### Section 3.5 - Staging

#### The Tyranny of the Rocket Equation

Launching a rocket presents us with a catch-22. The rocket and payload have mass, so we need fuel to launch it. But the fuel itself has mass, so we need more fuel to launch the fuel, and more fuel to launch that fuel... how do we get around this?

The solution is to build rocket stages, each containing its own rocket engines and propellant. A stage is jettisoned when it runs out of propellant, saving the weight of the depleted section of the rocket, while the next stage fires its engine and continues accelerating the now-smaller rocket The LEGO model of the Saturn V clearly displays its stages

#### The Tsiolkovsky Rocket Equation

Also known as the ideal rocket equation, it relates a rocket's maximum change in velocity with the effective exhaust velocity and the initial and final masses of the rocket

$\Delta v = v_e \cdot ln(\frac{m_0}{m_f})$

$\Delta v =$ maximum change in velocity
$v_e =$ effective exhaust velocity
$m_0 =$ initial mass including propellant
$m_f =$ final mass without propellant (aka dry mass)

Note: effective exhaust velocity is basically the same as actual exhaust velocity, corrected for losses such as friction and non-axially directed flow (ie not straight down) as well as pressure differences between the rocket and the atmosphere. It's calculated as simply as $v_e = I_{sp} \cdot g_0$

The equation can be rearranged into the following form by dividing both sides by $v_e$ and exponentiating everything by $e$ to give:

$\frac{m_0}{m_f} = e^{\Delta v / v_e}$

Since the difference between $m_0$ and $m_f$ is the propellant that was combusted and expelled from the rocket, this equation is handy because it can tell us what proportion of the rocket needs to be propellant. We'll apply this formula to solve two related problems that will mathematically prove the advantage of rocket staging (adapted from Wikipedia)

#### Example 1

Assume a single stage rocket produces effective exhaust velocity of 4,500 m/s and needs to reach a delta-v of 9,700 m/s in order to reach Low Earth Orbit. What percentage of the initial mass needs to be propellant?

$\frac{m_0}{m_f} = e^{9700/4500} = 8.633$

But we're looking for $\frac{m_f}{m_0}$ since the final mass over the initial mass tells us what proportion of the rocket is dry mass

$\frac{m_f}{m_0} = \frac{1}{8.633} = 11.6\%$

$1 - 11.6\% = 88.4\%$

88.4% of the initial mass of the rocket is propellant

The above steps demonstrate algebraically that $\frac{m_0 - m_f}{m_0} = 1-e^{-\Delta v / v_e}$. We'll go ahead and use this shorter formula for the next problem to keep it more concise

#### Example 2

Now assume the rocket has two stages. The first stage will get us to a delta-v of 5,000 m/s, and the second stage will get us the remaining 4,700 m/s. Also, assume that the dry mass of the first stage alone is 8% of the total mass. Exhaust velocity is still 4,500 m/s

$\frac{m_0 - m_f}{m_0} = 1-e^{-5000/4500} = 67.1\%$

The first stage is accelerating the rocket to 5,000 m/s, so 67.1% of the total rocket's mass is the propellant of the first stage

$100\% - 67.1\% - 8\% = 24.9\%$

67.1% of the rocket's total mass just got expelled as burned propellant. Then, when the first stage jettisoned, it's dry mass (given as 8%) is also dropped, so we know that the second stage is equal to 24.9% of the total mass of the rocket

$\frac{m_0 - m_f}{m_0} = 1-e^{-4700/4500} = 64.8\%$

$64.8\% \cdot 24.9\% = 16.1\%$

64.8% of the remaining weight (second stage only) must be propellant, which is 16.1% the weight of the original rocket before the first stage was jettisoned. We could also determine that the dry mass of the second stage is equal to 8.8% of the original weight of the rocket ($35.2\% \cdot 24.9\%$)

$67.1\% + 16.1\% = 83.2\%$

83.2% of the mass of the original rocket must be propellant, a 5.2% improvement over the 88.4% needed for the single stage rocket

#### Max-$Q$

Max-$Q$ refers to the point during a rocket's ascent into orbit at which aerodynamic stress on the vehicle is at its highest. Aerodynamic stress is given by the formula:

$q = \frac{1}{2} pv^2$

$p$ is the local air density
$v$ is velocity

If it looks like the formula for kinetic energy, $KE = \frac{1}{2} mv^2$, that's because aerodynamic pressure can be thought of as the kinetic energy of the air with respect to the vehicle. Consider this:

• $Q$ is $0$ at liftoff because although the air is thick at the surface, the rocket is stationary ($v=0$).
• $Q$ is also $0$ by the time the rocket is in space because although the rocket is flying super fast, there's no air in the vacuum of space ($p=0$).

So if we were to plot $Q$ over the course of launch, we’d know that at some point during the flight there must be a maximum. Rockets engines are designed to throttle back a bit as max-$Q$ is reached to avoid putting too much stress on the vehicle and reduce the risk of structural failure. Max-$Q$ occurred for the Space Shuttle at an altitude of about 11 km. For the Saturn V, it occurred around 13 km

In videos of rocket launches, they'll often announce when max-$Q$ is reached. In the above video, it occurs at 1:25 ("vehicle is passing through max-$Q$")

#### Extra: Deriving the Tsiolkovsky Rocket Equation

This part is optional, but I try not to accept equations at face value without understanding the derivation (although sometimes the derivation of formulas is so complex I have no choice :( my finance degree isn't very useful for STEM). But I looked into the Tsiolkovsky Rocket Equation and was able to figure it out! (adapted from Wikipedia)

Consider the following system:

$V =$ velocity of the rocket at $t = 0$
$V + \Delta V =$ velocity of the rocket at $t = \Delta t$
$V_e =$ velocity of the mass added to the exhaust (and lost by the rocket) during time $\Delta t$
$m + \Delta m =$ mass of the rocket at $t = 0$
$m =$ mass of the rocket at $t = \Delta t$

$\Sigma F_i = \lim_{\Delta t \to 0} \frac{P_2 - P_1}{\Delta t}$

Newton's second law relates all external forces to the change in linear momentum of the whole system. We can recognize that if $F = ma$, $p = mv$, and $a = \frac{dv}{dt}$, then we can conclude $F = \frac{dp}{dt}$ (ie force is a change in momentum over time)

$P_1 = (m + \Delta m)V$          Momentum of the rocket at $t = 0$

$P_2 = m(V + \Delta V) + \Delta mV_e$          Momentum of the rocket at $t = \Delta t$

We're also going to want to make note of this equation:

$V_e = V - v_e$

Velocity of the exhaust in the observer (ie inertial frame of reference), $V_e$, is equal to the velocity of the rocket $V$ minus the apparent velocity of the exhaust from the point of view of the rocket, $v_e$

For example, for a rocket moving forward at 100 mph, if the exhaust appears to be traveling backward at 150 mph from the point of view of the rocket, then an observer would record an exhaust velocity of -50 mph (50 mph in the negative direction)

Now that we've laid the groundwork, we can move forward:

$P_2 - P_1 = m\Delta V - \Delta m v_e$

Subtract the formulas of $P_2$ and $P_1$. The $mV$ terms cancel out, and the $\Delta m v_e$ term arises from the substitution of $v_e = V - V_e$

$\Sigma F_i = m\frac{dV}{dt} + v_e \frac{dm}{dt}$

Substitute the formula for $P_2 - P_1$ back into the original equation. Also, keep in mind that $dm = -\Delta m$ since ejecting positive $\Delta m$ results in a decrease in mass

$m \frac{dV}{dt} = -v_e \frac{dm}{dt}$

If there are no external forces, then $\Sigma F_i = 0$

$$\int_{0}^{\Delta t} \frac{dV}{dt} = \int_{0}^{\Delta t} -v_e\frac{1}{m}\frac{dm}{dt}$$

Divide both sides by m and integrate with respect to t

$\Delta V = v_e \cdot ln(\frac{m_0}{m_1})$

Assuming $v_e$ is a constant