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>
> "Here we assume a "solar constant" of 1367 W m^-2 (Hartmann 1994),
> and because the incoming solar radiation is one-quarter of this, that
> is, 342 W m^-2., a planetary albedo of 31% is implied."
>
> Stripping this sentence to bare bones we are told that-- 'Because
> the incoming solar radiation is one-quarter of the "solar constant" a
> planetary albedo of 31% is implied', which is of course nonsense. So
> one has to ask what the authors might have had in mind when they wrote
> this.
>
Hi again:
I think that I have this sorted out, and I think part of it is that the
original statement is worded poorly. I think what it should read is:
> Here we assume a "solar constant" of 1367 W m^-2 (Hartmann 1994), and
> because the incoming solar radiation is one-quarter of this, that is,
> 342 W m^-2. A planetary albedo of 31% is used.
There key thing about the solar constant (1367 W per square metrr) is
that it is for a plane that is perpendicular to the sunlight
For the first point, the total power available to the Earth would be
1367 W per square metre multiplied by the circular area of the Earth.
which is pi x radius x radius; or 1367 x PI x r X r.
Obviously not all of the Earth's surface is perpendicular to sunlight.
The total area on which the amount just calculates impinges is the
total surface area of the Earth facing the Sun, which = 4 X PI X radius
x radius.
Therefore on average, the each square metre of the Earth will receive:
1367 x PI x r x r
------------------------------------------------
4 x PI x r x r
= 342 Watts per square metre.
Now for the second point. The The Earth's albedo (or its reflectivity),
as measured by satellites is 0.31, that is 31% of all incoming
radiation is reflected away by clouds, ice, water, etc. In Figure 7,
31% of the 342 Watts per square metre would be 106 Watts per square
metre. The left side of the figure shows 77 W/m^2 being reflected by
the atmosphere, and 30 W/m^2 by the ground for a total of 107 W/m^2
> i.e. no allowance is made for the greater surface area at the top of
> the atmosphere than at the surface of the earth.
The difference thickness of the atmosphere is so small compared to the
radius of the Earth that it can be ignored. If you take 20 km to be the
thickness of the atmosphere, and the radius of the Earth to be 6378 km*
then the surface are at the top of the atmosphere will be increased by
a factor of (6398/6378)^2 or 0.6%
Given that I don't really see any problem unless there is something
that I missed.
Pat
* From the 2007 RASC Observer's Handbook, available from Atlantic News
and The Book Room or on-line at http://store.rasc.ca. I would recommend
this book even if I wasn't the new editor ;-)
========================================================================
==
Patrick Kelly
Director of Computer Facilities
========================================================================
==
Faculty of Architecture and Planning
Dalhousie University
========================================================================
==
PO Box 1000 Stn Central 5410 Spring Garden Road
Halifax, Nova Scotia B3J 2X4 Halifax, Nova Scotia B3J 2X4
Canada Canada
========================================================================
==
Phone:(902) 494-3294 FAX:(902) 423-6672 E-mail:patrick.kelly@dal.ca
========================================================================
==
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<excerpt>
"Here we assume a "solar constant" of 1367 W m^-2 (Hartmann 1994),
and because the incoming solar radiation is one-quarter of this, that
is, 342 W m^-2., a planetary albedo of 31% is implied."
Stripping this sentence to bare bones we are told that-- 'Because
the incoming solar radiation is one-quarter of the "solar constant" a
planetary albedo of 31% is implied', which is of course nonsense. So
one has to ask what the authors might have had in mind when they wrote
this.
</excerpt>
Hi again:
I think that I have this sorted out, and I think part of it is that
the original statement is worded poorly. I think what it should read
is:
<excerpt>Here we assume a "solar constant" of 1367 W m^-2 (Hartmann
1994), and because the incoming solar radiation is one-quarter of
this, that is, 342 W m^-2. A planetary albedo of 31% is used.
</excerpt>
There key thing about the solar constant (1367 W per square metrr) is
that it is for a plane that is perpendicular to the sunlight
For the first point, the total power available to the Earth would be
1367 W per square metre multiplied by the circular area of the Earth.
which is pi x radius x radius; or 1367 x PI x r X r.
Obviously not all of the Earth's surface is perpendicular to sunlight.
The total area on which the amount just calculates impinges is the
total surface area of the Earth facing the Sun, which = 4 X PI X
radius x radius.
Therefore on average, the each square metre of the Earth will receive:
<fontfamily><param>Courier</param> 1367 x PI x r x r
------------------------------------------------
4 x PI x r x r<bold>
</bold></fontfamily>= 342 Watts per square metre.
Now for the second point. The The Earth's albedo (or its
reflectivity), as measured by satellites is 0.31, that is 31% of all
incoming radiation is reflected away by clouds, ice, water, etc. In
Figure 7, 31% of the 342 Watts per square metre would be 106 Watts per
square metre. The left side of the figure shows 77 W/m^2 being
reflected by the atmosphere, and 30 W/m^2 by the ground for a total of
107 W/m^2
<excerpt>i.e. no allowance is made for the greater surface area at the
top of the atmosphere than at the surface of the earth.<color><param>0000,6363,1212</param>
</color></excerpt>
The difference thickness of the atmosphere is so small compared to the
radius of the Earth that it can be ignored. If you take 20 km to be
the thickness of the atmosphere, and the radius of the Earth to be
6378 km* then the surface are at the top of the atmosphere will be
increased by a factor of (6398/6378)^2 or 0.6%
Given that I don't really see any problem unless there is something
that I missed.
Pat
* From the 2007 RASC Observer's Handbook, available from Atlantic News
and The Book Room or on-line at http://store.rasc.ca. I would
recommend this book even if I wasn't the new editor ;-)
<fontfamily><param>Courier</param>
==========================================================================
Patrick Kelly
Director of Computer Facilities
===============================================================