A satellite is in an orbit with a period \(T=205\) minutes and eccentricity \(e=0.40\) about the Earth. When the true anomaly of the satellite is \(f=70\) degrees, find the time \(t-\tau \) since perigee passage, in minutes.
Answer
\[ n\left ( t-\tau \right ) =E-e\sin E \]
But \(n=\frac{2\pi }{T}\) hence
\begin{equation} t-\tau =\frac{E-e\sin E}{n}=\frac{T\left ( E-e\sin E\right ) }{2\pi } \tag{1} \end{equation}
But \(\tan \left ( \frac{f}{2}\right ) =\sqrt{\frac{1+e}{1-e}}\tan \left ( \frac{E}{2}\right ) \), hence \(E\) can be found. Substituting it in the above, solves for \(t-\tau \)
\begin{align*} \tan \left ( \frac{70\pi }{2\left ( 180\right ) }\right ) & =\sqrt{\frac{1+0.4}{1-0.4}}\tan \left ( \frac{E}{2}\right ) \\ 0.70021 & =1.5275\tan \left ( \frac{E}{2}\right ) \\ \tan \left ( \frac{E}{2}\right ) & =\frac{0.70021}{1.5275}=0.4584\\ \frac{E}{2} & =\arctan \left ( 0.4584\right ) =0.42982\\ E & =0.85964 \end{align*}
Hence from Eq (1)
\begin{align*} t-\tau & =\frac{205\left ( 0.85964-0.40\sin \left ( 0.85964\right ) \right ) }{2\pi }\\ & =18.16\text{ min} \end{align*}
A satellite is in an orbit with a period \(T=205\) minutes and eccentricity \(e=0.40\) about the Earth. Find the true anomaly of the satellite, in degrees, when it is \(50\) minutes past perigee passage.
Answer
\begin{align*} n\left ( t-\tau \right ) & =E-e\sin E\\ \frac{2\pi }{T}\left ( t-\tau \right ) & =E-e\sin E\\ \frac{2\pi }{205}\left ( 50\right ) & =E-0.4\sin E\\ 1.5325 & =E-0.4\sin \left ( E\right ) \end{align*}
Solving for \(E\)
\[ E=1.9097\text{ rad}\]
Hence
\begin{align*} \tan \left ( \frac{f}{2}\right ) & =\sqrt{\frac{1+e}{1-e}}\tan \left ( \frac{E}{2}\right ) \\ \tan \left ( \frac{f}{2}\right ) & =\sqrt{\frac{1+0.4}{1-0.4}}\tan \left ( \frac{1.9097}{2}\right ) \\ & =2.1581 \end{align*}
Hence
\begin{align*} \frac{f}{2} & =\arctan \left ( 2.1581\right ) =1.1369\\ f & =\left ( 1.1369\right ) 2=2.2738\\ & =2.2738\left ( \frac{180}{\pi }\right ) \\ & =130.28\text{ deg} \end{align*}
A spaceship in a circular orbit above the Earth at an altitude of \(300\) km. At time \(t=0\), it retrofires its engine, reducing its speed by \(500\) m/s. How long (in minutes) does it take to impact the Earth? Neglect atmospheric drag.
Answer
\[ \mu =3.986\times 10^{5}\text{ \ km}^{3}\text{/s}^{2}\]
But \[ \Delta V=V_{2}-V_{1}\]
Where \(V_{1}=\sqrt{\frac{\mu }{r_{a}}}=\sqrt{\frac{\mu }{r_{E}+alt}}\) where \(r_{E}\) is earth radius and \(alt\) is the altitude at \(t=0\) when the spaceship was in circular orbit. Hence \(V_{1}=\sqrt{\frac{3.986\times 10^{5}}{6378+300}}=7.7258\) km/s hence \(V_{2}=V_{1}-500\times 10^{-3}=7.7258-0.5=\) \(7.2258\) km/sec. This is the speed at apogee for the new orbit.
\[ V_{a}=7.2258\text{ km/sec}\] But \begin{align} V_{a} & =\sqrt{\frac{\mu }{a}\left ( \frac{1-e}{1+e}\right ) }\nonumber \\ 7.2258 & =\sqrt{\frac{3.986\times 10^{5}}{a}\left ( \frac{1-e}{1+e}\right ) } \tag{1} \end{align}
But also we know that \(r_{a}=a\left ( 1+e\right ) \), hence
\begin{align} 6378+300 & =a\left ( 1+e\right ) \nonumber \\ a & =\frac{6678}{1+e} \tag{2} \end{align}
Substitute (2) in (1) \begin{align*} 7.2258 & =\sqrt{\frac{3.986\times 10^{5}}{6678}\left ( 1-e\right ) }\\ 52.212 & =\frac{3.986\times 10^{5}}{6678}\left ( 1-e\right ) \\ \frac{\left ( 52.212\right ) \left ( 6678\right ) }{3.986\times 10^{5}} & =1-e\\ 0.87474 & =1-e\\ e & =1-0.87474\\ & =0.12526 \end{align*}
Hence from (2) we find \(a\)
\[ a=\frac{6678}{1+0.12526}=5934.6 \]
Hence \(n\) the mean speed is \begin{align*} n & =\sqrt{\frac{\mu }{a^{3}}}\\ & =\sqrt{\frac{3.986\times 10^{5}}{5934.6^{3}}}\\ & =1.381\times 10^{-3}rad/s \end{align*}
At impact \(r=r_{E}\), hence
\begin{align*} r_{E} & =a\left ( 1-e\cos E\right ) \\ 6378 & =5934.6\left ( 1-0.12526\cos E\right ) \\ \frac{6378}{5934.6} & =1-0.12526\cos E\\ \cos E & =\frac{1-\frac{6378}{5934.6}}{0.12526}=-0.596\,47\\ E & =\arccos \left ( -0.596\,47\right ) \\ E & =2.2099 \end{align*}
Solving this equation gives \(E=126^{0}\), but we want to use the \(E\) shown in the diagram. Hence remember to do \(E_{actual}=2\pi -E\) to obtain \(E_{actual}=233^{o}\) and that is the \(E\) to use in \(n\left ( t-\tau \right ) =E-e\sin E.\)Hence, measured from perigee, \[ E=2\pi -2.2099 \]
Using Kepler equation
\begin{align*} n\left ( t-\tau \right ) & =E-e\sin E\\ 1.381\times 10^{-3}\left ( t-\tau \right ) & =\left ( 2\pi -2.2099\right ) -0.12526\sin \left ( 2\pi -2.2099\right ) \\ \left ( t-\tau \right ) & =\frac{\left ( 2\pi -2.2099\right ) -0.12526\sin \left ( 2\pi -2.2099\right ) }{1.381\times 10^{-3}}\\ & =3022.\text{ sec}\\ & =\allowbreak 50.37\text{ min} \end{align*}
But the period is \(T=2\pi \sqrt{\frac{a^{3}}{\mu }}\) \(=2\pi \frac{1}{n}=2\pi \frac{1}{1.381\times 10^{-3}}=4549.7\) sec \(=75.828\) min
Hence the time to impact is
\[ 50.37-\frac{75.828}{2}=12.456\text{ min}\]
Russians use Molniya orbits for their communications satellites. A typical Molniya orbit has a perigee altitude of 500 km and a period of 12 hr.
What is the eccentricity of a Molniya orbit?
Answer
\begin{align*} T & =2\pi \sqrt{\frac{a^{3}}{\mu }}\\ 12\times 60\times 60 & =2\pi \sqrt{\frac{a^{3}}{3.986\times 10^{5}}}\\ \left ( 12\times 60\times 60\right ) ^{2} & =\left ( 2\pi \right ) ^{2}\frac{a^{3}}{3.986\times 10^{5}}\\ a^{3} & =\frac{\left ( 12\times 60\times 60\right ) ^{2}\left ( 3.986\times 10^{5}\right ) }{\left ( 2\pi \right ) ^{2}}=1.884\,3\times 10^{13}\\ a & =\left ( 1.8843\times 10^{13}\right ) ^{1/3}=26610\text{ km} \end{align*}
We are given that \(r_{p}=6378+500=6878\), but \(r_{p}=\frac{a\left ( 1-e^{2}\right ) }{1+e}=a\left ( 1-e\right ) \), hence
\begin{align*} e & =1-\frac{r_{p}}{a}\\ & =1-\frac{6878}{26610}\\ & =0.741\,53 \end{align*}
What is the apogee radius of a Molniya orbit, in km?
Answer
\begin{align*} r_{p} & =a\left ( 1+e\right ) \\ & =26610\left ( 1+0.741\,53\right ) \\ & =46342\text{ km} \end{align*}
Determine the time, in hours, that a satellite on a Molniya orbit has a true anomaly greater than 135\(^{o}\) and less than 225\(^{o}\)
Answer
Let \(\theta _{1},\theta _{2}\) be the true anomaly angles at position \(1\) and \(2\), and let \(E_{1},E_{2}\) be the corresponding circular angles. We first find \(E_{1},E_{2}\)
\begin{align*} \tan \left ( \frac{E_{1}}{2}\right ) & =\tan \left ( \frac{\theta _{1}}{2}\right ) \sqrt{\frac{1-e}{1+e}}\\ & =\tan \left ( \frac{135\pi }{2\times 180}\right ) \sqrt{\frac{1-0.741\,53}{1+0.741\,53}}=0.930\,07\\ \frac{E_{1}}{2} & =\arctan \left ( 0.93007\right ) =0.749\,18\\ E_{1} & =0.749\,18\times 2\\ & =1.498\,4\text{ rad}\\ & =1.498\,4\times \left ( \frac{180}{\pi }\right ) =85.85^{0} \end{align*}
Similarly
\begin{align*} \tan \left ( \frac{E_{2}}{2}\right ) & =\tan \left ( \frac{\theta _{2}}{2}\right ) \sqrt{\frac{1-e}{1+e}}\\ & =\tan \left ( \frac{225\pi }{2\times 180}\right ) \sqrt{\frac{1-0.741\,53}{1+0.741\,53}}=-0.930\,07\\ \frac{E_{1}}{2} & =\arctan \left ( -0.930\,07\right ) =-0.749\,18\\ E_{1} & =-0.749\,18\times 2\\ & =-1.498\,4\text{ rad} \end{align*}
Hence \(E_{2}=-1.49836\) rad or \(-85.75^{o}\), Measured anticlockwise from perigee, it becomes \(E_{2}=360-85.75=274.15^{o}\)
Now the time to reach point \(1\), is
\begin{align*} n\left ( t_{1}\right ) & =E_{1}-e\sin E_{1}\\ t_{1} & =\frac{E_{1}-e\sin E_{1}}{\sqrt{\frac{\mu }{a^{3}}}}=\frac{1.498\,4-0.741\,53\sin \left ( 1.498\,4\right ) }{\sqrt{\frac{3.986\times 10^{5}}{26610^{3}}}}=5217.1\text{ sec} \end{align*}
and
\begin{align*} n\left ( t_{2}\right ) & =E_{2}-e\sin E_{2}\\ t_{1} & =\frac{\left ( 2\pi -1.49836\right ) -0.741\,53\sin \left ( 2\pi -1.49836\right ) }{\sqrt{\frac{3.986\times 10^{5}}{26610^{3}}}}=37983\text{ sec} \end{align*}
Hence the difference is \(37983\) \(-5217.1=32766\) sec or \(\frac{32766}{60\times 60}=9.1017\) hr