SOLUTION:
From the vis-viva relation \begin{equation} v_{perigee}=\sqrt{\frac{\alpha }{m}\left ( \frac{2}{r_{p}}-\frac{1}{a}\right ) } \tag{1} \end{equation} Where \(m\) is the reduced mass and \(\alpha =GM_{earth}m_{satt}\), which reduces to \(GM_{earth}\) and known constant called the Standard gravitational parameter which for earth is (From table)\[ \frac{\alpha }{m}=398600\text{ km}^{3}/s^{2}\] And\begin{align*} r_{p} & =220+6378\\ & =6598\text{ km} \end{align*}
Where \(6378\) is the equatorial radius of earth. And \(v_{perigee}=28070\) km/h. Therefore, we use (1) to solve for \(a\), the length of the semimajor axes of the elliptical orbit of the satellite around the earth. From (1), by squaring both sides \begin{align*} v_{p}^{2} & =\frac{\alpha }{m}\left ( \frac{2}{r_{p}}-\frac{1}{a}\right ) \\ \left ( \frac{28070}{60\times 60}\right ) ^{2} & =398600\left ( \frac{2}{220+6378}-\frac{1}{a}\right ) \end{align*}
Solving for \(a\) gives\[ a=6640\text{ km}\] Hence the apogee distance is \[ 2a=13280\text{ km}\]
We can also find \begin{align*} r_{a} & =2a-r_{p}\\ & =13280-6598\\ & =6682\text{ km} \end{align*}
When the satellite is at the apogee, it will be above the earth at height of\begin{align*} h_{a} & =r_{a}-r_{earth}\\ & =6682-6378\\ & =304\text{ km} \end{align*}
The period \(T\) is given by\begin{align*} T & =2\pi \sqrt{\frac{a^{3}}{\frac{\alpha }{m}}}\\ & =2\pi \sqrt{\frac{6640^{3}}{398600}}\\ & =5385\text{ sec}\\ & =\frac{5385}{60\times 60}=1.496\text{ hr} \end{align*}
SOLUTION:
The total energy is\[ E=\frac{1}{2}m\dot{r}^{2}+U_{effective}\] The escape velocity is when \(U_{effective}=0\) , therefore\[ 0=-U+\frac{l^{2}}{2mr^{2}}\] But angular momentum \(l=mrv\) and \(U=\frac{GM_{e}m}{r}\), hence the above becomes\begin{align} 0 & =-\frac{GM_{e}m}{r}+\frac{m^{2}r^{2}v^{2}}{2mr^{2}}\nonumber \\ & =-\frac{GM_{e}m}{r}+\frac{mv^{2}}{2}\nonumber \\ & =-\frac{GM_{e}}{r}+\frac{v^{2}}{2} \tag{1} \end{align}
Now we are given that the satellite was at \(r=200+6378=6578\) km (this is \(r_{p}\) for the new orbit as well). Using \(GM_{e}=398600\) km\(^{3}/s^{2}\) from tables then we solve now for \(v\) in (1), which will be the new velocity. Hence\begin{align*} 0 & =-\frac{398600}{6578}+\frac{v^{2}}{2}\\ v & =11.009\text{ km/sec} \end{align*}
Before this, the spacecraft was in circular orbit. So its speed was \begin{align*} v_{c} & =\sqrt{\frac{\alpha }{m}\frac{1}{r}}\\ & =\sqrt{\frac{398600}{6578}}\\ & =7.784\text{ km/sec } \end{align*}
The difference is the minimum speed kick needed, which is\[ 11.009-7.784=3.225\text{ km/sec}\] This orbit is parabolic since \(U_{effective}=0\) as seen on the \(U_{effective}\) vs. \(r\) graph. parabolic is the first orbit beyond elliptic that do not contain turn points. The next orbit is hyperbolic.
SOLUTION:
Eccentricity is defined as (for all conic sections)\begin{equation} e=\sqrt{1+\frac{2El^{2}}{m\alpha ^{2}}} \tag{1} \end{equation} Where \(\alpha =GM_{sun}m\) and \(l\) is the angular momentum \begin{align*} l & =m\left \vert \mathbf{r}\times \mathbf{v}\right \vert \\ & =mrv\sin \phi \end{align*}
Therefore (1) becomes\[ e=\sqrt{1+\frac{2E\left ( rv\sin \phi \right ) ^{2}}{m\left ( GM_{sun}\right ) ^{2}}}\] The energy of the comet is given by \(E=\frac{1}{2}mv^{2}-\frac{GM_{sun}m}{r}\), then the above becomes\begin{align*} e & =\sqrt{1+\frac{2\left ( \frac{1}{2}mv^{2}-\frac{GM_{sun}m}{r}\right ) \left ( rv\sin \phi \right ) ^{2}}{m\left ( GM_{sun}\right ) ^{2}}}\\ & =\sqrt{1+\left ( \frac{2\left ( \frac{1}{2}mv^{2}-\frac{GM_{sun}m}{r}\right ) }{m}\right ) \left ( \frac{rv\sin \phi }{GM_{sun}}\right ) ^{2}}\\ & =\sqrt{1+\left ( v^{2}-\frac{2GM_{sun}}{r}\right ) \left ( \frac{rv\sin \phi }{GM_{sun}}\right ) ^{2}} \end{align*}
Let \(v=qv_{e}\) where \(v_{e}\) is earth velocity around the sun and let \(r=dr_{e}\) where \(r_{e}\) is the astronomical unit (the distance between the earth and sun) then result of part (1) becomes\begin{equation} e=\sqrt{1+\left ( \left ( qv_{e}\right ) ^{2}-\frac{2GM_{sun}}{dr_{e}}\right ) \left ( \frac{dr_{e}qv_{e}\sin \phi }{GM_{sun}}\right ) ^{2}} \tag{2} \end{equation} Looking at the earth/sun system, we know that \begin{align*} \frac{GM_{sun}m_{earth}}{r_{e}^{2}} & =\frac{m_{earth}v_{e}^{2}}{r_{e}}\\ \frac{GM_{sun}}{r_{e}} & =v_{e}^{2}\\ GM_{sun} & =r_{e}v_{e}^{2} \end{align*}
Replacing \(GM_{sun}\) in (2) by the above result gives\begin{align*} e & =\sqrt{1+\left ( \left ( qv_{e}\right ) ^{2}-\frac{2r_{e}v_{e}^{2}}{dr_{e}}\right ) \left ( \frac{dr_{e}qv_{e}\sin \phi }{r_{e}v_{e}^{2}}\right ) ^{2}}\\ & =\sqrt{1+\left ( \left ( qv_{e}\right ) ^{2}-\frac{2v_{e}^{2}}{d}\right ) \left ( \frac{dq\sin \phi }{v_{e}}\right ) ^{2}}\\ & =\sqrt{1+\left ( q^{2}-\frac{2}{d}\right ) \left ( dq\sin \phi \right ) ^{2}}\\ & =\sqrt{1+\left ( \frac{q^{2}d-2}{d}\right ) \left ( dq\sin \phi \right ) ^{2}} \end{align*}
We are now ready to answer the final part. If \(q^{2}d=2\) then \(e=1\) which means it is parabolic. If \(q^{2}d>2\) then \(\left ( \frac{q^{2}d-2}{d}\right ) \) is positive and the expression inside \(\sqrt{.}\) is larger than one, and hence \(e>1\), which means the orbit is hyperbolic. Finally, if \(q^{2}d<2\) then \(\left ( \frac{q^{2}d-2}{d}\right ) \) is negative, and the expression inside \(\sqrt{.}\) is less than one, which means \(e<1\) and hence the orbit is elliptic.
SOLUTION:
The angular momentum \(l\) is constant. At perigee, where the speed is maximum, we have\[ l_{p}=mv_{\max }r_{p}\] And at apogee, where the speed is minimum, we have\[ l_{a}=mv_{\min }r_{a}\] Since \(l\) is constant, then\begin{align} mv_{\max }r_{p} & =mv_{\min }r_{a}\nonumber \\ v_{\max }r_{p} & =v_{\min }r_{a}\tag{1} \end{align}
But \begin{align*} r_{a} & =a\left ( 1+e\right ) \\ r_{p} & =a\left ( 1-e\right ) \end{align*}
Hence (1) becomes\begin{align*} v_{\max }a\left ( 1-e\right ) & =v_{\min }a\left ( 1+e\right ) \\ v_{\max }\left ( 1-e\right ) & =v_{\min }\left ( 1+e\right ) \\ v_{\max }-ev_{\max } & =v_{\min }+ev_{\min }\\ v_{\max }-v_{\min } & =e\left ( v_{\min }+v_{\max }\right ) \\ e & =\frac{v_{\max }-v_{\min }}{v_{\min }+v_{\max }} \end{align*}
Replacing \(v_{\max }=v+v_{0}\) and \(v_{\min }=v-v_{0}\) gives\begin{align*} e & =\frac{\left ( v+v_{0}\right ) -\left ( v-v_{0}\right ) }{\left ( v+v_{0}\right ) +\left ( v-v_{0}\right ) }\\ & =\frac{2v_{0}}{2v}\\ & =\frac{v_{0}}{v} \end{align*}
SOLUTION:
In this calculation, the standard symbol \(\mu \) is used for \(GM_{earth}\) which is the Standard gravitational parameter (in class, we used \(\frac{\alpha }{m}\) for this same parameter). For earth \[ \mu =398600\text{ km}^{3}/s^{2}\] The first step is to find \(a\) for the transfer ellipse. This is given by\[ a=\frac{R_{LEO}+R_{GEO}}{2}\] Next, we first find \(V_{1}\), which is velocity in the LEO circular orbit just before initial kick to \(V_{2}\). Since this is circular, the speed is given by \[ V_{1}=\sqrt{\frac{\mu }{R_{LEO}}}\] Next step is to find \(V_{2}\), which is the speed at the perigee of the ellipse (the transfer orbit). This is given by the standard vis-viva relation \begin{equation} V_{2}=\sqrt{\mu \left ( \frac{2}{R_{LEO}}-\frac{1}{a}\right ) }\tag{1} \end{equation} Where \(R_{LEO}=r_{perigee}\) for the ellipse. Now that we found \(V_{2}\) and \(V_{1}\), then \begin{align*} \Delta V_{12} & =V_{2}-V_{1}\\ & =\sqrt{\mu \left ( \frac{2}{R_{LEO}}-\frac{1}{a}\right ) }-\sqrt{\frac{\mu }{R_{LEO}}} \end{align*}
When at the apogee of the transfer ellipse, the speed is given by\[ V_{3}=\sqrt{\mu \left ( \frac{2}{R_{GEO}}-\frac{1}{a}\right ) }\] We now want to be of GEO circular orbit, hence \[ V_{4}=\sqrt{\frac{\mu }{R_{GEO}}}\] And therefore, the speed boost is\begin{align*} \Delta V_{34} & =V_{4}-V_{3}\\ & =\sqrt{\frac{\mu }{R_{GEO}}}-\sqrt{\mu \left ( \frac{2}{R_{GEO}}-\frac{1}{a}\right ) } \end{align*}
