<p>I'm not sure what you mean by "heating effect". The amount another body is heated by a star with luminosity $L$ is quantified by its <a href="https://en.wikipedia.org/wiki/Effective_temperature" rel="nofollow noreferrer">effective temperature</a>:
$$T_{eff}=\left(\frac{L(1-a)}{16\pi\sigma D^2}\right)^{1/4}$$
where $a$ is the albedo, $D$ is the distance to the other body, and $\sigma$ is the Stefan-Boltzmann constant. If the luminosity is given, then there is no direct dependence on the size of the star.</p>

<p>The only thing that matters is the luminosity. You stated yourself that the luminosities are the same, which means that the red star in question is likely a <a href="https://en.wikipedia.org/wiki/Red_supergiant" rel="nofollow noreferrer">red supergiant</a> - a member of a group containing many of the largest stars in the universe.</p>

<p>Why, mathematically, is this the case? Well, stellar models can tell us that this is the case, but we can also figure this out via the <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law" rel="nofollow noreferrer">Stefan-Boltzmann law</a> by assuming that the stars are <a href="https://en.wikipedia.org/wiki/Black_body" rel="nofollow noreferrer">black bodies</a>. For an object of radius $R$ and temperature $T$, the luminosity is approximately
$$L=4\pi\sigma R^2T^4$$
If star 1 is a blue supergiant and star 2 is a red supergiant, then, setting the luminosities equal, we have
$$L_1=L_2\to R_1^2T_1^4=R_2^2T_2^4$$
We know that $T_2$ is much less than $T_1$ - possibly by an order of magnitude - so $R_2$ must be much larger for the luminosities to be the same. This matches what we observe and what models predict.</p>

CC BY-SA 3.0
2016-11-29T20:16:21.060Z
2016-12-10T15:29:44.563Z
2016-12-10T15:29:44.563Z
HDE 226868
<p>Because a star's temperature is an indication of the energy passing through each unit of its surface, it follows that for cool red and hot blue stars of identical luminosities, the cooler red star must be considerably smaller so that the energy escapes through a reduced surface area and has a far greater <strong><em>heating effect</strong>.</em> Hence, knowledge of a star's colour and luminosity can reveal its size.</p>

<p>In particular, I can't understand the relation between energy and the heating effect. Of course, I searched for the word "the heating effect" in my own language. Also, I Googled in English. </p>

<p>In this context, what is the heating effect? Could you explain to me? If you give me any help, it will be very useful to me.  </p>

CC BY-SA 3.0
2016-11-29T19:37:43.853Z
If a blue and red star have equal luminosity then (roughly) they give off the same amount of energy per second. Since the red star is cooler, less energy is emitted per unit area. If the total energy is the same the total area of the red star must be larger, not smaller as you state. I have difficulty understanding what you mean by "heating effect". Do you mean the power of the star (amount of energy per second)?
James K
<star><temperature><heat>
The relation between the energy of star and heating effect
Purple Rain Kim
# The relation between the energy of star and heating effect

(Asked by: username_0 on 2016-11-29)

Because a star's temperature is an indication of the energy passing through each unit of its surface, it follows that for cool red and hot blue stars of identical luminosities, the cooler red star must be considerably smaller so that the energy escapes through a reduced surface area and has a far greater ***heating effect***. Hence, knowledge of a star's colour and luminosity can reveal its size.

In particular, I can't understand the relation between energy and the heating effect. Of course, I searched for the word "the heating effect" in my own language. Also, I Googled in English.

In this context, what is the heating effect? Could you explain to me? If you give me any help, it will be very useful to me.

**Tags:** star, temperature, heat

> **Comments:**
> **username_1**: If a blue and red star have equal luminosity then (roughly) they give off the same amount of energy per second. Since the red star is cooler, less energy is emitted per unit area. If the total energy is the same the total area of the red star must be larger, not smaller as you state. I have difficulty understanding what you mean by "heating effect". Do you mean the power of the star (amount of energy per second)?  

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## 1 answers

### Answer by username_2 on 2016-11-29. Score: 5

I'm not sure what you mean by "heating effect". The amount another body is heated by a star with luminosity $L$ is quantified by its [effective temperature](https://en.wikipedia.org/wiki/Effective_temperature):
$$T\_{eff}=\left(\frac{L(1-a)}{16\pi\sigma D^2}\right)^{1/4}$$
where $a$ is the albedo, $D$ is the distance to the other body, and $\sigma$ is the Stefan-Boltzmann constant. If the luminosity is given, then there is no direct dependence on the size of the star.

The only thing that matters is the luminosity. You stated yourself that the luminosities are the same, which means that the red star in question is likely a [red supergiant](https://en.wikipedia.org/wiki/Red_supergiant) - a member of a group containing many of the largest stars in the universe.

Why, mathematically, is this the case? Well, stellar models can tell us that this is the case, but we can also figure this out via the [Stefan-Boltzmann law](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law) by assuming that the stars are [black bodies](https://en.wikipedia.org/wiki/Black_body). For an object of radius $R$ and temperature $T$, the luminosity is approximately
$$L=4\pi\sigma R^2T^4$$
If star 1 is a blue supergiant and star 2 is a red supergiant, then, setting the luminosities equal, we have
$$L\_1=L\_2\to R\_1^2T\_1^4=R\_2^2T\_2^4$$
We know that $T\_2$ is much less than $T\_1$ - possibly by an order of magnitude - so $R\_2$ must be much larger for the luminosities to be the same. This matches what we observe and what models predict.


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