$$b = \frac{L}{4\pi r^2}\quad[\text{Eq. (6.18)}]$$
Note: If you feel uncomfortable with working with variables, fractions, exponents or scientific notation, you should review the following tutorials before working through this tutorial:
Sections
Introduction
So-called inverse square laws appear numerous times in astronomy and physics. They all have the form of $$\text{something} = \frac{\text{stuff}}{r^2},$$ where \(r\) is the distance from “the stuff”. In fact, you have already encountered one inverse square law in Section 5.3: $$F_\text{gravity} = G\frac{Mm}{r^2},$$ which is Newton’s universal law of gravitation, Equation (5.7).
These types of laws are called inverse square laws simply because the quantity being squared, in this case distance as represented by \(r\), is in the denominator of a fraction. The term “inverse” is a mathematical term that means essentially that the inverse operation on a quantity “undoes” the original operation on that quantity. For example, for the number \(2^2 = 4\), its inverse is \(\frac{1}{2^2} = \frac{1}{4}\) so that \(2^2 \times \frac{1}{2^2} = 4 \times \frac{1}{4} = 1\), and the inverse of \(r^2\) is \(\frac{1}{r^2}\) so that \(r^2 \times \frac{1}{r^2} = \frac{r^2}{r^2} = 1\). The squaring a quantity, say \(2^2 = 4\) is undone by the inverse operation, taking the square root of that quantity; \(\sqrt{2^2} = \sqrt{4} = 2\), or \(\sqrt{x^2} = x\). [Don’t forget that variables, like \(r\) or \(x\), are just general placeholders for numbers, like \(2\). Writing equations using variables saves writing an infinite number of equations with all possible combinations of valid numbers. Algebra is just generalized arithmetic. BTW, setting \(r=-2\) is not valid since negative distances violate the (natural) law!]
All inverse square laws behave the same way in terms of distance from an object: doubling the distance reduces by a factor of four (\(2^2 = 4\)) the quantity being calculated \(\left[\frac{1}{(2r)^2} = \frac{1}{2^2r^2} = \frac{1}{4r^2}\right]\); tripling the distance reduces the calculated quantity by a factor of nine (\(3^2 = 9\)), or \(\frac{1}{(3r)^2} = \frac{1}{3^2r^2} = \frac{1}{9r^2}\); increasing the distance by a factor of five, reduces the calculated quantity by a factor of twenty-five, \(5^2 = 25\), or \(\frac{1}{(5r)^2} = \frac{1}{5^2r^2} = \frac{1}{25r^2}\); and increasing the distance by a factor of ten, reduces the calculated quantity by a factor one hundred, \(10^2 = 100\), or \(\frac{1}{(10r)^2} = \frac{1}{10^2r^2} = \frac{1}{100r^2}\).
The reason that the equation for brightness, Equation (6.18), is an inverse square law has to do with geometry; the surface area of a sphere is \(A_\text{sphere} = 4\pi r^2\), where \(r\) is the distance from the source of the light, say a star. As the distance from the light source increases, so does the surface area of the sphere centered on that light source. This means that the same amount of light from the source gets spread out over more area, but the size of the detector, say your eye or a telescope, remains the same size. As a result, less light comes into your eye or the telescope, and the object looks dimmer. The amount of light being emitted by the star is its luminosity, \(L\). Assuming that the star’s luminosity doesn’t change while we are looking at it, then its brightness, \(b\), is the star’s luminosity, \(L\), divided by the surface area of a giant sphere centered on the star with a radius of our distance from the star: $$\text{brightness} =\frac{\text{luminosity}}{\text{the area that the light is spread across}}$$ or $$\quad b=\frac{L}{4\pi r^2}.$$
Never forget that mathematics is the language of nature, and like all languages, it must be translated in order to understand what is being presented to you. Always try to read an equation like you would a sentence, with each part of the equation being like a component of that sentence: in this case, “brightness is the amount of light being emitted by a star spread out evenly over all of the surface area centered on the star.” As you become more proficient with the language of mathematics and its interpretation, what you are studying can guide the mathematics, and the mathematics can help describe what you are studying.
An Alternate Form of the Inverse Square Law for Light
The “trick” that we have used before of dividing the general case by a standard case can simplify comparing cases. One example of this “trick” is in the escape speed tutorial. Dividing one object at one distance by another object at a different distance gives $$\frac{b}{b_\text{standard}} = \frac{L/(4\pi r^2)}{L_\text{standard}/(4\pi r_\text{standard}^2)}.$$ Canceling the \(4\pi\) terms in the numerator and the denominator, and rearranging gives the equation$$b = b_\text{standard} \times \frac{L/L_\text{standard}}{(r/r_\text{standard})^2}.$$
Examples
1. Mars is approximately \(1.5\text{ au}\) from the Sun. How much dimmer does the Sun appear from Mars when compared with the Sun’s brightness from Earth?
Using the alternate form of the inverse square law for light that we just developed, \(L\) and \(L_\text{standard}\) both refer to the Sun, the Sun seen from Mars and the Sun seen from Earth, so \(L/L_\text{standard} = 1\). For the distances, \(r = 1.5\text{ au}\) and \(r_\text{standard}\) is the distance from Earth to the Sun, or \(r_\text{standard} = 1\text{ au}\). Therefore, $$b_{\,\text{from Mars}} = b_{\,\text{from Earth}}\times\frac{1}{(1.5/1)^2} = b_{\,\text{from Earth}}/2.25 = 0.44b_{\,\text{from Earth}}.$$ In other words, the Sun’s brightness from Mars is only 44% of its brightness from Earth.
2. The dwarf planet, Sedna, is in a very eccentric orbit. It is 76 au from the Sun during its closest approach (perihelion) and it is 890 au from the Sun at aphelion. By what factor does the Sun’s brightness change as seen from Sedna, during the dwarf planet’s orbit?
From the alternate form: $$b_\text{perihelion} = b_\text{aphelion} \times \frac{1}{(r_\text{perihelion}/r_\text{aphelion})^2} = b_\text{aphelion}\times\frac{1}{(76/890)^2} = b_\text{aphelion}\times \frac{1}{0.0073} = 137b_\text{aphelion}.$$ The Sun appears 137 times brighter when Sedna is at perihelion than it does when Sedna is at aphelion.
3. A planet is discovered orbiting a star that has twice the Sun’s luminosity. Furthermore, that exoplanet is 1 au from its parent star. How bright does the star appear from the exoplanet compared to the Sun as seen from Earth?
The star’s luminosity, \(L = 2\text{ L}_\text{Sun}\) and \(r = r_\text{Earth} = 1\text{ au}\). As a result, $$b_\text{star from exoplanet} = b_\text{Sun from Earth} \times \frac{2}{1^2} = 2b_\text{Sun from Earth}.$$
4. Suppose that the same star from Example 3 has another planet in orbit about it, but this one is 10 au from the star. How bright would the star appear from that exoplanet compared to our Sun from Earth?
In this case \(r/r_\text{Earth from Sun} = 10\) so \((r/r_\text{Earth from Sun})^2 = 100\). This means that $$b_\text{star from exoplanet} = b_\text{Sun from Earth} \times \frac{2}{100} = 0.02b_\text{Sun from Earth}.$$ The star as seen from its more distant exoplanet is only 2% of the Sun’s brightness as seen from Earth.
5. Suppose that a star with 5 times the Sun’s luminosity is 1 parsec (pc) from Earth, where \(1\text{ pc} = \text{206,265}\text{ au}\). How bright would that star appear from Earth compared to the Sun’s brightness from Earth.$$b = b_\text{Sun}\times\frac{5}{\text{206,265}^2} = b_\text{Sun}\times\frac{5}{4.25\times10^{10}} = (1.2\times 10^{-10})\times b_\text{Sun}$$. This is 120 trillionths of the brightness of the Sun. (There is no star that is remotely close to that bright anywhere near as close to the Sun as our hypothetical star. This is why when you look up at the sky at night you don’t see any star that is anything like as bright as the Sun.)
Quiz Yourself
Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.
(Answers are available below.)
- If the orbital radius of Earth suddenly (and miraculously) increased by a factor of 2, would the Sun’s brightness as seen from Earth increase or decrease, and by what factor?
- A very gifted magician suddenly increased the luminosity of the Sun by a factor of 16 while simultaneously increasing Earth’s orbital radius by a factor of 4. How much would the brightness of the Sun change, and by what factor?
- Mercury is 0.39 au from the Sun. Is the Sun brighter or dimmer when seen from Mercury than it is when seen from Earth? By what factor?
- An exoplanet is orbiting its parent star at a distance of 0.5 au and the star has one-fourth of the Sun’s luminosity. Compare the star’s brightness as seen from the exoplanet with our Sun’s brightness as seen from Earth. (You shouldn’t need to use a calculator for this one.)
- How much brighter does the Sun appear from Mercury, orbiting at a distance of 0.39 au compared to the Sun seen from Neptune at a distance of 30 au?
- The star Sirius A has a luminosity that is about 25 times greater than the luminosity of the Sun. Sirius A also has a small stellar companion, Sirius B, that is 7.5 au from Sirius A. How bright does Sirius A appear from Sirius B, relative to the Sun from Earth?
- The luminosity of Betelgeuse, a red supergiant star in the shoulder of Orion, is typically \(\text{126,000 L}_\text{Sun}\), although its luminosity does vary significantly, brightening and dimming semi-periodically over time. Betelgeuse is also 168 parsecs, or 34,670,000 au from Earth. Compare Betelgeuse’s brightness as seen from Earth with the Sun’s brightness as seen from Earth.
- You find yourself marooned on a moon dwarf planet that is orbiting the Sun. From your vantage point the Sun appears to be 0.227% as bright as it does from Earth. How far from the Sun are you? (FYI, it’s a long way home!)
Answers
- The Sun’s brightness would decrease by a factor of 4.
- The Sun’s brightness wouldn’t change at all.
- Brighter by a factor of 6.6.
- The two brightnesses are the same.
- 5900 times brighter from Mercury
- Sirius A is 44% as bright
- \(1\times 10^{-10}\) times the Sun’s brightness
- 47.7 au (the dwarf moon is Makemake)