How Does the Brightness and Temperature of l Carinae Change with Its Pulsations?

[d97]

Homework Statement

The Cepheid variable star l Carinae pulsates with a period of about 36 days. During each pulsation, the star becomes about twice as bright and its radius grows by about 25%.
(a) In its smaller state, the star has a surface temperature of about 5100 K. Treating the star as a blackbody, use Wein’s displacement law to compute the wavelength at which l Carinae is brightest. [2 marks]
(b) If l Carinae swells by 25% and becomes twice as bright, what is its surface temperature in its swollen state? At what wavelength is it brightest in this state? [4 marks]

Homework Equations

λ=2.898*10^-3/T
T=(L/16πσa^2)^0.25

The Attempt at a Solution

For (a) I think I have got the correct solution from the first equation which was 5.68*10^-7 m. My main problem is (b) because I'm unsure on how to work out luminosity with the variables I have been given. Do you work out 'a' in the equation by saying that P^2=a^3?, I'm unsure.
Any help would be much appreciated thank you [/B]

 

[TSny]

Hello.

For (b) use your second relevant equation to set up the ratio of the final temperature to the initial temperature. Simplify before plugging in numbers.

 

[d97]

Oh ok thank you I tried doing it that way and I got a temperature of 2.086*10^9 K, not sure if that is correct but it seems like its too big?

 

[TSny]

Oh ok thank you I tried doing it that way and I got a temperature of 2.086*10^9 K, not sure if that is correct but it seems like its too big?

Yes, that's too big. Can you show your calculation?

 

[d97]

Yeah I wasn't sure how to do it your way, but I tried like this: T2/T1=(2L/32π^2*σ^2*1.25*a^4)^0.5, T1=5100K, L=3.417*10^17, a=26625, σ=5.67*10^-8

 

[TSny]

The reason for setting up a ratio is that all constants will cancel out. So, you have made a mistake in simplifying the ratio. Also, I don't understand how you got the power of 0.5 rather than 0.25.

For practice, what is the ratio of (a*b)^0.25 and (a*c)^0.25? Recall that $\frac{x^b}{y^b} = \left( \frac{x}{y} \right)^b$

 

[d97]

Oh ok would it be (a*b*/a*c)^0.25?

 

[d97]

I re-calculated my temperature and I got 5507.4 K which seems more likely to be correct?

 

[TSny]

Oh ok would it be (a*b*/a*c)^0.25?

Yes, and note that you could simplify this by canceling the a.

 

[TSny]

I re-calculated my temperature and I got 5507.4 K which seems more likely to be correct?

I don't get this answer. What expression do you get for the ratio of the temperatures after you simplify by canceling whatever you can and before you plug in any numbers?

 

[d97]

So what I got was T2/T2=(2L/a^4)^0.25 but I think I've canceled too many constants?

 

[d97]

So what I got was T2/T2=(2L/a^4)^0.25 but I think I've canceled too many constants?

Sorry *T2/T1

 

[TSny]

So what I got was T2/T2=(2L/a^4)^0.25 but I think I've canceled too many constants?

This doesn't look right.

From T=(L/16πσa^2)^0.25, you have $$T_1 = \left( \frac{L_1}{16 \pi \sigma a_1^2} \right )^{0.25} $$ and $$T_2= \left( \frac{L_2}{16 \pi \sigma a_2^2} \right )^{0.25}$$ So, when you form the ratio $\frac{T_2}{T_1}$, you should get something that involves $L_1$, $L_2$, $a_1$ and $a_2$.

 

[d97]

This doesn't look right.

From T=(L/16πσa^2)^0.25, you have $$T_1 = \left( \frac{L_1}{16 \pi \sigma a_1^2} \right )^{0.25} $$ and $$T_2= \left( \frac{L_2}{16 \pi \sigma a_2^2} \right )^{0.25}$$ So, when you form the ratio $\frac{T_2}{T_1}$, you should get something that involves $L_1$, $L_2$, $a_1$ and $a_2$.

Oh ok so I tried it again and I got (L2a1^2/L1a2^2)^0.25 and I got a temperature of 5426.97 K

 

[TSny]

That looks good.

 

[d97]

That looks good.

Ok thank you for all your help