Discovery of Gliese 581C

Ast 4 Lecture 23 Notes J. E. Ybarra - Sacramento State

1 Extra Solar Planets - Discovery of Gliese 581c

Source: ESO

M dwarf stars M dwarfs are ideal stars to observe in order to ﬁnd small planets • Low mass - M dwarfs are low mass stars thus it is easier to detect lower mass planets around them • Low luminosity - habital zones for M dwarfs are closer (∼ 0.1 A.U.)

Gliese 581 Gliese 581 is a M dwarf star. • Low Activity: Gl 581 has low photospheric activity which means that there will be low radial velocity noise (= better detection)

• Mass ≈ 0.31 M

• Luminosity ≈ 0.013 L • Has a Neptune-mass planet Gl 581b Kepler’s 1st Law of Motion

Doppler Method

Gl 581c Using the radial-velocity doppler method detected a second planet around Gl 581 (Udry et al. 2007) • P = 12.931 days

• Wobble due to second planet V? = 3.01 m/s

2 Gl 581c - Kepler’s 3rd Law revisited Remember Modiﬁed Kepler’s 3rd Law 3 2 apl Ppl = Mtotal

where P is in years and a is in A.U.

Gl 581c - Kepler’s 3rd Law revisited

1 year P = 12.931 days = 0.0354 yr pl 365 days then a3 (0.0354)2 = 0.31 3.89 × 10−4 = a3 0.073 = a

Using Kepler’s 3rd law we can estimate the distance from GL 581c to its parent star to be a = 0.073 A.U.

Gl 581c Using the period and the semi-major axis of the orbit we can estimate the planets velocity. If we assume a circular orbit, then 2πa v = pl P 1.5 × 1011 m a = 0.073 A.U. = 1.1 × 1010 m 1 A.U. 3.15 × 107 s P = 0.0354 yr = 1.12 × 106 s 1 yr

Gl 581c

2π(1.1 × 1010 m) v = = 6.1 × 104 m/s pl 1.12 × 106 s

3 Gl 581c Kepler’s 1st law states that the star and the planet both orbit the center of mass. As a consequence, the relationship between the masses and velocities is

M?V? = mplvpl then 4 (0.31M?)(3.01 m/s) = mpl(6.1 × 10 m/s) −5 and we estimate the mass of the planet to be mpl = 1.5 × 10 M = 5.0M⊕

Gl 581c From theoretical models of planet formation for planets with masses 1 to 10 M⊕ (Valencia et al. 2006) the relationship between radius and mass is

0.267−0.272 rpl ∝ mpl

Using this model we can estimate that the radius of this planet is 1.5 R⊕

Gl 581c The acceleration due to gravity on the surface of Earth is

GM⊕ 2 = g R⊕

Then for Gl 581c with rpl = 1.5R⊕ and mpl = 5M⊕

G(5M⊕) GM⊕ 2 = 2.2 2 = 2.2g (1.5R⊕) R⊕ At the surface of the planet the force of gravity is about twice that of Earth. References • Udry, S., et al. 2007, A&A preprint (http://obswww.unige.ch/∼udry/udry preprint.pdf) • Valencia, D., OConnell, R. J., & Sasselov, D. 2006, Icarus, 181, 545