<p>The <span class="math-container">$\mathrm{\frac{S}{4}}$</span> represents the area- and time-averaged incident solar flux and the whole term on the LHS represents solar flux emitted by the planet.</p>
<blockquote>
<p>The factor of 1/4 comes from the fact that only a single hemisphere is lit at any moment in time (creates a factor of 1/2), and from integrating over angles of incident sunlight on the lit hemisphere (creating another factor of 1/2). (<a href="https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature" rel="nofollow noreferrer">Wikipedia</a>)</p>
</blockquote>

CC BY-SA 4.0
2022-01-28T06:09:35.587Z
2022-01-28T06:16:08.003Z
2022-01-28T06:16:08.003Z
Nilay Ghosh
<p>It's simple geometry:</p>
<p>A planet absorbs energy from its star with its geometric cross-section <span class="math-container">$\pi r^2$</span>.</p>
<p>The same planet re-radiates the energy via its complete surface <span class="math-container">$4\pi r^2$</span>.</p>
<p>In equilibrium, in the incoming energy equals the re-radiated energy:</p>
<p><span class="math-container">$$
E_{in} = E_{out}\\
\pi r^2 \cdot S \cdot (1-A) = 4\pi r^2 \sigma T^4
$$</span>
Divide by the surface area of the planet and you arrive at
<span class="math-container">$$
\frac{S}{4}(1-A) = \sigma T^4
$$</span></p>

CC BY-SA 4.0
2022-01-28T10:59:26.537Z
2022-01-28T10:59:26.537Z
planetmaker
<p>This is an equation my professor wrote to calculate the equilibrium surface temperature of a planet:<span class="math-container">$\frac{S}{4}(1-A) = \sigma T^4$</span>. (A is the albedo of the planet and S is the intensity of starlight at that distance). <br>Why is there a  factor of <span class="math-container">$\frac{1}{4}$</span> on the left hand side?</p>

<temperature><planetary-science>
Why does the factor of $\frac{1}{4}$ appear in this equation?
Ambica Govind
# Why does the factor of $\frac{1}{4}$ appear in this equation?

(Asked by: username_0 on 2022-01-28)

This is an equation my professor wrote to calculate the equilibrium surface temperature of a planet:$\frac{S}{4}(1-A) = \sigma T^4$. (A is the albedo of the planet and S is the intensity of starlight at that distance).   
Why is there a factor of $\frac{1}{4}$ on the left hand side?

**Tags:** temperature, planetary-science

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## 2 answers

### Answer by username_1 on 2022-01-28. Score: 3 | Accepted answer

The $\mathrm{\frac{S}{4}}$ represents the area- and time-averaged incident solar flux and the whole term on the LHS represents solar flux emitted by the planet.

> The factor of 1/4 comes from the fact that only a single hemisphere is lit at any moment in time (creates a factor of 1/2), and from integrating over angles of incident sunlight on the lit hemisphere (creating another factor of 1/2). ([Wikipedia](https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature))


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### Answer by username_2 on 2022-01-28. Score: 4

It's simple geometry:

A planet absorbs energy from its star with its geometric cross-section $\pi r^2$.

The same planet re-radiates the energy via its complete surface $4\pi r^2$.

In equilibrium, in the incoming energy equals the re-radiated energy:

$$
E\_{in} = E\_{out}\\
\pi r^2 \cdot S \cdot (1-A) = 4\pi r^2 \sigma T^4
$$
Divide by the surface area of the planet and you arrive at
$$
\frac{S}{4}(1-A) = \sigma T^4
$$


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