Energy conservation does not necessarily limit the maximum thrust possible.
The key is obtaining the highest *propulsion* "specific impulse" (m/s) possible.
i.e. the highest "exhaust velocity" (m/s) that is possible.

Back to the Internet to find some realistic numbers for that velocity.  EAM.

---- Original message ----
>Date: Mon, 20 Jun 2011 10:27:08 -0700 (PDT)
>From: "John M. Steele" <[email protected]>  
>Subject: Re: [USMA:50686] Solar Electric Propulsion  
>To: [email protected], "U.S. Metric Association" <[email protected]>
>
>   But that side effect may equal or exceed the ion
>   thrust.  The photons are either reflected or
>   absorbed, giving two values for light pressure
>   (reality is a mix based on percentage
>   reflectivity).  Some of the absorbed ones can be
>   electricity rather than thermal impact energy, but
>   the impulse is still there.
>    
>   The electricity can power an ion drive, but I think
>   the 1st Law ensures the ion thrust can't exceed the
>   photon thrust on the system.  The photon thrust is
>   the direction of the Poynting vector for absorbed
>   photons.  By Snell's law, some thrust vectoring can
>   be obtained for the reflected photons, and up to
>   twice the pressure since they leave at the speed of
>   light too.  However the "lightship" would have a
>   much smaller range of navigational opportunity than
>   a sailing vessel with a keel and the ability to
>   resist sideways thrust.  I doubt it can tack to
>   sunward, it would be more like a square-rigged ship.
>
>     ------------------------------------------------
>
>   From: "[email protected]" <[email protected]>
>   To: John M. Steele <[email protected]>;
>   U.S. Metric Association <[email protected]>
>   Sent: Mon, June 20, 2011 1:00:14 PM
>   Subject: Re: [USMA:50686] Solar Electric Propulsion
>   Yes, Ions are the masses which impart the propulsion
>   thrust,  The much less massive electrons need to be
>   discarded to maintain charge neutrality.
>
>   The forces of solar photons are a side effect in the
>   application of ion drives.
>
>   Although the velocities of the ions in the
>   propulsion jet stream are very high,
>   as close to c, (3 x 10^8 m/s), as possible, they are
>   still well below the speed
>   of light.  How are high are they in proposed (and
>   already flown) ion drives?
>   Let's search the Internet to find out.  EAM 
>
>   ---- Original message ----
>   >Date: Mon, 20 Jun 2011 09:15:04 -0700 (PDT)
>   >From: "John M. Steele"
>   <[email protected]> 
>   >Subject: Re: [USMA:50686] Solar Electric
>   Propulsion 
>   >To: [email protected], "U.S. Metric Association"
>   <[email protected]>
>   >
>   >  Ignoring relativity, the physical interpretation
>   is
>   >  the velocity at which you can throw fuel mass
>   >  "overboard."  If it is some kind of "ion drive" I
>   >  suppose you can't ignore relativity.
>   >   
>   >  The photons from the sun also act with direct
>   force
>   >  on the solar panels.  Past studies have looked at
>   >  solar sails, etc.  Converting some of the energy
>   to
>   >  an ion drive could be a way to "vector" the
>   thrust
>   >  from the light pressure and provide more control.
>   >
>   >   
>   ------------------------------------------------
>   >
>   >  From: "[email protected]"
>   <[email protected]>
>   >  To: U.S. Metric Association <[email protected]>
>   >  Sent: Mon, June 20, 2011 9:50:32 AM
>   >  Subject: [USMA:50686] Solar Electric Propulsion
>   >  The NASA John Glenn Center in Cleveland, Ohio is
>   >  requesting Proposals
>   >  for a demonstration of  Solar Electric
>   Propulsion, a
>   >  method of slowly thrusting
>   >  probes along spiral trajectories into the outer
>   >  regions of our solar system.
>   >
>   >  The "specific impulse" (m/s) required for such
>   >  missions can only be achieved by electric
>   propulsion
>   >  (not by chemical rockets) fueled by solar or
>   nuclear
>   >  energy.
>   >
>   >  Specific Impulse (thrust times time; divided by
>   the
>   >  mass ejected by the rocket motor)  (newtons x
>   >  seconds per kilogram) i.e. (kg x m/s^2) x s per
>   kg =
>   >  m/s.
>   >
>   >  In outdated studies, some engineers equated mass
>   >  with weight and got the less
>   >  acceptable result that pounds times seconds
>   divided
>   >  by pounds equals seconds.
>   >
>   >  Let's see if the winning proposal(s) can
>   correctly
>   >  distinguish mass and weight?
>   >
>   >  Gene Mechtly

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