AST 311: Homework III

Joshua W. Kern

Ast 311: Homework III

1) If the contrast of star to planet is 10^-10 then,

Contrast = Flux_planet /Flux_star ≈ Luminosity_planet /Luminosity_star = L_planet /L_star

               = (0.2)(pi*r_p²) L_star / (8*pi*D²) L_star

               = (0.2)(pi*r_p²) / (8*pi*D²)

Therefore,

           r_p = (Contrast*(8*pi*D²) / (0.2*pi))^1/2 ≈ 540,749 km; using D from problem 2.

So,

           r_p = 7.73 r_jupiter = 84.88 r_earth

2) If we observe the star at max seperation of 0.2 arcsec, then from the small angle formula,

                                                              Sin φ ≈ φ

So since,

                                         Distance to the star = Distance to the planet.

Then from geometry,

            D = Distance between planet and host star

                = Sin φ*Distance to the planet

                = φ*Distance to the star.

After converting to appropriate units

            D = 57.2 AU

3) If the probability to transit is given by

                     P = r_star / D = 0.5 r_sun / 57.2 AU

Then

                     P = 4.07 x 10^-5

4) From Kepler's 3^rd Law, the orbital period is given by

                     T = 2*pi*(D³/GM)^1/2

                        = 2*pi*(57.2 AU³/G(0.5 M_sun))^1/2

                        = 612.97 years

Then

                     Orbital velocity = 2*pi*D / T = 2.78 km/s

Therefore

                     Length of transit time = 2*r_star/ Orbital velocity

                                                         = 69.57 hours = 2.90 days

5) If transit depth is given by

                     δ = (r_planet / r_star )²

Then ,

                     δ = 2.41 %

This transit would be visible from the ground and from space.

6) The time for a full orbit is

                     T = 2*pi*(D³/GM)^1/2

                        = 2*pi*(57.2 AU³/G(0.5 M_sun))^1/2

                        = 612.97 years

7) Assuming the planet is at maximum separation means that if we want to view a primary or secondary transit then we would have to wait 25% or 75% of the orbital period for the planet to align with the star (assuming the inclination angle is 0).

Thus we would have to wait either 153.24 years or 459.73 years to see a transit

8) If

                M_starV_star = M_planetV_planet

And we know M_star, V_star , and V_planet

Then,

                M_planet= M_starV_star / V_planet

                            = 1.78 x 10²⁶ kg

Therefore,

                M_planet= 29.83 M_earth = 0.094 M_Jupiter