Hocus-Pocus Conditional Physics-101

In sci.physics,

wrote
on 17 Mar 2007 12:32:37 -0700
.com:
Apparently our moon's L1 is still fully hocus-pocus rated. However,
moving our moon into Earth's L1 simply doesn't get any crazier, does
it?

It would take some doing. The Moon is currently in a
tilted elliptical orbit about 3.85 * 10^8 m out, and is
moving relative to Earth at about 1 km/s. The Moon's L1
point (distance depends on mass of the bodies) is
2.4 * 10^10 m inward, approximately, if one uses
the Hill sphere approximation:

r_L1/2 =~ d_SE * cuberoot(m_M / (3 * m_E))

where

r_L1/2 is (approximately) the distance between the Earth
and either L1 or L2,
d_SE = distance between Sun and Earth
m_M = Moon mass
m_E = Earth mass

Were this a rocket problem, two burns would be required: one to get the
Moon into an orbit whose apogee is at the right radius, and one to stop
the Moon in its tracks, precisely positioning it at the L1 point.

You'll need to divert at least two astronomical masses to hit the Moon
at exactly the right point and velocity.


Of what's totally crazy is educating common folks into appreciating
those regular laws of physics, and thus accepting upon the best
available science, as such would become the ultimate demise of most
religions, as well as having uncovered our mutually perpetrated cold-
wars, and we simply can't have that sort of outcome, can we.

Asking silly questions, like any of these following, is usually asking
for yet another Usenet butt-load of trouble, or simply lethal loads of
mainstream flak being tossed your way without so much as a stitch of
remorse.

How much h2o does our atmosphere hold? (I've heard it stated as worth
50e12 tonnes)

The atmosphere can be approximated by a spherical shell of air at
uniform pressure about 9 km thick. (The 9 km is called "scale height",
and is equal to kT / (mg). k is Boltzmann's Constant, T is the
temperature, m is presumably molecular mass, and g is the acceleration.)

Given that, I can do the volume easily enough; that's
simply 4 * Pi * (6.378*10^6)^2 * 9000 = 4.6 * 10^18 m^3.
At 41 moles per cubic meter, or 1.2 kg/m^3, that gives a total
atmospheric mass of 5.5 * 10^18 kg. At 1%-4% water vapor that
suggests 5.5 * 10^16 kg or 55 trillion metric tonnes at a minimum, and
220 trillion metric tonnes maximum. So you're a bit dry, but not too
far off.

(The 41 moles/m^3 is a bit magic but can be derived using
the Ideal Gas law.)


How much applied energy is our atmospheric gravity-forced tide worth
these days?

Unknown.


Is our badly soot polluted atmospheric pressure going up, or is it
going down?


It is probably increasing slightly. Hard to say, though.

Wouldn't those nifty little diatoms (if given a chance) once again
manage to save our CO2/GW day?

Depends on a fair number of factors. The diatoms, after all, require
sunlight to photosynthesize.


Unfortunately, most Usenet folks (including myself) are not happy
campers, especially those of us having since realized just how badly
snookered and summarily dumbfounded we've all been taken through the
infomercial cleaners, of having our clocks cleaned by way of whatever
the mainstream status quo had within their silly mindset to do to the
rest of us village idiots.

[rest snipped]

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