Astrodynamics/Basic Rocketry

This section of the introduction will cover the basic ideas and theory of how rockets fly and leave the atmosphere as well as introduce the rocket equation.

The Ideal Rocket Equation

Rockets are momentum exchange devices the function by expelling some fluid (usually very hot gas or plasma) which "pushes" the rocket via Newtons third law of motion. Rocket are technically all around us ranging from simple water bottle rocket, to fireworks, to more sophisticated rockets like the Saturn V. The Tsiolkovsky Rocket Equation, derived below, is basic and fundamental principle of how rockets fly and shows the maximum change in velocity, $\Delta V$ (Delta V), that can be achieved by a rocket provided no external forces act on it.

$\Delta V=v_{e}\ln {\left({\frac {m_{0}}{m_{f}}}\right)}=I_{SP}g_{0}\ln {\left({\frac {m_{0}}{m_{f}}}\right)}$

Where,

• $v_{e}$ is the effective exhaust velocity (out of the nozzle) equal to $I_{SP}g_{0}$.
• $I_{SP}$ is the specific impulse of measuring in seconds. It's a measure of solid rocket fuel efficiency specifically the impulse created per unit weight (on Earth) of propellant.
• $g_{0}$ is the standard gravity on Earth near ground level.
• $m_{0}$ is the mass of the rocket before firing it's engines. Commonly the initial total mass is used (wet mass).
• $m_{f}$ is the mass of the rocket after the engines stop burning. Commonly the final total mass is used (dry mass).

The Rocket Equation states that with a faster exhaust velocity the greater finally velocity of the rocket; however, due to the natural logarithm there is an exponential increase in the initial mass of the rocket. Therefore, it is not beneficial to increase the mass of rocket as it will have negligible return and instead the uses of stages (multiple rocket stacked on top of each other) is more beneficial. Further, it's important to state that it might be misleading thinking that more efficient rockets (higher $I_{SP}$) are better at launch, since the rocket may not lift of the ground if the thrust force propelling it up is too small. This is due to the fact that the equation functions with no external forces acting on it.

Picture a rocket or a mass in space of mass, $m$, traveling at a velocity, $v$. If we take a small lump of the mass, $dm$, and throw it, as a speed of $v_{e}$ relative to the rocket, opposite to the direction of the flight (anit-colinear with velocity). We can expect a small change in velocity, $dv$, of the mass in space. Since, there are no other forces acting on this the momentum of the mass before expelling of the smaller mass and the one after must be equal due to the conservation of momentum. Let $P_{1}$ and $P_{2}$ denote the linear momentum before and after respectively.

$P_{1}=mv$

$P_{2}=(m-dm)(v+dv)+dm(v+dv-v_{e})$

Note, that both moments are taken from a stationary observer not on either of the masses, hence why the dv is added to the small masses momentum. Setting these two equations equal to one another the equation becomes:

$mv=(m-dm)(v+dv)+dm(v+dv-v_{e})$

$mv=mv-vdm+mdv-dmdv+vdm+dmdv-v_{e}dm$

After cancelations,

$mdv=v_{e}dm$

It would be unfair to not mention here that this equation can create two useful equations in itself, an ODE (first one shown below) of a 1D rocket dynamics, and the Tsiolkovsky equation (second one shown below). These equations are achived by making the time between the two moments $P_{1}$ and $P_{2}$ smaller. In other words taking the limit as time approaches zero.

$m(t){\frac {dv}{dt}}={\dot {m}}v_{e}$

$\Delta v=v_{f}-v_{0}=v_{e}\ln {\left({\frac {m_{0}}{m_{f}}}\right)}$

Although the Rocket Equation has it's limitations it is a useful resources for explaining future topics such as transfers. However, to elaborate on reason why the efficiency might be misleading it's worth noticing that in the derivation we have found the $F=ma$ equation. Hence, the instantaneous thrust force on the rocket in 1D is shown below. As can be seen if the ISP is high and the rocket motor is efficient it will not mean much if the mass flowrate, ${\dot {m}}$, is low and the thrust cannot surpass gravity and aerodynamical drag.

$T=v_{e}{\dot {m}}=g_{0}I_{SP}{\dot {m}}$