The Mathematics of Star Trek

The Mathematics of Star Trek Lecture 11: Extra-Solar Planets

Outline • Finding Extra-Solar Planets • The Two-Body Model • A Celestial Cubic • Example-51-Pegasi

Finding Extra-Solar Planets • Recent discoveries of planets orbiting stars rely on a type of problem known as an inverse problem. • In the paper “A Celestial Cubic”, Charles Groetsch shows how the orbital radius and mass of an unseen planet circling a star can be obtained from the star’s spectral shift data, via the solution of a cubic equation!

The Two-Body Model • Assume a far-off star of mass M is orbited by a single planet of mass m<M, with a circular orbit of radius R. • The star and planet orbit a common center of mass (c.o.m.). • To an observer on Earth, the star will appear to wobble. • Think of a hammer thrower spinning around—the thrower is the star and the hammer is the planet!

The Two-Body Model (cont.) • On earth, we see this wobble as a Doppler shift in the wavelength of the light from the star. • As the star moves towards us, the light shifts towards the blue end of the spectrum. • As the star moves away from us, the light shifts towards the red end of the spectrum. • The magnitude of these shifts determine the radial velocity of the star relative to Earth. • The time between successive peaks in the wavelength shifts gives the orbital period T of the star and planet about their center of mass.

The Two-Body Model (cont.) • For our model, we assume the following: • The star orbits the center of mass in a circle of radius r with uniform linear speedv. • The Earth lies in the orbital plane of the star-planet system. • The distance D from the Earth to the center of mass of the star-planet system is much greater than r (D >> r).

The Two-Body Model (cont.) • Recall from trigonometry that v =  r, where  is the angular speed. • Also recall that  =/t, where  is the angle in radians traced out in t seconds by the star as it orbits around the center of mass. D c.o.m. Earth  r

The Two-Body Model (cont.) • Since v is constant, it follows that  is also constant, so when t = T,  = 2, and thus  = 2/T. • Using this fact, we can write the radial velocity, given by V(t) = d’(t), as follows: • Hence, V is sinusoidal, with amplitude equal to star’s linear speed v, and period equal to the star’s period T about the center of mass!

The Two-Body Model (cont.) • Measuring wavelength shifts in the star’s light over time, a graph for V(t) can be found, from which we can get values for v and T. • Then, knowing v and T, we can find the orbital radius r of the star about the center of mass: • Finally, the mass M of the star can be found by direct observation of the star’s luminosity.

The Celestial Cubic • At this point, we know M, v, T, and r. • We still want to find the radius R of the planet’s orbit about its star and the mass m of the planet. • From physics, the centripetal force on the star rotating around the c.o.m. is equal to the gravitational force between the planet and star.

The Celestial Cubic (cont.) • The centripetal force is given by • Parameterizing the star’s orbit about the center of mass, we find the planet’s position vector to be:

The Celestial Cubic (cont.) • Differentiating twice, we see that the acceleration of the star is given by: so the magnitude of the centripetal force on the star is

The Celestial Cubic (cont.) • The magnitude of the gravitational force is where G is the universal gravitation constant • Equating forces, we get

The Celestial Cubic (cont.) • We now have one equation that relates the unknown m and R. • To get another equation, we’ll use the idea of finding the balance point (center of mass) for a teeter-totter. • Archimedes discovered that the balance point (center of mass) for a board with masses m1 and m2 at each end satisfiesm1r=m2r2 (Law of the Lever). Balance Point m2 m1 r1 r2

The Celestial Cubic (cont.) c.o.m. r R-r • Thinking of the planet and star as masses on a teeter-totter, the Law of the Lever implies, • Solving (2) for R and substituting into (1), we find

The Celestial Cubic (cont.)

The Celestial Cubic (cont.) • Dividing (3) by M2, and setting and we find that x and  satisfy the following cubic equation:

Measured wavelength shifts of light from the star 51-Pegasi show that v = 53 m/s, T = 4.15 days, and M = 1.99 x 1030 kg. Use Mathematica to find r, , x, and m by finding the roots of (4) directly. Example-51-Pegasi

Example-51-Pegasi (cont.) • Repeat, using a fixed-point method to solve the following equation which is equivalent to (4): • Groetsch argues that equation (4) can be solved by iteration of (5), via • Try this with Mathematica and compare to the solution above.

References • C.W. Groetsch, “A Celestial Cubic”, Mathematics Magazine, Vol. 74, No. 2, April 2001, pp. 145 - 152. • C.P. McKeague, Trigonometry (2cd ed), Harcourt Brace, 1988. • J. Stewart, Calculus: Early Transcendentals (5th ed), Brooks - Cole, 2003. • http://zebu.uoregon.edu/51peg.html

The Mathematics of Star Trek