Re: Venus's atmosphere/afterthought

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

I'm confused again.

From: <http://hyperphysics.phy-astr.gsu.edu/hbase/solar/venusenv.html>
> "The mass of the Venus atmosphere is about 90 times that of the Earth's

> atmosphere. 90% of the Earth's atmosphere is within 10 km of the 
> surface, whereas you have to go to 50 km to capture 90% of the 
> atmosphere of Venus."

A factor of five?  How do they know this?

Since the molecular weight on Venus is about 1.5 times that on Earth,
g is 0.9 times that on Earth and the air temperature is about 2.5 times
that on Earth*, I'd guess that the height on Venus should be multiplied
by a factor of 2.5/(0.9*1.5) or about 2, not 5.

To check my logic, I checked the pressure profile for Venus.
<http://www.ebicom.net/~rsf1/vel/1918vpt.htm> shows
that at 50 km, Venus' atmosphere has a pressure of about 1 bar.
Since the surface pressure is 91 bars (according to
<http://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_british.htm
l>)
This implies that 99%, not 90%, of the atmosphere is within 50 km
of the surface.

To double-check my logic, I checked the height at which the
pressure on Venus is 10% of its surface value.  That appears to be
about 30 km.  On Earth, the height is about 16 km.  This is a factor
of 2.

Hmmm...am I doing something wrong or the factor of 5 incorrect?

*Note: I initially estimated the temperature factor by taking the
Average temperature between the surface and 50 km (Venus) or 10 km
(Earth). I get (300+750)/2 for Venus and (223+288)/2 for Earth, or
a factor of 2.  However, I wanted to err on the side of a large
factor and so I upped it to 2.5.

____________________________________________________
Robert Cohen; 570-422-3428; www.esu.edu/~bbq
East Stroudsburg University; E. Stroudsburg, PA 18301