Question

jupiter is 7.78x10^8 km from the sun and has a planetary albedo of about 0.4. calculate jupiters radiative temperature. recall that the solar constant in 1367 W/m^2 a Earth's orbit which is 1.50 x10^8 km from the sun

Answer 1

The formula to find the temperature of the planet is

where A is the reflectivity (albedo) of the planet, L is the luminosity of its star in multiples of the sun's power, and D is the distance between the planet and the star in Astronomical Units (AU), where 1 AU is the distance from Earth to the sun (150 million km)

Now, we have to find Luminosity from solar constant.

The formula for that is

where (Lsun) is solar luminosity and (Dsun) is 1 AU

hence, Luminosity L = I * 4 * pi * D^2 = 1367 * 4 * pi * D^2 = 17178.23 * D^2

D(sun - jupiter) = 7.78x10^8 km / 1.50 x10^8 (in AUs)

= 5.19 * D

Now, radiative temperature of jupiter (T) = 273*[ {(1 - 0.4) * 17178.23 * D^2 }/ D^2 ]^(1/4)

= 273 * [0.6 * 17178.23]^(1 / 4)

= 2750.71 K