User:Azjps/sandbox/gov school

Trajectory and Calculations

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

[asy] size(150); 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] size(150); 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.)

For the sake of simplicity of calculations, we let the orbits be (1) coplanar (2) concentric (3) circular. Our path will have three distinct segments: A circular parking orbit about the Earth, an elliptical transfer from the Earth to the orbit of Mars, and then a circular orbit about Mars. Data for Mars, Earth, and the Sun can be found in Appendix A, along with specific calculations.

[asy] size(120); 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] size(120); 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] size(120); 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.)

The satellite has a height of $h = 300\ km$ above the Earth. From the law of conservation of energy, the satellite must convert some of its kinetic energy into gravitational potential energy to reach this altitude; from this we find that $V = 2370\ \frac ms\quad\mathrm{I}$ is necessary to propel a shuttle into orbit. For a satellite to keep this altitude, the force of gravity upon the satellite must provide sufficient centripetal force for revolution about the Earth, with a tangential speed of $V_{E\ orbit} = 7726\ \frac ms$. The escape velocity of the spacecraft from the orbit in space at a distance of $r_E + h$ from the center of the Earth can also be computed using the transformation of kinetic energy into potential energy: $10,926\ \frac ms$.

The next step involves an elliptical transfer (known as the Hohmann transfer) from the orbit of Earth to that of Mars, which relies upon gravity from the Sun. With known distances, the speeds of the spacecraft at the periapsis and apapsis (closest and farthest points, respectively) are found to be $V_{ES} = 32729 \frac ms, \quad V_{MS} = 21475 \frac ms$.

To enter our desired elliptical orbit, we use a hyperbolic escape trajectory, known as the Oberth hyperbola. This allows the spacecraft to immediately enter the elliptical orbit, leaving a residual speed of $V_{\infty} = V_{ES} - V_E = 2940\ \frac ms$ at a point far from the Earth. This escape requires a speed of $11,314\ \frac ms \end{align*}$ (Error compiling LaTeX. Unknown error_msg), or a thrusting of $\Delta V = 11,314 - V_{E\ orbit} = 3,560\ \frac ms \quad(\mathrm{II})$. We can also compute the angle of the spacecraft's trajectory with respect to the motion of the Earth. To find this, we note that the spacecraft's velocity should be parallel to the Earth's motion when the spacecraft exits the region affected by Earth's gravity (this region is known as the sphere of influence of the Earth). For Earth against the Sun, the radius of this sphere is $145\times$ the radius of the Earth. The angle turns out to be $\nu_{\infty} = 106.2^{\circ}$.

[asy] size(200); 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((2,-2)--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); [/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, by Kepler's Laws, is $P = 8.5\ \mathrm{months}$. Mars and Earth are aligned up at February 2031, March 2033 (see Timeline of launch), so the spacecraft must be launched 8.5 months ahead, respectively at May 2030, July 2032.

When the spacecraft reaches the apapsis of the orbit, the relative velocity of the spacecraft to Mars is $|V_{\infty}| = 24,140 - V_M = 2665 \frac ms.$ 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 easy communication. This Martianosynchronous orbital period is given by Kepler's Laws to be at a radius of $r = 20,437,500\ m$. Using conservation of energy, the spacecraft reaches a speed of $V_p = 3363\ \frac ms,$ where $V_p$ is the speed of the spacecraft at the closest point of 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 $V_{M\ orbit} = 1450\ \frac ms$. Thus, as the spacecraft makes the flyby, it must decrease its speed by $\Delta V = V_p - V_{M\ orbit} = 1913\ \frac ms \quad(\mathrm{III})$. In conclusion, the velocity budget for this journey is $\Delta V_T = \mathrm{(I) + (II) + (III)} = 2370 + 3560 + 1913 = \boxed{7,843\ \frac ms}$.

We would like to launch our lander into the Martian atmosphere. We can approximate the speed of the lander using the conservation of energy. 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.