Calculating exoatmosphereic flux and figuring out amount of SW Earth will absorb at an albedo of 0.3? 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
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 $W/m^2$.