Alternative to Rockets

In sci.space.tech Jim Logajan wrote:
Sander Vesik wrote:
Strictly speaking, the solar sail too throws mass (photons) out in one
direction and moves in the other direction. Where the mass being
thrown out comes from is not too important.

I'm sure someone else will also point this out but photons are massless. A

I doubt. Photons are not massless - they don't have rest mass, but that is
not at all the same as photons not having mass. E = m*c^2, so each photon has
a mass determined by how energetic it is.

solar sail basically redirects the momentum vector of some of the incident
photons. In order for momentum to be conserved the solar sail must acquire
the vector difference.


If there was no mass in the form of photons, precicely where would the vector
of momentum be coming from ?

--
Sander

+++ Out of cheese error +++

Gregory L. Hansen wrote:
In article ,
nafod40 wrote:

Mike Miller wrote:

2) Mass drivers/railguns/coilgun launchers. Unfortunately, the big
electromagnetic launchers that can fling a spaceship into orbit
without turning passengers to goo is really, really long, like 600-700
miles for a 3G launch.

Not that it would be a "make or break" factor, but they did centrifuge
tests way back where the subjects were immersed in water, then spun up.
The water pressure outside the body opposed the internal pressures, and
the subjects/victims were able to carry on conversations at 12+ G's.

I've thought that if you need to carry water on missions, for shielding
and all of the other reasons, might as well put it to work for you.

Just did a quick google...the Swiss have a suit called the Libelle that
is based on the principle, lets you chit chat at 12 G's sustained. It
would be higher laying back in a couch.


Cool! If they're chit chatting and, well, breathing, I assume the head
doesn't need additional protection at 12 G's?

I think the standard flight suit applies regular pressure on the legs and
torso, but as I recall a pilot can only go to about 8 G's before blacking
out in a turn. What's different about that?

The big diff was that with the water suits, it was as if they weren't
even at 12 Gs. Just a day at the office. Using standard G-suits, you
have to really strain to keep the blood in the noggin. The G-suit only
provides 1-2 G improvement alone, but it also lets you push against it
to raise the tolerance. Very exhausting.

In article , (Greg) writes:
wrote in message ...
While it never quits, it goes asymptotically to zero (1/r^2) with
growing distance to the light source. It is a rather simple exercise
to estimate to limiting velocity a solar sail with a given mass/area
ratio will achieve, upon being launched from a given distance from the
sun. While it'll beat chemical rockets, in the long haul, it is still
far from adequate for really long trips.


Its far from trivial!!If you include finding a optimal trajectory.

The angular momentum reversal manoeuvre can get huge velocity's, and
while sail weight remains a very critical parameter you can get more
speed by using a sail that can be used closer to the sun/star in
question. Also there is nothing to say that some other trajectory
won't provide higher velocity.

While figuring out optimal trajectory is far from trivial, limit
estimates are straightforward.

The force generated by the light pressure is (assuming full
reflection), 2*P*a/c, where P is power per unit area, a is the area
and c is the speed of light. The power drops like 1/r2 and it is
convenient, for what follows, to represent it as

P(r) = P_s*r_s^2/r^2

where r_s is the radius of the sun and P_s is the energy flux (power
per unit area) at the surface of the sun. Checking the numbers you
get (rounding up a bit)

P_s*r_s^2 = 3.15*10^25 Watt

The constant we really care about is

K = 2*P_s*r_s^2/c = 2.1*10^17 N.

and using this, the light generated force on the sail is simly

F = K*a/r^2

Note that this has the same form as gravity, only repulsive instead of
attractive, so you can associate with it an 1/r potential energy (with
positive, instead of negative, sign). Thus, startin tracking the sail
from some moment when it has a velocity v and is at distance r from
the sun, you can write for its energy

E = m( v^2/2 - GM/r + K/r*(a/m))

where m is the mass of the sail (and all that's attached to it). Now, the
first two terms are just the standard energy, kinetic and potential (M here
is the mass of the sun). What they're, depends on the past history of
the sail but, as long as you're in a bound orbit, the sum of the first
two terms is negative. So, assume that the point at which we start
tracking is the perihelion. Furthermore, assume that before you got
there, you went through whatever maneuvers to maximize your energy.
Such as, using the sail (or anything else) to take you ver far from
the sun (in numerous passes if needed) then killing your excess
angular momentum, folding the sail and dropping to as low an orbit
around the sun as survivable. The sum of the first two terms will
still be negative but playing the game right you can bring this sum
close to zero. So, lets assume the *limiting* case that you got it
exactly to zero. And now you unfurl the sail and go out, full blast.

OK, so taking the sum of the first two terms as zero, you've initial
energy of

E_i = m*K*(a/m)*1/r_i

Where r_i is the starting distance from the Sun, of course. The finel
energy (at infinity) is

E_f = m/2*v_f^2

where v_f is the "speed at infinity". equating the two you get

v_f = sqrt(2*K*(a/m)*1/r_i) = 6.5*10^8*sqrt((a/m)*1/r_i) (m/s)

and it depends just on the starting radial distance and the mass to
area ratio, m/a. OK, what are reasonable values.

m/a of the order of 10^-3 kg/m^2 is (optimistically) achievable. For
the sail alone, this would correspond to thickness of hte order of a
micron, for reasonable materials (note, that you really should count
not just the mass of the sail but the probe and the cabling as well,
but I ignore this at the moment). 10^-4 kg/m^2 is beyond realistic,
being both too fragile and not significantly reflective. As for r_i,
it depends how much heat and radiation you can take. 10 million
kilometer may be possible, 1 million, I don't think so.

So, taking the reasonable figures of a/m = 10^3 m^2/kg, r_i = 10^10 m
and substituting in the formula above we get a limiting velocity of

v_f = 2.1*10^5 m/s = 21 km/s

(I rounded up in the above). Taking the insanely optimistic values of
1/m = 10^4 and r_i = 10^9 you'll gain an order of magnitude in the
velocity. Two hundred something km/s, still below 0.001 c.

The max speeds available do look promising for interstellar probes if
your patient(ie don't mind waiting a long time).

A very long time, as you can see from the above. Thousands of years.

Braking at the other
planet could be done with the sail and the destination star.

Depending on the star. Alpha or Beta Centauri should work fine.
Proxima, won't.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

In article , Sander Vesik writes:
In sci.space.tech Jim Logajan wrote:
Sander Vesik wrote:
Strictly speaking, the solar sail too throws mass (photons) out in one
direction and moves in the other direction. Where the mass being
thrown out comes from is not too important.

I'm sure someone else will also point this out but photons are massless. A

I doubt. Photons are not massless - they don't have rest mass, but that is
not at all the same as photons not having mass. E = m*c^2, so each photon has
a mass determined by how energetic it is.

Mass is defined through

E^2 = m^2*c^4 + p^2*c^2

Photons are massless.

solar sail basically redirects the momentum vector of some of the incident
photons. In order for momentum to be conserved the solar sail must acquire
the vector difference.


If there was no mass in the form of photons, precicely where would the vector
of momentum be coming from ?

Momentum is not necessarily associated with mass. p = mv is *not* the
definition of momentum. It is just a special case.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

In article ,
Sander Vesik wrote:
In sci.space.tech Jim Logajan wrote:
Sander Vesik wrote:
Strictly speaking, the solar sail too throws mass (photons) out in one
direction and moves in the other direction. Where the mass being
thrown out comes from is not too important.

I'm sure someone else will also point this out but photons are massless. A

I doubt. Photons are not massless - they don't have rest mass, but that is
not at all the same as photons not having mass. E = m*c^2, so each photon has
a mass determined by how energetic it is.

You're thinking of E = m_r*c^2. But when it's said that photons are
massless, people aren't talking about relativistic mass. They're talking
about the mass parameter that would appear in the equations of motion
(Maxwell's equations), i.e. rest mass.


solar sail basically redirects the momentum vector of some of the incident
photons. In order for momentum to be conserved the solar sail must acquire
the vector difference.


If there was no mass in the form of photons, precicely where would the vector
of momentum be coming from ?

Electromagnetic fields have momentum. That's been known since the 19th
century on purely classical reasoning. The argument comes from the
conservation of linear momentum. Lorentz's law gives

F = q(E + 1/c vxB)

F is a dp/dt. If p is transferred to a particle, then p must be
transferred to the field to conserve momentum. The result is that when
the field has energy E, it will have momentum p=E/c, which is also what
you get from the relativistic equation for energy

E^2 = m^2 c^4 + p^2 c^2

with m=0.


--
"What are the possibilities of small but movable machines? They may or
may not be useful, but they surely would be fun to make."
-- Richard P. Feynman, 1959

Sander Vesik writes:

In sci.space.tech Jim Logajan wrote:
Sander Vesik wrote:
Strictly speaking, the solar sail too throws mass (photons) out in one
direction and moves in the other direction. Where the mass being
thrown out comes from is not too important.

I'm sure someone else will also point this out but photons are massless.

I doubt. Photons are not massless - they don't have rest mass, but that
is not at all the same as photons not having mass. E = m*c^2, so each
photon has a mass determined by how energetic it is.

You have just committed a very common conceptual error: The famous formula
you quote is only a _SPECIAL CASE_ of a more general formula, which is only
valid for _MASSIVE_ particles _AT REST_. The correct formula valid in all
cases is:

E = \sqrt[ p^2 + (m_0*c^2)^2 ],

which for photons with m_0 = 0 yields "E = p*c." The more famous formula
should be written "E = m_0 * c^2, iff p = 0," in order to avoid the conceptual
error you make. See the Physics FAQ entries for a more detailed discussion:

http://www.math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html
http://www.math.ucr.edu/home/baez/physics/Relativity/SR/mass.html


solar sail basically redirects the momentum vector of some of the incident
photons. In order for momentum to be conserved the solar sail must acquire
the vector difference.

If there was no mass in the form of photons, precicely where would the vector
of momentum be coming from ?

The fact that photons have a velocity and carry energy, and the fact that
in relativity, energy and momentum are unified into a single four-vector,
whose four-dimensional norm is the so-called "rest mass." (The "rest mass"
is now more correctly called the "proper mass," since photons can never be
at rest). The "proper mass" is the only physically meaningful concept of "mass"
in relativity --- and the proper mass of a photon is _IDENTICALLY ZERO_.


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'

In article , writes:
In article , (Greg) writes:
wrote in message ...
While it never quits, it goes asymptotically to zero (1/r^2) with
growing distance to the light source. It is a rather simple exercise
to estimate to limiting velocity a solar sail with a given mass/area
ratio will achieve, upon being launched from a given distance from the
sun. While it'll beat chemical rockets, in the long haul, it is still
far from adequate for really long trips.


Its far from trivial!!If you include finding a optimal trajectory.

The angular momentum reversal manoeuvre can get huge velocity's, and
while sail weight remains a very critical parameter you can get more
speed by using a sail that can be used closer to the sun/star in
question. Also there is nothing to say that some other trajectory
won't provide higher velocity.

While figuring out optimal trajectory is far from trivial, limit
estimates are straightforward.

The force generated by the light pressure is (assuming full
reflection), 2*P*a/c, where P is power per unit area, a is the area
and c is the speed of light. The power drops like 1/r2 and it is
convenient, for what follows, to represent it as

P(r) = P_s*r_s^2/r^2

where r_s is the radius of the sun and P_s is the energy flux (power
per unit area) at the surface of the sun. Checking the numbers you
get (rounding up a bit)

P_s*r_s^2 = 3.15*10^25 Watt

The constant we really care about is

K = 2*P_s*r_s^2/c = 2.1*10^17 N.

and using this, the light generated force on the sail is simly

F = K*a/r^2

Note that this has the same form as gravity, only repulsive instead of
attractive, so you can associate with it an 1/r potential energy (with
positive, instead of negative, sign). Thus, startin tracking the sail
from some moment when it has a velocity v and is at distance r from
the sun, you can write for its energy

E = m( v^2/2 - GM/r + K/r*(a/m))

where m is the mass of the sail (and all that's attached to it). Now, the
first two terms are just the standard energy, kinetic and potential (M here
is the mass of the sun). What they're, depends on the past history of
the sail but, as long as you're in a bound orbit, the sum of the first
two terms is negative. So, assume that the point at which we start
tracking is the perihelion. Furthermore, assume that before you got
there, you went through whatever maneuvers to maximize your energy.
Such as, using the sail (or anything else) to take you ver far from
the sun (in numerous passes if needed) then killing your excess
angular momentum, folding the sail and dropping to as low an orbit
around the sun as survivable. The sum of the first two terms will
still be negative but playing the game right you can bring this sum
close to zero. So, lets assume the *limiting* case that you got it
exactly to zero. And now you unfurl the sail and go out, full blast.

OK, so taking the sum of the first two terms as zero, you've initial
energy of

E_i = m*K*(a/m)*1/r_i

Where r_i is the starting distance from the Sun, of course. The finel
energy (at infinity) is

E_f = m/2*v_f^2

where v_f is the "speed at infinity". equating the two you get

v_f = sqrt(2*K*(a/m)*1/r_i) = 6.5*10^8*sqrt((a/m)*1/r_i) (m/s)

and it depends just on the starting radial distance and the mass to
area ratio, m/a. OK, what are reasonable values.

m/a of the order of 10^-3 kg/m^2 is (optimistically) achievable. For
the sail alone, this would correspond to thickness of hte order of a
micron, for reasonable materials (note, that you really should count
not just the mass of the sail but the probe and the cabling as well,
but I ignore this at the moment). 10^-4 kg/m^2 is beyond realistic,
being both too fragile and not significantly reflective. As for r_i,
it depends how much heat and radiation you can take. 10 million
kilometer may be possible, 1 million, I don't think so.

So, taking the reasonable figures of a/m = 10^3 m^2/kg, r_i = 10^10 m
and substituting in the formula above we get a limiting velocity of

v_f = 2.1*10^5 m/s = 21 km/s

Ooops sorry, lost a zero there. That's 210 km/s, of course. Still
not much.

(I rounded up in the above). Taking the insanely optimistic values of
1/m = 10^4 and r_i = 10^9 you'll gain an order of magnitude in the
velocity. Two hundred something km/s, still below 0.001 c.

The max speeds available do look promising for interstellar probes if
your patient(ie don't mind waiting a long time).

A very long time, as you can see from the above. Thousands of years.

Braking at the other
planet could be done with the sail and the destination star.

Depending on the star. Alpha or Beta Centauri should work fine.
Proxima, won't.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

In article ,
Uncle Al wrote:
On another point, the sail itself may have low mass, but something is
going to have to hold it open and taught.

Radiation pressure plus the solar wind will balloon it out...

Actually, no, it won't. The solar wind is insignificant. And light
pressure is *not* like air in a parachute -- it will not hold a curved
canopy open. Solar sails must be rigid, via either internal structure or
centrifugal force.

1) Steering in general. How do you tack?

You don't, not because of steering difficulties, but because there is no
equivalent of a sailboat's keel. There is, however, a loose equivalent:
reduce your orbital velocity and let the Sun's gravity pull you down.

Attitude control for a solar sail is straightforward in principle,
although details are tricky. You can have movable vanes, say on the tips
of the booms that hold the sail rigid, or you can move the payload around,
shifting the center of mass away from the center of pressure to induce a
torque.

2) The curved solar sail - huge area - will have a focus or
caustics.

Since sails have to be rigid, in general they don't curve much.
--
MOST launched 30 June; science observations running | Henry Spencer
since Oct; first surprises seen; papers pending. |

In article ,
Olli Wilkman wrote:
This isn't exactly in response to the question, but what is the
current opinion on laser rockets? The idea is fairly simple, but I get
the impression that currently it is not a very strong candidate.

The idea is not ridiculous, but advanced propulsion in general is very
poorly funded. The laser rocket suffers from needing a very large laser,
which is a big up-front capital expense. For near-term applications, it
is not obvious that the money needed for that wouldn't be better spent on
improved application of conventional rocket technology.

This seems to imply that no propellant is used as such, but the news
article mentions a platic coating...

It does need reaction mass of some kind; the laser is heating the mass,
not exerting thrust directly. (The laser *does* exert a very small thrust
by light pressure, but that's insignificant by comparison.)

The simplest and most straightforward laser rockets are pure rockets,
which make no attempt to use the surrounding air as reaction mass. Using
the air is theoretically appealing but in practice is a nasty swamp of
technical problems.
--
MOST launched 30 June; science observations running | Henry Spencer
since Oct; first surprises seen; papers pending. |

wrote in message
and using this, the light generated force on the sail is simly

F = K*a/r^2

Note that this has the same form as gravity, only repulsive instead of
attractive, so you can associate with it an 1/r potential energy (with
positive, instead of negative, sign). Thus, startin tracking the sail
from some moment when it has a velocity v and is at distance r from
the sun, you can write for its energy

E = m( v^2/2 - GM/r + K/r*(a/m))

where m is the mass of the sail (and all that's attached to it).

This is not correct because the *direction* of the force is *not* away
from the sun in general. In fact its *not* constant in general. It's
not a conservative "feild", that is the path the sail takes and its
relative angle history all effect the final velocity.

heres a ref:
Vulpetti, G., Sailcraft at High Speed by Orbital Angular Momentum
Reversal, Acta Astronautica, Vol. 40, No. 10, pp. 733-758, 1997

Greg