title | date |
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Operating |
2016-07-17 |
Let's face it, space is a risky business. I always considered every launch a barely controlled explosion. -- Aaron Cohen , Deputy Administrator of NASA
Consider a spacecraft
Now, this velocity change is done relatively to
The goal of rendez-vous is to make the position and velocity of a
spacecraft
Same position and same velocity means a similar orbit. Once both satellites are revolving around the same body, the first correction is that of inclination: one will perform an inclination change (either (anti-)normal burn, straight burn, or bi-elliptical). When they are orbiting in the same plane, they will have to consort their orbits, and synchronize on the final orbit.
A simple way to do this is to have the orbits joining in one point (say, the periapsis) while their period is significantly different. After a few revolutions, the two spacecraft will get in the common point at the same time and will be able to join their orbit. It is handy to correct the orbits before the last revolution to get a more accurate rendez-vous.
\begin{tikzpicture} \def\peri{2} \def\apoa{3} \def\apob{5} \node[point=O] (O) at (0,0) {}; \node[loint=X] (X) at (-\peri,0) {}; \orbit[blue]{O}{\apoa}{\peri}{0}{360}; \orbit[red] {O}{\apob}{\peri}{0}{360}; \orbitpoint[point=P1]{O}{\apoa}{\peri}{ 40}{P1}{}; \orbitpoint[point=P2]{O}{\apob}{\peri}{110}{P1}{}; \end{tikzpicture} $\posit{P2}$ will reach $\posit{X}$ ahead of $\posit{P1}$; however, the red orbit has a higher apoapsis and thus a longer period: after some time, $\posit{P1}$ will catch up with $\posit{P2}$Assume we want to set up a satellite network around a body
We are interested in two values: the closest approach of the lines-of-sight to
What we have to choose is the number of satellites
and
Thus, with the requirement that
Conversely, if we want to know the the necessary value of
Similar information is available on the KSP wiki @coverage.
The curvature of Kerbin makes that a satellite sees less than half the surface; the closer it is the surface, the less it can see.
\begin{tikzpicture} \def\R{2} \def\r{5} \node[boint=O] (O) at (0,0) {}; \node[point=P] (P) at (\r,0) {}; % tangents \pgfmathsetmacro{\be}{acos(\R/\r)} \node[point=X1] (X1) at ( \be:\R) {}; \node[point=X2] (X2) at (-\be:\R) {}; \draw (P) -- (X1); \draw (P) -- (X2); % visible cone \fill[grey!20] (P) -- (-\be:\R) arc (-\be:\be:\R) -- cycle; % values \draw (O) --node[above left]{$\dist{R}$} (X1); \draw (O) --node[below] {$\dist{r}$} (P); \markangle{X1}{P}{O}{$\alpha$}{1} \markangle{X1}{O}{P}{$\beta$} {1} % show covered area \draw ( \be:\R) arc ( \be:360-\be:\R); \draw[red] (-\be:\R) arc (-\be:\be :\R); \end{tikzpicture} $\posit{X1}$ and $\posit{X2}$ are the farthest point that $\posit{P}$ can see from this distance; the line $(PX1)$ is said to be tangent to the circle, and there is a right angle in $\posit{X1}$ This property is used on the surface: the crow's nest is an observation spot located high in the masts of a ship to see significantly farther away. The distance you can see from altitude $\dist{a}$ is $\dist{R} \arccos \frac {\dist{R}} {\dist{R} + \dist{a}}$.We quickly see that
Consider the typical case of a satellite running on photovoltaic cells. Most of the time, it can relies on the sunlight emitted by the local star. However, for some time during each orbit, it will pass behind its primary (unless it is orbiting the star itself). During that duration, it will thus have to run on batteries. To know the needed electrical capacity, we will need to know how long an episode lasts.
\begin{tikzpicture} % inspired from http://www.texample.net/tikz/examples/earth-orbit/ \def\R{1.0} \def\ra{5.0} \def\rp{1.5} \node[point=F] (F) at (5,5) {}; \node[point=O,right = ((\ra-\rp)/2) of F] (O) {}; % shadow \draw[yellow!20!black!80] (F) +(0,-\R) -- +(5,-\R); \draw[yellow!20!black!80] (F) +(0,+\R) -- +(5,+\R); % planet \draw[brown] (F) circle (\R); % arrow \draw[black,->,thick] (O) +(-3,2) --node[below]{star} +(-4,2); % orbit \orbit[red]{F}{\ra}{\rp}{0}{360}; \begin{scope} \clip (F) +(0,-\R) rectangle +(5,\R); \orbit[blue]{F}{\ra}{\rp}{0}{360}; \end{scope} \orbitpoint[roint=S]{F}{\ra}{\rp}{320}{S}{} \pgfmathsetmacro{\halfnightangle}{asin(\R / \ra)} % notable points \orbitpoint[loint=Pe]{F}{\ra}{\rp}{180}{Pe}{} \orbitpoint[roint=Ap]{F}{\ra}{\rp}{0}{Ap}{} \orbitpoint[point=N_e]{F}{\ra}{\rp}{+\halfnightangle}{N_e}{} \orbitpoint[boint=N_s]{F}{\ra}{\rp}{-\halfnightangle}{N_s}{} \coordinate[boint=H] (H) at ($(F)!(N_s)!(Ap)$); % show angles \draw[dashed] (O) -- (Ap) (O) -- (Pe) (O) -- (N_s) (O) -- (N_e) (O) -- (S) (N_e) -- (H); \markangle{Ap}{O}{N_s}{$\theta$}{0.7} \markangle{Ap}{O}{N_e}{$\theta$}{0.7} \markangle{Pe}{O}{S}{$E$}{0.5} \draw (H) +(0,.2) -- +(-.2,.2) -- +(-.2,0); \end{tikzpicture} A schematic representing the situation with some points, lines and angles that will turn useful.As shown in the schematic above, we can see that
From this, we can determine that the dark time start with eccentric anomaly
\begin{align*} \delay{\Delta t_{\text{dark}}} &= \delay{t}(\angle{E}=\angle{E_e}) - \delay{t}(\angle{E}=\angle{E_s}) \ &= \frac 1 n (M(\angle{E}=\angle{\pi}+\angle{\theta}) - M(\angle{E}=\angle{\pi}-\angle{\theta})) \ &= \frac 1 n ( (\strike[red]{\angle{\pi}}+\angle{\theta} - e \sin(\angle{\pi}+\angle{\theta})) - (\strike[red]{\angle{\pi}}-\angle{\theta} - e \sin(\angle{\pi}-\angle{\theta}) ) \ &= \frac 2 n (\angle{\theta} + e \sin \angle{\theta}) \ &= \frac 2 n \left(\sin^{-1}\left(\frac R b\right) + e \frac R b\right) \end{align*}
TODO