The Mathematics of Star Trek

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# The Mathematics of Star Trek - PowerPoint PPT Presentation

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.

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## The Mathematics of Star Trek

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### 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.)
• Dividing (3) by M2, and setting

and

we find that x and  satisfy the following cubic equation:

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