I stumbled at this point. The Moon is so low-mass (relatively) and
close, and the distance to the Earth-Sun L1 point is so
large, that a trajectory from the Moon's surface to L1 would be
essentially the same as a trajectory from the Earth to L1. Nobody
has proposed the latter, AFAIK.
As for a stable orbit at Earth-Sun L1, all the debris in the Solar
System has had 4.5 billion years to drift there and stay, if that
were true. Nothing's visible.
Comments?
>> Does anyone know if a ballistic trajectory exists directly from the
>> Moon’s surface to the Earth-Sun L1 Lagrange point?
>
>I stumbled at this point. The Moon is so low-mass (relatively) and
>close, and the distance to the Earth-Sun L1 point is so
>large, that a trajectory from the Moon's surface to L1 would be
>essentially the same as a trajectory from the Earth to L1. Nobody
>has proposed the latter, AFAIK.
The trajectory is similar.
But, from the Moon, one is starting from a place with a quarter of the
radius, a sixth of the surface gravity, no atmosphere, no residents, and
a sufficient supply of non-volatile material there for the picking.
Anything arriving at or near any Lagrange point with a non-negligible
speed has a strong tendency to depart with a similar speed. Anything
arriving with a negligible speed either has propulsion or takes too long
to get there. Approximately.
--
(c) John Stockton, nr London, UK. ?@merlyn.demon.co.uk Turnpike v6.05 MIME.
Web <URL:http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.
Nature has recently provided a sort of yard stick as to dust quantity
effects on climate,
http://expo.edu.ph/pinatubo/index.html
That seriously affects the DVI, (Dust Veil Index),
http://www.geo.mtu.edu/volcanoes/vc_web/overview/o_sc_volcano_climate.html
Personal experience: In 1991 we were wintering in central
Ontario, and normally used ~6 cords of wood / year, with
1 cord back-up and 250 gallons of heating oil emergency
reserve if the woodstove malfunctioned.
We ended up burning 9 cords that winter.
Daily high's stayed below -20F to 0F for weeks.
By putting a fist at arm's length in front of the day-time Sun,
the dust was quite apparent in the sky region around it.
Regards
Ken
Oneandonly1 wrote:
> I stumbled at this point. The Moon is so low-mass (relatively) and
> close, and the distance to the Earth-Sun L1 point is so
> large, that a trajectory from the Moon's surface to L1 would be
> essentially the same as a trajectory from the Earth to L1. Nobody
> has proposed the latter, AFAIK.
>
> As for a stable orbit at Earth-Sun L1, all the debris in the Solar
> System has had 4.5 billion years to drift there and stay, if that
> were true. Nothing's visible.
>
> Comments?
>
Must stop hand... must stop hand from getting near the keyboard...
"No precious, we _wants_ to post to the moderated newsgroup, _don'ts we_?"
L1, L2, and L3 are unstable; L4 and L5 are theoretically stable, so
things sent there will remain in the same spot, rotating in a small
orbit around the center of the L point as the Sun-Earth-Moon geometry
changes during any given month.
Pat
I understand L1 is unstable, in fact I’m depending on it, since I
intend the process to be reversible and this requires that the dust
disperse over time. Ejecting dust for global cooling would require an
ongoing process until warming was fixed by other means. Ideally the
dust would then disperse and avoid the risk of creating an ice-age.
Dispersal would involve dust falling to the Sun, Earth and Moon (and
perhaps the inner planets). A small quantity might find stable orbits
around the Earth and Moon (though not L1) but I suspect even this
would disperse under the action of the solar wind.
In addition to having low gravity the Moon also has an orbital
velocity sufficient to resist the Earth’s gravity, so avoiding the
dust falling directly to Earth should simply be a question of pointing
the ejection nozzle in the right direction.
The key issue concerns how long dust can be kept between the Sun and
Earth in a free trajectory originating from the Moon’s surface. If
the period is too short then the amount that needs to be ejected will
be impossibly large. If the period is too long then we risk losing
control and over-cooling the planet. The chances of a trajectory
existing that is just right is perhaps slim and of course I don’t know
what “just right” is at this stage. But I still think the question is
worth asking, does anyone know the answer?
John Hampson
It's actually solar radiation pressure that's relevant for dust.
Note the difference between comet dust tails and ion tails. How big
an effect radiation pressure is, relative to gravity, depends on
particle size.
It occurs to me that by making the particles the right size, one
might be able to launch the material in a cannister on a fast
trajectory, disperse it at L1, and allow radiation pressure to slow
the dispersed dust down so it remains in place for a useful
duration. I have no quantitative idea what this would involve, and
maybe it's silly.
Depending on the quantity of dust needed -- and I haven't calculated
that -- it isn't obvious why launching from the Moon would be better
than launching from Earth. To get the benefit of the lower gravity
well, you have to create a massive infrastructure on the lunar
surface.
--
Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)
You're right : it is.
If radiation pressure is enough to retard particles to a stop, it is
also sufficient to accelerate them back out. Unless your particles are
more-or-less disc shaped, travel in the axial direction, and turn
sideways on arrival - which more or less defeats the object of the
exercise - or are clever enough to be reflective when slowing, and
merely slightly dispersive when stopped.
Note : there's no need to absorb the light which would have hit the
Earth; it just needs to be deflected a little. A thin converging
Fresnel lens with a focal length a fraction of the Earth-lens distance
will do that, and the refraction will tend to keep the lens stretched
out.
For radiation pressure to match solar gravity, the mass of a perfectly-
reflecting disc-shaped particle would have to be about 1.6 grams per
square metre, i.e. rather thin. Neglecting perturbations, a reflecting
body of that thickness would be able to hover over the Sun, remaining on
a line of fixed absolute direction from the Sun, at any altitude. See
below.
If radiation pressure is significant, the point of unstable equilibrium
is no longer at L1. Since both radiation and gravity are inverse-
square, the radiation in effect reduces big G for the sun-particle
interaction; the particle has to be correspondingly nearer to the sun in
order to be in a 365.242...-day orbit and remain between Sun and Earth.
Indeed. Particles will have to be replenished on the deceleration
time scale, whatever that is. I won't be surprised if that makes the
idea unfeasible, but I haven't done the calculation.
> For radiation pressure to match solar gravity, the mass of a perfectly-
> reflecting disc-shaped particle would have to be about 1.6 grams per
> square metre,
If you want particles to remain at L1 for a long time, you want
radiation force to be negligible compared with gravitational force.
As you wrote, that may conflict with using radiation pressure for
deceleration.
Lets throw some numbers at this problem:
* How to get from lunar surface to earth-sun L1.. trivial
A linear boost of some 2.6 km/s will do the job, taking about a month to
get there, at very low resultant velocity.
* How much dust will you need?
To block just ONE PERCENT of sunlight, you need to block some 1.35
million square kilometers.
Assuming you grind down your regolith to about cement-dust scale (10 nm
grain size), you will need some 53 million metric tons. In addition, at
this dust size, solar radiation pressure is quite significant, imparting
some 250 m/s of acceleration per day. If you plan your trajectory well,
you could have your dust linger in a usefull region for up to a week, so
you only need to be able to sling 200 million tons of material delivered
per month. Not too practical.
Increasing the dust particle size fixes your solar pressure problem, but
increases the initial mass required as the cube of the particle size..
1mm grains would be stabile, but mass required is some 4.6e12kg
For comparison: the cooling effect caused by mount Pinatubo's eruption
ejected some 500 million tons of dust and, more significant to the
cooling issue, some 20 million tons of SO2. The earth's
gravity+atmosphere neatly contains this material from scattering, until
normal weather effects wash them from the air, which takes some months.
Conclusion: If you want to block the sunlight from reaching earth
surface, you need to keep your solution close to earth. On or just above
the surface, like volcanoes do, or at most in LEO where you have control
over its location. And given the sheer mass required for any meaningfull
shading, space is just not an option.
Good post. I think your conclusions are correct.
Though the idea of using lunar materials propelled
from it's surface is always interesting, such as,
http://en.wikipedia.org/wiki/Mass_driver
Part of the idea floated in Project Constellation is
to use the Moon as a springboard to Mars and
beyond, and I've read a few ideas on the subject
that seem borderline sci-fi, or economically harsh.
Backup to 1975, and consider the state of the
PC (personal computer), and there was a lot of
hand waving and maybe's, then 25 years later,
from a toy to a mammoth industry.
The Moon might be like that.
Ken
OK. (This assumes a grain blocks its geometric area, but it actually
blocks a factor of two more because of scatttering, but we won't
worry about that.)
> Assuming you grind down your regolith to about cement-dust scale (10 nm
> grain size), you will need some 53 million metric tons.
But I don't get this. The area is 1.35E12 m^2, and the area of one
grain is 8E-17 m^2, so the number of grains needed is 1.7E28. Each
grain has a volume of 5E-25 m^3 and (at density = 3 g cm^-3) mass of
1.6E-27 kg for total mass 27000 tonnes. What am I doing wrong?
Whether this (if correct) is small enough to make the idea feasible,
I don't know.