Surface temperature of a planet revolving a sun

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1. Apr 30, 2015

1. The problem statement, all variables and given/known data

Find the surface temperature of a small planet having circular orbit around the sun with time period T,assuming sun and planet to be black bodies. Take radius of sun = R, its mass = M and its surface temperature as $\theta_0$.

2. Relevant equations

$P=eA\sigma T^4$
Total energy of planet is $\frac{GMm}{2R}$
$I=\frac{P}{4\pi d^2}$

3. The attempt at a solution
Power emitted by sun is $P=eA\sigma\theta^4_0$.
The energy of planet is the sum of its kinetic and potential enegies (E) + the energy recieved from sun.
From these relations, I am not able to find the planet's surface temperature.
Since the planet is also a black body, the energy absorbed by planet = the energy imcident on the planet. I can find the energy it recieved during one complete revolution.
But I am not able to get any usefull relation.

2. Apr 30, 2015

Orodruin

Staff Emeritus
You do not need to know the energy during one revolution. The energy loss rate needs to equal the energy gain from absorbing sunlight.

3. Apr 30, 2015

Quantum Defect

You need to calculate the power incident on the planet from the sun. This can be approximated by the ratio of the cross sectional area presented to the sun (pi*r_planet^2) divided by the total area (solid angle) 4*pi*r_orbit^2 times the power output by the sun (assumes isotropic radiation). At equilibrium, the power deposited into the planet from the sunlight must be equal to the power radiated by the planet to space. This can be used to calculate the black body temperature for the planet. So, to figure all of this out, you need to find out what the orbital radius is.

4. May 1, 2015

$T=\frac{2\pi R_0}{v}$
And $v=\sqrt{\frac{GM}{R}}$
So $T=2\pi R_0\sqrt{\frac{R}{GM}}$
So $R_0=\frac{T}{2\pi}\sqrt{\frac{GM}{R}}$
But radius of planet is not given.

5. May 1, 2015

Orodruin

Staff Emeritus
Assume a radius r and see what comes out. Do not enter a value for it.

Edit: Also, v is irrelevant.

6. May 1, 2015

Its $R_0^{3/2}=\frac{T}{2\pi}(GM)^{1/2}$
The power received by planet is $P=\pi r^2\frac{P}{4\pi R_0^2}$
Now P emitted=$4\pi r^2\sigma\theta^4$