<p>Agree with <a href="https://earthscience.stackexchange.com/a/14088/6298">Ash</a>, I think it's a mistake.  Dimensionally:</p>

<p>$$\frac{S}{4}(1−A)$$ is going to have the units of $S$ has because $A$, <a href="http://www.cocorahs-albedo.org/about/" rel="nofollow noreferrer">albedo, is a unitless ratio</a>.</p>

<p>Insolation comes in over an area, so is measured in $W/m^2$ as you've got.  The only way I can see you'd get units of $W/m$ is by multiplying by a distance somewhere, perhaps by integrating its values over a latitudinal distance or something... but that's not really insolation (or exoatmosphereic flux as you call it!) anymore.</p>

<p>No, all values of S should be $W/m^2$.</p>

CC BY-SA 4.0
2018-05-10T15:35:36.877Z
2018-05-10T15:35:36.877Z
JeopardyTempest
<p>Looking at your equations and assuming you've plugged in the right numbers everything looks right, what <em>doesn't</em> look right is question b.; Watts per metre has no place in such an equation it just doesn't fit the question, <strong>all</strong> the units should be expressed and calculated in square metres, and the answer you have in the last section is.</p>

CC BY-SA 4.0
2018-05-10T00:07:43.653Z
Do think the answer would then be 247.17K? and then maybe the Wm-1 was an error?
Foo Fighter
CC BY-SA 4.0
2018-05-11T17:49:32.887Z
@FooFighter energy wouldn't be in units of Kelvin.  A subsequent question could be "what temperature will this create", but question B, despite the error in units, does not appear to be asking this at all.
JeopardyTempest
CC BY-SA 4.0
2018-05-09T18:23:54.277Z
2018-05-09T18:23:54.277Z
Ash
<p>Am working through two problems which I am not sure if I am handling correctly.   Here they are:</p>

<blockquote>
  <p>a. If the Earth decreases its distance from the sun by 0.03
  Astronomical Units (AU), what will be the resulting exoatmospheric
  flux?<br>
  Recall that at present the exoatmospheric flux = 1,370 $Wm^{-2}$</p>
</blockquote>

<p>My solution (use inverse square law):  </p>

<p>Inverse-square law:
$S = s_0(r_0/r)^2$</p>

<p>$S = 1370 \; (1\;AU/0.97\;AU)^2 = 1412.37\;W/m^{-2}$</p>

<blockquote>
  <p>b. Assuming   an  albedo  of  0.3, how much incoming shortwave energy will
  the   Earth absorb on average in $Wm^{-1}$?<br>
  Recall    that    Energy  absorbed    =   $S/4(1-A) Wm^{-2}$</p>
</blockquote>

<p>This confuses me, the difference between $Wm^{-1}$ and $Wm^{-2}$.  What's the differences between Watts per meter to the first and watts per meter squared? Not sure how to combine the two to make the math/metrics come out right.</p>

<p>Going off what I have for part a:</p>

<p>Energy absorbed = $S/4(1-A)Wm^{-2}$<br>
Energy absorbed = $1412.37/4(1-0.3)Wm^{-2} = 247.17$ (is this in $Wm^{-2}$ or $Wm^{-1}$?)</p>

<p>Thank you</p>

CC BY-SA 4.0
2018-05-10T15:20:35.550Z
I tried to cleanup everything to make it more readable ([Mathjax formatting](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) is unfortunately not intuitive to most!).  One thing that you should make sure you're being clear on is that if you use the negative exponent for meters, you don't need the fraction... so your answer for S would either be 1412.37 $W/m^2$ or 1412.37 $Wm^{-2}$, but not 1412.37 $W/m^{-2}$!  Maybe just been a mistake due to the frustrations of having to type all that again and again, but wanted to make sure :-D
JeopardyTempest
<atmosphere><climate-change>
Calculating exoatmosphereic flux and figuring out amount of SW Earth will absorb at an albedo of 0.3?
Foo Fighter
# Calculating exoatmosphereic flux and figuring out amount of SW Earth will absorb at an albedo of 0.3?

(Asked by: username_0 on 2018-05-09)

Am working through two problems which I am not sure if I am handling correctly. Here they are:

> a. If the Earth decreases its distance from the sun by 0.03
> Astronomical Units (AU), what will be the resulting exoatmospheric
> flux?  
> Recall that at present the exoatmospheric flux = 1,370 $Wm^{-2}$

My solution (use inverse square law):

Inverse-square law:
$S = s\_0(r\_0/r)^2$

$S = 1370 \; (1\;AU/0.97\;AU)^2 = 1412.37\;W/m^{-2}$

> b. Assuming an albedo of 0.3, how much incoming shortwave energy will
> the Earth absorb on average in $Wm^{-1}$?  
> Recall that Energy absorbed = $S/4(1-A) Wm^{-2}$

This confuses me, the difference between $Wm^{-1}$ and $Wm^{-2}$. What's the differences between Watts per meter to the first and watts per meter squared? Not sure how to combine the two to make the math/metrics come out right.

Going off what I have for part a:

Energy absorbed = $S/4(1-A)Wm^{-2}$  
Energy absorbed = $1412.37/4(1-0.3)Wm^{-2} = 247.17$ (is this in $Wm^{-2}$ or $Wm^{-1}$?)

Thank you

**Tags:** atmosphere, climate-change

> **Comments:**
> **username_1**: I tried to cleanup everything to make it more readable ([Mathjax formatting](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) is unfortunately not intuitive to most!).  One thing that you should make sure you're being clear on is that if you use the negative exponent for meters, you don't need the fraction... so your answer for S would either be 1412.37 $W/m^2$ or 1412.37 $Wm^{-2}$, but not 1412.37 $W/m^{-2}$!  Maybe just been a mistake due to the frustrations of having to type all that again and again, but wanted to make sure :-D  

---

## 2 answers

### Answer by username_1 on 2018-05-10. Score: 1 | Accepted answer

Agree with [Ash](https://earthscience.stackexchange.com/a/14088/6298), I think it's a mistake. Dimensionally:

$$\frac{S}{4}(1−A)$$ is going to have the units of $S$ has because $A$, [albedo, is a unitless ratio](http://www.cocorahs-albedo.org/about/).

Insolation comes in over an area, so is measured in $W/m^2$ as you've got. The only way I can see you'd get units of $W/m$ is by multiplying by a distance somewhere, perhaps by integrating its values over a latitudinal distance or something... but that's not really insolation (or exoatmosphereic flux as you call it!) anymore.

No, all values of S should be $W/m^2$.


---

### Answer by username_2 on 2018-05-09. Score: 0

Looking at your equations and assuming you've plugged in the right numbers everything looks right, what *doesn't* look right is question b.; Watts per metre has no place in such an equation it just doesn't fit the question, **all** the units should be expressed and calculated in square metres, and the answer you have in the last section is.

> **Comments:**
> **username_0**: Do think the answer would then be 247.17K? and then maybe the Wm-1 was an error?  
> **username_1**: @username_0 energy wouldn't be in units of Kelvin.  A subsequent question could be "what temperature will this create", but question B, despite the error in units, does not appear to be asking this at all.  

---

