Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ames!pasteur!ucbvax!decwrl!sun!pitstop!sundc!seismo!uunet!nih-csl!jim
From: jim@nih-csl.UUCP (jim sullivan)
Newsgroups: sci.space
Subject: Re: approaching "C"
Message-ID: <901@nih-csl.UUCP>
Date: 3 Feb 89 16:01:08 GMT
References: <Added.UXtbE9y00Ui38CnU9S@andrew.cmu.edu> <EXtpzNy00WB9E0jnx8@andrew.cmu.edu> <10116@ut-emx.UUCP>
Reply-To: jim@nih-csl.UUCP (jim sullivan)
Organization: NIH-CSL, Bethesda, MD
Lines: 29

In article <10116@ut-emx.UUCP> ethan@ut-emx.UUCP (Ethan Tecumseh Vishniac) writes:
>In article <EXtpzNy00WB9E0jnx8@andrew.cmu.edu>, rg20+@andrew.cmu.edu (Rick Francis Golembiewski) writes:
>> Here's a question I've alaways wondered about relativity:
>> Suppose there are two space ships, one going at .6 C (an attainable
>> theoretical velocity)  and one going .6 in the opposite direction,
>> what would observers inside each see?  
>
>Each one would see the other receding at a velocity of 1.2/1.36 times
>c.  Velocities do not add in the simple way you are used to.  The 
>correct formula to apply involves dividing the sum by one plus the
>product of the velocities (as fractions of c). 
>

	Last night I was reading Calder's book "Einstein's Universe"
	which I've said in past postings (maybe too many times) is
	a good laymans book on this subject.  It gave an example
	which I must share:

	If you are stationary and ship A is moving at .75*c 
	away from you and ship B is doing the same but in the 
	opposite direction, you would think that ship A would
	see ship B moving at 1.5*c.  But you can communicate
	with both ship A and ship B.  Therefore, both ships should be
	able to communicate with one another and they can because
	as shown in the previous posting, each ship's relative speed
	to the other would be (1.5/(1+(.75*.75)))*c = (1.5/1.56)*c = ~.96*c

	Just had to share this...   Jim
				    jim@alw.nih.gov