Planet Cooling Homework: Calculating Time to Cool Down from 100K to 50K

  • Thread starter Thread starter ma18
  • Start date Start date
  • Tags Tags
    Cooling
Click For Summary

Homework Help Overview

The discussion revolves around calculating the time it takes for a planet with a surface temperature of 100K and a radius of 500 km to cool down to 50K, while ignoring solar heating. The problem involves concepts from thermodynamics and blackbody radiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to set up a differential equation due to the non-constant power associated with the cooling process. There are attempts to express thermal energy in terms of temperature and to relate energy loss to temperature changes. Questions arise about the assumptions made regarding the planet's shape and material properties, particularly in calculating the number of particles.

Discussion Status

Participants are actively exploring different approaches to the problem, including setting up integrals and differential equations. Some guidance has been offered regarding the formulation of the energy equations and the assumptions about the planet's geometry. There is no explicit consensus, but various lines of reasoning are being investigated.

Contextual Notes

There are discussions about the assumptions of the planet being spherical and made of iron, which may affect calculations of mass and particle number. Participants express uncertainty about the correctness of their approaches and results, including concerns about negative time values in their calculations.

ma18
Messages
93
Reaction score
1

Homework Statement



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.

Homework Equations



E_thermal = 3/2 NkT

Power emitted/Area = σT^4

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:
Physics news on Phys.org
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.
 
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?
 
ma18 said:
Yes it supposed to be a differential equation but I tried doing that and I didn't get anywhere.

I get

dE/dt = σT^4 * A

then what?

Well, what is E(T)? That will allow you to express everything in terms of T.
 
ma18 said:
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?

Oh, you edited your post. Yes. I think that is the right approach.
 
  • Like
Likes   Reactions: 1 person
Thanks!
 
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 calculate the mass of the planet and then get the number of particles N.

Is there any other way to get N?
 
ma18 said:
Is it reasonable to assume that the planet is a sphere?

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

ma18 said:
I am given that the planet is made of iron (know the density and mass of iron). Using p = m/V I would calculate the mass of the planet and then get the number of particles N.

Is there any other way to get N?

This sounds like the way to get N.
 
I get time to be negative :(Tf
∫dt/T^4
Ti

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

Since Ti>Tf
 
  • #10
Plus I get a timescale of 10^42 years
 
  • #11
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.
 
  • Like
Likes   Reactions: 1 person
  • #12
I got it, thanks for all your help. I like this method, much better than other places which just give you the answer.
 
  • #13
Just out of curiosity, what did you get as the cooling time to 50 K?
 

Similar threads

Replies
6
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
5
Views
4K
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
11K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K