0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I tried to cleanup everything to make it more readable (Mathjax formatting 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 $\endgroup$ – JeopardyTempest May 10 '18 at 15:20
1
$\begingroup$

Agree with Ash, 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.

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 $W/m^2$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do think the answer would then be 247.17K? and then maybe the Wm-1 was an error? $\endgroup$ – Foo Fighter May 10 '18 at 0:07
  • 1
    $\begingroup$ @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. $\endgroup$ – JeopardyTempest May 11 '18 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.