Path: utzoo!mnetor!uunet!husc6!tut.cis.ohio-state.edu!bloom-beacon!gatech!uflorida!codas!cpsc6a!atl2!akgua!mtunx!mtune!mtgzz!bba
From: bba@mtgzz.UUCP (XMRJ40000[pjc]-b.banerjee)
Newsgroups: comp.sys.amiga
Subject: Re: JET
Message-ID: <3759@mtgzz.UUCP>
Date: 22 Mar 88 17:17:06 GMT
References: <4578@garfield.UUCP> <9280@sunybcs.UUCP> <4110@dayton.UUCP>
Reply-To: bba@mtgzz.UUCP (XMRJ40000[pjc]-b.banerjee)
Organization: AT&T Information Systems, Middletown NJ
Lines: 64


I can't believe that no one else here was a physics wienie before
becoming a computer wienie. Makes me feel really old.

G's.
---

G   stands  for   the  universal   gravitational  constant.   The
acceleration due to gravity is denoted  by 'g', and has the value
of approx.  32 ft/sec^2.  I guess the  fighter pilots  decided it
sounded sexier to capitalise it.

Escape Velocity.
---------------

This  can  be  calculated  quite easily.  We  use  the  following
equations:

a.	Work = Force * Distance over which the force is exerted.

b.	Kinetic Energy = 0.5 * Mass * Velocity^2

c.	Gravitational Attraction (Force) = GMm/r^2 (a la Newton)

The work required  to move a body distance dr  against the Earths
Gravitational attraction is:

	GMm dr
	---
	r^2

The work required  to escape the gravitational  attraction of the
earth is

	infinity
	 /
	 | GMm dr
	 / --
	 R r^2

where R is the radius of the earth. This simplifies to

	- GMm | infinity
	  --- |
	  r   | R

	= GMm/R	(Where R = radius of the earth).
	= mgR	(g = accel. due to gravity)

Now this must equal the initial kinetic energy, so

	0.5*m*v^2 = mgR

or	v = sqrt(2*g*R).

This is the escape velocity.  The figures are available in any almanac
and can be figured out easily enough.

Regards,


	Binayak Banerjee
	{ihnp4,allegra}!mtgzz!bba
	bba@mtgzz.ATT.COM