Transcript Extra-Solar Planets

The Mathematics of Star
Trek
Lecture 11: Extra-Solar
Planets
Outline
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Finding Extra-Solar Planets
The Two-Body Model
A Celestial Cubic
Example-51-Pegasi
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Finding Extra-Solar Planets
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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!
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The Two-Body Model
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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.
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Think of a hammer thrower spinning around—the
thrower is the star and the hammer is the planet!
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The Two-Body Model (cont.)
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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.
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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.)
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For our model, we assume the following:
• The star orbits the center of mass in a circle
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of radius r with uniform linear speed v.
The Earth lies in the orbital plane of the starplanet 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.)
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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.)
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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!
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The Two-Body Model (cont.)
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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.
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The Celestial Cubic
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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.
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The Celestial Cubic (cont.)
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The centripetal force is given by
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Parameterizing the star’s orbit about the
center of mass, we find the planet’s
position vector to be:
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The Celestial Cubic (cont.)
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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.)
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The magnitude of the gravitational force is
where G is the universal gravitation constant
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Equating forces, we get
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The Celestial Cubic (cont.)
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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 satisfies
m1r1=m2r2 (Law of the Lever).
Balance Point
m1
m2
r1
r2
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The Celestial Cubic (cont.)
c.o.m.
R-r
r
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Thinking of the planet and star as masses on a teetertotter, the Law of the Lever implies,
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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.)
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Dividing (3) by M2, and setting
and
we find that x and  satisfy the following cubic equation:
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Example-51-Pegasi
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Measured wavelength
shifts of light from the
star 51-Pegasi show that
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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.)
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Repeat, using a fixed-point method to solve the
following equation which is equivalent to (4):
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Groetsch argues that equation (4) can be
solved by iteration of (5), via
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Try this with Mathematica and compare to the
solution above.
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References
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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
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