title | date |
---|---|
Launch to orbit |
2016-07-24 |
Can it be that you have come from outer space?
— As a matter of fact, I have! -- Yuri Gagarin, first human being in space, to people near his landing site
From the Tsiolkovsky equation (\ref{eql:thrust}), we know that:
If we now assume that
\begin{align*} \dist{\Delta x_t} &= - \speed{v_e} \times \frac 1 {- \frac {\dot m} {\mass{m}(\delay{0})}} \left( \left(1 - \frac {\dot m} {\mass{m}(\delay{0})} \delay{t}\right) \ln\left(1 - \frac {\dot m} {\mass{m}(\delay{0})} \delay{t}\right) + \frac {\dot m} {\mass{m}(\delay{0})} \delay{t} \right) \ & = \speed{v_e} \left(\frac {\mass{m}(\delay{0})} {\dot m} - \delay{t}\right) \ln\left(1 - \frac {\dot m} {\mass{m}(\delay{0})} \delay{t}\right) + \speed{v_e} \delay{t} \end{align*}
This gives us the displacement done by the thrust of the engines.
Let us notate
Then, integrating twice:
This is the displacement due to gravity.
We now compare the graphs for the different approaches (ignoring gravity, constant gravity, or numerical approximation). To get actual values, we simulate a basic rocket made of the following parts:
- Command Pod Mk1
- FL-T800 Fuel Tank
- LV-T30 Liquid Fuel Engine
An object moving through a fluid (gaz or liquid) pushes matter around; by the principle of action-reaction, the object will be subjected to an opposing force. We are only interested in two effects of this: drag, which is the prograde component (along the velocity vector), and lift, a normal component.
For now, we will assume an axial symmetry around the velocity vector. In other words, we make it so as not to generate lift. It has been measured that the force of drag was of the form:
where
Let us consider a falling object subjected only to gravitation and atmospheric drag. Gravitation is a constant force oriented downwards; here, drag is a force oriented upwards and growing with speed. There will hthus be a point where gravitation and drag balances and speed becomes constant. This speed is named terminal velocity.
When terminal velocity is reached,
\begin{align*} \accel{\ddot z} = 0 &\Leftrightarrow \accel{a_g} = \accel{a_d} \ &\Leftrightarrow \force{F_g} = \force{F_d} \ &\Leftrightarrow \frac 1 2 \rho \times C_d \area{A} \times \speed{v_{\text{term}}}^2 = \mass{m} \accel{g} \ &\Leftrightarrow \speed{v_{\text{term}}} = \sqrt{\frac {2 \mass{m} \accel{g}} {\rho C_d \area{A}}} \end{align*}
In Kerbal Space Program pre-1.0, \area{A} used to be computed as $\frac 1 {125} ~\area{m^2}/\mass{kg} \times \mass{m}$. This means that the terminal velocity only depended on the air density $\rho$ and the coefficient of drag $C_d$ (usually about 0.2).This is not realistic since, according to the square-cube law, mass tends to grow cubically while area tends to grow quadratically: larger ships should experience proportionally less drag.
The new aerodynamic model ("drag cubes") avoids this issue.
Since drag grows with speed, we face the problem of aiming for an ascent speed that minimizes the time of ascent (implying a higher speed) as well as the drag (implying a lower speed).
Ultimately, we are trying to minimize the cost in propellant, measured in
\begin{align*} \speed{\Delta v} &= (\accel{a_g} + \accel{a_d}) \delay{\Delta t} \ &= \left(\frac {\force{F_d}} {\mass{m}} + \accel{g}\right) \delay{\Delta t} \ &= \left(\frac 1 2 \rho C_d \times \area{A} \times \speed{v}^2 + \mass{m} \accel{g} \right) \frac 1 {\mass{m}} \frac {\dist{\Delta z}} {\speed{v}} \ &= \underbrace{\left( \frac 1 2 \rho C_d \area{A} \speed{v} + \mass{m} \accel{g} \frac 1 {\speed{v}} \right) }_{f(\speed{v})} \dist{\Delta z} \end{align*}
To find the minimum of
\begin{align*} f(\speed{v}) \text{~min} &\Rightarrow f'(\speed{v}) = 0 \ &\Rightarrow \frac 1 2 \rho C_d \area{A} - \mass{m} \accel{g} \frac 1 {\speed{v}^2} = 0 \ &\Rightarrow \frac 1 2 \rho C_d \area{A} - \mass{m} \accel{g} \frac 1 {\speed{v}^2} = 0 \ &\Leftrightarrow \speed{v} = \sqrt{\frac {2 \mass{m} \accel{g}} {\rho C_d \area{A}}} \ &\Leftrightarrow \speed{v} = \speed{v_{\text{term}}} \end{align*}
From this, we see that the optimal vertical ascent speed is the terminal velocity.
Assume the atmosphere is in a stable state and consider an infenitesimal volume
The resultant force of pressure on
The value
where
To solve it, we need to find a primitive. We will assume that the gravitation
is constant and that the temprature decrease linearly with altitude (rate
Thus:
Using the closed expression for the pressure, we can now derive the density and then the terminal velocity depending on the altitude:
\begin{tikzpicture} \pgfmathsetmacro{\Pz}{101325} \pgfmathsetmacro{\L}{0.0065} \pgfmathsetmacro{\Tz}{288.15} \pgfmathsetmacro{\g}{9.80665} \pgfmathsetmacro{\M}{0.0289644} \pgfmathsetmacro{\R}{8.31447} \pgfmathsetmacro{\Rsp}{287.058} \pgfmathsetmacro{\H}{\R*\Tz/\g/\M} \pgfmathsetmacro{\m}{1000} \pgfmathsetmacro{\A}{0.008*\m} \pgfmathsetmacro{\Cd}{0.2} \pgfmathdeclarefunction{p}{1}{\pgfmathparse{\Pz*(1 - \L*#1/\Tz)^(\g*\M/\R/\L)}} %\pgfmathdeclarefunction{p}{1}{\pgfmathparse{\Pz*exp(-#1/\H)}} \pgfmathdeclarefunction{T}{1}{\pgfmathparse{\Tz - \L * #1}} \pgfmathdeclarefunction{rho}{1}{\pgfmathparse{p(#1)/\Rsp/T(#1)}} \pgfmathdeclarefunction{vterm}{1}{\pgfmathparse{sqrt(2*\g*\m/rho(#1)/\A/\Cd)}} \begin{axis}[ domain=0:30000, samples=\samples, no markers, axis lines=left, xlabel=$\dist{h}$, x unit=m, ylabel=$\rho$, y unit=kg/m\^3, legend style={ cells={anchor=west}, legend pos=north west, }, scaled ticks=false, yticklabel style={/pgf/number format/fixed}, ] \addplot[black] {p(x)}; \end{axis} \begin{axis}[ domain=0:30000, samples=\samples, no markers, axis lines=left, axis y line=right, ylabel=$\speed{v_{\mathrm{term}}}$, y unit=m/s, scaled ticks=false, yticklabel style={/pgf/number format/fixed}, ] \addplot[speed] {vterm(x)}; \end{axis} \end{tikzpicture} Terminal velocity increases as air density decreases; notice that the terminal velocity at the surface is about \speed{100~m/s} in this exampleOnce the spacecraft has left the atmosphere, it is not subject to air drag anymore and can set its orbit. For this, it simply need to raise its orbit from the center of the body to above the atmosphere. It is more convenient for further maneuvers to have a circular orbit.
\begin{tikzpicture} \def\R{2} \def\h{0.5} \def\ap{3} \def\pea{0.1} \def\peb{3} \def\an{-7} \node[loint=O] (O) at (0,0) {}; \path[ right color=blue!5, left color=blue!5!black!10, ] (O) circle (\R+\h); \orbit[red]{O}{\ap}{\pea}{-180}{\an}; \path[ right color=blue!20, left color=blue!20!black, ] (O) circle (\R); \orbitpoint[boint=P]{O}{\ap}{\pea}{\an}{}{}; \orbit[red,dotted]{O}{\ap}{\pea}{\an}{360+\an}; \orbit[blue]{O}{\ap}{\peb}{-180}{180}; \orbitpoint[roint=B]{O}{\ap}{\peb}{0}{}{}; \end{tikzpicture} When $\posit{P}$ leaves the atmosphere and burns in $\posit{B}$, it can change from the \textcolor{red}{red} “orbit” to the \textcolor{blue}{blue} orbit.Using the vis-viva equation (\ref{eql:visviva}), we can compute the requisite speed for orbiting:
\frac {\mathcal G \mass{M}} {\dist{r_a}} $$
For example, on Kerbin, you need to reach an horizontal velocity of:
\sqrt{\frac {6.67 \times 10^{11}m^3/kg/s^2 \times \mass{5.29 \times 10^{22}kg}} {\dist{600km} + \dist{80km}}}
\speed{2279~m/s} $$
To save $\speed{\Delta v}$, launches are done close to the equator to go with the movement of the surface due to the body's rotation. On Kerbin, we save ($\delay{T}$ is the orbital period):\speed{174.5~m/s} $$