# Temperature of blackbody (star)

• pat666
In summary, two stars, behaving like black bodies, radiate the same total energy per second. The cooler star has a surface temperature, T, and is 3.0 times the diameter of the hotter star. In terms of T, the temperature of the hotter star is 1/sqrt(3)T, and the ratio of the peak intensity wavelength of the hotter star to that of the cooler star is 1/sqrt(3). This is based on the fact that P is proportional to AT^4 and A1 T^4 is equal to A2 T^4, which leads to the solution for T2 as 1/sqrt(3)T1.
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?

## The Attempt at a Solution

Calling the cooler star 1 and the hotter 2
A1=4$$\pi$$r12
A2=4$$\pi$$(3r1)2
P1=P2
P=$$\epsilon$$$$\sigma$$AT4
so P$$\propto$$AT 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?

pat666 said:

## The Attempt at a Solution

Calling the cooler star 1 and the hotter 2
A1=4$$\pi$$r12
A2=4$$\pi$$(3r1)2
P1=P2
P=$$\epsilon$$$$\sigma$$AT4
so P$$\propto$$AT 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

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

pat666 said:
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.

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

pat666 said:
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

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

Looks good, that's what I get too.

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

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

## 1. What is a blackbody star?

A blackbody star is an idealized astronomical object that absorbs and emits all wavelengths of electromagnetic radiation with perfect efficiency. It is often used as a theoretical standard to compare the temperature of other stars.

## 2. How does the temperature of a blackbody star affect its appearance?

The temperature of a blackbody star determines the peak wavelength of light it emits, which in turn affects its color. Higher temperatures emit shorter, bluer wavelengths while lower temperatures emit longer, redder wavelengths.

## 3. What is the relationship between a blackbody star's temperature and its luminosity?

The temperature of a blackbody star directly affects its luminosity, or brightness. The hotter the star, the more energy it emits, making it appear brighter. This relationship is known as the Stefan-Boltzmann law.

## 4. How is the temperature of a blackbody star measured?

The temperature of a blackbody star can be estimated using its spectral class and peak wavelength of light emitted. It can also be calculated using its luminosity and radius through the Stefan-Boltzmann law.

## 5. How do scientists use the temperature of blackbody stars in their research?

Scientists use the temperature of blackbody stars to study the properties and evolution of stars. By comparing the temperature of different stars, they can determine their age, mass, and other characteristics. It also helps in understanding the formation and evolution of galaxies and the universe as a whole.

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