Everyone knows about the famous Tsiolkovsky rocket equation. However, he formulated the equation neglecting many things. So, here, I am trying to find the real equation.
My thinkings:
Suppose, the rocket has mass $M_i$ (including the fuel) at time $t_0$. At time $t_f$, it will have mass $M_f$.
Now, we have to calculate the velocity that it will have at that point when time is $t_f$.
Suppose,at time $t$, the rocket has mass $m$ having velocity $v$. After time interval $dt$ , it has mass $m – dm$ now having velocity $v + dv$ and the gas of $dm$ has relative velocity $u$ . So, after doing a little work of algebra and calculus, then applying 2nd law, we get,
$$\sum{F_\text{ext}} + F_{\text{thrust}} = m.\dfrac{dv}{dt}$$
Now, since height changes, $\dfrac{dg}{dt} = J \implies dg = J.dt \implies g_t = J.t + g_0$
Now, velocity changed during $dt$ :
$$ dv = g.dt + \dfrac{f_\text{drag}}{m} + \dfrac{u\frac{dm}{dt}}{m} .$$
Now, we have to integrate this as:
$$ \int_{0}^{v_f} dv = \int_{t_0}^{t_f} g(t).dt + \int_{M_i}^{M_f} \dfrac{f_\text{drag}}{m} + \int_{M_i}^{M_f} \dfrac{\frac{dm}{dt}}{m}$$ where $F_\text{ext}$ is the net external force; $F_\text{thrust}$ is the thrust force;$f_\text{drag}$ is the drag force of the surroundings. Hmmm… Here, I am in dilemma, say it because of my limited knowledge of calculus( I am still learning,17 only!), due to change in height $g$ changes; therefore we have to include the rate of change of $g$ i.e. Jerk, air thickness changes as a result the drag also changes; and also, the relative velocity and the rate of ejection of gases change along with the change of main mass. So, how to integrate the equation considering all these changings?? Please help me in formulating the final equation.
Best Answer
I do what I can to take questions at face-value, but there's a flaw in the formulation here. That's okay, because rocket science isn't intuitive. Your question comes down to
Here are the factors which have so-far been considered:
This misses gravity drag. What is gravity drag? I say it's the loss due to imperfect vectoring of an orbit change. To do a perfect orbit-change maneuver, you will ideally burn your rockets in the opposite direction of your motion (and at perigee, a factor I ignore here). But as a Earth-dwelling individual, you have a practical constraint because you can't go below r = {the radius of Earth}. So you're stuck spending a lot of your fuel dealing with the problem of not crashing into the ground. Imagine a jet pack. You fire your rockets, however for a minute, and land right where you started (for fun). That's 100% gravity drag.
For some accounting, I would read this question on Space Stack Exchange. Different answers will give you a few attempts at a break-down of how much different factors count for. This is a relatively common topic, but I it's hard to find examples where people get the specifics right. The bottom line is the gravity drag dominates among the different factors. Air drag is somewhat significant, but it's not the dominant one. In fact, air drag is mostly a problem because it compounds the impact from gravity drag. You have to move slower through the atmosphere, and that causes you to spend more fuel - NOT just because you're pushing to move through the air, but because it causes you to hover at sub-orbital speed, costing you propellant via gravity drag. So a perfect accounting is really like gravity drag, air-drag induced gravity drag, air drag, and then altitude climb. These are inter-related, mutually compounding each other, so it's hard to nearly organize them into different numbers. Perhaps that's why it's so hard to find good lessons on this.
Implementation
So let me answer your question. Dealing with tidal effects combined with air drag will result in an unsolvable equation. Even dealing with air drag alone will be unsolvable. You just can't do that because of the gravity turn - the process where the rocket starts mostly vertical and then turns sideways. An algebraic equation is hopeless for these cases.
But let's not be so gloomy. The largest factor is gravity drag, after all. So instead of revising the rocket equation for a laundry list of factors, perhaps we can make a more accurate expression by accounting for the most major factor. I can believe this is possible for gravity drag.
In my humble opinion, you would have the best shot at getting a solution by assuming that the altitude remains the same throughout the entire orbital launch. You could (somewhat validly) assume a constant thrust or constant acceleration. That means that your rocket will turn based on the need to remain at a constant radius. For a launch on an airless world, this wouldn't be so bad. It might also get you a large fraction of the way between Earth's 7.8 km/s orbital velocity and the 9 or 10 km/s that is required to launch a rocket to orbit.
Off the top of my head, I'm not sure if this scenario has been solved before. I'm not even sure if it's possible. But if you wanted to go at the problem, this is the most promising route.
Caveats
If you solved the above scenario, it wouldn't be a new form of the "rocket equation" because it only applies to trips from a stationary point on the surface to a low orbit. If you were transferring from one orbit to another with a Holman transfer, you would have to do it all over again. This fact is one reason that the rocket equation has so many fans - it applies generally.