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
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

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

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:

The second integration:

Hence the result.