Burnout Height of Upward Rocket under Constant Gravity

Theorem

Let $R$ be a rocket whose structural mass is $m_1$.

Let $R$ contain fuel of initial mass $m_2$.

Let $R$ be fired straight up from the surface of a planet whose gravitational field exerts an Acceleration Due to Gravity of $g$, assumed constant.

Let $R$ burn fuel at a constant rate $a$, with a constant exhaust velocity $b$ relative to $R$.

Let all forces on $R$ except that due to the gravitational field be neglected.

Then the burnout height of $R$ is given by:

$h_b = -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a + \dfrac {b m_1} a \ln \dfrac {m_1} {m_1 + m_2}$

Proof

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The total mass of the rocket at time $t$ is given by

$m = m_1 + m_2 - at$

The time $t_b$ at which the rocket runs out of fuel is given by:

$t_b = \dfrac {m_2} a$

Now we can calculate the total force on the rocket to be:

$F = b a - mg$

where:

$b a$ is the rocket's thrust as calculated from Newton's Second Law of Motion

$-mg$ is the force of gravity on the rocket

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In particular: Not obvious why thrust is $b a$
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We can now calculate the acceleration of the rocket as a function of time using our calculated total force in Newton's First Law of Motion to obtain

$\ds \ddot h$

$=$

$\ds \dfrac 1 m \paren {b a - m g}$

$\ds $

$=$

$\ds \dfrac {b a} {m_1 + m_2 - a t} - g$

where $\ddot h = \dfrac {\d^2 h} {\d t^2}$ is the second time derivative of the height of the rocket.

Integrating the acceleration function twice with respect to time will yield the height of the rocket as a function of time.

Performing the first integration:

$\ds \dot h$

$=$

$\ds \int_0^t \paren {\dfrac {b a} {m_1 + m_2 - at'} - g} \rd t'$

$\ds $

$=$

$\ds -b \ln \paren {m_1 + m_2 - a t} + b \map \ln {m_1 + m_2} - g t$

The second integration:

$\ds \map h t$

$=$

$\ds \int_0^t \paren {b \map \ln {m_1 + m_2} - b \map \ln {m_1 + m_2 - a t'} - g t'} \rd t'$

$\ds $

$=$

$\ds -\dfrac 1 2 g t^2 + b t \map \ln {m_1 + m_2} - b \ds \int_0^t \paren {\map \ln {m_1 + m_2 - at'} } \rd t$

$\ds $

$=$

$\ds -\dfrac 1 2 g t^2 + b t \map \ln {m_1 + m_2} + \dfrac b a \paren {\paren {m_1 + m_2 - a t} \map \ln {m_1 + m_2 - a t} + a t - \paren {m_1 + m_2} \map \ln {m_1 + m_2} }$

$\ds \leadsto \ \ $

$\ds \map h {t_b}$

$=$

$\ds -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a \map \ln {m_1 + m_2} + \dfrac b a \paren {m_1 \ln m_1 + m_2 - \paren {m_1 + m_2} \map \ln {m_1 + m_2} }$

substituting $t_b = \dfrac {m_2} a$ for $t$

$\ds $

$=$

$\ds -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a + {b m_1} a \map \ln {\dfrac {m_1} {m_1 + m_2} }$

simplifying

Hence the result.

$\blacksquare$

Sources

1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $29$