Note: If you feel uncomfortable with working with fractions, variables, or exponents you should review the following tutorials before working through this tutorial:
Sections
Introduction
Kepler’s total mass equation
\[m_\text{total, Sun} = m_\text{1, Sun} + m_\text{2, Sun} = \frac{a_\text{au}^3}{P_\text{y}^2}\quad[\text{Eq. (15.6)}]\]
Kepler’s total mass equation [Eq. (15.6)] is a rewritten version of Equation (5.10) on p. 174 that is solved for total mass, and \(M_\text{Sun}\) and \(m_\text{Sun}\) are placed by \(m_\text{1, Sun}\) and \(m_\text{2, Sun}\), respectively. The notational changes were only made so that the masses are consistent with the masses in Equations (15.7)-(15.10). The semimajor axis, \(a_\text{au}\), is the sum of the semimajor axes of the two objects orbiting their mutual center of mass, as described by Equation (15.8) seen in the next section of this tutorial. This mean that Equation (15.6) could also be written as \[m_\text{total, Sun} = m_\text{1, Sun} + m_\text{2, Sun} = \frac{(a_\text{1, au} + a_\text{2, au})^3}{P_\text{y}^2}\quad[\text{Eq. (15.6) modified}]\].
Kepler’s total mass equation provides one piece of information that allows astronomers to determine individual masses of two orbiting objects such as two stars in a binary star system. The equation also works for moons orbiting planets, asteroids orbiting other asteroids, or planets, asteroids, or comets orbiting the Sun. The equation even works for stars orbiting the centers of galaxies or two galaxies orbiting one another. This is an example of the universality of physical laws (they work everywhere and for all time).
Center of mass and determining individual masses
\[m_1\,a_1 = m_2\,a_2\quad\text{[Eq. (15.7)]}\quad\quad a=a_1+a_2\quad\text{[Eq. (15.8)]}\]
\[m_\text{1, Sun} = \left(\frac{a_2}{a}\right)\times m_\text{total, Sun}\quad\text{[Eq. (15.9)]}\quad\quad m_\text{2, Sun} = \left(\frac{a_1}{a}\right)\times m_\text{total, Sun}\quad\text{[Eq. (15.10)]}\]
Equations (15.7) and (15.8) describe the relative separations of two objects from their mutual center of mass based on the relative masses of those objects. You are certainly intuitively familiar with the relationship. If you try to balance two objects with identical masses, intuitively you realize that the balance point must half-way between them. You can try this at home if you have a ruler for example. Assuming that the mass distribution of the ruler is evenly distributed throughout the ruler, you can balance it on an outstretched finger in front of you by placing the midpoint of the ruler on your finger. In this way exactly one-half of the ruler’s mass is on the right side of you finger and the other half of the ruler’s mass is on the left side of your finder. The same would hold true if you set two identical masses on the ruler, one on the left side and the other on the right side, as long as both masses were the same distance from the center of your finger. However, if you were to replace one of the two objects with a mass that is twice as great, you would need to shift the balance point of the ruler, the center of mass, closer to the object of greater mass. The greater the difference in masses of the two objects, the closer the center of mass is to the object of greater mass. If the greater mass were extremely massive compared to the less massive object, the center of mass would be nearly at the center of the more massive object. This behavior is what Equations (15.7) and (15.8) are all about.
The figure below illustrates how distances from the center of mass depend on the relative masses of the objects. For each example, call the masses on the left-hand side \(m_1\) and call the masses on the right-hand side \(m_2\). Similarly, the separations of the masses from the center of mass are \(a_1\) on the left-hand side and \(a_2\) on the right-hand side.
- The center-of-mass in each example is the location of the fulcrum (triangle).
- According to Equation (15.7), in the first example \(m_1\,a_1 = m_2\,a_2\) leads to \(1\times 1/2 = 1\times 1/2=1/2\) .
- The distance between the single mass on the left-hand side and the multiple masses on the right-hand side is always taken to be 1. This means that the length to the left of fulcrum added to the length on the right-hand side always equals \(1\); for example, \(1/2+1/2=2/2=1\) and \(2/3+1/3 = 3/3 = 1\). This is Equation (15.8): \(a = a_1+a_2\).
Examples
Now let’s apply all of the theory and abstract mathematics discussed in the introduction to some examples:
- In the diagrams above, assume that the blocks are actually stars orbiting their mutual center of mas, the masses listed are in terms of the mass of the Sun, and the distances from the center of mass are in astronomical units. Use Equations (15.9) and (15.10) to show that the bottom distances and masses are consistent.
- Let \(m_\text{1, Sun} = 1\text{ M}_\text{Sun}\) and \(m_\text{2, Sun} = 20\text{ M}_\text{Sun}\). This means that the total mass is \(m_\text{1, Sun} +m_\text{2, Sun} = 21\text{ M}_\text{Sun}\). That value corresponds to the denominators in the distances given. The value of the first mass based on its distance from the center of mass is given by \(m_\text{1, Sun} = \left(\frac{a_2}{a}\right)\times m_\text{total, Sun}\) where \(a_2 = (1/21)\text{ au}\). Substituting, \(m_\text{1, Sun} = (1/21)\times 21\text{ M}_\text{Sun}= 1\text{ M}_\text{Sun}\) The value of the second mass also works by using Equation (15.10).
- Suppose one star has a mass of \(1\text{ M}_\text{Sun}\) and its binary companion has a mass of \(3\text{ M}_\text{Sun}\). If the total separation of the two stars from each other is \(1\text{ au}\), what are the distances of the two stars from their mutual center of mass?
- Following the pattern of the diagrams above, \(a_\text{1, au} = (3/4)\text{ au}\) and \(a_\text{2, au} = (1/4)\text{ au}\). The same results would be obtained by using Equations (15.9) and (15.10).
- Suppose one star orbits \(6\text{ au}\) from a binary system’s center of mass and its binary companion is \(2\text{ au}\) from the center of mass. Based on Kepler’s total mass equation, astronomers find that the total mass of the system is \(2\text{ M}_\text{Sun}\). What are the individual masses of the two stars?
- The total separation of the two stars from one another is \(a_\text{au} = 6\text{ au} + 2\text{ au} = 8\text{ au}.\) Assuming that the star farthest from the center of mass is Star 1 and the other star is Star 2, from Equations (15.9) and (15.10), \(m_\text{1, Sun} = (a_\text{2, au}/a_\text{au})\times m_\text{total, Sun} = (2/6)\times 2\text{ M}_\text{Sun} = 2/3\text{ M}_\text{Sun} .\) Since the total mass of the two stars is \(2\text{ M}_\text{Sun}\), the mass of the second star must be \(m_\text{2, Sun} = 4/3\text{ M}_\text{Sun} .\)
- What is the orbital period of the two stars in Example 3?
- Kepler’s total mass equation must be solved for the orbital period squared, and then the square root of the result must be taken to get the final result. Solving for \(P^2\) gives \[P_\text{y}^2 = \frac{a_\text{au}^3}{m_\text{total, Sun}} = \frac{8^3}{2} \text{ y}^2= \frac{512}{2}\text{ y}^2 = 256\text{ y}^2.\] Taking the square root, \(P_\text{y} = \sqrt{256}\text{ y} = 16\text{ y}.\)
- Two stars in a binary star system are found to have an orbital period of \(2\) years and they are separated by a distance of \(4\) astronomical units. What is the sum of the masses of the two stars?
- In Equation (15.6), \(P_\text{y} = 2\) and \(a_\text{au} = 4\). Substituting into Kepler’s total mass equation we find \[m_\text{total, Sun} = \frac{4^3}{2^2}\text{ M}_\text{Sun} = \frac{64}{4}\text{ M}_\text{Sun} = 16\text{ M}_\text{Sun}.\] The combined mass of the two stars is \(16\text{ M}_\text{Sun}\).
- One of the two stars, call it Star 1 in Example 5 is found to be orbiting the system’s center of mass at a distance of \(3\text{ au}\). How far is the second star from the center of mass.
- The total distance of the two stars from one another is \(a_\text{au} = 4\text{ au}\) and Star 1 is \(a_\text{1, au} = 3\text{ au}\) from their mutual center of mass. Since \(a_\text{total} = a_\text{1, au} + a_\text{2, au}\), algebraically this means that \(a_\text{2, au} = a_\text{total} – a_\text{1, au} = 4\text{ au}-3\text{ au} = 1\text{ au}\). In other words, if the separation between the two stars is \(4\text{ au}\) and Star 1 is \(3\text{ au}\) from the center of mass, then Star 2 must be \(1\text{ au}\) from the center of mass so that \(3\text{ au} + 1\text{ au} = 4\text{ au}\).
- From Example 5 we know that the total mass of the two stars is \(16\text{ M}_\text{Sun}\) and from Example 6 we know the individual distances of the two stars from their mutual center of mass, \(3\text{ au}\) and \(1\text{ au}\) for Star 1 and Star 2, respectively. What are the individual masses of the two stars?
- From Equation (15.9), \[m_\text{1, Sun} = \frac{a_\text{2, au}}{a_\text{au}}\times m_\text{total, Sun} = \frac{1 \text{ au}}{4\text{ au}}\times 16\text{ M}_\text{Sun} = 4\text{ M} _\text{Sun}.\]
- The mass of Star 2 can be found from either Equation (15.10) or \(m_\text{total, Sun} = m_\text{1, Sun} + m_\text{2, Sun}\); in other words, the total mass is just the sum of the two masses. Since the total mass is \(16\text{ M}_\text{Sun}\) and the mass of Star 1 is \(4\text{ M}_\text{Sun}\), the mass of Star 2 must be \(12\text{ M}_\text{Sun}\). Note: You can check your answer by solving Equation (15.10).
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 to the problems are given below)
- The orbital period of a binary star system is 4 years and the separation between the two stars is 4 astronomical units. What is the total mass of the system?
- The orbital period of a binary star system is 27 years and the separation between the two stars is 9 astronomical units. What is the total mass of the system?
- One of the stars in the system from Example 2 is 3 astronomical units from the center of mass. What are the masses of the two stars?
- The orbital period of the Procyon binary star system is 40.8 years and the distance between Procyon A and Procyon B is 15.2 au. What is the total mass of the Procyon system?
- (Advanced) The ratio of the distances of Procyon A to Procyon B from the center of mass is \(a_\text{A}/a_\text{B} = 2/5\). This ratio can be used to determine the ratio of the stars’ masses from Equation (15.7). Solving for the mass ratio gives \(m_\text{B}/m_\text{A} = a_\text{A}/a_\text{B}\). Further solving for the mass of Procyon B in terms of Procyon A gives \(m_\text{B} = (a_\text{A}/a_\text{B})\times m_\text{A} = (2/5)\times m_\text{A}. \) Finally, inserting that relationship into Equation (15.6) and solving for the mass of Procyon A gives \[m_\text{A} = \left(\frac{5}{7}\right)\times \frac{a_\text{au}^3}{P_\text{y}^2}.\] What is the mass of Procyon A? What is the mass of Procyon B?
- Jupiter’s mass is about \(1/1000\) the mass of the Sun. Jupiter’s orbital distance from the center of mass of the Sun-Jupiter system is 5.2 au and its orbital period is 11.86 years. What is the Sun’s orbital distance from their mutual center of mass? The Sun’s radius, expressed in astronomical units, is \(0.0046\text{ au}\). Where is the system’s center of mass relative to the Sun?
Answers
- 4 solar masses
- 1 solar mass
- The star that is 3 au from the center of mass has a mass of \(2/3\text{ M}_\text{Sun}\) and the other star’s mass is \(1/3\text{ M}_\text{Sun}\).
- 2.1 solar masses
- \(m_\text{A} = 1.5\text{ M}_\text{Sun}\), \(m_\text{B} = 0.6\text{ M}_\text{Sun}\)
- \(a_\text{Sun} = 0.005\text{ au}\). The center of mass is just beyond the surface of the Sun. Neglecting the other objects in the Solar System, the Sun would wobble around that point every 5.2 years. Including all of the other objects makes the situation more complex.