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The Mathematics of Star TrekPowerPoint Presentation

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.

The Mathematics of Star Trek

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The Mathematics of Star Trek

Lecture 11: Extra-Solar Planets

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

- 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!

- 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!

- 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.

- 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).

- 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

- 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!

- 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.

- 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 centripetal force is given by
- Parameterizing the star’s orbit about the center of mass, we find the planet’s position vector to be:

- 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 magnitude of the gravitational force is
where G is the universal gravitation constant

- Equating forces, we get

- 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

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

- 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.

- 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.
- http://zebu.uoregon.edu/51peg.html