The Mathematics of Star Trek

1
The Mathematics of Star Trek

• Lecture 11 Extra-Solar Planets

2
Outline

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

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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 stars spectral shift data, via the
  solution of a cubic equation!

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The Two-Body Model

• Assume a far-off star of mass M is orbited by a
  single planet of mass mof 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 aroundthe
  thrower is the star and the hammer is the planet!

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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.

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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 speed v.
  
• 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).

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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
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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
  stars linear speed v, and period equal to the
  stars period T about the center of mass!

9
The Two-Body Model (cont.)

• Measuring wavelength shifts in the stars 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 stars luminosity.

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The Celestial Cubic

• At this point, we know M, v, T, and r.
• We still want to find the radius R of the
  planets 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.

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The Celestial Cubic (cont.)

• The centripetal force is given by
• Parameterizing the stars orbit about the center
  of mass, we find the planets position vector to
  be
  

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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

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The Celestial Cubic (cont.)

• The magnitude of the gravitational force is
• where G is the universal gravitation constant
• Equating forces, we get

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The Celestial Cubic (cont.)

• We now have one equation that relates the unknown
  m and R.
  
• To get another equation, well 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 satisfies m1rm2r2 (Law of the
  Lever).

Balance Point
m2
m1
r1
r2
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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

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The Celestial Cubic (cont.)
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The Celestial Cubic (cont.)

• Dividing (3) by M2, and setting
• and
• we find that x and ? satisfy the following cubic
  equation

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Example-51-Pegasi

• 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.

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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.

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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