User talk:Azjps/sandbox/gov school

Timeline and trajectory

Most of the materials to be sent to Mars will spend time on a parking orbit around the Earth, which is an orbit about the Earth which we have set to $300\ km$. This gives a greater number of opportunities for launch. The spacecraft itself will be kept in orbit about Mars, while a lander and return craft will allow transportation from the Martian surface to the spacecraft.

Timeline of launch

A base constructed in parking orbit will be sent to Mars first. This will precede the actual launch by approximately two years, and will be timed to open and self-develop at the correct time. The base will enter the atmosphere of Mars and will use powerful retro-rocket firings to dramatically decrease the impact of collision.

The spacecraft will be constructed in parking orbit similarly. The landing craft will be constructed on Earth and then shuttled to the spacecraft. The spacecraft then launches itself from the Earth and reaches the orbit of Mars, and enters a parking orbit about Mars. It then deploys the landing craft.

These launch windows will occur respectively at February 2031 and March 2033 (Citation: Wikipedia). Mars and Earth line up with respect to each other every $2.135$ years (this period is known as the synodic period). For example, launch windows occured August 2005 and October 2007, which we can write in decimal format as $2007 \frac{8}{12} \approx 2007.67$. Since the period is $2.135$ years, then other launch windows can be written in the form $2007.67 + 2.135n$ for integers $n$; for consecutive $n = 11,12$, we get February $2031$, March $2033$, which will be our launch dates for the base and the spacecraft, respectively.

The journey to Mars will take approximately six months; details are located under Trajectory and calculations. The astronauts will stay on Mars for a period of a year before returning.

Trajectory and Calculations

The elliptical orbits of Mars and the Earth look as follows:

[asy] pointpen = black; pathpen = black + linewidth(0.7); pen s = fontsize(10); pair S = (0,0); D(MP("\mathrm{Sun}",S,s),orange); D(yscale(0.99)*CR(S-(.0167,0),1)); D(rotate(2)*yscale(0.98)*CR(S-(.1422,0),1.523)); pair E = 0.99*expi(0), M=1.57* expi(0.7 * pi); D(MP("\mathrm{Earth}",E,SE,s),blue); D(MP("\mathrm{Mars}",M,NW,s),red);  [/asy]     [asy]  pointpen = black; pathpen = black + linewidth(0.7); pen s = fontsize(10); pair S = (0,0); D(MP("\mathrm{Sun}",S,s),orange); D(CR(S,1)); D(CR(S,1.524)); pair E = expi(0), M=1.524* expi(0.7 * pi); D(MP("\mathrm{Earth}",E,SE,s),blue); D(MP("\mathrm{Mars}",M,NW,s),red); [/asy]

(Figure 1(a,b). The first diagram shows the elliptical orbits; the orbits will be approximated as circles for calculational ease. The scale ratio here is approximately 0.75 inches : 1AU.)

However, for the sake of simplicity of calculations, we let the orbits be (1) coplanar (2) concentric (3) circular, with the Sun at the center. We describe the error caused by these assumptions under the Calculational Errors section.

Our path will take three distinct segments: An escape by the spacecraft into a circular parking orbit about the Earth, an elliptical transfer from the orbit of the Earth about the Sun to the orbit of Mars about the Sun, and then a landing from the orbit about Mars to the surface of Mars. Data for Mars, Earth, and the Sun can be found in Appendix A.

[asy]  pointpen = black; pathpen = black + linewidth(0.7); pen s = fontsize(10); pair E = (0,0);  path sphere = CR(MP("\mathrm{Earth}",E,(0,0),s),1); D(sphere); fill(sphere,rgb(0.8,1,1)); D(CR(E,1.1),MidArrow(6));  real r = 20;  D(arc((-r,0),r+1.1,0,6),EndArrow(6));  [/asy]           [asy]  pointpen = black; pathpen = black + linewidth(0.7); pen s = fontsize(10); pair S = (0,0); D(MP("\mathrm{Sun}",S,s),orange); D(CR(S,1)); D(CR(S,1.524));  pair E = expi(0), M=1.524* expi(0.3 * pi); D(MP("\mathrm{Earth}",E,SE,s),blue); D(MP("\mathrm{Mars\ at\ launch}",M,(-1,3),s),red); D(CR(E,0.05)); D(CR(MP("\mathrm{Mars}",D((-1.524,0),red),SE,s),0.05)); D(yscale(0.9782)*CR((-.262,0),1.312),dashed+linewidth(0.7),Arrow(6,Relative(0.3)));  [/asy]           [asy]  pointpen = black; pathpen = black + linewidth(0.7); pen s = fontsize(10); pair M = (0,0);  path sphere = CR(MP("\mathrm{Mars}",M,(0,0),s),0.75); D(sphere); fill(sphere,rgb(1,0.9,0.9)); D(CR(M,1.2),EndArrow(6)); D(CR(M,4),dashed); MP("\mathrm{SOI}",4*expi(12*pi/7),SE); MP("\mathrm{Hyperbolic\ flyby}",2.4*expi(4*pi/7),E,s); MP("\mathrm{Ellipse}",4.3*expi(2.8*pi/7),E,s); MP("\mathrm{Circular\ orbit}",expi(3*pi/2),S,s); real r = 3;  D(arc((r,0),r+1.2,110,180),EndArrow(6));  [/asy]

(Figure 2 (a,b,c). The three diagrams, respectively, show the transfer from a circular orbit about the Earth to the ellipse, the to and return journey of the spacecraft, and the transfer from the elliptical transfer to an orbit about Mars. All three are exaggerated to show detail.)

We consider an orbit of a satellite at a height of $h = 300\ km = 3 \times 10^5\ m$ (arbitrarily picked to be close to the orbits of other satellites) about the surface of the Earth. In order for an object to reach this altitude (in our case, we are shuttling the necessary parts/supplies), the conservation of energy from kinetic to potential applies: $\frac 12 mV^2 = \frac{GMm}{r_E} - \frac{GMm}{r_E+h} = \frac{hm\mu}{r_E(r_E+h)}$, where $\mu = GM$ is the gravitational parameter of an astronomical body. It follows that an initial shuttle thrusting of $V = 2370\ \frac ms \quad(\mathrm{I})$ is necessary.

For a satellite to maintain this altitude, the force of gravity upon the satellite must provide sufficient centripetal force for revolution about the Earth, or $\frac{V^2}{r} = \frac{\mu}{r^2} \Longrightarrow V_{orbit} = \sqrt{\frac{\mu_E}{r_E+h}} = 7726\ \frac ms \tag{(1)}$ (Error compiling LaTeX. ! Package amsmath Error: \tag not allowed here.). The escape velocity of the spacecraft from the orbit in space at a distance of $r_E + h$ can be computed using the transformation of kinetic energy into potential energy: $\frac 12mV_{esc}^2 = \frac{\mu m}{r} \Longrightarrow V_{esc} = \sqrt{\frac{2\mu_E}{r_E + h}} = 10,926\ \frac ms \tag{(2)}$ (Error compiling LaTeX. ! Package amsmath Error: \tag not allowed here.).

The next step involves the elliptical transfer (known as the Hohmann transfer) from the orbit of Earth to that of Mars, which relies upon the gravitation provided by the Sun. Thus the sun is at one of the foci of the ellipse. We can now calculate some general properties of the ellipse (shown in Figure y(b); variables are explained in Appendix A):


2a &= a_E + a_M \Longrightarrow a = 1.888 \times 10^{11}\ m \\ c &= a - a_{E} = 3.92 \times 10^{10}\ m \\ \epsilon &= \frac ca = 0.2076 \\ p &= a(1 - \epsilon^2) = 1.807 \times 10^{11}\ m

\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

The specific angular momentum of the orbit is

$\begin{align*}H &= a_{ES}v_{ES} = a_{MS}v_{MS} = \sqrt{p\mu_{S}} = 4.896 \times 10^{15} \\ &\Longrightarrow V_{ES} = 32729 \frac ms, \quad V_{MS} = 21475 \frac ms$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

To enter our desired elliptical orbit, we have two options:

  • Use a parabolic escape trajectory to exit the Earth orbit, and then make a thrust to enter the elliptical orbit. Note that a parabolic escape trajectory leaves a $V_{\infty} = 0 \frac ms$, where an infinite distance from the Earth is defined at the points where the gravity of the Earth no longer has a strong effect upon the spacecraft.
  • Use a hyperbolic escape trajectory, which will cause the spacecraft to automatically enter the elliptical orbit. This leaves a finite $V_{\infty} = V_{ES} - V_E = 2940\ \frac ms$.

The latter option requires less energy, and is known as the Oberth hyperbola. Indeed,


E &= \frac{V_{\infty}^2}{2} = \frac{V_r^2}{2} - \frac {\mu}r \\ \Longrightarrow V &= \sqrt{V_{esc}^2 + V_{\infty}^2} = \sqrt{\frac{2\mu}{r_E + h} + (V_{ES} - V_E)^2}. \\

&= \sqrt{10926^2 + 2940^2} = 11,314\ \frac ms \end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

To achieve a speed of $V$, we only need to fire a thrust in addition to the $V_{orbit}$. Thus $\Delta V = 11,314 - 7726 = 3,560\ \frac ms \quad(\mathrm{II})$. This firing will be sufficient to carry the spacecraft from the Earth to Mars.

Another point to take into account is the angle with which the satellite should be launched with respect to the tangential velocity of the Earth about the Sun. The sphere of influence of a certain body is defined as the region in which that body has greater gravitational influence than any other bodies. Using $F = \frac{-Gm_1m_2}{r^2}$, the ratio of two gravitational forces upon a single object is $\frac{F_{m_2}}{F_{m_1}} = \left(\frac{r_1}{r_2}\right)^2 \frac{m_2}{m_1}$. For the sake of convention, we define the sphere of influence of a mass $m_2$ to be the region at which $r_{SOI} = r\left(\frac{m_2}{m_1}\right)^{0.4}$. For Earth against the Sun, this value is $9.25 \times 10^{8}\ m$, or $\approx 145\times$ the radius of the Earth. At the end of the Oberth hyperbola, which is at the edge of this sphere of influence, our satellite should be aligned with the direction of the Earth about the Sun ($V_{pl}$). From $\theta = \cos^{-1} \left[\frac 1{\epsilon}\left(\frac{p}{r}-1\right)\right]$ and the approximation that $r_{\infty} >> p$, we find that $\theta \approx \cos^{-1} \left(1/\epsilon\right)$.

[asy] size(320); real h = 5, eqn_d = 0.5, eqn_xoffset = -1.5, eqn_yoffset = -1.5; pointpen = black; pathpen = black+linewidth(0.7); pen d = dashed, eqn = fontsize(12); pair E = (0,0); D(E); path sphere = CR(E, 1), hyperbola = arc((-1.2*h,1.2*h),(1.2*h)*2^.5 + 1,315,360); pair P = IP(hyperbola, (0,h)--(10,h));  D(sphere); D(CR(E,0.8),linewidth(1)+green); fill(CR(E,0.8),rgb(0.8,1,1)); D((-2,h)--(4,h)); D((3,-3)--E--(0,1.2*h),d); D((-1,0)--(-1,h),d); D(E--(0,h*.6),EndArrow(6)); D((1/2^.5,-1/2^.5)--(1/2^.5,-1/2^.5)+(2,2),EndArrow(6));  D(hyperbola,EndArrow(6)); D(rightanglemark(E,expi(7*pi/4),(2^.5,0))); D(rightanglemark((0,10), (0,h), (10,h))); D(rightanglemark(P+(0,2), P, (10,h))); D(anglemark(P-(0.2,2), P, (10,h)));  MP("b",(1.6,h),N); MP("V_{\infty}",(-1.2*h+(1.2*h+1)*2^.5-0.5,1.1*h),NW); MP("\mathrm{SOI}",(4,h),(1,0)); MP("V_{pl}",(0,2),W); D(anglemark((1,-1),(0,0),(0,1))); MP("\nu_{\infty}",(0,0),(2,2)); MP("r_{orbit}",(0,0.5),W); MP("\phi",P,(2,-2)); MP("r_{SOI}",(-1,h/2),W); MP("V_i",(1/2^.5,-1/2^.5)+(1,1),SE); MP("E = \frac{v_i^2}{2} - \frac{\mu}{r} = \frac{-\mu}{2a}",(eqn_xoffset,eqn_yoffset),NE,eqn); MP("\Longrightarrow a = 3.74 \times 10^6\ m",(eqn_xoffset,eqn_yoffset-eqn_d),NE,eqn); MP("\epsilon = \frac ca = 1 + \frac ra = 3.575",(eqn_xoffset,eqn_yoffset-2*eqn_d),NE,eqn); MP("\nu_{\infty} = \cos^{-1}\left(\frac{-1}{\epsilon}\right) = 106.2^{\circ}",(eqn_xoffset,eqn_yoffset-3*eqn_d),NE,eqn); [/asy]

(Figure 3. Asymptotic overview showing the hyperbolic path of the spacecraft parallel to the Earth's orbit.)

We know that the period of this journey is half of the period of the elliptical orbit, and by Kepler's Laws,

$P = \pi\sqrt{\frac{a^3}{\mu_S}} = 2.237 \times 10^7\ s = 8.5\ \mathrm{months}$

Mars and Earth are lined up at February 2031, March 2033 (see Timeline of launch). Thus, the spacecraft must be launched 8.5 months ahead, respectively at May 2030, July 2032. At this point, the spacecraft will enter the sphere of influence (SOI) of Mars. Since Mars is traveling at $24,140 \frac ms$ about the Sun, the relative velocity of the spacecraft at the periapsis for the Hohmann ellipse is

$|V_{\infty}| = 24,140 - V_M = 2665 \frac ms.$

To determine whether the spacecraft will fly about Mars and get captured by its gravity or if it will crash into Mars, we define the impact parameter, which is a condition that determines whether or not a flying object will collide with another due to gravity. Our goal is to have the spacecraft revolve in a circular orbit, where the spacecraft hovers over the same spot above Mars at all times, which would allow quick communication. This Martianosynchronous orbital period is given by Kepler's Laws to be $P = 2\pi\sqrt{\frac{r^3}{\mu_M}} \Longrightarrow r = \sqrt[3]{\mu_M\left(\frac{P}{2\pi}\right)^2} = 20,437,500\ m$. Using conservation of energy, the spacecraft reaches a speed of


\frac{V_\infty^2}{2} &= \frac{V_p^2}{2} - \frac{\mu_p}{r_p} \\ \Longrightarrow V_p^2 &= \frac{2\mu_M}{r} + V_{\infty}^2 \Longrightarrow V_p = 3363\ \frac ms,

\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

where $V_p$ is the speed of the spacecraft at the periapsis (closest point) on the hyperbolic flyby of Mars.

Since our goal is to achieve a permanent circular orbit of the spacecraft about Mars, we require that the gravitation force by Mars provides a sufficient centripetal force, or $\frac{V^2}{r} = \frac{\mu}{r^2} \Longrightarrow V_{M\ orbit} = \sqrt{\frac{\mu_M}{r}} = 1450\ \frac ms$. Thus, as the spacecraft makes the flyby, it must decrease its speed by $\Delta V = V_p - V_{M\ orbit} = 3363 - 1450 = 1913\ \frac ms \quad(\mathrm{III})$.

In conclusion, the velocity budget for this journey is $\Delta V_T = \mathrm{(I) + (II) + (III)} = 2370 + 3560 + 1913 = \boxed{7,843\ \frac ms}$.

Finally, we would like to launch our lander into the Martian atmosphere. We can approximate the speed of the lander using the conservation of energy: $\frac 12mV_{M\ orbit}^2 - \frac 12mV^2 = \frac{m\mu}{r} - \frac{m\mu}{r_M} \Longrightarrow V = 6745\ \frac ms$. The lander uses a system of backwards thrusts to dramatically reduce its speed upon entering the atmosphere, and then uses a parachute to reach a low terminal velocity. A system of inflatable balloons will soften the impact of the lander on the Martian surface.

The return craft, which goes with the base on the first trip, will be used to return the crew back to the orbiting spacecraft. The return journey will follow a similar elliptical orbit, except the order is now backwards: the journey starts with leaving from Mars, and reaches the Earth. Specifically, the same orbital speeds about Mars and the Earth and the same elliptical orbit applies, so the journey will be of approximately the same duration and have a similar shape.

Calculational Errors

Outside of natural error introduced by measurement problems, certain errors are introduced with several of the assumptions used to perform these calculations. The plane of Mars with respect to that of Earth is tilted at $1.85^{\circ}$, and Mars and Earth have eccentricities respectively of $\approx 0.0933$ and $0.01671$. Earth's orbit, at extremes, differs only about $2\%$ from its average radius (1 AU), while the Martian orbit differs by about $10\%$. The problems we discussed above at all times used a "patched-conic" solution, during which we switch between different conic section orbits instantaneously (such as the sharp bound at the sphere of influence). They also assume that the masses are point sources, and that there are at most 2 bodies affecting the spacecraft's trajectory at any point. Certain design specifications, such as that the acceleration of a rocket occurs instantaneously, can be assumed because of the relative distances. In all of these cases, the percentage differences are fairly small, yet all extremely significant.

Design and Materials

Our rotating room is in the form of annulus, with an inner radius of 64 feet and an outer radius of 74 feet. Using the equation from the previous section, $w^2 = 32/74 \Longrightarrow w = 0.658\, \mathrm{rad/sec}$. Approximating the human as six-feet tall, the centripetal force at the head is $29.4\ N$, a difference of 8% between the forces on the head and feet of the astronaut, which is tolerable.

Appendix A

$\begin{tabular}{|l||l|} \hline \textbf{Earth} & \\ \hline Gravitation parameter (\mu_E) & 3.986 \times 10^{14}\ m^3/s^2 \\ Semi-major axis (a_E) & 1 AU = 1.496 \times 10^{11}\ m \\ Orbital velocity (V_E) & 29,790\ m/s\\ Radius (r_E) & 6.378 \times 10^6\ m \\ Sphere of influence & 9.245 \times 10^8\ m \\ Surface escape velocity & 5,027\ m/s \\ \hline \textbf{Mars} & \\ \hline Gravitation parameter (\mu_M) & 4.297 \times 10^{13}\ m^3/s^2 \\ Semi-major axis (a_M) & 1.524 AU = 2.280 \times 10^{11}\ m \\ Orbital velocity (V_M) & 24,140\ m/s \\ Radius (r_M) & 3.393 \times 10^6\ m \\ Sphere of influence & 5.781 \times 10^8\ m \\ Surface escape velocity & 11,186\ m/s \\ \hline \textbf{Sun} & \\ \hline Gravitation parameter (\mu_S)& 1.327 \times 10^{20}\ m^3/s^2 \\ \hline \end{tabular}$ (Error compiling LaTeX. ! Missing $ inserted.)

Relevant Equations

About Physics:

Forces ($F=ma$, Newton's Second Law) - Gravitational: $F_G = \frac{GM_1M_2}{r^2}$ where $G$ is the universal gravitation constant, Centripetal: $F_C = m\frac{V^2}{r}$.

Energy ($W = E$) - Kinetic: $K = \frac 12 mv^2$, Gravitational potential: $U = \frac{-GM_1M_2}{r}$

About Conic Sections:

[asy] pen d = dashed; pair O = (0,0), F1 = (6, 0), F2 = (-6, 0); path ellipse = yscale(0.8) * CR(O, 10); D(O); pair P = IP((6,0)--(6,0)+10*expi(2 * pi / 7),ellipse), F3 = IP((-6,0)--(-6,10),ellipse); D(ellipse); D((-10,0) -- (10,0), d); D(F1); D(F2); D((0,-8) -- (0,8), d); D(F2 -- F3, d); D(F1 -- P);  MP("p",(F2+F3)/2,E); MP("a",(5,0),S); MP("c",(-3,0),S); MP("b",(0,4),E);  MP("r",(F1+P)/2,NW); MP("\theta",F1,(3,1.5));  [/asy]

(Figure 4: An example of an ellipse with $a = 10,\ \epsilon = 0.6$ )

The shape of a conic section is determined by a certain dimension and an eccentricity, $\epsilon$, that defines out circular the section is. An ellipse has an oval shape (see dashed trajectory in Figure Y for example), with a major axis spanning its longer length and a minor axis spanning its shorter length. The semi-major axis has length $a$, and the distance from the center of an ellipse to a focus is $c$. Then $\epsilon = \frac ca$. The equation of an ellipse can be written in the form $r = \frac{a(1-\epsilon^2)}{1+\epsilon \cos \theta} \Longleftrightarrow \theta = \cos^{-1} \left[\frac 1{\epsilon}\left(\frac{p}{r}-1\right)\right]$ where $\theta, r$ parameterize the points of the ellipse in terms of the distance from one focus. The quantity $a(1-\epsilon^2) = p$ is the semi-latus rectum of the ellipse. Some equations that come from physics regarding an object in an elliptical orbit include:

$H = rV\cos \phi = r_AV_A = r_PV_P = \sqrt{p \mu},\quad E = \frac{V^2}{2} -  \frac{\mu}{r} = \frac{-\mu}{2a}$

where $H$ is the specific angular momentum, $E$ is the energy. A circle is a specific ellipse that has an $\epsilon = 0$. A hyperbola has $\epsilon > 1$, but similar equations for ellipses apply to a hyperbola.

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