Laboratory Exercise II - On the Use of Kepler's Third Law
Isaac Newton expanded Kepler's version of the 3rd Law of Planetary Motion which simply stated is that the cube of the semimajor axis of a planet's elliptical orbit is proportional to the square of its period of revolution. Newton's expansion of the 3rd Law can be written in the following simplified form:
(M1 + M2) = a3 / P2
Where M1 and M2 are the masses of the two bodies (say the Sun as mass #1 and the Earth as mass #2) with one in orbit about the other, a is the semimajor axis of the orbit (1 AU for the Earth orbiting the Sun), and P is the period of motion (1 year for the Earth). This simple form of the law holds when the masses are in units of the Sun's Mass called solar masses (solar masses have nothing to do with church or with King Loui XIV of France!), the semimajor axis is in units of Astronomical Units (1 AU = 1.49 x 1011 m), and the period is in units of years (1 year = 365.25 mean solar days). The basis of this law is Newton's Law of Gravitation. It can be applied in this form to satellites orbiting planets, binary star systems, clusters of stars, and even symmetrical galaxies, whether ellipsoidal or spiral.
I. Calculating the Masses of Binary Stars
Double stars are stars which are close enough and move slowly enough that they orbit each other. Each star is located at the focus of the ellipse of its orbit around the other star.
1. Consider a binary star system which has a semimajor axis of 6.1" arc and a period of 87.3 years. The annual parallax of the stars, p, is 0.192"arc.
We call the measure of the angular separation of the two stars, a. [remember that 1 degree is divided into 60 'arc (read this as 60 minutes of arc) and each 1'arc is subdivided into 60"arc (read this as 60 seconds of arc)]. The distance to the binarystar system is calculated from its parallax , p, of 0.192"arc, which has been measured carefully over a period of the last 92 years. First we must calculate the distance to the binary system:
D = 1/p where p is the parallax in seconds of arc giving D in parsecs.
The distance, D = 1/p = ________ pc
How many light years does this correspond to? (remember that 1 pc = 3.26 lt yr)
D (in light years) =
2. This angular measure of the semimajor axis of the orbit can be converted to a linear measure of the distance by multiplying the distance, d, to the binary system (measured in parsecs) by the angular separation (measured in seconds of arc - make sure you don't use the parallax!!) so:
abinary = a D = ( "arc) ( pc) = ( AU)
3. Now, since we know that the period, P, of the binary stars is 87.3 years, we can now calculate the combined masses of the stars (designated 1 and 2 below) using Kepler's Third Law:
(M1 + M2) = a3 / P2 = ( AU)3 / ( yr)2 = ( _________ MS)
4. Binary stars are important because they provide the only direct method of estimating the masses of other stars. From the amplitude of the motion of the stars around their center of mass (called the barycenter) we can estimate the portion of the total mass which is in each star. If the stars were of equal mass, they would be located equal distances from the center of mass. If one star is twice as massive as the other star, the less massive star is located at twice the distance from the center of mass to the more massive star. These variations are also evident in the variations in the radial velocity of the two stars which can be measured using the Doppler Effect on the spectra of the binary star components. The least massive star moves faster relative to the center of mass of the system, the more massive star moves slower. If one of the stars is found to have a mass of 1.0 MS, what is the mass of the second star?
5. Consider now another binary star example. If our Sun had another star of identical size and mass orbiting it in the orbit of the Earth (in other words, the semimajor axis of its orbit is 1AU), what would be the period of their orbit?
Whyis this period (greater than) / (equal to) / (less than) the period of 1 year which is the period of the Earth's orbit?
II. Calculating the Mass of the Milky Way Galaxy
Isaac Newton showed that if a mass is distributed in a uniform spherical shells (so that the density is constant in each spherical shell), we can calculate the attraction of that distribution at a point some distance from the center of the distribution. When the attraction of shells further away from the center are calculated at the point we find that they cancel. That is, there is no attraction from shells further from the center of the spherical distribution than the point at which the attraction is calculated. What this means in the case of a galaxy which is an elliptical or spiral galaxy is that only matter between the center of a galaxy and a star will attract that star. In this section, we will use this effect to calculate the mass of our galaxy.
1. Consider our Sun - it is in orbit around the center of our Milky Way Galaxy. The velocity of the Sun in its orbit is about 250 km/s. The distance to the center of the galaxy is about 9.1 kpc (kiloparsecs). We can use Kepler's third law to calculate the mass of the galaxy interior to the Sun's orbit. We assume that the orbit is circular so that the semimajor axis is just the radius of the circular orbit = 9.1 kpc. First we need to calculate the number of AU's in 9.1 kpc. (Note that 1 kpc = 1000 pc = 3260 lt yrs and 1 pc = 206,265 AU. )
a = rSun = 9.1 kpc = (9.1 kpc)(1000 pc/1kpc)(206,265 AU/1 pc) = ______AU
2. Next calculate the period of the Sun's orbit around the center of the galaxy by calculating the distance the Sun travels in one period divided by the Sun's orbital velocity. But first, the circumferance of the Sun's orbit:
So the distance the Sun travels = 2 p r = (2)(3.14)( AU) = ________ AU
Calculate the Sun's velocity in AU/yr:
Number seconds/yr = (60 s/1 min)(60 min/1 hr)(24 hr/1 day)(365.25 days/1 yr) =
VSun = (250 km/s) (1 AU/1.49 x 108 km) ( _______ s/1 yr) = _________ AU/yr
The Period of the Sun's Orbit = PSun = circumference/VSun= ( _____ AU) / ( ______ AU/yr)
= _____________ yr.
3. Calculate the combined mass of the Sun and the portion of the Milky Way inside the Sun's orbit:
(M1 + M2) = a3 / P2 = ( ____ AU)3 / ( ____ yr)2 = ( _____ MS)
4. How many billions of Suns is that?
III. The Missing Mass Problem
Measurements of the orbital speed of stars in the Milky Way and other galaxies give an estimated mass of the Milky Way and other galaxies which are in contrast with mass estimates based on stars, star clusters, gas, and dust which are visible in these galaxies. The contrast between these kinds of measurements has lead astronomers to infer that there must be a great amount of non-luminous matter in the outer portions of most galaxies.
Assume that the orbital speed of stars is measured at a distance of 15 kpc from the center of our galaxy. If the rotational speed is 250 km/s, calculate the mass inside the orbit of such stars.
1. First calculate the semimajor axis length in AU's for a circular orbit with radius 15 kpc.
a = (15 kpc) ( ______ pc/1 kpc) ( ___________ AU / 1 pc) = ___________ AU
2. And calculating the circumference of the orbit:
circumference of orbit = 2 p a = (2)(3.14)( AU) = ____________ AU
3. Now calculate the mass of such a star and the portion of the Milky Way inside of this orbit:
VStar = (250 km/s) (1 AU/1.49 x 108 km) ( _______ s/1 yr) = _________ AU/yr
4. The Period of the Star's Orbit = PStar = circumference/VStar= ( _____ AU) / ( _____ AU/yr)
= _____________ yr.
5. Calculate the combined mass of the Star and the portion of the Milky Way inside the Star's orbit:
(Mstar + Mgalaxy) = a3 / P2 = ( _______ AU)3 / ( _______ yr)2 = ( ________ MS)
6. What fraction of the mass inside an orbit of 15 kpc is within the Sun's orbit (9.1 kpc)?
7. What conclusions do you draw regarding the mass distribution of the Milky Way Galaxy?
IV. Conclusions
Describe biefly what you find most interesting about how Kepler's 3rd Law can be used beyond the Solar System.
