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.
  • #1
halcyon_zomboid
2
0
TL;DR Summary
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.

$$\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v(1-v_\mathrm{ex}\frac{v}{c^2})}{(1-\frac{v^2}{c^2})(-v_\mathrm{ex}\frac{v^2}{c^2}+(1-a)v+av_\mathrm{ex})}
$$
This is purportedly the relativistic rocket equation when accounting for annihilation to usable exhaust. To quote Wikipedia, " ##a## is the fraction of the original (on board) propellant mass (non-relativistic) remaining after annihilation (i.e., ##a=0.22## for the charged pions)."

I can see that this reduces to the usual relativistic rocket equation when ##a=0##:
$$
\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v}{v_\mathrm{ex}(1-\frac{v^2}{c^2})}
$$

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 ##M+\mathrm{d}M##. Its final mass is ##M## and its velocity in this frame is ##\mathrm{d}v'##. Let's say the exhaust has mass ##\mathrm{d}m## and velocity ##v_\mathrm{ex}##, so if all the exhaust were usable, energy conservation would give
$$(M+\mathrm{d}M)c^2-Mc^2=\frac{c^2\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}
$$
(to first order in ##\mathrm{d}v'##). However, in reality, not all of the exhaust is usable. Let's say only a fraction ##\alpha## of the energy loss ##c^2\,\mathrm{d}M## is captured by beamed core exhaust pions:
$$
\alpha c^2\,\mathrm{d}M=\frac{c^2\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}.
$$

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 ##M\,\mathrm{d}v'## (to first order in ##\mathrm{d}v'##), and that of the exhaust is
$$
-\frac{v_\mathrm{ex}\,\mathrm{d}m}{\sqrt{1-\frac{v_\mathrm{ex}^2}{c^2}}}
=-\alpha v_\mathrm{ex}\,\mathrm{d}M
$$

But doesn't this then mean that

$$
\frac{\mathrm{d}M}{M}=-\frac{\mathrm{d}v'}{\alpha v_\mathrm{ex}}=-\frac{\mathrm{d}v}{\alpha v_\mathrm{ex}(1-\frac{v^2}{c^2})}
$$
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 ##v_{ex}## and ##v## in your formula ? I don' t find the former in Wikipedia you referred.
 

FAQ: Deriving the relativistic rocket equation with exhaust efficiency

What is the relativistic rocket equation?

The relativistic rocket equation describes the relationship between the velocity of a rocket and the mass of the fuel it expels, taking into account the effects of special relativity. It extends the classical Tsiolkovsky rocket equation to velocities approaching the speed of light.

How do you account for exhaust efficiency in the relativistic rocket equation?

Exhaust efficiency can be incorporated into the relativistic rocket equation by introducing a parameter that represents the fraction of the exhaust velocity that contributes to the rocket's thrust. This parameter, often denoted by η (eta), modifies the effective exhaust velocity in the equation.

What assumptions are made when deriving the relativistic rocket equation?

Key assumptions include that the rocket expels mass at a constant rate, the exhaust velocity is constant in the rocket's instantaneous rest frame, and the effects of gravity and other external forces are negligible. Additionally, the derivation assumes the principles of special relativity apply.

Can you provide the final form of the relativistic rocket equation with exhaust efficiency?

The relativistic rocket equation with exhaust efficiency can be expressed as:\[ \frac{v}{c} = \tanh\left(\frac{\eta v_e}{c} \ln \frac{m_0}{m_f}\right) \]where \(v\) is the final velocity of the rocket, \(c\) is the speed of light, \(v_e\) is the exhaust velocity, \(\eta\) is the exhaust efficiency, \(m_0\) is the initial mass, and \(m_f\) is the final mass of the rocket.

Why is exhaust efficiency important in the context of relativistic rockets?

Exhaust efficiency is crucial because it determines how effectively the rocket converts the energy of the expelled fuel into thrust. Higher exhaust efficiency means more of the fuel's energy is used to accelerate the rocket, which is especially important when dealing with relativistic velocities where energy requirements are significantly higher.

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