Temperature of blackbody (star)

[pat666]

Homework Statement

b) Two stars, both of which behave like black bodies, radiate the same total energy per second. The cooler star has a surface temperature, T, and 3.0 times the diameter of the hotter star.
i) What is the temperature of the hotter star in terms of T?
ii) What is the ratio of the peak intensity wavelength of the hotter star to that of the cooler star?

Homework Equations

The Attempt at a Solution

Calling the cooler star 1 and the hotter 2
A₁=4\pir₁²
A₂=4\pi(3r₁)²
P₁=P₂
P=\epsilon\sigmaAT⁴
so P\proptoAT and that I think means T \propto 1/A
Here is where I get stuck, not sure if anything I am doing is right (basically just playing to see what happens) and not sure where to go now?

 

[Redbelly98]

The Attempt at a Solution

Calling the cooler star 1 and the hotter 2
A₁=4\pir₁²
A₂=4\pi(3r₁)²
P₁=P₂
P=\epsilon\sigmaAT⁴
so P\proptoAT and that I think means T \propto 1/A
Here is where I get stuck, not sure if anything I am doing is right (basically just playing to see what happens) and not sure where to go now?

You're pretty much on track, except that your T has lost it's ⁴ exponent.

P α A T ⁴​

 

[pat666]

So A_2 is 12 times the size of A_1 and T is proportional to 1/A^4
T=12T_1?

 

[Redbelly98]

So A_2 is 12 times the size of A_1

No, it isn't. How did you calculate that? What is A2/A1, given the correct expressions for A1 and A2 that you wrote earlier?

and T is proportional to 1/A^4

No, you need to use the fact that A·T⁴ is a constant to get the proportionality relation.

T=12T_1?

No.

 

[pat666]

A2/A1=9
P is proportional to AT^4
still not sure what to do here?

 

[Redbelly98]

A2/A1=9

Yes.

P is proportional to AT^4
still not sure what to do here?

Yes, and since P is the same for both stars,

A₁ T₁⁴ = A₂ T₂⁴​

Solve for T₂

 

[pat666]

I get T2=1/sqrt(3)T1? since the indices on T cancel out.

 

[Redbelly98]

Looks good, that's what I get too.

(I assume you mean T1/sqrt(3), and not 1/[T1*sqrt(3)] )

 

[pat666]

yeah I meant the 1st one. Thanks for helping me to the answer.