BAAO 2020, Теория — Задача 1. Apollo 11

• a. (i) F = 3.30 × 10⁷ N; a = 1.14 g0 at lift-off and 4.26 g0 when the S-IC finishes its burn

• a. (ii) a = 8.35 m s⁻² (= 0.85 g0); m = 126.8 tonnes

• a. (iii) S-IC: 1.14 g0 to 4.26 g0 (0–168 s); S-II: 0.74 g0 to 2.31 g0 (168–552 s); S-IVB: 0.85 g0 (552–699 s).

• a. (iv) 11430 m/s

• b. (i) 192 km

• b. (ii) 1.73

  revolutions

• c. a = 2.01 × 10⁸ m; duration 124 hr 54 mins

• d. (i) x = 3.55 × 10⁸ m; a_AC = 1.81 × 10⁸ m; a_BC = 2.06 × 10⁷ m

• d. (ii) v_A = 10.8 km s⁻¹ (99.1% of v_esc,A); v_B = 2.22 km s⁻¹ (97.7% of v_esc,B); the eccentricities must be very close to 1

• d. (iii) 71 hr 33 mins (2.58 × 10⁵ s)

Q1 – Apollo 11 [35 marks]

a. (i)

$F = -I_{sp} g_0 \dot{m} = -263 \times 9.81 \times \frac{(135.6-2283.9)\times10^{3}}{168} = 3.30 \times 10^{7}\ \mathrm{N}$ [1] [1]

[Since this was a ‘show that’ question, students should have at least given to 3 s.f. Allow the mark whether they’ve done the change in the mass of S-IC (as shown) or in the change in the mass of the whole rocket (i.e. the numbers shown below for $m_{\mathrm{tot}}$)]

At lift off, $m_{\mathrm{tot}} = 2283.9 + 483.7 + 121.0 + 49.7 = 2938.3$ tonnes

$\therefore a = \frac{F}{m_{\mathrm{tot}}} = \frac{3.30\times10^{7}}{2938.3\times10^{3}} = 11.23\ \mathrm{m\,s^{-2}} = 1.14\ g_0$ [1] [1]

At the end of S-IC, , $m_{\mathrm{tot}} = 135.6 + 483.7 + 121.0 + 49.7 = 790.0$ tonnes

$\therefore a = \frac{F}{m_{\mathrm{tot}}} = \frac{3.30\times10^{7}}{790.0\times10^{3}} = 41.76\ \mathrm{m\,s^{-2}} = 4.26\ g_0$ [1] [1]

[Using only the mass of the S-IC (giving accelerations at lift off and end of S-IC of 1.47 $g_0$ and 24.8 $g_0$ respectively) loses both of these marks, but this conceptual error is only penalised here with full ecf allowed in the rest of the question]

a. (ii)

Combining the given thrust equation with Newton’s 2nd Law to determine m = f(t),

$-I_{sp} g_0 \dot{m} = ma \quad \therefore \dot{m} = -\frac{a}{I_{sp} g_0} m \quad \therefore m = m_0 \exp\left(-\frac{a}{I_{sp} g_0} t\right)$ [1]

Rearranging and solving for the initial and final masses during the whole S-IV burn,

$a = \frac{\ln\left(\frac{m_0}{m}\right) I_{sp} g_0}{t} = \frac{\ln\left(\frac{(121.0+49.7)}{(13.2+49.7)}\right)\times421\times9.81}{(147+347)} = 8.35\ \mathrm{m\,s^{-2}}\ (= 0.85\ g_0)$ [1] [2]

Using the exponential equation to calculate the final mass at the end of the first burn,

$m = (121.0 + 49.7)\exp\left(-\frac{8.35}{421\times9.81}\times147\right) = 126.8$ tonnes [1] [1]

[Full ecf available for ignoring the mass of the Apollo spacecraft (49.7 t), assuming the same conceptual error was penalised earlier, leading to a = 18.5 m s$^{-2}$ and m = 62.6 t]

[This is close to the 118 tonnes that the Saturn V was designed to be able to carry into Low Earth Orbit, the largest of any rocket that has ever flown (although planned rockets launched in the coming decade will exceed this)]

a. (iii)

So far we have for S-IC $a_{\mathrm{init}} = 1.14\ g_0$ and $a_{\mathrm{final}} = 4.26\ g_0$, plus for S-IV $a = 0.85\ g_0$

Doing the same thing for S-II gives,

$F = 4.77 \times 10^{6}\ \mathrm{N} \quad \therefore a_{\mathrm{init}} = 0.74\ g_0$ and $a_{\mathrm{final}} = 2.31\ g_0$ [2]

Plotting all of this information,

Graph has horizontal line for S-IV and only goes up to the end of first burn [1]

Graph has straight, diagonal lines for S-IC and S-II [1]

Appropriate axes labels (accel in $g_0$) [1] [5]

[The values of a (as summarised above) and t (0 s, 168 s, 552 s and 699 s) should either be listed in the working or clearly indicated in the sketch (since few students will try and do it to scale, unlike the graph above). Full ecf for incorrectly calculating the mass of the rocket gives for S-IC $a_{\mathrm{init}} = 1.47\ g_0$ and $a_{\mathrm{final}} = 24.8\ g_0$, for S-II $a_{\mathrm{init}} = 1.01\ g_0$ and $a_{\mathrm{final}} = 12.2\ g_0$, and for S-IV $a = 1.89\ g_0$. The lines are straight since we are ignoring the changing weight]

a. (iv)

The area under the graph is the change in velocity of the rocket [1]

Working it out for each stage and adding them together,

$\Delta v = 4451 + 5752 + 1227 = 11430\ \mathrm{m\,s^{-1}}$ [2] [3]

[Full ecf from the incorrect rocket masses leads to ~49250 m s$^{-1}$]

[It is worth noting that in under 12 minutes astronauts are in Earth orbit – staggeringly quick if you think about it – and they have to train to endure several seconds at an acceleration of about 4 $g_0$ (deeply uncomfortable!). If you integrate this graph twice you get the total change in displacement of the rocket at 4517 km – at this height above sea level and the calculated speed the orbit would not be bound (i.e. $E_{\mathrm{tot}} > 0$), meaning that the effects of air resistance and weight of the rocket make a big difference, as does the fact the rocket rapidly rotates onto its side rather than following a radial path]

b. (i)

For a circular orbit,

$v = \sqrt{\frac{GM}{r}} \therefore r = \frac{GM}{v^2} = \frac{6.67\times10^{-11}\times5.97\times10^{24}}{(7.79\times10^{3})^2} = 6.56 \times 10^{6}\ \mathrm{m}$ [1]

$\therefore h = r - R_{\oplus} = 6.56 \times 10^{3} - 6.37 \times 10^{3} = 192\ \mathrm{km}$ [1] [2]

[Must be in km for the final mark]

b. (ii)

From Kepler’s third law, [Students may just use v and the orbital circumference to find T]

$T^2 = \frac{4\pi^2}{GM} r^3 \quad \therefore T = \sqrt{\frac{4\pi^2}{6.67\times10^{-11}\times5.97\times10^{24}} \times (6.56 \times 10^{6})^3}$ [1]

$= 5293\ \mathrm{s}$ [1]

$\therefore n = \frac{(2\times3600)+(32\times60)+27}{5293} = \frac{9147}{5293} = 1.73$ revolutions [1] [3]

c.

The semi-major axis is half the total distance from A to B,

$a_{Hoh} = \frac{1}{2}\left(h_A + R_{\oplus} + d_{E\to M} + R_M + h_B\right)$

$= \frac{1}{2}(334 + 6370 + 3.94 \times 10^{5} + 1740 + 161) \times 10^{3} = 2.01 \times 10^{8}\ \mathrm{m}$ [1] [1]

From Kepler’s third law,

$T^2 = \frac{4\pi^2}{GM} a_{Hoh}^3 \quad \therefore T = \sqrt{\frac{4\pi^2}{6.67\times10^{-11}\times5.97\times10^{24}} \times (2.01 \times 10^{8})^3}$ [1]

$= 8.99 \times 10^{5}\ \mathrm{s}$ [1]

We will only travel half of the ellipse, so the time we want is half the period,

$T_{Hoh} = \frac{1}{2} T = 4.50 \times 10^{5}$ s = 124 hr 54 mins (9 secs) [must be in hr and min] [1] [3]

d. (i)

At point C, the gravitational forces from both bodies are equal,

$\frac{GM_{\oplus}}{x^2} = \frac{GM_M}{(d_{E\to M}-x)^2} \therefore \frac{d_{E\to M}-x}{x} = \sqrt{\frac{M_M}{M_{\oplus}}} \quad \therefore x = \left(1 + \sqrt{\frac{M_M}{M_{\oplus}}}\right)^{-1} d_{E\to M}$ [1]

$\therefore x = \left(1 + \sqrt{\frac{7.35\times10^{22}}{5.97\times10^{24}}}\right)^{-1} \times 3.94 \times 10^{8} = 3.55 \times 10^{8}\ \mathrm{m}$ [1] [2]

Using a similar approach to the previous part of the question,

$a_{AC} = \frac{1}{2}\left(h_A + R_{\oplus} + x\right) = 1.81 \times 10^{8}\ \mathrm{m}$ [1] [1]

$a_{BC} = \frac{1}{2}\left(h_B + R_M + x\right) = 2.06 \times 10^{7}\ \mathrm{m}$ [1] [1]

[Allow the use of a first order binomial expansion for x giving $x = \left(1 - \sqrt{\frac{M_M}{M_{\oplus}}}\right) d_{E\to M}$ leading to $x = 3.50 \times 10^{8}\ \mathrm{m}$, $a_{AC} = 1.78 \times 10^{8}\ \mathrm{m}$ and $a_{BC} = 2.28 \times 10^{7}\ \mathrm{m}$.]

d. (ii)

Using the vis-viva equation given at the beginning of the paper,

$v_A = \sqrt{GM_{\oplus}\left(\frac{2}{R_{\oplus}+h_A} - \frac{1}{a_{AC}}\right)}$

$= \sqrt{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times \left(\frac{2}{(6370+334)\times10^{3}} - \frac{1}{1.81\times10^{8}}\right)}$

$= 10.8\ \mathrm{km\,s^{-1}}$ [1] [1]

As a percentage of the escape velocity,

$v_{\mathrm{esc},A} = \sqrt{\frac{2GM_{\oplus}}{R_{\oplus}+h_A}} = 10.9\ \mathrm{km\,s^{-1}} \therefore v_A = 99.1\%\ v_{\mathrm{esc},A}$ [1] [1]

Similarly,

$v_B = 2.22\ \mathrm{km\,s^{-1}}$ [1] [1]

$v_{\mathrm{esc},B} = 2.27\ \mathrm{km\,s^{-1}} \therefore v_B = 97.7\%\ v_{\mathrm{esc},B}$ [1] [1]

Since these periapsis velocities are very close to the escape velocity, the eccentricities must be very close to 1 [1] [1]

[Do not award the marks for the percentage of the escape velocity if they have calculated the escape velocity at the surface of the body, rather than at the given altitude (i.e. $R_x$ rather than ($R_x$ + $h_x$)). The use of the binomial expansion in the previous part hardly affects these values (the only one changing at 3 s.f. is $v_B = 97.9\%\ v_{esc,B}$)]

d. (iii)

Using a similar approach to part c. where we only want half of the full ellipse,

$T_{AC} = \frac{1}{2}\sqrt{\frac{4\pi^2}{6.67\times10^{-11}\times5.97\times10^{24}} \times (1.81 \times 10^{8})^3} = 3.82 \times 10^{5}\ \mathrm{s}$ [1]

$T_{BC} = \frac{1}{2}\sqrt{\frac{4\pi^2}{6.67\times10^{-11}\times7.35\times10^{22}} \times (2.06 \times 10^{7})^3} = 1.33 \times 10^{5}\ \mathrm{s}$ [1]

The best estimate is exactly half of the sum of these two durations,

$T_{\mathrm{Apollo}} = \frac{1}{2}\left(T_{AC} + T_{BC}\right) = 2.58 \times 10^{5}$ s = 71 hr 33 mins (47 secs) [1] [3]

[Must be in hr and min for the final mark. Use of the binomial expansion earlier gives 73 hr 36 mins 38 s]

[The real duration of the Apollo 11 translunar coast was 73 hr 5 mins 35 secs, so our second model of patched conics has done rather well compared to the Hohmann transfer. The main reasons for the differences are that the moon is not in a perfectly circular orbit around the Earth (so its gravitational influence varies during the coast more than our model allowed for) and we are assuming that the coast and the orbit are coplanar, which is not generally true. In later Apollo missions they used a manoeuvre during the coast to change the trajectory away from a simple translunar free-return trajectory, allowing for a greater range of landing sites – this means this simple model will only get close to the real durations for the early missions and for the later missions the full (non-trivial) calculations must be done.]