# Planterary Cooling

1. Nov 7, 2013

### ma18

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

How much time does it take a planet with surface temperature of 100K and a radius of 500 km to cool down to 50K, ignoring solar heating. Assume E_thermal = 3/2 NkT and that the surface radiates as a blackbody. Calculate this by using the relationship between thermal energy and temperature and the relationship between the rate of energy loss dE/dt and temperature.

2. Relevant equations

E_thermal = 3/2 NkT

Power emitted/Area = σT^4

3. The attempt at a solution

Power = Energy/Time

Energy/Area*Time = σT^4

Energy = σT^4 * A*t

Subbing for thermal energy

3/2 NkT = σT^4 * A*t

T = (3NKt/2A)^(1/3)

I feel like I've missed something and the T should be a delta T but I'm not sure where I've going wrong.

Thanks for any help,

Last edited: Nov 7, 2013
2. Nov 7, 2013

### cepheid

Staff Emeritus
Something to consider: the power is not constant in this problem, because cooling causes the planet's surface temperature to get lower, which causes it to radiate less. So, you probably have to set up an integral here.

Edit: Er, actually, a differential equation, I think.

3. Nov 7, 2013

### ma18

I get

dE/dt = σT^4 * A

d(3/2 NkT)/dt = σT^4 * A

and take out the constant and solve using separation of variables?

4. Nov 7, 2013

### cepheid

Staff Emeritus
Well, what is E(T)? That will allow you to express everything in terms of T.

5. Nov 7, 2013

### cepheid

Staff Emeritus
Oh, you edited your post. Yes. I think that is the right approach.

6. Nov 7, 2013

Thanks!

7. Nov 7, 2013

### ma18

Is it reasonable to assume that the planet is a sphere?

I am given that the planet is made of iron (know the density and mass of iron). Using p = m/V I would calcluate the mass of the planet and then get the number of particles N.

Is there any other way to get N?

8. Nov 7, 2013

### cepheid

Staff Emeritus
Yes. In fact, since you've only been given one dimension, you can't really assume anything else.

This sounds like the way to get N.

9. Nov 8, 2013

### ma18

I get time to be negative :(

Tf
∫dt/T^4
Ti

[(1/Ti^3)-(1/Tf^3)]

Since Ti>Tf

10. Nov 8, 2013

### ma18

Plus I get a timescale of 10^42 years

11. Nov 8, 2013

### cepheid

Staff Emeritus
Regarding the first problem: you could just be out by a negative sign. Remember that the power radiated is equal to the rate of decrease in thermal energy.

Second problem: it is an algebraic or arithmetic error. Please post all of your work so far.

12. Nov 8, 2013

### ma18

I got it, thanks for all your help. I like this method, much better than other places which just give you the answer.

13. Nov 8, 2013

### cepheid

Staff Emeritus
Just out of curiosity, what did you get as the cooling time to 50 K?