Surface temperature of a planet revolving a sun

[AdityaDev]

Homework Statement

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 $\theta_0$.

Homework Equations

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

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 received 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 received during one complete revolution.
But I am not able to get any usefull relation.

 

[Orodruin]

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

 

[Quantum Defect]

Homework Statement

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 $\theta_0$.

Homework Equations

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

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 received 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 received during one complete revolution.
But I am not able to get any usefull relation.

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.

 

[AdityaDev]

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.

To find radius of orbit:
$T=\frac{2\pi R_0}{v}$
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.

 

[Orodruin]

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

Edit: Also, v is irrelevant.

 

[AdityaDev]

sorry, the value of Radius of orbit is wrong in post 4.
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=$4\pi r^2\sigma\theta^4$
So the r^^2 terms goes out and i got the correct answer.