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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.

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the mathematics of star trek

The Mathematics of Star Trek

Lecture 11: Extra-Solar Planets

  • Finding Extra-Solar Planets
  • The Two-Body Model
  • A Celestial Cubic
  • Example-51-Pegasi
finding extra solar planets
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
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
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 cont6
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 cont7
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.





the two body model cont8
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 cont9
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
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 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 cont12
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 cont13
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 cont14
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





the celestial cubic cont15
The Celestial Cubic (cont.)




  • 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 cont17
The Celestial Cubic (cont.)
  • Dividing (3) by M2, and setting


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

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.

example 51 pegasi cont
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.
  • 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.