<p>I might as well write up my comment into an answer.</p>

<p>One of my new favorite tools in my equations toolkit is the formula for <a href="https://en.wikipedia.org/wiki/Effective_temperature" rel="nofollow">effective temperature</a>:
$$T=\left( \frac{L(1-a)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$
where $T$ is the temperature of the planet, $L$ is the star's luminosity, $a$ is the planet's albedo, $\sigma$ is the <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant" rel="nofollow">Stefan-Boltzmann constant</a>, and $D$ is the distance between the star and the planet. To find the star's luminosity, simply solve for $L$:
$$L=\frac{T^4 16 \pi \sigma D^2}{1-a}$$
where all the terms are for the first planet (Mars). Now, knowing $D$ for Earth, solve for $T$ on Earth.</p>

<p>All you have to know for this that <em>isn't</em> in the question is albedo. That could be tricky, because it's not given. You <em>could</em> assume that it's the same for both planets - and so solve not for $L$ but for $L(1-a)$ - but that might not be better than an approximation.</p>

CC BY-SA 3.0
2015-01-18T18:55:26.943Z
2015-01-18T18:55:26.943Z
HDE 226868
<p>If you use no additional data you assume that in all respects other than distance from the star the planets are the same. Then assuming thermal equilibrium the energy radiated bt the planet is proportional to the stellar intensity at the planets distance. Also assume that the Stefan–Boltzmann law applies to the energy radiated from the planets (so it's proportional to the fourth power of the temperature). The intensity of the stellar radiation at the planets distance is inversely proportional to the square of the distance. </p>

<p>This gives us the equation:
$$
\frac{T_1^4}{T_2^4}=\frac{I_1}{I_2}=\frac{R_2^2}{R_1^2}
$$
which simplifies to:
$$
T_1=\sqrt{\frac{R_2}{R_1}}T_2
$$</p>

<p>Now using the data provided we get:
$$
T_1=\sqrt{1.524}\times 210 \approx 259\ \text{K}
$$</p>

<p>Which is of course much less than the average temperature of the Earth since not all else is the same in reality.</p>

<p>One of the significant differences is the Earth has a more massive atmoshpere than Mars, and it has a significant grenhouse effect on the temprature of the Earth. Also the Earth has a higher albedo than Mars which would reduce the equilibrium temperature from that calculated. I expect there are other factors that also contribute but I will leave those to others to mention.</p>

CC BY-SA 3.0
2015-01-21T23:49:30.343Z
yeah, the result is just the same as from the formula for luminosity. thank you very much!
Gerda
CC BY-SA 3.0
2015-01-20T12:22:56.580Z
2015-01-20T12:59:47.380Z
2015-01-20T12:59:47.380Z
Conrad Turner
<p>I am to compute the temperature of planet--in fact I've already found out it's Earth--knowing only: </p>

<ol>
<li>the surface temperature of Mars (210 K) and its distance from Sun (1.524 AU)</li>
<li>and of course distance of Earth from Sun (1 AU). </li>
</ol>

<p>I have no data concerning the temperature of Sun, surfaces of the planet, etc. What formula should I use? Thank you.</p>

CC BY-SA 3.0
2015-01-18T18:06:00.563Z
So you don't know albedo? With it, you could use the [effective temperature](https://en.wikipedia.org/wiki/Effective_temperature) formula.
HDE 226868
CC BY-SA 3.0
2015-01-18T18:29:58.943Z
the problem is that I don't have luminosity. I can assume that it's a blackbody
Gerda
CC BY-SA 3.0
2015-01-18T18:31:31.370Z
You can calculate the luminosity because you have the temperature on the planet already. You sort of go in reverse.
HDE 226868
CC BY-SA 3.0
2015-01-18T18:44:11.907Z
oh yes, you're right. I haven't seen that. thanks very much and have a nice evening!
Gerda
<solar-system><temperature>
Finding the temperature of Earth from temperature of Mars and its distance from Sun
Gerda
# Finding the temperature of Earth from temperature of Mars and its distance from Sun

(Asked by: username_0 on 2015-01-18)

I am to compute the temperature of planet--in fact I've already found out it's Earth--knowing only:

1. the surface temperature of Mars (210 K) and its distance from Sun (1.524 AU)
2. and of course distance of Earth from Sun (1 AU).

I have no data concerning the temperature of Sun, surfaces of the planet, etc. What formula should I use? Thank you.

**Tags:** solar-system, temperature
> **Comments:**
> **username_1**: So you don't know albedo? With it, you could use the [effective temperature](https://en.wikipedia.org/wiki/Effective_temperature) formula.  
> **username_0**: the problem is that I don't have luminosity. I can assume that it's a blackbody  
> **username_1**: You can calculate the luminosity because you have the temperature on the planet already. You sort of go in reverse.  
> **username_0**: oh yes, you're right. I haven't seen that. thanks very much and have a nice evening!  

---

## 2 answers

### Answer by username_1 on 2015-01-18. Score: 2 | Accepted answer

I might as well write up my comment into an answer.

One of my new favorite tools in my equations toolkit is the formula for [effective temperature](https://en.wikipedia.org/wiki/Effective_temperature):
$$T=\left( \frac{L(1-a)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$
where $T$ is the temperature of the planet, $L$ is the star's luminosity, $a$ is the planet's albedo, $\sigma$ is the [Stefan-Boltzmann constant](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant), and $D$ is the distance between the star and the planet. To find the star's luminosity, simply solve for $L$:
$$L=\frac{T^4 16 \pi \sigma D^2}{1-a}$$
where all the terms are for the first planet (Mars). Now, knowing $D$ for Earth, solve for $T$ on Earth.

All you have to know for this that *isn't* in the question is albedo. That could be tricky, because it's not given. You *could* assume that it's the same for both planets - and so solve not for $L$ but for $L(1-a)$ - but that might not be better than an approximation.


---

### Answer by username_2 on 2015-01-20. Score: 1

If you use no additional data you assume that in all respects other than distance from the star the planets are the same. Then assuming thermal equilibrium the energy radiated bt the planet is proportional to the stellar intensity at the planets distance. Also assume that the Stefan–Boltzmann law applies to the energy radiated from the planets (so it's proportional to the fourth power of the temperature). The intensity of the stellar radiation at the planets distance is inversely proportional to the square of the distance.

This gives us the equation:
$$
\frac{T\_1^4}{T\_2^4}=\frac{I\_1}{I\_2}=\frac{R\_2^2}{R\_1^2}
$$
which simplifies to:
$$
T\_1=\sqrt{\frac{R\_2}{R\_1}}T\_2
$$

Now using the data provided we get:
$$
T\_1=\sqrt{1.524}\times 210 \approx 259\ \text{K}
$$

Which is of course much less than the average temperature of the Earth since not all else is the same in reality.

One of the significant differences is the Earth has a more massive atmoshpere than Mars, and it has a significant grenhouse effect on the temprature of the Earth. Also the Earth has a higher albedo than Mars which would reduce the equilibrium temperature from that calculated. I expect there are other factors that also contribute but I will leave those to others to mention.

> **Comments:**
> **username_0**: yeah, the result is just the same as from the formula for luminosity. thank you very much!  

---

