Tuesday

January 17, 2017
Posted by **Mark** on Wednesday, March 4, 2009 at 4:19am.

There are currently 340 known extrasolar planets, i.e., planets around other suns. Most of them we infer only by their gravitational pull on their host stars, and a handful appear to dim the light of their host star as they transit in front of the star. In November of last year, however, astronomers announced the detection of a planet which not only was not only resolved but whose orbital motion was also conﬁrmed by direct imaging. In other words, in two pictures made by Hubble Space Telescope, the planet was observed to have shifted position.

This Jupiter-like planet is called Fomalhaut b, orbiting a star Fomalhaut, which is 25 light years from Earth. The host star has a mass of 2.1 M and its radius is 1.8 R. Its surface temperature is 8700 K. The planet orbits with the semimajor axis of 115 AU around the star, has small eccentricity, and the plane of the orbit is in the plane of the sky (in other words we are seeing the orbit face-on).

a) Determine the angular separation between the star and the planet, in arc seconds.

b) Determine the period of the planet, in years. Hubble imaged the system in 2004 and 2006. What is the distance that the planet traveled between the images, in AU? (Hint: if you are reaching for your lecture notes to ﬁnd the formula for the velocity of

the planet, you are doing it the hard way.)

c) What is the angular separation between the position of the planet in 2004 and 2006, in arc seconds? Can the Hubble telescope, with the angular resolution of 0.04 arc seconds, detect this motion?

d) What is the equilibrium temperature on Fomalhaut b? You can ignore the eﬀects of albedo and greenhouse eﬀect. Can liquid water exist on Fomalhaut b?

e) At what distance from the star Fomalhaut would the planet have the same equilibrium temperature as on Earth? Again, ignore albedo and greenhouse eﬀects.

Thanks!

- Astronomy (algebra-based physics) -
**drwls**, Wednesday, March 4, 2009 at 9:17amWe don't give "detailed solutions" to long-winded problems here. Here is how to go about solving them yourself. If you want further help, show your work.

a) The ratio of the orbital distance to the distance of the star (in the same units) is the angle in radians. Convert that to degrees and then arc seconds.

b) Use Kepler's third law in the form

M = R^3/T^2

where M is the mass of Fomalhgaut in "suns", R is the orbit radius in AU and T is the period in years. In two years, the planet moves 2/T of the orbit circumference. Compute that distance (X)

c) X/(star distance) in radians

Convert that to arc seconds

d) Set the received energy from the star (one side only, using the projected area pi r^2) equal to the energy that the planet radiates away (using 4 pi r^2 for the surface area). The planet's radius, r, cancels out. So does sigma. Use the Stefan Boltzmann law for both star and planet and solve for the planet temperature. The planet's received radiated power is

4 pi*R^2(sigma)*Tstar^4 * pi r^2/(4 pi d^2) and that equals

4 pi r^2 (sigma)*Tplanet^4

(the energy raidated away by the planet).

d is the distance of the planet from the star. Solve for Tplanet

pi*(R/d)^2 *Tstar^4 = Tplanet^4

You can compare Tplanet with the freezing point of water, which is not very sensitive to the pressure. Whether it exists as a liqud or not will depend upon atmospheric pressure, which you don't know.

(e) Use the formula I just derived to compute Tstar to make Tplanet = 280 K (about the equilibrium T of earth)