A comet is on a parabolic orbit about the Sun. At its point of closest approach, the distance between the comet and the center of the Sun is 5 million km.
What is the speed of the comet, in km/s, relative to the Sun at its point of closest approach?
Answer
\begin{align*} v_{p} & =\sqrt{\frac{2\mu }{r_{p}}}=\sqrt{\frac{2\left ( 1.327\times 10^{11}\right ) }{5\times 10^{6}}}\\ & =230.39\left [ km/\sec \right ] \end{align*}
How long is the comet within 150 million km of the Sun?
Answer
\[ r=\frac{p}{1+\cos \theta }=\frac{2r_{p}}{1+\cos \theta }\]
\(p=2r_{p}=2\times 5\times 10^{6}=10\times 10^{6}\). Hence\begin{align*} \cos \theta & =\frac{p}{r}-1\\ & =\frac{10\times 10^{6}}{150\times 10^{6}}-1=-0.933\,33 \end{align*}
The above can also be found using
\[ r=\frac{\frac{h^{2}}{\mu }}{1+\cos \theta }\]
Where \(h=r_{p}v_{p}=5\times 10^{6}\times 230.39=1.152\,0\times 10^{9}\left [ km^{2}/s\right ] \)
Hence\begin{align*} \cos \theta & =\frac{h^{2}}{r\mu }-1\\ & =\frac{\left ( 1.152\,0\times 10^{9}\right ) ^{2}}{150\times 10^{6}\times 1.327\times 10^{11}}-1=-0.933\,33 \end{align*}
Therefore, \(\theta =\arccos \left ( -0.933\,33\right ) =2.774\,4\left [ rad\right ] =158.96^{o}.\) Now, from \begin{align*} 2\sqrt{\frac{\mu }{p^{3}}}\left ( t-\tau \right ) & =\tan \left ( \frac{\theta }{2}\right ) +\frac{1}{3}\left ( \tan \left ( \frac{\theta }{2}\right ) \right ) ^{3}\\ \left ( t-\tau \right ) & =\frac{\tan \left ( \frac{-2.7744}{2}\right ) +\frac{1}{3}\left ( \tan \left ( \frac{-2.7744}{2}\right ) \right ) ^{3}}{2\sqrt{\frac{1.327\times 10^{11}}{\left ( 10\times 10^{6}\right ) ^{3}}}}\\ & =2.493\,5\times 10^{6}\left [ \sec \right ] \\ & =\frac{2.493\,5\times 10^{6}}{60\times 60\times 24}\\ & =28.8558\left [ day\right ] \end{align*}
To account for both sides of the trajectory, then number of days is doubled, hence \(28.8558\times 2=57.712\left [ days\right ] \)
A spaceship is in a circular orbit about the Earth at an altitude of 700 km. It fires its rocket engine for a short time to instantaneously increase its speed by 75% and boost the spaceship to a hyperbolic orbit.
What is the speed increase (del V) of the spaceship in km/s as a result of the rocket burn?
Answer:
\[ V_{cir}=\sqrt{\frac{\mu }{r}}=\sqrt{\frac{3.986\times 10^{5}}{6378+700}}=7.5044\left [ km/s\right ] \]
Hence
\begin{align*} V_{2} & =V_{1}+\Delta V\\ 1.75V_{1} & =V_{1}+\Delta V\\ \Delta V & =0.75V_{1}\\ & =0.75\left ( 7.5044\right ) \\ & =5.6283\left [ km/s\right ] \end{align*}
What is the semimajor axis a of the resulting hyperbolic orbit in km?
Answer:
The new speed at the point of the firing is \(V=V_{1}+\Delta V=7. 5044+5.6283=13.133\left [ km/s\right ] \)
But \begin{align*} V & =\sqrt{\mu \left ( \frac{2}{r}+\frac{1}{a}\right ) }\\ V^{2} & =\mu \left ( \frac{2}{r}+\frac{1}{a}\right ) \\ \frac{1}{a} & =\frac{V^{2}}{\mu }-\frac{2}{r}\\ a & =\frac{1}{\frac{V^{2}}{\mu }-\frac{2}{r}}=\frac{1}{\frac{ 13.133^{2}}{3.988\times 10^{5}}-\frac{2}{6378+700}}\\ & =6670.2\left [ km\right ] \end{align*}
What is the eccentricity \(e\) of the resulting hyperbolic orbit?
Answer
These are 3 ways to find \(e\), the first is using \(r=\frac{a\left ( e^{2}-1\right ) }{1+e\cos \theta }\), where we can use that \(\theta =0\) at the time of firing since that is when \(r=r_{p}\) for the hyperbolic orbit. This is always the case, when an orbit changes to new orbit, we use the point of firing as perigee of the new orbit, and the true anamoly is hence zero at that point. This means \(r_{p}=\frac{a\left ( e^{2}-1\right ) }{1+e}\) and since we know \(a\) and \(r_{p}\) we can solve for \(e\)\begin{align*} 6378+700 & =\frac{6670.2\left ( e^{2}-1\right ) }{1+e}\\ 7078.0 & =\frac{6670.2\left ( e^{2}-1\right ) }{1+e}\\ 7078.0+7078.0e & =6670.2e^{2}-6670.2\\ 6670.2e^{2}-6670.2-7078.0-7078.0e & =0\\ 6670.\,2e^{2}-7078.0e-13748 & =0 \end{align*}
Hence \(e=\) \(2.0614\) or \(e=-1\), and since \(e\) is positive, we use \(e=2.0614\) as the solution.
Another way, is to note that since \(e=\frac{c}{a}\) and \(c=r_{p}+a\), hence \[ e=\frac{\left ( 6378+700\right ) +6670.2}{6670.2}=\allowbreak 2.061\,1 \]
Another way to find \(e\) is using \(e=\sqrt{1+\frac{2\mathcal{E}h^{2}}{\mu ^{2}}}\) where Energy \(\mathcal{E=}\frac{v^{2}}{2}-\frac{\mu }{r}\) and \(h=rv\), hence
\begin{align*} e & =\sqrt{1+\frac{2\left ( \frac{v^{2}}{2}-\frac{\mu }{r}\right ) \left ( rv\right ) ^{2}}{\mu ^{2}}}\\ & =\sqrt{1+\frac{2\left ( \frac{13.133^{2}}{2}-\frac{3.988\times 10^{5}}{6378+700}\right ) \left ( \left ( 6378+700\right ) 13.133\right ) ^{2}}{\left ( 3.988\times 10^{5}\right ) ^{2}}}\\ & =2.0614 \end{align*}
How long (in hours) does it take the spacecraft to reach the Moon’s orbit, a distance of 384,000 km from the center of the Earth?
Answer
\[ r_{2}=384000\left [ km\right ] \]
Using
\[ \sqrt{\frac{\mu }{a^{3}}}\left ( t-\mu \right ) =e\sinh \left ( F\right ) -F \]
Where \(F\) is found from \begin{align*} r & =a\left ( e\cosh \left ( F\right ) -1\right ) \\ 384000 & =6670.2\left ( 2.0611\cosh \left ( F\right ) -1\right ) \\ \cosh \left ( F\right ) & =\frac{\frac{384000}{6670.}+1}{2.0611}=28.417\\ F & =4.03983 \end{align*}
Hence
\begin{align*} \sqrt{\frac{\mu }{a^{3}}}\left ( t-\tau \right ) & =e\sinh \left ( F\right ) -F\\ \sqrt{\frac{3.988\times 10^{5}}{6670.2^{3}}}\left ( t-\tau \right ) & =2.0611\sinh \left ( 4.03983\right ) -4.03983\\ \left ( t-\tau \right ) & =\frac{2.0611\sinh \left ( 4.03983\right ) -4.03983}{\sqrt{\frac{3.988\times 10^{5}}{6670.2^{3}}}}\\ & =47009\left [ \sec \right ] \\ & =\frac{47009}{60\times 60}=13.058\left [ hr\right ] \end{align*}
Using Matlab, EES, Mathcad, Maple or similar software, create a program to calculate the position and velocity components of a satellite in an \(x, y, z\) coordinate system given its classical orbital elements (a, e, i, GAMMA, OMEGA, f). Use the examples in the course notes to test your program, then apply it to the set of elements below. (Save your program somewhere you can find it again; you will need it later in the semester.)
a: 9000 km
e: 0.02
i: 28.5 degrees
GAMMA: 50 degrees
OMEGA: 20 degrees
f: 40 degrees
x = Answer km
y = Answer km
z = Answer km
vx = Answer km/s
vy = Answer km/s
vz = Answer km/s