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# Simple Rocket Science
[[!toc levels=4]]
## Getting Started
Real spaceflight dynamics are complex. There are only a few basic forces involved, but some of them (namely air resistance) are nonconservative and tricky. In order to make a useful, accurate model of spaceflight you have to build a complex numerical approximation. However there is no need to be too complicated too quickly. A great place to start is to find an analytical solution for a rocket, solving for the height it will travel. Put everything in the 'up' axis and ignoring a ton of things will result in an easy place to start.
### The Simplest Case
The true simplest case is a pointmass rocket that is in free space with no gravity, airresistance and has a simplified model of propulsion. Describing this situation mathematically should result in the classic
[Tsiolkovsky rocket equation](http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation).
#### Rocket in free space
Consider a rocket in free space. There are no external forces. We will be nice and constrain the motion of the rocket in one direction. At some time the rocket starts to eject mass out of one end of the rocket. Through conservation of linear momentum this will propel the rocket forward.
##### Derivation
The motion of the rocket in our simple case will be determined by Newton's 2<sup>nd</sup> law. Usually written as **F** = m**a** since we have a variable mass system we use the more general form of
[[!teximg code="\vec{F} = \vec{\dot{p}}" ]]
We can safely ignore the vector part because all motion is in one dimension. Also we can rewrite pdot as dp/dt because we are going to use some arguments about infinitesimals to finish our derivation.
[[!teximg code="F = \frac{dp}{dt}" ]]
We can write an equation for dp from the diagram below because we know the momentums involved. From the definition of infinitesimal values
[[!teximg code="dp = p(t + dt)  p(t)" ]]
The momentum at some time plus dt will be the momentum of the rocket plus the momentum of the exhaust gas
[[2.2.png]]
In this diagram you can see we have a rocket of mass m moving at some velocity relative to an inertial reference frame v. At some time t + dt it will release an exhaust mass dm' at some velocity relative to the rocket v<sub>e</sub>. Many textbooks refer to v<sub>e</sub> as u. In a real rocket is an ideal "effective velocity" of the exit gasses and directly proportional to efficiency of the engine. It is related to the more common term I<sub>sp</sub>.
[[!teximg code="\mbox{mass of rocket at } t + dt = m  dm^\prime" ]]
[[!teximg code="\mbox{velocity of rocket at } t + dt = v + dv" ]]
[[!teximg code="\mbox{mass of exhaust mass} = dm^\prime" ]]
[[!teximg code="\mbox{velocity exhaust mass} = (v + dv)  v_e" ]]
Therefore
[[!teximg code="dp(t + dt) = (m  dm^\prime)(v + dv) + (dm^\prime)(v + dv  v_e)" ]]
[[!teximg code="p(t) = mv" ]]
[[!teximg code="dp = (m  dm^\prime)(v + dv) + (dm^\prime)(v + dv  v_e)  mv" ]]
Simplifying and throwing out higher order terms will reveal
[[!teximg code="dp = m dv  v_e dm^\prime" ]]
A positive exhaust mass is a mass lost from the rocket so we say that
[[!teximg code="dm^\prime =  dm" ]]
[[!teximg code="dp = m dv + v_e dm" ]]
In this simple case we have no external forces.
[[!teximg code="F dt = dp = 0" ]]
[[!teximg code="0 = m dv + v_e dm" ]]
Solving for dv gives
[[!teximg code="dv =  \frac{v_e}{m}dm" ]]
We can integrate both sides to find
[[!teximg code="\int^{v_f}_{v_0}dv =  v_e\int^{m_f}_{m_0}\frac{1}{m}dm" ]]
[[!teximg code="v_f  v_0 =  (v_e \ln (m_f)  v_e\ln (m_0))" ]]
Simplified this becomes the
##### Tsiolkovsky rocket equation
[[!teximg code="\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)" ]]
Since it is hard to look up values of v<sub>e</sub> in books about engines we can use its relation to I<sub>sp</sub>
[[!teximg code="v_e = g_0 I_{sp}" ]]
Where g<sub>0</sub> is the acceleration due to gravity at the surface of the Earth and I<sub>sp</sub> is the specific impulse in seconds.
[[!teximg code="\Delta v = g_0 I_{sp} \ln\left(\frac{m_0}{m_f}\right)" ]]
### Adding Gravity
Starting with our previous result of
[[!teximg code="Fdt = m dv + v_e dm" ]]
But now having an external force of gravity, mg
[[!teximg code=" mgdt = m dv + v_e dm" ]]
Solving for dv gives
[[!teximg code="dv =  \frac{v_e}{m}dm  g dt" ]]
Or as an equation of motion
[[!teximg code="\ddot x =  \frac{v_e}{m} \dot m  g"]]
Integrating over v, t and m (Where t<sub>bo</sub> is the burntime of the rocket)
[[!teximg code="\int^{v_{bo}}_{v_0}dv =  v_e\int^{m_f}_{m_0}\frac{1}{m}dm  g\int^{t_{bo}}_{0}dt" ]]
Gives an equation for the velocity at burnout
#### Burnout Velocity
[[!teximg code="v_{bo} = v_e \ln\left(\frac{m_0}{m_f}\right)  gt_{bo} + v_0" ]]
The term gt is referred to as gravity loss. This represents the losses endured by launching in a gravity well. To maximize burnout velocity you want to minimize gravity loss, which means burning the fuel as fast as possible. This makes sense because when you spend a long time burning fuel you are wasting energy lifting unburnt fuel to a higher altitude rather than your payload.
In the real world usually rocket motors are categorized by thrust and not burn time. So it becomes useful to rewrite this equation in terms of a new parameter, thrusttoweight ratio. We define thrusttoweight ratio, Ψ, as the thrust (which we assume is constant) divided by the weight at liftoff, m<sub>0</sub>g.
[[!teximg code="\mbox{ThrusttoWeight Ratio } \equiv \Psi = \frac{F_{thrust}}{m_0g}" ]]
If we realize that thrust is
[[!teximg code="F_{thrust} = v_e \dot m" ]]
then
[[!teximg code="\Psi = \frac{v_e \dot m}{m_0g}" ]]
We can find a relation to the time it takes to burn through the fuel. If we take the burn rate to be constant (again, not a bad assumption) then the time is to burn out is
[[!teximg code="\dot m = \frac{dm}{dt}" ]]
[[!teximg code="dt = \frac{dm}{\dot m}" ]]
[[!teximg code="\int^{t_{bo}}_{0} dt = \frac{1}{\dot m} \int^{m_f}_{m_0} dm" ]]
[[!teximg code="t_{bo} = \frac{m_f  m_0}{\dot m}" ]]
However we see that m<sub>f</sub>  m<sub>0</sub> is negative! This okay because mdot is secretly also negative. What we actually want is the fuel mass divided by a (positive) burn rate.
[[!teximg code="t_{bo} = \frac{m_0  m_f}{\dot m}" ]]
We also want this in terms of mass ratio and thrusttoweight ratio. So we do some clever math, first pulling out m<sub>0</sub>
[[!teximg code="t_{bo} = \frac{m_0}{\dot m} \left( 1  \frac{m_f}{m_0} \right)" ]]
Now if we multiply by "one" we get
[[!teximg code="t_{bo} = \frac{v_e}{g} \frac{g m_0}{v_e \dot m} \left( 1  \frac{m_f}{m_0} \right)" ]]
We recognize the term [[!teximg code="\frac{g m_0}{v_e \dot m}"]] as the inverse of Ψ Thus we have the burnout time in reasonably terms
[[!teximg code="t_{bo} = \frac{v_e}{g} \frac{1}{\Psi} \left( 1  \frac{m_f}{m_0} \right)" ]]
Plugging this in for t in our burnout velocity equation gets
[[!teximg code="v_{bo} = v_e \left[\ln\left(\frac{m_0}{m_f}\right)  \frac{1}{\Psi} \left( 1  \frac{m_f}{m_0} \right)\right] + v_0" ]]
Now we can also introduce the symbol μ for the mass ratio (for brevity's sake) and replace v<sub>e</sub> with gI<sub>sp</sub>
[[!teximg code="v_{bo} = g I_{sp} \left[\ln(\mu)  \frac{1}{\Psi} \left( 1  \frac{1}{\mu} \right)\right] + v_0" ]]
We see now that we have burnout velocity that is in general a function of three parameters: μ, Ψ and I<sub>sp</sub>
[[!teximg code="v_{bo} (\mu, \Psi, I_{sp})" ]]
Three variables and 3 dimensions is just begging for some graphing. Here is a graph showing the burnout velocity of a vertical rocket at an I<sub>sp</sub> of 250 [s]
[[!img v_bo.png size="500x500" caption="Burnout Velocities for an Isp of 250. Click to enlarge"]]
You can see that there a heavy dependence on mass ratio. But also to avoid gravity loss it is beneficial to have Ψ as high as you can too. A typical thrusttoweight ratio is somewhere between 2 and 3 for large commercial rockets.
We can also talk about what effect changing I<sub>sp</sub> has on the burnout velocity. If we take the partial with respect to I<sub>sp</sub> we get
[[!teximg code="\frac{\partial v_{bo}}{\partial I_{sp}} = g \left[\ln(\mu)  \frac{1}{\Psi} \left( 1  \frac{1}{\mu} \right)\right]" ]]
For a rocket with a μ of 10 and a Ψ of 2 we see that adding an extra second of I<sub>sp</sub> results in of 18.2 m/s of extra velocity.
[[!teximg code="\frac{\partial v_{bo}}{\partial I_{sp}} = 9.80665 [m/s^2] \left[\ln(10)  \frac{1}{2} \left( 1  \frac{1}{10} \right)\right] = 18.2 [m/s/s]" ]]
#### Burnout Height
Now that we have done some work to find the burnout velocity we can find the burnout height much the same way. Starting with our differential equation for rocket motion we double integrate to find range
[[!teximg code="dv =  \frac{v_e}{m}dm  g dt" ]]
[[!teximg code="\iint^{h_{bo}}_{h_0} dv =  v_e \iint^{m_f}_{m_0}\frac{1}{m}dm  \iint^{t_{bo}}_{0}g dt" ]]
I had to look this one up in a table of integrals but the answer should be
[[!teximg code="h_{bo} = v_e \left[ m_f \ln(m_f)  m_f  m_0 \ln(m_0)  m_0 \right]  \frac{1}{2} g t_{bo}^2 + h_0"]]
This can be rearranged to look like
[[!teximg code="h_{bo} = v_e \frac{m_0}{\dot m} \left( 1  \frac{m_f}{m_0} \left( \ln\left(\frac{m_0}{m_f}\right) + 1 \right)\right)  \frac{1}{2} g t_{bo}^2 + h_0"]]
Doing the same substitution of t and using μ we get
[[!teximg code="h_{bo} = v_e \frac{m_0}{\dot m} \left( 1  \frac{1}{\mu} ( \ln(\mu) + 1)\right)  \frac{1}{2} g \left( \frac{v_e}{g\Psi} \left( 1  \frac{1}{\mu} \right)\right)^2 + h_0"]]
[[!teximg code="h_{bo} = \frac{v_e^2}{\Psi g} \left( 1  \frac{1}{\mu} ( \ln(\mu) + 1)\right)  \frac{v_e^2}{2g\Psi}\left(1  \frac{1}{\mu}\right)^2 + h_0" ]]
Now making our I<sub>sp</sub> substitution
[[!teximg code="h_{bo} = \frac{g I_{sp}^2}{\Psi} \left[\left( 1  \frac{\ln(\mu) + 1}{\mu}\right)  \frac{1}{2}\left(1  \frac{1}{\mu}\right)^2\right] + h_0" ]]
## Ballistic Trajectory
After burnout the rocket will continue to move upwards on a ballistic trajectory. Since we are considering gravity as a constant and ignoring air resistance we can use energy arguments to easily figure out the final height the rocket will go
[[!teximg code="KE_{bo} = PE_{final}"]]
[[!teximg code="\frac{1}{2} m_f v_{bo}^2 = m_f g \Delta h"]]
Solving for the change in height
[[!teximg code="\Delta h = \frac{v_{bo}^2}{2g}"]]
[[!teximg code="h_f = \frac{v_{bo}^2}{2g} + h_{bo}"]]
## Bringing it all together
So if we really like math we can combine the ballistic and burnout equations to find the total height a rocket under constant gravity with no air resistance will travel (as a function of μ, Ψ and I<sub>sp</sub>)
[[!teximg code="h_f = g I_{sp}^2 \left[\frac{1}{2} \left[\ln(\mu)  \frac{1}{\Psi} \left( 1  \frac{1}{\mu} \right)\right]^2 + \frac{1}{\Psi} \left[\left( 1  \frac{\ln(\mu) + 1}{\mu}\right)  \frac{1}{2}\left(1  \frac{1}{\mu}\right)^2\right]\right]"]]
## References
[1] Turner, Martin J.L. _Rocket and Spacecraft Propulsion._ Springer Praxis Books. [New York]: Praxis Publishing Ltd, Chichester, UK, 2005.
[2] Thomson, William Tyrrell. _Introduction to Space Dynamics._ New York: Wiley, 1961.
