os.popen3 with windows; help?

[Russell E. Owen]

I'm trying to launch an application from Python 2.3 on Windows. The 
application is "ds9" (an image viewer), and is installed in C:\Program 
Files\ds9\ds9

On unix I just do:
os.popen3("ds9")
and close the returned files and all is good. (I'm not trying to 
communicate with the program via popen3 and so had been using 
os.spawnlp, but that doesn't exist on Windows.)

On Windows os.popen3("ds9") does nothing, and the stderr returned from 
os.popen3 has a message saying the program is unknown.

So I tried being specific:
os.popen3("C:\\Program Files\\ds9\\ds9")
This also fails and the program that is not found is "C:\Program", 
suggesting that the space in "Program Files" is causing the problem. The 
following failed in exactly the same way:
os.popen3("C:\\Program\ Files\\ds9\\ds9")

Any suggestions?

-- Russell

P.S. I installed the app using its Windows binary installer, which 
unpacks a few files in C:\\Program files\. Windows doesn't seem to know 
the program exists; it's not in the task bar, for example. But one can 
double-click it to run it.