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Date: Mon, 8 Aug 88 01:04:01 PDT
From: Ted Anderson <ota@angband.s1.gov>
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To: Space@angband.s1.gov
Reply-To: Space@angband.s1.gov
Subject: SPACE Digest V8 #318
SPACE Digest Volume 8 : Issue 318
Today's Topics:
Re: Are these postings useful?
Contractor selected for solid-fueled rocket motor bondline study (Forwarded)
From article <1068@cfa.cfa.harvard.EDU>, by wyatt@cfa.harvard.EDU (Bill Wyatt):
> The above is correct except I think you have to emphasize that the
> escape velocity varies with the square root of the distance from the
> Earth's (or other gravity well's) center. Thus, if you have sufficient
> thrust to move away from the Earth at 55 mph, I calculate that
> somewhere in the vicinity of Jupiter's orbit you'll be at the escape
> velocity of Earth. After that point, even stopping the thrust will not
> let you fall back to Earth (assuming no perturbations, etc.), although
> you'd presumeably orbit or fall into the Sun!
Thanks for the clarification, Bill. You're quite right, at 9 AU ( about
the orbit of Saturn) 55 mph is indeed escape velocity from Earth. But
what happens if the thrust stops at that point? You're not travelling at
55 mph relative to the Sun..
Either: the problem implies one travels in a straight line with constant
velocity relative to an inertial frame instantaneously fixed in the
center of the Earth at the moment of takeoff (remember that at 55 mph
the Earth has made 1700 revolutions of the Sun by the time you get to
the point in question!) In this case one is no longer travelling at 55
mph relative to the Earth, but relative to where the Earth was.
Or: You keep a constant 55 mph velocity relative to the center of the
Earth at all times. This is very tricky as the Earth is accelerating
around the Sun. In either case, you do not have the local orbital
velocity relative to the Sun, you have the Earth's Keplerian orbital
velocity.
Since the escape velocity at a given point is root 2 times the orbital
velocity, and both decrease with the root of the distance from the
central mass, you just have to double your distance from the central
mass while keeping the same velocity to achieve escape; so you have
escape velocity relative to the Sun by the time you get to 2 AU.
You're headed out into the Great Unknown.
Of course, the point you make is that we shouldn't be trying to keep our
velocity low anyway, since that's not what costs..it's only relative.
The trick is to keep your energy requirements low, since that's what you
pay for.
Jonathan McDowell.
------------------------------
Date: Sun, 31 Jul 88 14:09:31 CDT
From: sedspace@doc.cc.utexas.edu (Steve Abrams)
Posted-Date: Sun, 31 Jul 88 14:09:31 CDT
Subject: Re: Solar Sails
In V8 #306, Eric "TheBoo" Bazan (eric@shorty.cs.wisc.edu) writes:
>My question is this: just how does the sun 'push' against the sail? Is it the
>solar wind of charged particles (protons and electrons), or the actual
>photonic flux, or both?
Since the wave and particle paradigms for electromagnetic radiation are
equivalent, you can certainly think of "material" photons bouncing off a sail
film and, thereby, transferring momentum to the sail. However, I assume you
want to view it from the wave perspective.
With the propagation of electromagnetic radiation (light energy) there
is a flow of energy. However, you've probably also heard that mass and energy
are equivalent (E=mc^2). Therefore, you have a mass-equivalent entity traveling
with a velocity...the product of mass and velocity is linear momentum. More mathematically, the "momentum density", g, is equal to a constant (the product of
the permeability and permittivity of free space) times the "Poynting vector"
(since we're in a vacuum). The Poynting vector is simply a vector defining the
energy flow of the light (EM radiation). The value for this Poynting vector
is just the solar constant you mention, 1.353 kW/m^2.
The physical process is (loosely) that the photon is "absorbed" by an
atom in the sail film. The energy is used to raise an electron's energy level.
That electron then spontaneously decays thereby emitting another photon *with
the same energy* as the original photon but in the opposite direction. In each
process of absorption and re-emission, momentum is transferred in the same
direction to the sail film. The energy of the photon doesn't change (a scalar
quantity) and the magnitude of the velocity of the photon doesn't change *BUT*
the direction changes. By Newton's Third Law (Action/Reaction), momentum is
transferred to the sail from the photon.
I realize that this won't satisfy purists, but it should be sufficient
for someone with a poor physics background.
Incidentally, the book Eric is probably referring to is _Starsailing_
by Louis Friedman. It is an excellent introductory, survey-type book on
the subject. Does anyone know if a more technical, mathematically-oriented
sequel is in the works?
Ad Astra,
Steve Abrams
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