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[parent] single stage rocket burnout height (Example)

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

$\displaystyle y_{max} = -\frac{g t^2_b}{2} + \frac{v_e m_f t_b}{m_i - m_f} ln \left (\frac{m_f}{m_i} \right ) + v_e t_b + \frac{v^2_b}{2 g}$ (1)

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

$\displaystyle F = ma = -mg$ (2)

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

$\displaystyle F = \frac{d}{dt}\left (m v \right)$ (3)

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

$\displaystyle -mg = \frac{dm}{dt} v_e + m \frac{dv}{dt} $

multiply by dt

$\displaystyle -mg dt = v_e dm + m dv$ (4)

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

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

so

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

plug this into (4) to get

$\displaystyle -\frac{mg}{k} dm = v_e dm + m dv $

Separate variables to setup the integration

$\displaystyle \left ( \frac{mg}{k} - v_e \right ) dm = m dv $

divide by m

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

integrate

$\displaystyle \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

$\displaystyle 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

$\displaystyle v(m) = \frac{g}{k} \left ( m - m_i \right) - v_e ln(\frac{m}{m_i})$ (5)

It is time to take care of the constant k.

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

rearrange

$\displaystyle dm = -k dt $

integrate

$\displaystyle \int_{m_i}^{m} dm = -k \int_{0}^{t} dt $

$\displaystyle m - m_i = -kt $

which gives

$\displaystyle 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).

$\displaystyle v(t) = \frac{g \left(m - m_i \right) t}{\left( m_i - m \right)} - v_e ln \left( \frac{m}{m_i} \right )$

cancel terms to get

$\displaystyle 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

$\displaystyle \frac{dy}{dt} = v(t) $

$\displaystyle \int_0^y dy = -g \int_0^{t_b} t dt - 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

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

$\displaystyle \int_0^y dy = -g \int_0^{t_b} t dt + \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.

$\displaystyle \int_0^y dy = -g \int_0^{t_b} t dt + \frac{v_e}{k} \int_{m_i}^{m_f} ln \left( \frac{m}{m_i} \right) dm $

$\displaystyle y = -\frac{g t_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

$\displaystyle w = \frac{m}{m_i} $

$\displaystyle dw = \frac{dm}{m_i} $

$\displaystyle dm = m_i dw $

substituting this in leaves us with

$\displaystyle \frac{v_e m_i}{k} \int ln(w) dw $

Integrating the logarithm function is done through integration by parts

$\displaystyle \int u dv = uv - \int v du$ (6)

setting

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

then differentiating and integrating yields

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

plugging these into (6) gives

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

which is equal to

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

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

$\displaystyle y = -\frac{g t_b^2}{2} + \frac{v_e m_i}{k} \left ( \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 $ln(1) = 0$

$\displaystyle y = -\frac{g t_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

$\displaystyle y = -\frac{g t_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_f - m_i \right) t_b}{m_i - m_f} $

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

$\displaystyle y = -\frac{g t_b^2}{2} + \frac{v_e m_f t_b}{m_i - m_f}ln \left( \frac{m_f}{m_i} \right) + v_e t_b$ (7)

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

$\displaystyle F = ma $

with a constant force due to gravity

$\displaystyle F = -mg = ma $

$\displaystyle a = -g $

$\displaystyle \frac{dv}{dt} = -g $

$\displaystyle dv = -g dt $

integrate to get velocity

$\displaystyle \int_{v_b}^0 dv = - \int_0^t g dt $

$\displaystyle 0 - v_b = -gt $

$\displaystyle v_b = gt $

integrate again to get distance traveled

$\displaystyle \frac{dy}{dt} = gt $

$\displaystyle dy = gt \, dt $

$\displaystyle \int_0^y dy = \int_0^t gt \,dt $

so the distance traveled after burnout is

$\displaystyle y = \frac{g t^2}{2} $

combining this with the distance traveled during the burn (7), yields equation (1), the total distance rocket traveled

$\displaystyle y_{max} = -\frac{g t^2_b}{2} + \frac{v_e m_f t_b}{m_i - m_f} ln \left (\frac{m_f}{m_i} \right ) + v_e t_b + \frac{v^2_b}{2 g}$ (8)

References

[1] Ellis, R., Gulick, D. "Calculus" Harcourt Brace Jovanovich, Inc., Orlando, FL, 1991.

[2] Etgen, G. "Calculus" John Wiley & Sons, New York, 1999.

[3] Marion, J., Thornton, S. "Classical dynamics of Particles and systems" Fourth Edition, Harcourt College Publishers, Fort Worth, 1995.

[4] Ketsdever, A. "Launch Vehicle Analysis", lecture notes, University of Colorado at Colorado Springs, Spring 2006.



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Cross-references: systems, dynamics, position, function, velocity, acceleration, mass, motion, Newton's laws

This is version 13 of single stage rocket burnout height, born on 2006-06-18, modified 2006-10-26.
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Physics Classification: 45.50.Dd (General motion)
 45.50.Pk (Celestial mechanics )
 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
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