Derivation of Planet Size to Account for WD2359-434 Variation

Peak-to-peak variation = 10.3 mmag (ratio = 1.00953).

Adopt star mass = 0.94 x sun. If sun had this mass, & Earth was at 1 a.u., period would be 377 days since P ~ 1/sqrt(m).

Moving Earth closer so that P = 2.7 hrs requires orbit radius = 0.0045 a.u., since ai/ao = (Pi/Po)^2/3 & Pi/Po = 2.7/(24*377) = 1/3350. Orbit radius = 667,000 km.

Adopt star radius = 0.93 x R_earth. R_star = 5932 km.

Ratio of solid angle of star (as seen from planet) to a hemisphere is 1 - cos(alpha), where alpha = angular size of star radius as viewed from planet = atan (R_star / a ) = 0.508 deg. Ratio solid angles = 3.93x10^-5. This is ratio of surface brightness of planet to star.

Ratio of apparent solid angle of planet to star = (R_p / R_star )^2.

Brightness ratio of system when planet is fully illuminated to not illuminated (assuming planet albedo of 1) = 1 + {(R_p / R_star )^2} x {3.93x10^-5}

Set above ratio equal to peak-to-peak variation ratio, & solve for R_p:

{(R_p / R_star )^2} x {3.93x10^-5} = 0.00953
(R_p / R_star )^2 = 0.00953 / 3.93x10^-5 = 242
R_p / R_star = 15.6
R_p = 92,300 km

Since R_jup = 69,100 km (average of equatorial & polar), R_p / R_jup = 1.34 (assuming planet albedo = 1)

If planet albedo = 1/2, then R_p / R_jup = 1.9.

We need accurate RV measurements!