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- Newsgroups: sci.astro
- Subject: Re: Deflecting P/S-T
- Message-ID: <warhil-171192094827@warrenhill.ntu.edu.au>
- From: warhil@post.ntu.edu.au (Warren Hill)
- Date: 17 Nov 92 09:54:16 +0900
- Followup-To: sci.astro
- References: <1992Nov10.145458.25432@cam-orl.co.uk> <10NOV92.17034867@wl.aecl.ca>
- Distribution: world
- Organization: Northern Territory University
- Nntp-Posting-Host: warrenhill.ntu.edu.au
- Lines: 36
-
- In article <10NOV92.17034867@wl.aecl.ca>, dyckg@wl.aecl.ca wrote:
- >
- > dg@cam-orl.co.uk (Dave Garnett) writes:
- > >If you applied a constant thrust perpendicular to the
- > >track for 1 year (31e6 seconds), then the acceleration
- > >required from Newton is:-
- > >
- > > a = 2 . 8000 . 1000 / (31e6)^2
- > > = 16e-9 m/s^2
- > >
- > >Hence the force required is = mass . accel
- > >
- > > = 26e6 kgf
- > > = 2600 tonnes !
- >
-
- {SOME STUFF FROM GARY DELETED HERE}
-
- > Gary
- >
-
- I missed the original posting, but would like to offer the following.
- If mass = kg, and acceleration = m/s^2, then from the SI units we
- derive F = m.a as newtons (N), not kgf !
-
- Then, if this is what was intended, the answer should read 26e6 newtons,
- or, 26 meganewtons (MN) of thrust. As a comparison, I think that one 727
- engine can deliver about 400 kN of thrust ( leading to the somewhat silly
- conclusion that the earth could be "saved" by 22 Boeing 727s ). Maybe
- someone could give us a comparison with the thrust generated by the shuttle
- (I don't have these figures at my finger tips).
-
- Regards
-
- Warren Hill
- Northern Territory University
-