Light Powered Spacecraft

JRS: In article , dated
Tue, 25 Apr 2006 20:56:08 remote, seen in news:sci.space.policy,
jonathan posted :

"Erik Max Francis" wrote in message
...
Dr John Stockton wrote:

Remember that the laser will be impelled with roughly the same force as
the craft. Since you are powering the craft 1300 AU, and Mercury is at
about 0.3 AU from Sol, then, unless the station outweighs the craft by
more that 5000 times, you have the possibility of ramming the station
into Sol.



I think the laser would be propelled by half the force of the
sail. Doesn't the sail get one unit of force for stopping the
light, and another for returning it?

Yes, that's why I put "roughly". Photons that bounce off the sail count
double, those absorbed count singly, those that miss don't count at all.
The average will be roughly one. It cannot exceed two; and less than
half would seem a bad design.

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JRS: In article , dated Wed, 26 Apr 2006
13:58:45 remote, seen in news:sci.space.policy, Henry Spencer
posted :
In article ,
Dr John Stockton wrote:
Remember that the laser will be impelled with roughly the same force as
the craft. Since you are powering the craft 1300 AU, and Mercury is at
about 0.3 AU from Sol, then, unless the station outweighs the craft by
more that 5000 times, you have the possibility of ramming the station
into Sol.

The obvious approach is to build the laser in an orbit around Mercury,
with the orbital plane perpendicular to the craft's departure direction,
and then as it fires up, let the recoil shift the orbit out of plane until
Mercury's gravity stops it. The laser will then be hovering above its
nominal orbital plane on photon thrust. This essentially transfers the
recoil thrust to Mercury, which is heavy enough to take it without any
noticeable effect.

Provided that the laser system is heavy enough for there to be
sufficient binding force between laser and planet.

But is that argument obvious to persons of ordinary intelligence who
happen not to have read "Dragonfly"?


The major force component on the laser station must be the incoming
sunlight, assuming efficiency under 100%; the orbital plane must be more
or less perpendicular to the Mercury-Sun line.


If the entire energy of the Sun for a second, 4E9 kg, were beamed in one
direction from Mercury, that's enough to give the planet a speed change
of 4E9 * 3E8 / 0.3E24 m/s = 4E-6 m/s; if done for half a Mercury year,
about 4E6 seconds, that would give a speed change of 16 m/s, small
compared with system escape speed from Mercury orbit. So, under even
such propulsion, Mercury remains bound to Sol, and the effective
launcher mass is Sol's.

--
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JRS: In article .com
, dated Wed, 26 Apr 2006 11:04:26 remote, seen in news:sci.space.policy,
Stefan D. posted :
Can't you just connect the laser with the solar powersats feeding it,
the light pressure on the powersats counteracting the impulse of the
laser?


This is a Usenet newsgroup, to which Google have provided their own
peculiar form of Web interface. Experienced users of Usenet follow
long-established conventions.

If you find that, when you start a News reply, Google does not provide
the previous article in quoted form, note what Keith Thompson wrote in
comp.lang.c, message ID :-
If you want to post a followup via groups.google.com, don't use
the "Reply" link at the bottom of the article. Click on "show
options" at the top of the article, then click on the "Reply" at
the bottom of the article headers.

Since that is what the experts in this newsgroup prefer to read, it will
be to your advantage to comply.




To counteract fully, the laser beam must be generated with 100%
efficiency and heading directly away from the Sun (presuming the
receiving structure to be small compared with its distance from the
Sun).

If the laser, etc., is orbiting the Sun (whether or not it is also
orbiting Mercury), then it will have to be aimed at the (current) craft.
That means that in general it will be redirecting the momentum that it
delivers, and will feel a sideways force.

To be really clever, keep everything in a straight line, and make the
light collector plus laser system of such a low areal density that its
tendency to fall into the Sun is balanced by the net push from that
proportion of the intercepted light that is absorbed (with a few
corrections of detail). That means that it must be under a gram per
square metre, with an average "solid" thickness under a micron ...

--
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In article ,
Dr John Stockton wrote:
The obvious approach is to build the laser in an orbit around Mercury,
with the orbital plane perpendicular to the craft's departure direction,
and then as it fires up, let the recoil shift the orbit out of plane until
Mercury's gravity stops it. The laser will then be hovering above its
nominal orbital plane on photon thrust. This essentially transfers the
recoil thrust to Mercury...

Provided that the laser system is heavy enough for there to be
sufficient binding force between laser and planet.

The mass of the laser system is irrelevant -- it drops out of the math
completely. All that matters is that the recoil force not be too large
compared to the force exerted by Mercury's gravity at that altitude.

The major force component on the laser station must be the incoming
sunlight, assuming efficiency under 100%; the orbital plane must be more
or less perpendicular to the Mercury-Sun line.

That constrains you to launching more or less along the Sun-Mercury axis.
It's probably better to decouple the power system and the laser system by
doing power beaming between them.
--
spsystems.net is temporarily off the air; | Henry Spencer
mail to henry at zoo.utoronto.ca instead. |

: (Henry Spencer)
: The mass of the laser system is irrelevant -- it drops out of the math
: completely. All that matters is that the recoil force not be too large
: compared to the force exerted by Mercury's gravity at that altitude.

I don't quite follow this. If the force provided by gravity is what
matters, then the mass of the laser is important, since the force
is proportional to that mass. The less massive the laser, the
smaller the force exerted by Mercury's gravity.

I realize that the *acceleration* provided by mercury is independent
of the laser's mass. But if we're comparing accleration instead of
force, then for a given acceleration provided to the starship, the
acceleration of the laser due to recoil is inversely proportional to
it's mass, and again, mass enters into it.

In short, either the (gravitational) mass has to be large so the force
provided by mercury is large enough, or the (inertial) mass has to be
large enough so the acceleration due to recoil is small enough. No matter
how you look at it, the mass of the laser remains in the picture.

So... where am I going wrong? Aside from the simplifying assumptions
that the laser isn't large compared to mercury, and the starship's
change in velocity over the duration involved doesn't red-shift the
acceleration lower, and like that, which don't seem to signify.


Wayne Throop http://sheol.org/throopw

JRS: In article . com,
dated Wed, 26 Apr 2006 16:05:17 remote, seen in news:sci.space.policy,
Alex Terrell posted :

thought of 3. - Kept me awake -

3. The light reflected back would rely on a perfectly flat mirror, at
least for any long distance reflection (like reflecting back from
beyond Pluto orbit). Howeverm it will be virtually impossible to
refelct back in a coherent manner across a 100km diameter mirror. Such
a mirror cannot be flat.


Pish - that's thinking of the early 21st century and earlier. Engineers
of the 23rd century will be able to do better.

Note, for example, that rather than returning the energy by simple
reflection, the energy might be collected and used to power a laser.

--
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Henry Spencer wrote:
Dr John Stockton wrote:
Remember that the laser will be impelled with roughly the same
force as the craft. Since you are powering the craft 1300 AU,
and Mercury is at about 0.3 AU from Sol, then, unless the station
outweighs the craft by more that 5000 times, you have the
possibility of ramming the station into Sol.

The obvious approach is to build the laser in an orbit around
Mercury, with the orbital plane perpendicular to the craft's
departure direction, and then as it fires up, let the recoil
shift the orbit out of plane until Mercury's gravity stops it.

I'm surprised at you. The laser is presumably powered by sunlight
(otherwise why place it so close to the sun?), so the force of the
light from the sun on the laser's solar panels would necessarily be
at least as great as the force of the light emitted by the laser.

Does this qualify for a t-shirt?
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.

Dr John Stockton wrote:
JRS: In article . com,
dated Wed, 26 Apr 2006 16:05:17 remote, seen in news:sci.space.policy,
Alex Terrell posted :

thought of 3. - Kept me awake -

3. The light reflected back would rely on a perfectly flat mirror, at
least for any long distance reflection (like reflecting back from
beyond Pluto orbit). Howeverm it will be virtually impossible to
refelct back in a coherent manner across a 100km diameter mirror. Such
a mirror cannot be flat.


Pish - that's thinking of the early 21st century and earlier. Engineers
of the 23rd century will be able to do better.

Note, for example, that rather than returning the energy by simple
reflection, the energy might be collected and used to power a laser.

Using some reasonable guestimates, I get about 1500 tons for the sail,
assuming a radius of 100 km. Wouldn't it be a more effective use of
this much mass to make a smaller sail and use the collected energy to
accelerate propellant?

Actually, I should take this space to promote my favored method. Use a
magsail, but instead of the solar wind, how about a solar-powered mass
driver zipping conductive bits of foil through the center? You get the
braking for free then, and you can still remotely power the system (no
need for a big onboard power supply). Best of all, the mass of the
sail scales linearly (roughly) with the diameter.

JRS: In article , dated Thu, 27 Apr 2006
17:18:36 remote, seen in news:rec.arts.sf.science, Henry Spencer
posted :
In article ,
Dr John Stockton wrote:
The obvious approach is to build the laser in an orbit around Mercury,
with the orbital plane perpendicular to the craft's departure direction,
and then as it fires up, let the recoil shift the orbit out of plane until
Mercury's gravity stops it. The laser will then be hovering above its
nominal orbital plane on photon thrust. This essentially transfers the
recoil thrust to Mercury...

Provided that the laser system is heavy enough for there to be
sufficient binding force between laser and planet.

The mass of the laser system is irrelevant -- it drops out of the math
completely. All that matters is that the recoil force not be too large
compared to the force exerted by Mercury's gravity at that altitude.

The mass is important, since that is what the gravity pulls.


The major force component on the laser station must be the incoming
sunlight, assuming efficiency under 100%; the orbital plane must be more
or less perpendicular to the Mercury-Sun line.

That constrains you to launching more or less along the Sun-Mercury axis.

Which moves at over four degrees per day.


It's probably better to decouple the power system and the laser system by
doing power beaming between them.

That gives more freedom in the design.

But the sunlight-catchers are now diverting EM energy, and so have a
sideways force to deal with; and the drive laser is also redirecting.

If the catchers orbit Mercury in a plane parallel to the terminator, but
are light enough to be displaced enough to see the centre of the night
side, the laser could be at the centre of the night side, and used for a
month or so at a time on a given craft. Unfortunately, now that Mercury
is rotating, the laser would have to be on wheels ...


Each orbiting component needs an auxiliary mirror, for station-keeping.

--
© John Stockton, Surrey, UK. Turnpike v4.00 MIME. ©
Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links;
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In article ,
Keith F. Lynch wrote:
...let the recoil
shift the orbit out of plane until Mercury's gravity stops it.

I'm surprised at you. The laser is presumably powered by sunlight
(otherwise why place it so close to the sun?), so the force of the
light from the sun on the laser's solar panels would necessarily be
at least as great as the force of the light emitted by the laser.

As I noted earlier, the solar collectors and the laser aren't necessarily
in the same place. Although if you're doing power beaming between them,
the same issue arises for the beam receiver.

To get the full picture, in fact you have to consider three forces: the
recoil from the laser beam, the force from the absorbed sunlight or power
beam... and the photon thrust from the waste-heat radiators. The lasers
won't be perfectly efficient; indeed, most current lasers are pretty
inefficient. (Diode lasers and some of their relatives are an exception.)
So for a laser that powerful, you're going to have to get rid of quite a
lot of waste heat, and `F = P/c' applies just as much to it as to the
laser beam.

A further complication: if you're operating in Mercury orbit to use
Mercury as an anchor, that may limit which directions you can radiate in.
Unless your radiators (and hence your lasers) run *very* hot, you won't
want them to have a view of the planet, which is itself (on its day side)
quite hot and radiating furiously.

There might well be a journal paper to be had here, just looking at the
force balances and assessing what's feasible. Don't think I've seen an
analysis of it.
--
spsystems.net is temporarily off the air; | Henry Spencer
mail to henry at zoo.utoronto.ca instead. |