2 A B C D E F G H J K L M N O P R S T Y

Rocket Equation

by | Jun 28, 2020

 

A large fraction (typically 90%) of the mass of a rocket is propellant.  Since such a large portion of the rocket mass is propellant, it is important to consider the change in mass of the vehicle as it accelerates.

 Schematic for the application of the momentum theorem.
Image fig5RocketFBD_web

There are several ways to do this by applying the conservation of momentum. Apply the momentum theorem differentially by considering a small mass, $ dm$ , expelled from the rocket during time $ dt$ , (see above figure)

The initial momentum of the mass in the control volume (the vehicle) is $ m_v u$ . The final momentum of mass in the control volume (the vehicle and the mass expelled, $ dm$ ) is

$\displaystyle (m_v - dm)(u + du) + dm(u - u_e) = m_v u + m_v du - udm - du dm +u dm -u_e dm.$


The change in momentum during the interval $ dt$ is

$\displaystyle \textrm{change in momentum}$ $\displaystyle = \textrm{momentum}_{\textrm{final}} - \textrm{momentum}_{\textrm{initial}}$
$\displaystyle = m_v du - u_e dm,$


since $ du dm$ is a higher order term. Now consider the forces acting on the system which is composed of the masses $ m$ (the rocket), and $ dm$ (the small amount of propellant expelled from the rocket during time $ dt$ ):

$\displaystyle \sum F = (p_e - p_0)A_e - D - m g \cos\theta.$


Applying conservation of momentum, the resulting impulse, $ \sum F dt$ , must balance the change in momentum of the system:

$\displaystyle m_v du - dm u_e = [(p_e - p_0) A_e - D - m_v g \cos\theta]dt.$


Then since

$\displaystyle dm = \dot{m} dt = -\frac{dm_v}{dt} dt,$


where $ \dot{m}$ is the propellant mass flow rate, we have

$\displaystyle m_v du = [(p_e - p_0)A_e + \dot{m}u_e - D - m_v g \cos\theta]dt,$


or, for $ p_e = p_0$ ,

$\displaystyle du = -\frac{u_e dm_v}{m_v} - \frac{D}{m_v} dt - g \cos \theta dt.$ (ROCKET EQUATION)


Equation 14.1 is known as The Rocket Equation. It can be integrated as a function of time to determine the velocity of the rocket.

If we set $ u_e = \textrm{constant}$ , assume that at $ t=0$ , $ u=0$ , neglect drag, and set $ \theta = 0$ , then we can simplify the rocket equation to

$\displaystyle du = -u_e \frac{dm_v}{m_v} - gdt,$


which can be integrated to give

$\displaystyle u = -u_e\ln\left(\frac{m_v}{m_{v_0}}\right) - gt,$


where $ m_{v_0}$ is the initial mass of the rocket. We can also write this result as

$\displaystyle u = g \left[\textrm{Isp} \ln\left(\frac{m_{v_0}}{m_v}\right)-t\right].$


We can view this equation as being similar to the Breguet Range Equation for aircraft. It presents the overall dependence of the principal performance parameter for a rocket (velocity, $ u$ ), on the efficiency of the propulsion system (Isp), and the structural design (ratio of total mass to structural mass, since the initial mass is the fuel mass plus the structural mass and the final mass is only the structural mass).

Assuming the rate of fuel consumption is constant, the mass of the rocket varies over time as

$\displaystyle m_v(t) = m_{v_0} - (m_{v_0} - m_{v,\textrm{final}})\frac{t}{t_b},$


where $ t_b$ is the time at which all of the propellant is used. This expression can be substituted into the equation for velocity and then integrated to find the height at the end of burnout:

$\displaystyle h_b = \int_0^{t_b} u dt,$


which for a single-stage sounding rocket with no drag and constant gravity yields

$\displaystyle h_b = g \left[-t_b \textrm{Isp}\cfrac{\ln\left(\cfrac{m_{v_0}}{m_... ...}{m_{v,\textrm{final}}}-1\right)} + t_b \textrm{Isp} - \frac{1}{2}t_b^2\right].$


The final height of the rocket can then be determined by equating the kinetic energy of the vehicle at burnout with its change in potential energy between that point and the maximum height.

The image below shows the most powerful rocket built and successfully launched, the NASA Saturn V rocket.  The Saturn V launched 13 times to send people to the moon during the Apollo program and also launch America’s first space station, Skylab.

 The Saturn V rocket stood 365 feet tall and had 5 stages. It produced over 7.5 million pounds of thrust at liftoff (NASA, 1969).
Image fig5NASASaturnV_web

A large portion of the theoretical material here came from MIT’s website and was customized for use here. https://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node103.html

additional resources:

https://www.grc.nasa.gov/WWW/K-12/rocket/rktpow.html