**E178: Addition of Drag and Gravity to the 1-D Rocket Equation** |[home](index.md.html)|[syllabus](syllabus.md.html)|[assignments](assignments.md.html)|[labs](labs.md.html)|[final project](finalproject.md.html)|[flight data](FlightData.md.html)|[getting certified](RocketryCertification.md.html)| # Calculation of $\Delta v$ We’ll take as given the derivation at the end of the Rocket Equation video \begin{equation} \label{firsteq} m \frac{dv}{dt}(=ma)=\frac{dm}{dt}v_e \end{equation} where $v_e$, the exit velocity of the propellant, is a vector quantity in the direction opposite the rocket. We’ll assume that the rocket stays vertical and flies under thrust in the positive $x$ (or $z$) direction, so vector quantities are specified as the magnitude and a positive or negative sign to indicate up or down. We’ll first do a quick derivation of the relationship between specific impulse and exit velocity. Starting from the definition that \begin{equation} T=v_e \frac{dm}{dt}, \end{equation} and for a given motor $v_e$ is constant (not quite true but true enough), we have that \begin{equation*} dm = - \frac{1}{|v_e|}T dt \end{equation*} \begin{equation} \int^m_{m_0} dm = - \frac{1}{|v_e|} \int^t_{t_0}T dt \end{equation} \begin{equation*} \Delta m = - \frac{1}{|v_e|} I = -m_p \end{equation*} resulting in the impulse $I$ being given by \begin{equation} I = m_p |v_e|, \end{equation} where $m_p$ is the mass of the propellant. Rearranging, we have the exit velocity is \begin{equation} |v_e|= \frac{I}{m_p}, \end{equation} There are two definitions for specific impulse, $I_{\rm{sp}}$. The first is the impulse divided by the propellant mass, $I/m_p$, with SI units of \begin{equation} \rm{ \frac{Ns}{kg} = \frac{m}{s}}, \end{equation} and the second is impulse divided by the propellant weight in standard gravity, $I/(m_p g_0)$ , with SI units of \begin{equation} \rm{ \frac{Ns}{kg\frac{m}{s^2}} = s}, \end{equation} You can tell whether you need to multiply by or not to get by checking the units of $I_{\rm{sp}}$. With the rocket equation in this form we can derive the relationship between the change in mass and the change in velocity in the absence of other forces. Starting from \begin{equation} \label{dvdt} \frac{dv}{dt}=\frac{dm}{dt}\frac{v_e}{m} \end{equation} where $v_e$ still has a sign opposite to $v$, we can multiply through by $dt$ \begin{equation} dv=v_e\frac{dm}{m} \end{equation} and integrate from the starting velocity and mass, $v_0$ and $m_0$ to the current velocity and mass, $v$ and $m$. \begin{equation} \int^v_{v_0}dv=v_e \int^m_{m_0} \frac{dm}{m} \end{equation} \begin{equation} v - v_0 = \Delta v = v_e \log \frac{m}{m_0} = |v_e| \log \frac{m_0}{m} = I_{\rm{sp}} \log \frac{m_0}{m} \end{equation} This equation relates the change in velocity to the change in mass. It finds extensive use in estimating the needed propellant for a given final velocity. # Adding Gravity and Drag Substituting in thrust for $\dot{m}v_e$ in Equation [firsteq], we have \begin{equation} m \frac{dv}{dt}(=ma)=T \end{equation} At this point, we can simply add the gravity force, $–mg$ (with $m$ a function of time), and the drag force, $F_D = - \frac{1}{2}C_D A_p \frac{\rho}{m} v |v|$, to the rocket equation. \begin{equation} a=\dot{v}=\frac{T}{m}-g- \frac{1}{2}C_D A_p \frac{\rho}{m} v |v| \end{equation} There are several analytical solutions to the rocket equation, depending on what you assume about the thrust curve, the importance of drag, and whether to include gravity or not. In general, when you include a non-constant thrust curve, drag, and properties that change with altitude, you need to numerically integrate two coupled differential equations. We include integration for altitude or position as a third equation. The three are \begin{equation} \dot{v}=\frac{T}{m}-g- \frac{1}{2}C_D A_p \frac{\rho}{m} v |v| \end{equation} \begin{equation} \dot{m}=\frac{T}{v_e} \end{equation} and \begin{equation} \dot{x} = v \end{equation} The full 3-D equations of motion are given by these equations with $a$, $v$, $x$, $v_e$, and $g$ specified as vectors. This system of equations can be integrated with any handy D.E. solver, such as the Runge-Kutta 4th-order method. The method requires some way to read in the thrust curve, and parameter values or functions of velocity or altitude for $C_D$, $\rho$, and $g$. ## 1-D with constant gravity A useful analytical form is derived by assuming vertical flight and constant $g$. This model is Equation [dvdt] with $g$ added \begin{equation} \frac{dv}{dt}=\frac{dm}{dt}\frac{v_e}{m} - g \end{equation} Multiplying through by $dt$ we have \begin{equation} dv=v_e\frac{dm}{m} - g dt \end{equation} \begin{equation} \int^v_{v_0}dv=v_e \int^m_{m_0} \frac{dm}{m}-g \int^t_{t_0} dt \end{equation} \begin{equation} \Delta v = v_e \log \frac{m}{m_0} - g \Delta t \end{equation} This form is very useful for getting approximate parameters for setting up the full numerical integration. # From Burnout to Apogee for 1-D The rocket equation is useful for modeling flight up to burnout. Calculating the flight from burnout up to apogee can be approximated if the rocket is above the densest part of the atmosphere by summing the change in kinetic and potential energies from burnout to apogee. \begin{equation} E_k = \frac{1}{2} m v^2 \end{equation} \begin{equation} E_p = -GMm(r+x) \end{equation} where $E_k$ is the kinetic energy, $E_p$ is the potential energy, $G$ is Newton's gravitational constant, $M$ is the mass of the earth, $r$ is the radius of the earth, and $x$ is the AGL altitude. We will sum the changes from burnout (subscript $b$) to apogee (subscript $a$) \begin{equation} \Delta E_k + \Delta E_p = 0 = \frac{1}{2}m (v_a^2 - v_b^2)-GMm\left(\frac{1}{r+x_a}-\frac{1}{r+x_b}\right) \end{equation} Solving for $x_a$ we have \begin{equation} x_a=\frac{r^2 v_b^2 + 2 G M x_b+r v_b^2 x_b}{2 G M - v_b^2(r+x_b)} \end{equation} Again, this equation is useful for setting up the parameters for a numerical model.