Create a program to calculate the classical orbital elements (\(a,e,i,\Omega ,\omega ,f\)) of a satellite given its Cartesian position and velocity components (\(x,y,z,v_{x},v_{y},v_{z}\)). Use the examples in the course notes to test your program, then apply it to the state vector below. (Save your program somewhere you can find it again; you will need it later in the semester.)
\begin{align*} x & =-3000\text{ km}\\ y & =-6000\text{ km}\\ z & =4000\text{ km}\\ v_{x} & =6\text{ km/s}\\ vy & =-1\text{ km/s}\\ vz & =-3\text{ km/s} \end{align*}
Answer is
\begin{align*} a & =7108.84\text{ km}\\ e & =0.4615\text{ km}\\ i & =34.32^{o}\\ \Omega & =124.287^{o}\\ \omega & =242.65^{o}\\ f & =232.07^{o} \end{align*}
What combination of launch latitude and azimuth angle will allow a spacecraft to be launched directly into an equatorial geostationary orbit about the Earth?
Since \[ \cos i=\sin A_{z}\cos \phi \] Where \(i\) is inclination and \(A_{z}\) is the azimath and \(\phi \) is the latitude. Then for \(i=0^{0}\)
Latitude: \(0^{0}\)
Azimuth: \(90^{0}\)
Can a spacecraft be launched directly into an equatorial geostationary orbit about the Earth from the ETR (Eastern Test Range, Cape Canaveral)? No Since \(i\) is not zero.
Can a spacecraft be launched directly into an equatorial geostationary orbit about the Earth from the WTR (Western Test Range, Vandenburg AFB)? No, same reason.
A satellite is initially in a circular orbit about the Earth at an altitude of 200 km. Its target orbit is a circular orbit in the same plane with a radius of 130,000 km. Calculate the total \(\Delta V\) and transfer time (in hours) required to complete each of the orbit transfers below.
\begin{align*} a & =\frac{r_{1}+r_{2}}{2}=\frac{200+6378+130000}{2}=68289\text{ km}\\ V_{1} & =\sqrt{\frac{\mu }{r_{1}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{200+6378}}=7.7843\text{ km/s}\\ V_{2} & =\sqrt{\mu \left ( \frac{2}{r_{1}}-\frac{1}{a}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{200+6378}-\frac{1}{68289}\right ) }=10.74\text{ km/s}\\ \Delta V_{1} & =V_{2}-V_{1}=10.74-7.7843=2.9557\text{ km/s}\\ V_{3} & =\sqrt{\mu \left ( \frac{2}{r_{2}}-\frac{1}{a}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{130000}-\frac{1}{68289}\right ) }=0.54346\text{ km/s}\\ V_{4} & =\sqrt{\frac{\mu }{r_{2}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{130000}}=1.751\text{ km/s}\\ \Delta V_{2} & =V_{4}-V_{3}=1.751-0.54346=1.2075\\ \Delta V & =\left \vert \Delta V_{1}\right \vert +\left \vert \Delta V_{2}\right \vert =2.9557+1.2075=4.1632\text{ km/s} \end{align*}
Time of transfer\begin{align*} T & =\pi \sqrt{\frac{a^{3}}{\mu }}\\ & =\pi \sqrt{\frac{68289^{3}}{3.986\left ( 10^{5}\right ) }}=88799\text{ sec}\\ & =\frac{88799}{60\times 60}=24.666\text{ hr} \end{align*}
with an intermediate transfer radius of 200,000 km
\(r_{b}=200000\) km, \(r_{1}=200+6378\) km\(,r_{2}=130000\) km
\begin{align*} a_{1} & =\frac{r_{1}+r_{b}}{2}=\frac{200+6378+200000}{2}=1.0329\times 10^{5}\text{ km}\\ a_{2} & =\frac{r_{2}+r_{b}}{2}=\frac{130000+200000}{2}=1.65\times 10^{5}\text{ km}\\ V_{1} & =\sqrt{\frac{\mu }{r_{1}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{200+6378}}=7.7843\text{ km/s}\\ V_{2} & =\sqrt{\mu \left ( \frac{2}{r_{1}}-\frac{1}{a_{1}}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{200+6378}-\frac{1}{1.0329\times 10^{5}}\right ) }=10.832\text{ km/s}\\ \Delta V_{1} & =V_{2}-V_{1}=10.832-7.7843=3.0477\text{ km/s}\\ V_{3} & =\sqrt{\mu \left ( \frac{2}{r_{b}}-\frac{1}{a_{1}}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{200000}-\frac{1}{1.0329\times 10^{5}}\right ) }=0.35632\text{ km/s}\\ V_{4} & =\sqrt{\mu \left ( \frac{2}{r_{b}}-\frac{1}{a_{2}}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{200000}-\frac{1}{1.65\times 10^{5}}\right ) }=1.2531\text{ km/s}\\ \Delta V_{2} & =V_{4}-V_{3}=1.2531-0.35632=0.89678\text{ km/s}\\ V_{5} & =\sqrt{\mu \left ( \frac{2}{r_{2}}-\frac{1}{a_{2}}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{130000}-\frac{1}{1.65\times 10^{5}}\right ) }=1.9278\text{ km/s}\\ V_{6} & =\sqrt{\frac{\mu }{r_{2}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{130000}}=1.751\text{ km/s}\\ \Delta V_{3} & =V_{6}-V_{5}=1.751-1.9278=-0.1768\text{ km/s}\\ \Delta V & =\left \vert \Delta V_{1}\right \vert +\left \vert \Delta V_{2}\right \vert +\left \vert \Delta V_{3}\right \vert =3.0477+0.89678+0.1768=4.1213\text{ km/s} \end{align*}
Transfer time\begin{align*} T & =\pi \sqrt{\frac{a_{1}^{3}}{\mu }}+\pi \sqrt{\frac{a_{2}^{3}}{\mu }}\\ & =\pi \sqrt{\frac{\left ( 1.0329\times 10^{5}\right ) ^{3}}{3.986\left ( 10^{5}\right ) }}+\pi \sqrt{\frac{\left ( 1.65\times 10^{5}\right ) ^{3}}{3.986\left ( 10^{5}\right ) }}\\ & =4.9869\times 10^{5}\text{ sec}\\ & =\frac{4.9869\times 10^{5}\text{ }}{60\times 60}=138.53\text{ hr} \end{align*}
\(r_{b}=200000\) km, \(r_{1}=200+6378\) km\(,r_{2}=130000\) km\(,\)
\begin{align*} a & =\frac{r_{1}+r_{b}}{2}=\frac{200+6378+200000}{2}=1.0329\times 10^{5}\text{ km}\\ V_{1} & =\sqrt{\frac{\mu }{r_{1}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{200+6378}}=7.7843\text{ km/s}\\ V_{2} & =\sqrt{\mu \left ( \frac{2}{r_{1}}-\frac{1}{a}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{200+6378}-\frac{1}{1.0329\times 10^{5}}\right ) }=10.832\text{ km/s}\\ \Delta V_{1} & =V_{2}-V_{1}=10.832-7.7843=3.0477\text{ km/s}\\ V_{3} & =\sqrt{\mu \left ( \frac{2}{r_{2}}-\frac{1}{a}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{130000}-\frac{1}{1.0329\times 10^{5}}\right ) }=1.5077\\ V_{4} & =\sqrt{\frac{\mu }{r_{2}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{130000}}=1.751\text{ km/s}\\ e & =\frac{r_{b}-r_{1}}{r_{b}+r_{1}}=\frac{200000-\left ( 200+6378\right ) }{200000+\left ( 200+6378\right ) }=0.93631\\ \cos \gamma & =\sqrt{\frac{a^{2}\left ( 1-e^{2}\right ) }{r_{2}\left ( 2a-r_{2}\right ) }}=\sqrt{\frac{\left ( 1.0329\times 10^{5}\right ) ^{2}\left ( 1-0.93632^{2}\right ) }{130000\left ( 2\left ( 1.0329\times 10^{5}\right ) -130000\right ) }}=0.36351\\ \Delta V_{2} & =\sqrt{V_{4}^{2}+V_{3}^{2}-2V_{4}V_{3}\cos \gamma }=\sqrt{1.751^{2}+1.5077^{2}-2\left ( 1.751\right ) \left ( 1.5077\right ) \left ( 0.36351\right ) }=1.8493\\ \Delta V & =\left \vert \Delta V_{1}\right \vert +\left \vert \Delta V_{2}\right \vert =3.0477+1.8493=4.897 \end{align*}
To find transfer time, we first must find \(E\), which is found by solving \(r=a\left ( 1-e\cos E\right ) \) where \(r\) is the radius we want to find \(E\) at which is \(r_{2}\) in this case. Hence \begin{align*} r_{2} & =a\left ( 1-e\cos E\right ) \\ 130000 & =1.0329\times 10^{5}\left ( 1-0.93631\cos E\right ) \\ 0.93631\cos E & =1-\frac{130000}{1.0329\times 10^{5}}\\ 0.93631\cos E & =-0.25859\\ \cos E & =-0.27618\\ E & =\arccos \left ( -0.27618\right ) =1.8506\text{ rad}\\ n & =\sqrt{\frac{\mu }{a^{3}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{\left ( 1.0329\times 10^{5}\right ) ^{3}}}=1.9019\times 10^{-5}\\ \Delta t & =\frac{1}{n}\left ( E-e\sin E\right ) \\ \Delta t & =\frac{1}{1.9019\times 10^{-5}}\left ( 1.8506-\left ( 0.93632\right ) \sin \left ( 1.8506\right ) \right ) =49987\text{ sec}\\ & =\frac{49987}{60\times 60}=13.885\text{ hr} \end{align*}
with a transfer time half that required for a semi-tangential parabolic transfer
Semi-tangential parabolic transfer time: Answer hours
Semi-tangential hyperbolic transfer time: Answer hours
Semi-tangential hyperbolic total \(\Delta V\): Answer km/s
Answer:
\(r_{1}=200+6378\) km\(,r_{2}=130000\) km. For a parabolic orbit, the true anamoly \(\theta \) is found when \(r=r_{2}\). From\begin{align*} r_{2} & =\frac{2r_{p}}{1+\cos \theta }\\ \theta & =\arccos \left ( \frac{2r_{p}}{r_{2}}-1\right ) \\ & =\arccos \left ( \frac{2r_{p}}{r_{2}}-1\right ) \end{align*}
But \(r_{p}=r1\) hence\begin{align*} \theta & =\arccos \left ( \frac{2r_{1}}{r_{2}}-1\right ) \\ & =\arccos \left ( \frac{2\left ( 200+6378\right ) }{130000}-1\right ) \\ & =2.6878\text{ rad}\\ & =154^{0} \end{align*}
So the time for transfer if we are using a parabolic orbit is\begin{align*} \Delta t & =\frac{\tan \left ( \frac{\theta }{2}\right ) +\frac{1}{3}\left ( \tan \left ( \frac{\theta }{2}\right ) \right ) ^{3}}{2\sqrt{\frac{\mu }{2r_{1}}}}\\ & =\frac{\tan \left ( \frac{2.6878}{2}\right ) +\frac{1}{3}\left ( \tan \left ( \frac{2.6878}{2}\right ) \right ) ^{3}}{2\sqrt{\frac{3.986\left ( 10^{5}\right ) }{\left ( 2\left ( 200+6378\right ) \right ) ^{3}}}}\\ & =37547\text{ sec}\\ & =\frac{37547}{60\times 60}=10.430\text{ hr} \end{align*}
Hence required time for hyperbolic is\[ \Delta t_{hyper}=\frac{1}{2}\left ( 10.430\right ) =5.215\text{ hr}\] Now to obtain \(\Delta V\) for hyperbolic orbit.
If we know \(r_{1},r_{2}\) on the orbit, and know the travel time between these 2 points then \(a,e,F\) can be found by numerically solving these equations \begin{align*} r_{1} & =a\left ( e-1\right ) \\ r_{2} & =a\left ( e\cosh F-1\right ) \\ \Delta t & =\sqrt{\frac{a^{3}}{\mu }}\left ( e\sinh \left ( F\right ) -F\right ) \end{align*}
The above are 3 equations with 3 unknowns\begin{align*} 200+6378 & =a\left ( e-1\right ) \\ \sqrt{\frac{3.986\left ( 10^{5}\right ) }{a^{3}}}\left ( 5.215\times 60\times 60\right ) & =e\sinh \left ( F\right ) -F\\ 130000 & =a\left ( e\cosh F-1\right ) \end{align*}
Solving gives\begin{align*} e & =1.5468\\ a & =12029.4\text{ km}\\ F & =2.7213\text{ }rad \end{align*}
Hence\begin{align*} a & =12029.4\text{ km}\\ V_{1} & =\sqrt{\frac{\mu }{r_{1}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{200+6378}}=7.7843\text{ km/s}\\ V_{2} & =\sqrt{\mu \left ( \frac{2}{r_{1}}+\frac{1}{a}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{200+6378}+\frac{1}{12029.4}\right ) }=12.423\text{ km/s}\\ \Delta V_{1} & =V_{2}-V_{1}=12.423-7.7843=4.6387\text{ km/s}\\ V_{3} & =\sqrt{\mu \left ( \frac{2}{r_{2}}+\frac{1}{a}\right ) }=\sqrt{3.986\left ( 10^{5}\right ) \left ( \frac{2}{130000}+\frac{1}{12029.4}\right ) }=6.2664\\ V_{4} & =\sqrt{\frac{\mu }{r_{2}}}=\sqrt{\frac{3.986\left ( 10^{5}\right ) }{130000}}=1.751\text{ km/s}\\ \cos \gamma & =\sqrt{\frac{a^{2}\left ( e^{2}-1\right ) }{r_{2}\left ( 2a+r_{2}\right ) }}=\sqrt{\frac{\left ( 12029.4\right ) ^{2}\left ( 1.5468^{2}-1\right ) }{130000\left ( 2\left ( 12029.4\text{ }\right ) +130000\right ) }}=0.10031\\ \Delta V_{2} & =\sqrt{V_{4}^{2}+V_{3}^{2}-2V_{4}V_{3}\cos \gamma }=\sqrt{1.751^{2}+6.2664^{2}-2\left ( 1.751\right ) \left ( 6.2664\right ) \left ( 0.10031\right ) }=6.335\\ \Delta V & =\left \vert \Delta V_{1}\right \vert +\left \vert \Delta V_{2}\right \vert =4.6387+6.335=10.974 \end{align*}