Kepler’s Third Law

\[\left(M_\text{Sun} + m_\text{Sun}\right)P_\text{y}^2 = a_\text{au}^3\quad[\text{Eq. (5.10)}]\]

Note: If you feel uncomfortable with working with fractions, exponents, or variables you should review the following tutorials before working through this tutorial:

Sections

Introduction

Kepler’s third law is fundamentally important in advancing our understanding of objects in the universe. Derived from Newton’s laws and his universal law of gravitation, Kepler’s third law still remains the most direct way to determine the masses of two objects orbiting each other: two stars orbiting each other, planets orbiting stars, moons orbiting planets, or asteroids orbiting each other.

Kepler’s third law in the form of Equation (5.10) is really a way to compare two objects orbiting each other with Earth’s orbit around the Sun. The form derived directly from Newton’s laws and his universal law of gravitation uses SI units of kilograms, seconds, and meters, but it also involves constants, \(4\pi^2\) and the universal gravitational constant, \(G\). To use units that are characteristic of orbits (mass of the Sun, years, and astronomical units), and to avoid the complication of using the constants, the original general equation was divided by the specific equation for Earth’s orbit around the Sun. Dividing one equation by the other cancels the constants. This means that you must express the two objects’ masses in terms of the Sun’s mass \(\text{M}_\text{Sun}\)), their mutual orbital period in years (\(\text{y}\), and their average separation in astronomical units (\(\text{au}\)).

A very subtle point: If you look very carefully you will see that the masses in Equation (5.10) are variables, indicated by the fact that they are in math type (italicized) and they have Sun subscripts, such as \(M_\text{Sun}\). This means that the number that goes in the mass placeholder needs to be in terms of the mass of the Sun. For example, if the star has a mass that is two times the mass of the Sun, then \(M_\text{Sun}\) gets replaced by 2 in this instance. But there is also a mass unit that is not italicized, \(\text{M}_\text{Sun}\), that is a shortcut way of representing \(1.99\times 10^{30}~\text{kg}\), This means that if a star’s mass is twice the mass of the Sun, then its mass is \(3.98\times 10^{30}~\text{kg}\). To simplify writing the mass, it is common in astronomy to write \(2~\text{M}_\text{Sun}\), meaning \(2\times1.99\times 10^{30}~\text{kg}\). This is just like writing the amount of time it takes to complete one orbit in terms of years, where \(\text{1}~\text{y}=3.156 \times 10^7~\text{s}\); an orbital period of two years would equal \(6.312\times 10^7~\text{s}\), but it is a lot easier, and more understandable if the orbital period is simply written as \(2~\text{y}\). Using the units required in Equation (5.10) is both easier to express and easier to understand, given that they compare directly to things we know.

Examples

Let’s take a look at a couple of examples:

1. Sirius A and Sirius B orbit each other with a period, in years, of \(P_\text{y} = 50.13\), and they orbit each other at a distance, in astronomical units, of \(a_\text{au} = 19.78\). The sum of their two masses, in terms of the mass of the Sun, is \(M_\text{Sun} + m_\text{Sun}\) = 3.08\). Verify that Kepler’s third law is valid in this instance.

The left-hand side of the equation is \(\left(M_\text{Sun}+m_\text{Sun}\right)P_\text{y}^2 = 3.08 \times 50.13^2 = 3.08 \times 2513 = 7740\). The right-hand side is \(a_\text{au}^3 = 19.78^3 = 7739\). The fact that the two sides aren’t exactly equal is simply because of slight round-off errors in the calculations.

2. Titan orbits the planet Saturn with an orbital period of \(15.95~\text{days} = 0.0437~\text{y}\). Titan’s distance from Saturn is \(0.008\,167~\text{au}\), and the sum of Saturn’s mass and Titan’s mass is \(0.000\,2857~\text{M}_\text{Sun}\). Verify that Kepler’s third law is valid for the orbit of Titan around Saturn.

For the left-hand side: \(0.000\,2857 \times 0.0437^2 = 0.000\,2857 \times 0.00191 = 5.46\times 10^{-7}\). For the right-hand side:\(0.008\,167^3 = 5.48\times 10^{-7}\). Once again the results are very close, differing only by round-off errors.

Using Kepler’s Third Law to Calculate Masses

The most important application of Kepler’s third law is to calculate the sum of the masses of two objects orbiting one another. From Equation (5.10), solve for \(M_\text{Sun} + m_\text{Sun}\) by dividing both sides of the equation by \(P_\text{Sun}^2\). Canceling the period squared in both the numerator and the denominator on the left-hand side results in \[M_\text{Sun} + m_\text{Sun} = \frac{a_\text{au}^3}{P_\text{y}^2}\]. By knowing the semimajor axis of the orbit and the orbital period, the sum of the masses can be determined.

More Examples

3. An exoplanet is discovered around a star. The planet has an orbital period of \(12~\text{days}\) and a semimajor axis of \(0.3~\text{au}\). What is the sum of the masses of the planet and the star?

Converting \(12~\text{days}\) to years: \(12~\text{days}/(365.256~\text{days}/\text{year}) = 0.0392~\text{y}\). The sum of the two masses is \[M_\text{Sun}+m_\text{Sun} = \frac{a_\text{au}^3}{P_\text{y}^2} = \frac{0.3^3}{0.0392^2} = 17.6.\] The sum of the masses is \(\text{17.6}\) times the mass of the Sun, or \(\text{17.6}~\text{M}_\text{Sun}\).

4. Jupiter’s moon, Io, orbits the planet in \(1.769~\text{days}\) and has an orbital radius of \(2.82\times10^{-3}~\text{au}\). What is the sum of the masses of Jupiter and Io?

Using the same conversion as the last example, Io’s orbital period is \(4.84\times 10^{-3}~\text{y}\). As a result, the sum of the masses is \[M_\text{Sun} + m_\text{Sun} = \frac{\left(2.82\times10^{-3}\right)^3}{\left(4.84\times10^{-3}\right)^2} = 9.57\times10^{-4} = 0.000\,957 = 1/1045\]. The sum of the masses is slightly less than \(0.1\%\) of the Sun’s mass. (Note: Almost all of the mass of the Jupiter-Io system is in Jupiter.)

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 available below.)

  1. Two stars orbit each other at a distance of \(2~\text{au}\) and with an orbital period of \(2~\text{y}\). What is the sum of the masses of the two stars. (Note: Don’t even think about using a calculator for this problem! Think about what exponents mean.)
  2. Two stars orbiting each other have an orbital period of \(25~\text{years}\) and a separation of \(5~\text{au}\). What is the sum of the stars masses?
  3. A student is working on an observing project by observing a binary star system. He determines that the orbital period is \(3.5~\text{y}\), the separation of the two stars is \(2~\text{au}\), and the sum of their masses is \(0.5~\text{M}_\text{Sun}\). Should he get an A for his project? Why or why not?
  4. Assuming that the orbital period and the separation of the two stars is correct, what is the total mass of the system the student was studying in the last problem?
  5. In the future, the James Webb Space Telescope discovers a rogue planet (a planet not orbiting any star) that has a small moon orbiting it. The moon orbits the planet once every \(35~\text{days}\) at a distance of \(0.033~\text{au}\). What is the mass of the planet plus its moon? Compare the mass to Example 4 above.
  6. Phobos, one of the two tiny moons of Mars, orbits the planet with an orbital radius of \(9378~\text{km}\) and an orbital period of \(0.3189~\text{d}\). Since Phobos is so small, the sum of the masses of Mars and Phobos is essentially the mass of Mars. Determine the mass of Mars from the orbital data provided. Express your answer in kilograms. Note: \(1~\text{au} = 1.5\times 10^{11}~\text{m}\), \(1~\text{y} = 365.256~\text{d}\), and \(1~\text{M}_\text{Sun} = 1.99 \times 10^{30}~\text{kg}\).

Answers

  1. \(2~\text{M}_\text{Sun}\)
  2. \(0.2~\text{M}_\text{Sun}\)
  3. No. The left-hand side of Equation (5.10) equals \(6.125\), but the right-hand side only equals \(8\). Since the two sides aren’t equal, his result can’t be correct.
  4. \(0.65~\text{M}_\text{Sun}\)
  5. \(0.0081~\text{M}_\text{Sun}\). The rogue planet and its small moon is about 8 times more massive than Jupiter and Io.
  6. \(6.4\times 10^{23}~\text{kg}\)
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