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The general thrust equation is $F = \frac{dm}{dt}\cdot v$, where $m$ is propellant's mass and $v$ is the exhaust velocity, is the equation right? What if the propellant is highly relativistic? One gram of propellant expelled at $290,000,000\ \text{m/s}$ should produce $290,000\ \text{N}$, according to the above equation. However, this doesn't take relativistic change into account.

What I want to know is how to calculate thrust when relativistic exhaust is involved.

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  • $\begingroup$ Technically, the general thrust equation is $F=\dot{m}_ev_e+(p_e-p_0)A_e$ where $p_i$ are the exhaust & ambient pressures and $A_e$ the cross-sectional area of the nozzle. You've cited for the (ideal) case of $p_e=p_0$. $\endgroup$ – Kyle Kanos Jan 2 '16 at 15:02
  • $\begingroup$ Just as an aside: the $I_{sp}$ of that rocket motor would be around 29.4 million seconds (almost a whole year)! $\endgroup$ – pr1268 Apr 16 '17 at 13:15
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All you need to know is the rate of change of momentum, since that is equal to the force. For a relativistic body the momentum is given by:

$$ p = \gamma mv $$

where $\gamma$ is the Lorentz factor and $m$ is the rest mass of the body. So basically just multiply your non-relativistic calculation by $\gamma$.

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