Deriving the relativistic rocket equation with exhaust efficiency

In summary, the derivation of the relativistic rocket equation with exhaust efficiency involves applying the principles of special relativity to the traditional rocket equation. It incorporates the effects of relativistic mass increase and the exponential loss of mass as fuel is expelled. The equation accounts for the rocket's velocity, the effective exhaust velocity, and the efficiency of the fuel used, allowing for precise calculations of a rocket's performance at high speeds. This framework enables a better understanding of how relativistic effects impact the dynamics of space travel and propulsion systems.
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halcyon_zomboid
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My derivation produces a relativistic rocket equation with exhaust "efficiency" that disagrees with this one: https://en.wikipedia.org/wiki/Antimatter_rocket#Modified_relativistic_rocket_equation
Hi,

I'm looking at this relativistic rocket equation on Wikipedia. Something doesn't make sense here, and I can't find a derivation for this equation in the linked source, so I'm trying to derive it myself with limited success.


This is purportedly the relativistic rocket equation when accounting for annihilation to usable exhaust. To quote Wikipedia, " is the fraction of the original (on board) propellant mass (non-relativistic) remaining after annihilation (i.e., for the charged pions)."

I can see that this reduces to the usual relativistic rocket equation when :


However, I can't seem to derive the above equation. Here's my attempt (which succeeds for the normal relativistic rocket equation):

In the COM frame, the rocket's initial mass is . Its final mass is and its velocity in this frame is . Let's say the exhaust has mass and velocity , so if all the exhaust were usable, energy conservation would give

(to first order in ). However, in reality, not all of the exhaust is usable. Let's say only a fraction of the energy loss is captured by beamed core exhaust pions:


Okay, now let's look at momentum. For simplicity's sake, we'll assume all the useless energy is radiated symmetrically, so there's no effect on momentum. The final momentum of the rocket is (to first order in ), and that of the exhaust is


But doesn't this then mean that


when we go back to the lab frame? I get nothing resembling the equation at the start.

Is there something wrong with my derivation?
 
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  • #2
What are and in your formula ? I don' t find the former in Wikipedia you referred.
 

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