Understanding Relativistic Rockets: Exploring Energy and Momentum Conservation

[Mazimillion]

Hi. This one has been bugging me for a while. We briefly covered this in lectures getting just the main points with no derivations and I would like to fully understand this. I have a couple of questions and any help would be greatly appreciated.

Homework Statement

The main jist is this: A rocket is initially at rest, with respect to inertial frame K. Then the engines are fired and the rocket starts moving. Assume the exhaust gases are ejected at a constant rate v' in the opposite direction with respect to the rocket. Let m1 be the rest mass of the rocket that will, of course, change when material is being ejected.

The main goal is to calculate the velocity as a function of remaining mass m1. There are a couple of steps beforehand though.

Firstly, I note that the system rocket plus exhaust is isolated, thus total energy and momentum are conserved. Considering one infinitesimal step of rocket acceleration to the next, I let dm2 be the mass of a small chunk of exhausted material ejected within each infinitesimal step, v1 be the velocity of the rocket and v2 the velocity of the exhaust gas, as seen in K. Thus m1, v1 and v2 vary while dm2 is ejected.

Homework Equations

Energy conservation Law (given): $$d\left(\frac{m_{1}c^2}{\sqrt{1-\frac{v_{1}^2}{c^2}}}\right) + \frac{c^2 dm_{2}}{\sqrt{1-\frac{v_{2}^2}{c^2}}} = 0$$

Momentum Conservation (given): $$d\left(\frac{m_{1}v_{1}}{\sqrt{1-\frac{v_{1}^2}{c^2}}}\right) + \frac{v_{2} dm_{2}}{\sqrt{1-\frac{v_{2}^2}{c^2}}} = 0$$


Before I go any further, I am wondering why, in both conservation equations, we consider the differential of the entire first term, but only dm2 in the second term?

Secondly, I'm probably being really dense here, and I apologise for that, but if I wanted to calculate dm1 and dm2 as functions of m1, v1, dv1 and v2, how can this be done? It's clearly just a mathematical manipulation of the above expressions, but those differentials are putting me off - I'm probably just missing a trick.

Any help with this would be greatly appreciated, and should allow me to get a bit deeper into the problem.

Thanks in advance

 

[fantispug]

I'm having trouble with the line
"Assume the exhaust gases are ejected at a constant rate v' "
If this means the velocity of the ejected material is v' then you should be able to see why only the differential of m2 is taken in the second term.

If it means that dm/dt=v' then it's most certainly not clear, and I'd use total differentials.

In either case to help find the solution, can you think of any relation between dm1 and dm2?

 

[Mazimillion]

Ah, of course. If material is ejected at constant velocity, then its velocity does not depend on dm2. However, as the mass of the rocket m1 is decreasing, it velocity v1 is dependent on m1, hence the differential of the entire term.

Thus if dm2 is the mass of the ejected material, then dm1 must be the amount of material lost by the rocket, thus dm1 = dm2.

Right, that seems to make sense. Bearing that in mind, I'll work on expressing expressing dm1 and dm2 as functions of the other variables.

Thanks for your hint, fantispug

 

[Mazimillion]

Sorry I forgot to mention, your original assumption was correct. v' is the fixed speed at which the material is ejected.

 

[fantispug]

Actually, after doing a similar assignment myself I realize my advice was wrong, I'm really sorry to have given you rubbish advice.

Mass is not necessarily conserved, but energy is. The easiest way to see this is to think of a "light rocket" (e.g. a flashlight in space). It propels itself by releasing photons out of its back.
What is the mass of the emitted photons? What is the speed of the rocket? What is the mass of the rocket?
If you can answer these questions by conserving the appropriate quantities you're in good shape to tackle the more general problem.

The idea is to pick a good reference frame. As you say at each instant v2 is constant (but remember v' is relative to the rocket).

The key to solving this problem easily is picking a clever reference frame, one in which most of the terms are zero, and then transforming to the reference frame you're interested in.

Sorry again.