0
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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
    Commented Jan 2, 2016 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
    Commented Apr 16, 2017 at 13:15

1 Answer 1

Reset to default
3
$\begingroup$

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 $\gamma$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.