Aucbvax.6139
fa.space
utcsrgv!utzoo!decvax!ucbvax!space
Wed Feb 10 04:10:18 1982
SPACE Digest V2 #103
>From OTA@S1-A Wed Feb 10 03:43:54 1982
SPACE Digest Volume 2 : Issue 103
Today's Topics:
Re: half-time power from the moon
possible private funding for Shuttle #5
Re: Horseshoe (and other) orbits
Re: Re: sri-unix.707: Horseshoe Orbits
orbital speed
Orbital mechanix, please!
Re: sri-unix.707: Horseshoe Orbits
Re: Mooning Around
Mooning Around...
Mooning Around
Politics of Space
polar lunar solar power
----------------------------------------------------------------------
Date: Mon Feb 8 13:48:55 1982
To: Space at MIT-MC
From: ucbvax!decvax!watmath!jcwinterton at Berkeley
Subject: Re: half-time power from the moon
Source-Info: From (or Sender) name not authenticated.
I would expect that lunar power stations would go to low output
every two weeks (have to use earthshine only?).
------------------------------
Date: 9 Feb 1982 0959-EST
From: MPH at MIT-XX
Subject: possible private funding for Shuttle #5
To: space at mc
The February 12 issue of Science magazine states that the Space
Transportation Company, a group of investment bankers and venture
capitalists, is considering funding the fifth Shuttle. STC is considering
raising one billion dollars privately, and then turning the shuttle over to
NASA (or whomever is operating shuttles in 1986), in return for which STC
would become the sole ticketing and marketing agent for all commercial and
foreign users of the STS.
-------
------------------------------
From: MINSKY@MIT-AI
Date: 02/09/82 12:20:37
MINSKY@MIT-AI 02/09/82 12:20:37
To: space at MIT-MC
The lunar-polar 24 houd power station would be built on a mountain-top.
It is a question of fact: is there a peak with continuous sunlight,
through entire year? If not, how high a tower would it need?
------------------------------
Date: 9 Feb 1982 10:43 PST
From: Ciccarelli at PARC-MAXC
Subject: Re: Horseshoe (and other) orbits
In-reply-to: OTA's message of 07 Feb 1982 0302-PST
To: Space-Enthusiasts at MIT-MC
cc: Ciccarelli @ PARC-MAXC
About "horseshoe" orbits, actually about orbits in general... The higher orbit is
not faster (higher *velocity*) but of higher *energy*. As the trailing moon
(lower, faster, less energy) catches the leading moon (higher, slower, more
energy) it takes some of the leading moon's energy, swapping orbits. It's
counterintuitive -- higher energy does NOT imply higher orbital velocity. I dug
up the formulas for your reference...
---------------
The formula for orbital velocity (circular orbit approximation) is:
V = V0 * SQRT( Earth radius / Orbit radius), where
V0 = 7.86 KM/SEC ("Circular velocity at Earth's surface),
Earth radius = 6400 KM (approx.), and
SQRT is the Square-root operation (of course)
You can check the formula for the three familiar orbit radii; calculate the
velocity using the formula, then see that this velocity gives the right orbital
period [distance = 2 * PI * Orbit radius; divide by velocity to get the period].
1) near-earth: Orbit radius = Earth radius ==> V = V0 [gives 90 min. period]
2) geosynchronous: Orbit radius = 40,000 KM (approx.) [gives 24-hr period]
3) lunar: Orbit radius = 400,000 KM (approx.) [gives 28-day period]
This also gives the expected result of V=0 at very great distances.
---------------
The formula for orbit energy is:
Total Energy = - (G * M1 * M2 / 2 * Orbit Radius), where
G = Newton's gravitational constant, and
M1, M2 = masses (i.e. Earth and satellite)
Note that:
1) the total energy is negative (the physical interpretation is that the orbit is
"bound", i.e. the satellite has less energy than that required to escape).
2) Orbit radius appears in the denominator again, thus the total energy will
become greater (still negative, but closer to zero) as orbit radius increases. At
very great distance, the energy goes to zero (as expected).
[ Source: "Introductory Astronomy and Astrophysics", by Smith and Jacobs ]
/John
------------------------------
Date: 9 Feb 1982 11:17 PST
From: Lynn.ES at PARC-MAXC
Subject: Re: Re: sri-unix.707: Horseshoe Orbits
In-reply-to: ETC!dennis's message of Sun Feb 7 10:51:10 1982
To: Space-Enthusiasts at MIT-MC
cc: Lynn.es
Nice try on the orbital explanation, but it's unfortunately not right (even
ignoring the ground speed business). Both linear velocity and angular velocity
increase for satellites closer to the body they are orbiting. Take the moon (radius
238,000 miles, period 27+ days => 2200 mph, 0.04 rev/day) versus a low earth
orbit satellite (radius 4000 miles, period 1.5 hrs => 17,000 mph, 16 rev/day). The
angular velocity decreases with the 1.5 power of distance, the linear velocity
decreases with the 0.5 power (square root) of distance.
The actual explanation of why adding speed moves a satellite away from the
body is this: With additional speed, the satellite tends to go in more of a straight
line (has less time to fall toward the parent body), and increases its orbital
distance. While increasing its distance, it is slowed by gravitation. It reaches
equilibrium at a greater distance and a slower orbital speed than it originally
had.
This can be viewed another (equivalent) way: an orbiting body, given extra
speed, has too much kinetic energy for that orbit, so it exchanges some of its
kinetic energy for potential energy. The equilibrium is reached when the extra
speed we gave it and some of its original speed are exchanged, ending up quite a
bit higher and moving a little slower than originally.
Point of view is important in understanding the name "horseshoe". Imagine two
horseshoes, of slightly different size, mouth to mouth, on a plate. The plate
spins quickly. As seen by a viewer on the plate, the satellite on the inner orbit
is moving slowly counterclockwise along the smaller horseshoe, while the outer
one is moving slowly clockwise on the larger horseshoe. The spinning of the
plate represents the average rate of revolution of the two satellites, and so the
viewer on the plate sees only the DIFFERENCE between a satellite's orbital speed
and the average (plate's) speed. At encounter, we switch the sizes of the
horseshoes, and the satellites (still seen from our spinning point of reference)
seem to each reverse direction and traverse their (now slightly larger or smaller)
respective horseshoes in the opposite directions.
/Don Lynn
------------------------------
Date: 9 Feb 1982 21:51:13-EST
From: csin!cjh at CCA-UNIX
To: space at mit-mc
Subject: orbital speed
I suppose a lot of people will wake up on this one, but I might as
well put in my nickel's worth. High orbits \are/ slower in linear
velocity than low orbits. Local example (remember, earth orbits are
measured from the surface, so add ca. 3900 miles to these figures):
A satellite at LEO is ca. 150 miles up and has ca. 90-minute orbit;
orbital velocity ca. 140 miles per minute.
A satellite at GEO is ca. 22,300 miles up and has (by definition) a
24-hour orbit; orbital velocity ca. 57 miles per minute.
The moon is around .25e6 miles up, orbits in 28+ days; orbital velocity
ca. 19 miles per minute.
Need more data? Start with Pluto being at 39 AU (earth-orbit radii) with
a period of 200+ years.
All of these figures are out of my head, but date from a grade-school
infatuation with space and so are tolerably accurate. More precise figures
are welcome.
What happens when you add energy to an orbiting body is not that simple;
the only way the body can maintain a stable orbit is by turning all that
energy (and some of its own kinetic) into potential energy, i.e. take a
higher orbit. What the shuttle is doing when it turns around and blasts in
the direction it was going is throwing away enough KE that it can't keep
a stable orbit above the Earth's surface; if you got rid of the atmosphere
and dug a trench it could drop into a stable orbit below the net surface.
C'mon, guys, even Brunner got this right (and used it to make an effective
point in THE SHOCKWAVE RIDER).
------------------------------
Date: 9 February 1982 2232-EST (Tuesday)
From: David.Smith at CMU-10A (C410DS30)
To: space at mit-mc
Subject: Orbital mechanix, please!
Message-Id: <09FEB82 223233 DS30@CMU-10A>
In fact, higher orbits are slower any way you want to measure. But
due to the gravity well, they are still at a higher energy level.
Consider:
f = GMm/r^2
where f is the force exerted between two bodies, G is the gravitational
constant, M is the mass of the primary, m the mass of the satellite,
and r the radius between centers. Assume M >> m.
For a circular orbit, we use
f = ma (force, mass, acceleration)
a = rw^2 (acceleration, radius, angular velocity)
w = v/r (angular velocity, tangential velocity, radius)
Equating gravitational force on the satellite with the force required to
keep it in circular orbit,
GMm/r^2 = mrw^2 = mv^2/r
which produces
w = sqrt( GM/r^3 )
v = sqrt( GM/r )
which clearly shows angular and tangential velocity dropping as radius
rises.
Low earth satellites travel at nearly 18,000 mph; geosynchronous
satellites travel at around 6,000 mph; the moon travels at around
2,000 mph with respect to earth's center.
Consider the task of moving a satellite from LEO to GEO. (I'll
pull a few numbers out of my hat because I'm lazy, but the exact
numbers aren't the point.) Starting at 18,000 mph at 150 miles
up, you burn the rockets to accelerate it to 22000 mph. With
more than circular velocity, the satellite climbs, trading speed
for altitude, until it reaches apogee at geosynchronous height
(around 23,000 miles) with 2000 mph. Since circular velocity
there is 6000 mph, the satellite will drop back. It falls until
it reaches 22000 mph at its next perigee, 150 miles up. At next
apogee (23,000 miles, 2,000 mph), you burn the engines again to
raise the speed to 6,000 mph. This raises the perigee to put the
satellite into synchronous orbit. The satellite has gone from
a 18,000 mph circular orbit to a 6,000 mph one purely by firing
its rockets to increase speed.
- David Smith
------------------------------
Date: 10 February 1982 01:27-EST
From: Robert Elton Maas
Subject: Re: sri-unix.707: Horseshoe Orbits
To: HPLABS!MENLO70!UCBVAX!IHNSS!CBOSG!HARPO!CHICO!DUKE!PHS!DENNIS at MIT-MC
cc: SPACE at MIT-MC
You're mistaken (wrong).
Under inverse-square law, such as gravity, higher orbits actually
travel slower, not faster! If energy is added to a satellite, it rises
into a higher orbit, but more knetic energy is converted into potential
energy than was applied to make it rise to the new orbit and it has
less knetic energy (but much more potential energy) than when it was lower.
Earth-based common sense, if add energy an object travels faster, doesn't
apply in orbital mechanics; add energy and the object ends up in a new
slower orbit. Here's an example if you don't believe me. The moon is
about 225,000 miles from Earth, while geosynchronous satellites are
about 25,000 miles from Earth. Thus the moon has to travel about 10
times as far to get around, but takes about 29 times as long to do it
because it is traveling only about a third as fast in linear velocity.
------------------------------
Date: Tue Feb 9 23:02:53 1982
To: Space at MIT-MC
From: ucbvax!decvax!watmath!bstempleton at Berkeley
Subject: Re: Mooning Around
Source-Info: From (or Sender) name not authenticated.
Rick suggests that a geosynch sps would be in shadow half the time.
where do you get this from? If in shadow at all, it would
only be for a limited time due to bad design.
also, projecting to a station in lunar 'geo'synch orbit is a little silly,
if you think about what the distance of this orbit is.
(Hint: name an object that stays fixed in the lunar sky)
------------------------------
Date: Wed Feb 10 00:26:45 1982
To: Space at MIT-MC
From: ucbvax!decvax!watmath!pcmcgeer at Berkeley
Subject: Mooning Around...
Source-Info: From (or Sender) name not authenticated.
Oops, Brad's right. I goofed. The moon does rotate with respect to the
sun, but unfortunately the postion of a satellite that could take advantage of
that would be in the L5 position - rather further away than practicable.
However, I'm not so sanguine about solving the problem of half time
shadow by better design. The angular diameter of the Earth as seen by an
35,000 KM above the equator position is quite enough - comfortably enough -
to blot out the Sun. Remember, the Moon is large enough to cover the Sun,
and it's 400,000 KM away. Admittedly, the Earth wouldn't block the sun for
12 hours out of every day - six is actually more near correct. A solution would
be to put two SPS in high earth orbit, but this involves taking up not one but
two geostationary slots, which are pretty valuable. There are only 90 all told,
since there has to be a 4 degree separation between any two satellites due to
possible interference. Further, a quick glance at a globe will convince
the military that one of the SPSs will be over hostile territory permanently.
Rick.
------------------------------
Date: Tue Feb 9 20:35:00 1982
To: Space at MIT-MC
From: ucbvax!decvax!watmath!pcmcgeer at Berkeley
Subject: Mooning Around
Source-Info: From (or Sender) name not authenticated.
The Lunar SPS would, of course, deliver power only for half the time.
Of course, this could be said about a geostationary SPS, since it would be
in the shadow of the earth for 12 hours out of 24.
The Lunar SPS proposal has some merits, though :
1) We could keep it stable. Precisely how would we keep a free-floating SPS,
several square kilometers, from tumbling about a planar axis? And if we could,
precisely how do we stress something like that? These really are mundane
questions, but do we know how to do these things?
A lunar SPS, on the other hand, has no such problems. It's merely
a large, flat plain of collectors. We transmit from the surface (probably
Mare Crisium) to a sattelite in Geosynch orbit above the moon, which transmits
to a buddy in High Earth Orbit. The principal advantage is no large,
freestanding structure;
2) The materials are there, or at least we hope they are. Siliates are,
for sure.
3) (A cheap advantage, certainly) There would probably be much more
public support for a Lunar base than for one in High Earth Orbit. The Moon
has always had an emotional appeal that HEO doesn't share. The space program,
like all government programs, depend in the long run on their public support;
therefore, the chances are better that we will be able to build the Lunar
SPS, if it's technically feasible;
Which it might not be. The power-half-the-time problem can be
solved by putting another SPS over on Farside. Another, better question,
which I haven't got the foggiest idea about, is how we transmit the power -
even a laser spreads somewhat over 400,000 KM, and the satellites orbiting moon
and the earth, the ends of this game of celestial pitch - and - catch, will
have a velocity difference between them.
This is further complicated by the 3-sec feedback loop.
To another question - yes, the Moon does orbit in the plane of
the ecliptic, or near enough as to make no difference. This fact, plus
the low tug-of-war ratio for the moon (about .46, as against an empirical
minimum of 30.00 for a true satellite, led Asimov to speculate that the
Earth-Moon system is in fact a binary planet system..
Cheers,
Rick.
------------------------------
Date: 10 February 1982 04:39-EST
From: Jerry E. Pournelle
Subject: Politics of Space
To: Ward at USC-ISIF
cc: SPACE at MIT-MC
What's really needed is a skilled politician who'll undertake to
become the champion for space; someone to lead what is, after
all, the most fundamental revolution since the evoluton of
lungs.
At the L-5 Convention (Los Angeles Airport Hyatt, April
2-4 1982) we're going to try to generate some strategy; and we
might even have the politicians to help out.
------------------------------
Date: 10 February 1982 04:41-EST
From: Jerry E. Pournelle
Subject: polar lunar solar power
To: ucbvax!decvax!duke!cjp at UCB-C70
cc: SPACE at MIT-MC
Those really interested in direct power from the moon ought to
come to the L-5 Convention and hear Criswell on the subject.
Dave Criswell used to have the lunar rocks, and he is a lunar
fanatic.
------------------------------
End of SPACE Digest
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