Threading a moon
Logan Kearsley
but
the Jovian system makes for a useful example, so:
Would it be feasible to generate electricity on a Jovian moon by
running
conductive cables along lines between the near and far poles to take
advantage of the moon's motion between Jupiter's magnetic field?
And, if so, how much power could you generate on any particular moon?
-l.
------------------------------------
My inbox is a sacred shrine, none shall enter that are not worthy.
Nyrath the nearly wise
> Would it be feasible to generate electricity on a Jovian moon by
> running
> conductive cables along lines between the near and far poles to take
> advantage of the moon's motion between Jupiter's magnetic field?
> And, if so, how much power could you generate on any particular moon?
Dunno. In George Zebrowski's MACROLIFE, the colony
on Calliso generates energy with copper bars.
Each bar has a laser for power transmission.
It is boosted at Jupiter. As the bar cuts the
magnetic lines of force, electricity is
generated. This is converted into laser energy
and beamed back to a receptor at Callisto base.
Eventually it enters Jupiter's atmosphere and
is destroyed, but they are cheap to manufacture.
More are boosted to replace those destroyed.
Logan Kearsley
annoying.
On Oct 28, 4:49 pm, Nyrath the nearly wise
We have discussed that idea before; it seems that it's just a small
matter of engineering to keep the beam properly aligned to make it
work. But, a bar of copper and some beam-generating equipment has a
lot less gravitational potential energy to convert than a whole moon
does, and there are inevitable losses in the transmission. And what do
you do in a million years when you've run out of copper? Using the
mass of a whole moon for your powersat would provide a much more
permanent solution. Io's gravitational potential energy could supply
current world energy needs for over ten billion years, if you can
regulate the power right, and even a small, close-in moon like
Amalthea has quite a bit stored up. It Seems To Me that that would
make an ideal power source for a brown-dwarf colony, if it's really
feasible.
The biggest problem I can think of is that, if you stretch a cable
from front to back, you could get a very high voltage difference
between the two ends... but the leads would be a moon apart, and if
you stretch another cable to connect them to get a current flowing, it
will also be influenced by the primary's magnetic field. Could you
perhaps magnetically shield the return cable? Or maybe build enormous
field-emission electrodes to spew electrons into space on one side and
such them up on the other?
-l.
Mike Williams
Not having read Macrolife, I don't get the concept behind boosting more
bars to replace the old ones.
Generating electricity from a conductor moving through a magnetic field
doesn't generate energy for free. Generating a current causes a force on
the conductor which acts to reduce its speed, and you end up converting
its kinetic energy into electricity. Boosting up a new conductor to
replace one that's fallen into the atmosphere requires at least as much
energy as the kinetic energy that the old one converted into
electricity. So you end up with no more energy than you started with.
Attaching conductors to a moon means that the kinetic energy that gets
used is that of the moon itself, and a moon has a vast amount of KE, so
it won't run out for ages.
--
Mike Williams
Gentleman of Leisure
IsaacKuo
> Would it be feasible to generate electricity on a Jovian moon
> by running conductive cables along lines between the near and
> far poles to take advantage of the moon's motion between Jupiter's
> magnetic field?
As far as I can tell, yes. However, I think you can get much
better performance by putting the tethers in Jupiter orbit.
Put the tethers in Io's L1 or L2 point, to keep outside of
Io's "wake". You're still essentially getting energy from Io
due to the gravitational tugging, but you don't need to do
any tunneling and you can get a lot more exposure to
the ambient plasma.
> And, if so, how much power could you generate on any particular moon?
I don't know, and I'd really like to know.
Isaac Kuo
Logan Kearsley
Launch from the moon, not from Jupiter, so they start out high up with
lots of gravitational potential.
> Attaching conductors to a moon means that the kinetic energy that gets
> used is that of the moon itself, and a moon has a vast amount of KE, so
> it won't run out for ages.
-l.
Logan Kearsley
> On Oct 27, 8:35 pm, Logan Kearsley <chronosur...@gmail.com> wrote:
>
> > Would it be feasible to generate electricity on a Jovian moon
> > by running conductive cables along lines between the near and
> > far poles to take advantage of the moon's motion between Jupiter's
> > magnetic field?
>
> As far as I can tell, yes. However, I think you can get much
> better performance by putting the tethers in Jupiter orbit.
> Put the tethers in Io's L1 or L2 point, to keep outside of
> Io's "wake". You're still essentially getting energy from Io
> due to the gravitational tugging, but you don't need to do
> any tunneling and you can get a lot more exposure to
> the ambient plasma.
How would you keep it stable?
-l.
IsaacKuo
Tidal forces do the hard part for you--namely, keeping the
tether stretched taut and oriented perfectly up/down.
The more interesting part is staying in orbit around the L1
or L2 point. As I understand it, there are stable orbits
which circle around an L1 or L2 point. If you stand on the
moon and look at the lagrange point, the satellite appears
to keep the same distance as the lagrange point, but it
slowly circles around it.
Isaac Kuo
Erik Max Francis
> Generating electricity from a conductor moving through a magnetic field
> doesn't generate energy for free. Generating a current causes a force on
> the conductor which acts to reduce its speed, and you end up converting
> its kinetic energy into electricity. Boosting up a new conductor to
> replace one that's fallen into the atmosphere requires at least as much
> energy as the kinetic energy that the old one converted into
> electricity. So you end up with no more energy than you started with.
>
> Attaching conductors to a moon means that the kinetic energy that gets
> used is that of the moon itself, and a moon has a vast amount of KE, so
> it won't run out for ages.
That would be true if the conductors were being launched and then
electromagnetic interactions were causing them to fall back down.
Instead, what appears to be described is creating the conductors on
Callisto, launching them into Jupiter orbit (i.e., escape velocity from
Callisto), and then letting electromagnetic interactions to sap their
potential energy _with respect to Jupiter_, not Callisto.
Needless to say, escape energy from Callisto is a great deal less than
the potential energy of Jupiter near Callisto's orbit, so that generates
net energy.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM, Y!M erikmaxfrancis
I'll tell them that their daddy was / A good man
-- India Arie
Erik Max Francis
> The more interesting part is staying in orbit around the L1
> or L2 point. As I understand it, there are stable orbits
> which circle around an L1 or L2 point. If you stand on the
> moon and look at the lagrange point, the satellite appears
> to keep the same distance as the lagrange point, but it
> slowly circles around it.
Not sure what you're referring to here. L1 and L2 are never table; L4
and L5 _sometimes_ are.
Logan Kearsley
news:1193674323.1...@22g2000hsm.googlegroups.com...
Those orbits only work in the plane normal to the line between the
moon and
its primary. Magnetic drag sould cause the satelite to drop further
in
towards the primary (or get pushed farther out, if the magnetic field
is
rotating rapidly enough), after which it loses its orbit and goes
flying off
on its own.
I wonder if you might be able to set up a co-orbital arrangement, with
the
satelite starting out in a slightly lower, faster orbit as a
consequence of
magnetic drag, but then getting pulled up by the moon's gravity when
it
catches up with it from behind; periodic gravitational assists,
essentially,
to transfer KE from the moon to the powersat.
Even if that works, though, I still don't like transmission losses.
Erik Max Francis
> And, if so, how much power could you generate on any particular moon?
It's pretty hard to say. The upper limit would be a wire covering the
body's circumference, cutting through perpendicular magnetic field
lines, which would of course result in a ginormous current. The real
practicality would depend you rely on something else to close the
circuit, the moon's rotation, the makeup of the wire, etc.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM, Y!M erikmaxfrancis
I won't pretend / That I intend to stop living
-- Sade
John Schilling
<nyr...@projectrho.com.INVALID> wrote:
>Logan Kearsley wrote:
>> Would it be feasible to generate electricity on a Jovian moon by
>> running
>> conductive cables along lines between the near and far poles to take
>> advantage of the moon's motion between Jupiter's magnetic field?
>> And, if so, how much power could you generate on any particular moon?
>
> Dunno. In George Zebrowski's MACROLIFE, the colony
> on Calliso generates energy with copper bars.
>
> Each bar has a laser for power transmission.
> It is boosted at Jupiter. As the bar cuts the
> magnetic lines of force, electricity is
> generated.
Very little electricity, unless they've somehow got a current return path
in the system. Beaming the power out to Callisto, or wherever, may take
care of conservation of energy, but there's also conservation of charge
to worry about - and once you get any substantial accumulation of charge
at the ends of the bar, there isn't going to be any more current flowing.
The obvious solution of a second conductor parallel to the first, with
the current flowing in the antiparallel direction, won't work on account
of the induced EMF in the "return" conductor will oppose the necessary
return current. No arrangement of conductors moving at constant velocity
through a constant field will have the desired effect, and the scale of
the Jovian magnetic field is going to make it rather inconvenient to try
and sneak power out of the gradients.
The traditional approach to this sort of thing is to route the return
current through the local space plasma. That's easier said than done,
though, especially if you're trying to use a bare metal surface as your
cathode.
--
*John Schilling * "Anything worth doing, *
*Member:AIAA,NRA,ACLU,SAS,LP * is worth doing for money" *
*Chief Scientist & General Partner * -13th Rule of Acquisition *
*White Elephant Research, LLC * "There is no substitute *
*schi...@spock.usc.edu * for success" *
*661-951-9107 or 661-275-6795 * -58th Rule of Acquisition *
Logan Kearsley
news:mbudi3hpnulrmevkp...@4ax.com...
I always assumed that magnetic powersats would make use of vacuum-tube
tech
for that bit- feathery field-effect cathodes, or appropriately coated
thermionic emitters.
Can anything clever be done at the scale of a whole moon (like, maybe
taking
advantage of the varying magnetic field strength around the orbit to
induce
a circular current around the moon?) that couldn't be done with an
artificial satelite, or would you just have to have really huge vacuum
electrodes at each pole to exchange electrons with the local plasma?
Eivind Kjorstad
> That would be true if the conductors were being launched and then
> electromagnetic interactions were causing them to fall back down.
> Instead, what appears to be described is creating the conductors on
> Callisto, launching them into Jupiter orbit (i.e., escape velocity from
> Callisto), and then letting electromagnetic interactions to sap their
> potential energy _with respect to Jupiter_, not Callisto.
Yeah. But I can't imagine it's that impressive. Making copper-bars with
lasers on them will consume energy, as will launching them from Callisto.
Escape-velocity from jupiter is 60km/s or so, so at best, the energy
released is the same as the energy required to accelerate the
copper-bars to 60km/s. Minus the energy used for mining, transporting,
refining etc the copper. Minus the energy for making and installing the
laser. Times the efficiency of the laser and receiver.
One kg of copper could produce 500Kwh, if it fell from infinity and
everything was 100% efficient. In practice I think you'd be lucky to get
10% overall effiency out of this scheme.
At which point it makes sense if 50Kwh is worth more to you than a kg of
copper. Doesn't sound like a good deal to me, even assuming there where
huge amounts of easily accessible copper on Callisto.
*Certainly* doesn't sound anything at all like "huge amounts of
near-free energy".
I think someone was confused and thougth that the strong magnetic fields
of jupiter makes the energy huge, when in reality the energy is limited
not by that, but by the gravitational potential of whatever you drop at
jupiter.
Eivind Kjørstad
Erik Max Francis
> Yeah. But I can't imagine it's that impressive. Making copper-bars with
> lasers on them will consume energy, as will launching them from Callisto.
>
> Escape-velocity from jupiter is 60km/s or so, so at best, the energy
> released is the same as the energy required to accelerate the
> copper-bars to 60km/s. Minus the energy used for mining, transporting,
> refining etc the copper. Minus the energy for making and installing the
> laser. Times the efficiency of the laser and receiver.
60 km/s is a massic energy of 1.8 GJ/kg, which isn't anything to cough at.
> *Certainly* doesn't sound anything at all like "huge amounts of
> near-free energy".
Depends on your definition of "free," I suppose, and I haven't seen
anyone else in this thread use this phrase you've mentioned in quotes,
so I'm not sure whose argument you're actually trying to refute.
The energy released (even taking into account more realistic energy
estimates and efficiencies) exceeds the energy you have to put in,
that's for sure. Whether that's worth your use of copper is another
question.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM, Y!M erikmaxfrancis
You'll learn / Life is worth it / Watch the tables turn
-- TLC
IsaacKuo
> IsaacKuo wrote:
> > The more interesting part is staying in orbit around the L1
> > or L2 point. As I understand it, there are stable orbits
> > which circle around an L1 or L2 point. If you stand on the
> > moon and look at the lagrange point, the satellite appears
> > to keep the same distance as the lagrange point, but it
> > slowly circles around it.
> Not sure what you're referring to here. L1 and L2 are never table; L4
> and L5 _sometimes_ are.
I was thinking of "halo orbits", which actually aren't
stable. I recalled incorrectly. Still, it could be used
because orbital speed can be constantly adjusted
by adjusting the resistive load on the tether.
Different load leads to a different amount of drag.
Isaac Kuo
IsaacKuo
> "IsaacKuo" <mech...@yahoo.com> wrote in message
> > Tidal forces do the hard part for you--namely, keeping the
> > tether stretched taut and oriented perfectly up/down.
> > The more interesting part is staying in orbit around the L1
> > or L2 point. As I understand it, there are stable orbits
> > which circle around an L1 or L2 point. If you stand on the
> > moon and look at the lagrange point, the satellite appears
> > to keep the same distance as the lagrange point, but it
> > slowly circles around it.
> Those orbits only work in the plane normal to the line
> between the moon and its primary. Magnetic drag sould
> cause the satelite to drop further in towards the primary
> (or get pushed farther out, if the magnetic field
> is rotating rapidly enough), after which it loses its orbit
> and goes flying off on its own.
First off, whether the drag is inward or outward depends
on how fast the planet is spinning, not the magnetic field
(although obviously these may be related). The planet's
rotation rotates the surrounding plasma field along with
it (it's essentially really really upper atmosphere).
In Jupiter's case, the rotation rate is much faster than the
orbit of Io, so the drag will be outward.
Second, this drag would indeed pull a perfectly balanced
Lagrange point orbit out of balance, but you can instead
place the orbit slightly off center to counteract the drag.
Isaac Kuo
Eivind Kjorstad
> 60 km/s is a massic energy of 1.8 GJ/kg, which isn't anything to cough at.
It's 500KWh, what I said. It's hard to imagine more than a small
fraction of that ending up as usable power though.
>> *Certainly* doesn't sound anything at all like "huge amounts of
>> near-free energy".
>
> Depends on your definition of "free," I suppose, and I haven't seen
> anyone else in this thread use this phrase you've mentioned in quotes,
> so I'm not sure whose argument you're actually trying to refute.
Sorry should've made that clear. I was thinking of MACROLIFE, was
mentioned upstream as the source of this copper-bar-power idea. It had
large amounts of cheap power, though it also probably didn't say the
power was "free".
Seems unlikley to me that this mode of power-generation would be
economical. (or even practical, really) It's hard to imagine scenarios
where this would make more sense than, for example, nuclear.
Eivind Kjørstad
Logan Kearsley
> On Oct 29, 5:25 pm, Logan Kearsley <chronosur...@gmail.com> wrote:
>
> > "IsaacKuo" <mech...@yahoo.com> wrote in message
> > > Tidal forces do the hard part for you--namely, keeping the
> > > tether stretched taut and oriented perfectly up/down.
> > > The more interesting part is staying in orbit around the L1
> > > or L2 point. As I understand it, there are stable orbits
> > > which circle around an L1 or L2 point. If you stand on the
> > > moon and look at the lagrange point, the satellite appears
> > > to keep the same distance as the lagrange point, but it
> > > slowly circles around it.
> > Those orbits only work in the plane normal to the line
> > between the moon and its primary. Magnetic drag sould
> > cause the satelite to drop further in towards the primary
> > (or get pushed farther out, if the magnetic field
> > is rotating rapidly enough), after which it loses its orbit
> > and goes flying off on its own.
>
> First off, whether the drag is inward or outward depends
> on how fast the planet is spinning, not the magnetic field
> (although obviously these may be related). The planet's
> rotation rotates the surrounding plasma field along with
> it (it's essentially really really upper atmosphere).
Wah? What matters is the relative velocity between the conductor and
the local magnetic field lines, which depends on how fast the magnetic
field is rotating. That may or may not have anything to do with how
fast the planet is rotating (I don't know if it does or not in the
case of Jupiter). If we try for an alternating loop current induced by
variations in the local magnetic field strength, then that depends on
how fast the planet rotates, since that will determine the frequency
with which the magnetic poles sweep towards and away from you. I'm not
sure which direction drag is in in that case... I suspect it would be
outwards, with energy being drained from the planet's rotational KE.
>
> In Jupiter's case, the rotation rate is much faster than the
> orbit of Io, so the drag will be outward.
>
> Second, this drag would indeed pull a perfectly balanced
> Lagrange point orbit out of balance, but you can instead
> place the orbit slightly off center to counteract the drag.
OK, I think I get that now. If you go just a little bit off center of
the Lagrange point in the normal plane*, you get a small force back
towards the Lagrange point, from the projection of each body's gravity
into that plane. If the drag force exactly balances that force, you
can hold the satellite just off of the Lagrange point. And the
magnetic drag force should then be transferred by gravity to the moon,
so that it gets pulled along with the satellite very, very slowly. And
vice-versa.
Taking some inspiration from the calculating I did for magnetovores
around pulsars, we might be able to take advantage of alternating
currents induced by the changing voltage across the moon as it passes
through areas of different field strength. Assuming that the magnetic
field is close to spherical (which it certainly isn't for Jupiter,
which ought to make this option much more attractive, but should be a
good approximation for a not-severely-tilted brown dwarf),
dE = M*(2*pi*r/t - sqrt(Gm/r))*(sqrt(1+3*sin^2(x)) - sin(x))/(4pi*r^3)
over one quarter of a rotation, where M is the magnetic moment of the
primary in Wb*m, m is the mass of the primary, r is the orbital
radius, t is the primary's rotation time, and x is the angle between
the magnetic dipole and the primary's angular momentum. You get better
values for high tilts.
I suspect that might be worth looking into for Jupiter, since the
magnetic field is highly non-symmetrical. Around a brown dwarf,
though, which isn't having its magnetic field distorted by solar wind,
it's not going to produce very large electric field variations. But,
there's still loop inductance to consider.
Assuming I didn't screw something up, F =
A*M*sqrt(1+3*sin^2(l*e))*sin(l+e*(pi-l))/(4pi*r^3), where F is flux, M
is the magnetic moment, A is the cross-sectional area of the current
loop, r is the orbital radius, l is the angle between the dipole and
the primary's angular momentum, and e is a scaling parameter that
varies between 1 and 0 every quarter-rotation of the primary as seen
from the moon (a function of both the primary's rotational velocity
and the moon's orbital velocity). I'm not sure of the exact function
of how e varies with time, so I can't even attempt to take a
derivative of that to get exact voltages, but this is all BOTE anyway,
so I'll just assume that it's linear, in which case the voltage around
the loop will be
E = 4dF/t = 4dF/(1/(1/p-1/P)) = 4F(1/p-1/P), where p is the primary's
rotational period and P is the moon's rotational period. P =
2pi*sqrt(r^3/Gm), so
E = 4dF(1/p-sqrt(Gm/r^3)/2pi)
dF is A*M*sqrt(1+3*sin^2(l))/(4pi*r^3) - A*M*sin(l)/(4pi*r^3) =
A*M*(sqrt(1+3*sin^2(l))-sin(l))/(4pi*r^3)
So,
E = A*M*(sqrt(1+3*sin^2(l))-sin(l))*(1/p-sqrt(Gm/r^3)/2pi)/(pi*r^3)
Google calculator does not seem to want to evaluate that when I plug
in values for Jupiter and Io. Anyone care to check my math? Are the
units in order?
-l.
*But only a little bit. More than a little bit, and the plane
approximation becomes bad... it's a weird curved surface where the
pull towards one or the other body cancels out.
IsaacKuo
> On Oct 30, 7:07 am, IsaacKuo <mech...@yahoo.com> wrote:
> > First off, whether the drag is inward or outward depends
> > on how fast the planet is spinning, not the magnetic field
> > (although obviously these may be related). The planet's
> > rotation rotates the surrounding plasma field along with
> > it (it's essentially really really upper atmosphere).
> Wah? What matters is the relative velocity between the conductor and
> the local magnetic field lines, which depends on how fast the magnetic
> field is rotating.
No, in fact the relevant effect works just fine if the magnetic
field isn't rotating at all.
Essentially, you're using the exact same effect as an
MHD generator. There's a more or less constant
magnetic field. You have a couple electrodes as well
as plasma flowing past those electrodes. In this case,
the electrodes are the non-insulated ends of the tether.
The flowing plasma is the really thin exosphere of
the planet.
Lorentz force acts on the electrons in the plasma as
well as the electrons in the tether. Because they're
not moving at the same speed, the Lorentz force acts
differentially--and a current results.
The direction of this current depends on whether
the plasma is moving faster or the tether is moving
faster. In turn, this determines the direction of the
drag force. Essentially, the plasma will try to drag
the tether to its velocity and vice versa.
Isaac Kuo
Logan Kearsley
There are two relevant, and opposing (as far as being able to extract
energy), effects, then- voltage across the
conductor due to moving through the planet's magnetic field, and
voltage due
to the surrounding plasma moving through the field. I had not
considered
motion of the plasma through the planetary magnetic field before....
It seems to me that the net voltage, then, will depend on the
difference in
rotation rates between the planet's atmosphere and its magnetic field,
as
well as orbital speed. If they move in lock-step, you get no MHD
effect, and
all of the voltage is due to the motion of the conductor relative to
the
magnetic field.
John Schilling
The problem is, outer space isn't a vacuum. Or, rather, it is, and that's
a *good* thing because the whole concept depends on having that ambient
plasma around.
But cathodes are more than just electron sources; they are also ion sinks.
If there are ions around, and there will be, they are going to bombard the
cathode at high current and energy.
The sort of surface coatings you want for thermionic emission, don't much
like extensive ion bombardment - especially if you first heat them to any
temperature where you're going to get much thermionic emission in the
first place.
And the nice microstructured field-effect cathode "emitters", those get
worn down to smoothly-rounded non-emissive nubs in a right good hurry
when the ions start raining down upon them.
The best anybody has ever actually come up with for this sort of thing,
is a hollow-cathode discharge. And that just substitutes your own denser
but less energetic plasma for the one your cathode is trying to suck in,
which makes for a more complicated gadget than you were perhaps hoping
for and, possibly more importantly, a constant "fuel" flow requirment.
Keith F. Lynch
> It Seems To Me that that would make an ideal power source for a
> brown-dwarf colony, if it's really feasible.
I agree. Except the part about a colony on a brown-dwarf. It's too
hot, and the gravity is too high.
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.
Logan Kearsley
> Logan Kearsley <chronosur...@gmail.com> wrote:
> > It Seems To Me that that would make an ideal power source for a
> > brown-dwarf colony, if it's really feasible.
>
> I agree. Except the part about a colony on a brown-dwarf. It's too
> hot, and the gravity is too high.
brown
dwarf. :)
Logan Kearsley
> On Tue, 30 Oct 2007 03:35:23 -0000, Logan Kearsley
> >"John Schilling" <schil...@spock.usc.edu> wrote in message
If you're building on the scale of a moon, though, couldn't you start
to take advantage of those gradients?
When considering how to turn a moon into a giant powersat, I'm willing
to allow a little complication. The fuel requirement might be annoying
for a free-flying satellite (though I wonder if you could perhaps make
use of some of those Bussard ramjet designs to collect extra ambient
plasma for that purpose...), but an ice moon ought to provide plenty
of material for that.
If you've got cables a few thousands of kilometers long, though, and
potentially many square kilometers of emitter area, is the ion
bombardment as much of an issue anymore? For one thing, if the voltage
is high enough the microstructure for a field-effect cathode doesn't
need to be very micro anymore.
-l.