6)) = 100.4(−2.3 + 2.6) = 100.4×0.3 = 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.
Solution to problem 7: You want to know how much brighter Jupiter is, so make it “star” 1. m1 = −2.20 (Jupiter) and m2 = +0.71 (Mars). Again using formula (1),
B1/B2 = 100.4(0.71 − (−2.20)) = 100.4×2.91 = 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).
Solution to problem 8: Use formula (3):
m2 = m1 + 2.5 log(B1/B2)
m2 (to be found) is Mira’s magnitude at its brightest. m1 = 10 is its magnitude at its dimmest, and B1/B2 = 1/1600 is the ratio of brightness.
m2 = 10 + log(1/1600)
Since the brightness ration 1/1600 is less than 1, the logarithm is negative, so m2 < m1, which makes sense since m2 is the magnitude of Mira when it’s brighter. Punching the numbers into a calculator, you’ll get m2 = 1.9897, which is 2.0 after rounding.
Solution to problem 9: Use formula (2):
m2 − m1 = 2.5 log(B1/B2)
Here B1/B2 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.)
Solution to problem 11: You could use either formula (4a) or formula (4b), but I’ll use the second one for variety.
mC = −2.5 log(10−0.4m1 + 10−0.4m2)
mC = −2.5 log(10−0.4(0.01) + 10−0.4(1.33))
mC = −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!
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(dparsecs/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):
B1/B2 = 100.4(m2−m1)
B1/B2 = 100.4(4.83−4.37)
B1/B2 = 1.53
Alpha Centauri A is 1.53 times as bright as our Sun, or 53% brighter.
Solution to problem 14: Using formula (8a),
dparsecs = 101 + 0.2(m − M)
dparsecs = 101 + 0.2(−1.47 − 1.42)
dparsecs = 2.64
If you use formula (8b) instead, you get 8.62 light-years.
Updates and new info: https://BrownMath.com/bsci/