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   In outdated studies, some engineers equated mass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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