I've been trying to derive the rocket equation, and I've faced a minor confusion regarding a certain intuition. I've no problem deriving the equation mathematically though.
There are two ways, one can use momentum conservation to derive the rocket equation.
In both cases, you have initial momentum $p_i = mv$
As for final momentum, you have either $$p_f = (m+dm_r)(v+dv)-dm_r u$$
Or you can have $$p_f = (m-dm_g)(v+dv)+dm_gu$$
These are both correct in the sense that $m_g$ is mass of the released exhaust, and $dm_r$ is the mass of the rocket, or the mass of the gas in the rocket. And, from conservation of mass, we can easily say $dm_r = -dm_g$.
Hence we can solve the equation in two ways, either integrating over $dm_g $ or integrating over $dm_r$, noting that the former is positive while the latter is negative.
When we integrate over $dm_r$, the integral usually takes the form : $-v_{rel}\int_{m_i}^{m_f} \frac{dm_r}{m}$ where $m_f<m_i$.
Hence our final answer is positive, as the two negatives cancel each other out. The rocket accelerates, and this is what we expect.
Now, let us integrate over $dm_g$. This integral would be positive, and from the perspective of the exhausted gas, the relative velocity of the rocket is positive too. So, we'll have $+v_{rel}$ in front of the integral. This integral would take the form :
$v_{rel}\int_{m_f}^{m_i} \frac{dm_g}{m}$ and since $m_i>m_f$, our answer is exactly the same.
We have just flipped the sign of relative velocity, and the limits of the integral, and our result is now the same.
My question is, what exactly does this new integral represent?
In the old case, we had $\int_{m_i}^{m_f} \frac{dm_r}{m}$. This represented the change in mass of the rocket which is negative as the rocket is losing mass.
However, what does the new integral
$\int_{m_f}^{m_i} \frac{dm_g}{m}$ represent exactly? I've seen many posts describing it as the amount of mass of the exhaust that is removed from the rocket and added to the surrounding. Since we are looking from the perspective of this exhaust mass, and since it is increasing as it is leaving the rocket, the integration is positive, which is what we expect.
However, here comes the problem – the limits of the integral. In the previous case, there was some initial mass, and then there was some final mass, so the limits went from $m_i$ to $m_f$. However, in the case of the exhaust, there was no gas outside the rocket, but then gas was slowly added, as it escaped the rocket. Shouldn't the limits be from $0$ to some mass $m_i – m_f$, as that is the amount of gas that has been released from the rocket, and added to the outside?
This new integral doesn't make any physical sense to me, how do I exactly visualize it? We claim that the limits are $m_f$ to $m_i$, as this gives us the right answer. But the expression becomes intuitively non-sensical to me if I consider the integral to represent the addition of matter to the outside. How exactly do I make sense of this? Or should I just trust the math, regarding this?
So, my main problem is interpreting this second integral, if I'm integrating over $dm_g$, which is the mass being added to the surrounding. Shouldn't the limits go from $0$ to some mass, instead of initial mass inside the rocket to the final mass ?
Best Answer
There is nothing as integrating $dm_g$.
$dm_g$ is not defined in the problem, however, the initial and final masses of the main body are given. The change in mass of the main body is given by $dm_r$. We are integrating over the mass of the main body, and so, our variable of integration must be $dm_r$. Moreover, we know that $dm_g = -dm_r.$
Hence, when we change $dm_g$ into $dm_r,\space$ we plug in the limits of $dm_r$, and an extra minus sign in front, which interchanges the limit.
So, the integration $\int_{m_f}^{m_i} \frac{dm_g}{m}$ doesn't make any sense, firstly because $dm_g$ is not defined here and $m$ is the mass of the body, not the exhaust.
The real form is :
$$\int \frac{dm_g}{m} =-\int_{m_i}^{m_f} \frac{dm_r}{m}=\int_{m_f}^{m_i} \frac{dm_r}{m} \ne \int_{m_f}^{m_i} \frac{dm_g}{m}$$
It doesn't make sense to put any limits in the first one for obvious reasons.
The mistake that I made was, I correctly interchanged the limits, but didn't replace $dm_g$ with $dm_r$, which led me to believe that we were still integrating the mass of the exhaust.