PlanetPhysics/Single Stage Rocket Burnout Height

Applying Newton's laws to rocket motion is not only exciting, but also quite instructional. Problems involving rocket motion illustrate how to use Newton's 2nd law when mass is not constant. Here we will look at how high a single stage rocket will go under the influence of gravity. To make the problem manageable, a few simplifying assumptions are made:

Motion in the y direction only
Drag is neglected
Constant burn rate for rocket
force due to gravity is constant
Rocket does not escape Earth's gravity
Ideal rocket
Lots of other miscellaneous terms

To familiarize the reader with what is involved in this calculation, we will start with the answer and then derive the equation. So the max altitude the rocket will achieve is given by

$y_{max}=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{f}t_{b}}{m_{i}-m_{f}}}ln\left({\frac {m_{f}}{m_{i}}}\right)+v_{e}t_{b}+{\frac {v_{b}^{2}}{2g}}$

Description of variables:

$t_{b}$ : Time that the rocket is burning fuel (given) \\ $g$ : acceleration of Gravity (given) \\ $m_{f}$ : Final mass of rocket after burn (given) \\ $m_{i}$ : Initial mass of rocket (given) \\ $v_{b}$ : velocity of rocket at burnout (calculated) \\ $v_{e}$ : Exhaust velocity, velocity of the fuel as it is ejected out of the Rocket (given)\\

The problem is best approached by breaking it into two parts. First, we calculate the altitude that the rocket reaches when all its fuel is burned. After burnout, the rocket still climbs to a higher altitude until gravity finally brings its velocity to zero. Think of it like shooting a bullet into the sky, after the initial thrust of the gun, the bullet still goes higher (duh!). So the second calculation is to add the distance traveled after burnout.

The goal is to apply Newton's 2nd law, so let us start there. For our one dimensional case

$F=ma=-mg$

Since we do not have constant mass throughout the rocket burn, we also have

$F={\frac {d}{dt}}\left(mv\right)$

Using the chain rule and setting (2) equal to (3)

$-mg={\frac {dm}{dt}}v_{e}+m{\frac {dv}{dt}}$

multiply by dt

$-mgdt=v_{e}dm+mdv$

We still have three differentials, so we cannot directly integrate this equation. Since we are assuming a constant burn rate k and it is positive

${\frac {dm}{dt}}=-k$

so

$dt=-{\frac {dm}{k}}$

plug this into (4) to get

$-{\frac {mg}{k}}dm=v_{e}dm+mdv$

Separate variables to setup the integration

$\left({\frac {mg}{k}}-v_{e}\right)dm=mdv$

divide by m

$dv=\left({\frac {g}{k}}-{\frac {v_{e}}{m}}\right)dm$

integrate

$\int _{v_{i}}^{v}dv=\int _{m_{i}}^{m}{\frac {g}{k}}dm-\int _{m_{i}}^{m}{\frac {v_{e}}{m}}dm$

The initial velocity of the rocket is zero, so carrying out the integration gives us the velocity at a given mass

$v(m)={\frac {g}{k}}\left(m-m_{i}\right)-v_{e}ln(m)+v_{e}ln(m_{i})$

simlify using properties of the log function

$v(m)={\frac {g}{k}}\left(m-m_{i}\right)-v_{e}ln({\frac {m}{m_{i}}})$

It is time to take care of the constant k.

${\frac {dm}{dt}}=-k$

rearrange

$dm=-kdt$

integrate

$\int _{m_{i}}^{m}dm=-k\int _{0}^{t}dt$

$m-m_{i}=-kt$

which gives

$k={\frac {m_{i}-m}{t}}$

Note the sign, we have a positive k, since $m_{i}$ will always be bigger than m (the rocket is ejecting mass). Plug this into (5).

$v(t)={\frac {g\left(m-m_{i}\right)t}{\left(m_{i}-m\right)}}-v_{e}ln\left({\frac {m}{m_{i}}}\right)$

cancel terms to get

$v(t)=-gt-v_{e}ln\left({\frac {m}{m_{i}}}\right)$

While this equation gives us the velocity of the rocket at burnout, we also want altitude. As usual, integrate velocity to get position ${\frac {dy}{dt}}=v(t)$

$\int _{0}^{y}dy=-g\int _{0}^{t_{b}}tdt-v_{e}ln\left({\frac {m}{m_{i}}}\right)dt$

To integrate the last term, we need to replace dt with dm, since m is a function of t

$dt=-{\frac {dm}{k}}$

$\int _{0}^{y}dy=-g\int _{0}^{t_{b}}tdt+{\frac {v_{e}}{k}}\int _{m_{i}}^{m_{f}}ln\left({\frac {m}{m_{i}}}\right)dm$

All but the last term are simple. For the lazy, an integral table can be used to solve it.

$\int _{0}^{y}dy=-g\int _{0}^{t_{b}}tdt+{\frac {v_{e}}{k}}\int _{m_{i}}^{m_{f}}ln\left({\frac {m}{m_{i}}}\right)dm$

$y=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}}{k}}\int _{m_{i}}^{m_{f}}ln\left({\frac {m}{m_{i}}}\right)dm$

Use typical integral substitution technique for the last term, so set

$w={\frac {m}{m_{i}}}$

$dw={\frac {dm}{m_{i}}}$

$dm=m_{i}dw$

substituting this in leaves us with

${\frac {v_{e}m_{i}}{k}}\int ln(w)dw$

Integrating the logarithm function is done through integration by parts

$\int udv=uv-\int vdu$

setting

$u=ln(w)\,\,\,\,\,dv=1dw$

then differentiating and integrating yields

$du={\frac {1}{w}}dw\,\,\,\,\,v=w$

plugging these into (6) gives

$\int ln(w)dw=wln(w)-\int {\frac {w}{w}}dw$

which is equal to

$\int ln(w)dw=wln(w)-w$

so going back to the original integral and evaluating the limits we have

$y=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{i}}{k}}\left({\frac {m_{f}}{m_{i}}}ln\left({\frac {m_{f}}{m_{i}}}\right)-{\frac {m_{f}}{m_{i}}}-{\frac {m_{i}}{m_{i}}}ln\left({\frac {m_{i}}{m_{i}}}\right)+{\frac {m_{i}}{m_{i}}}\right)$

simplfying and using $ln(1)=0$

$y=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{i}}{k}}\left({\frac {m_{f}}{m_{i}}}ln\left({\frac {m_{f}}{m_{i}}}\right)-{\frac {m_{f}}{m_{i}}}+1\right)$

plug in k and expand

$y=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{f}t_{b}}{m_{i}-m_{f}}}ln\left({\frac {m_{f}}{m_{i}}}\right)+{\frac {v_{e}\left(m_{i}-m_{f}\right)t_{b}}{m_{i}-m_{f}}}$

finally, except for the last term, we get equation (1)

$y=-{\frac {gt_{b}^{2}}{2}}+{\frac {v_{e}m_{f}t_{b}}{m_{i}-m_{f}}}ln\left({\frac {m_{f}}{m_{i}}}\right)+v_{e}t_{b}$

To find how much more altitude is gained after burnout, we note that there is no more thrust and rocket is under constant acceleration, g, so we go back to Newton's 2nd law

$F=ma$