IOM 2020, Теория — Задача 2. To the Sun!

• 3) p = 2r(R_E + h)/(r + R_E + h) ≈ 11522 km; ε = (r − R_E − h)/(r + R_E + h) ≈ 0,727; |E₀/E₁| ≈ 5,49

  r ≈ 42222 km is the GSO radius (official solution).

• 4) Δv₂ ≈ 1,47 km/s (1,4665 km/s); m₂ ≈ 0,52260·m₀, i.e. about 52 %

• 5) summer, noon; v_A ≈ 2,83 km/s (2,8273 km/s); ṽ₂ ≈ 28,73 km/s (28,7267 km/s)

• 6) Δm ≈ 0,21264·m₀ (about 21 %); m̃₂ ≈ 0,02·m₀ (about 2 %)

• 7) $t = \frac{1}{2}\left(\frac{r_A + R_s}{r_A + r_P}\right)^{3/2}T_0$ ≈ $66.7$ days

  Official solution: t ≈ 66,7 days.

• 8) α_max ≈ 4°; no

Problem 2 (proposed solution)

1. Let us determine the radius of GSO and the ship velocity in this orbit ($r$ and $v$, respectively). To do this we use the equation for the centripetal component of the ship acceleration in the GSO: $m\dfrac{v^2}{r} = \dfrac{GM_Em}{r^2} = \dfrac{gR_E^2m}{r^2}$, hence $v = R_E\sqrt{\dfrac{g}{r}}$. On the other hand, the orbital period of the ship must be the same as the period of Earth’s rotation around its axis, i.e. $\dfrac{2\pi r}{v} = T$, therefore $r = \sqrt[3]{\dfrac{gT^2R_E^2}{4\pi^2}} \approx 42222$ km and $v = \sqrt[3]{\dfrac{2\pi gR_E^2}{T}} \approx 3{,}0705$ km/s. According to the orbit equation, one has for the intermediate orbit $R_E + h = \dfrac{p}{1 + \varepsilon}$ and $r = \dfrac{p}{1 - \varepsilon}$. These equations yield: $p = \dfrac{2r(R_E + h)}{r + R_E + h} \approx 11522$ km and $\varepsilon = \dfrac{r - R_E - h}{r + R_E + h} \approx 0{,}727$.

Below the symbols $a$ and $p$ stand for the quantities related to aphelion and perihelion of a ship orbit around Earth and symbols $A$ and $P$ relate to aphelion and perihelion of an orbit around the Sun. For an elliptical orbit of the ship both the total mechanical energy and angular momentum are conserved and can be expressed via the given distances from the Earth’s center to the aphelion and the perihelion. Therefore, $\dfrac{mv_a^2}{2} - \dfrac{GM_Em}{r_a} = \dfrac{mv_p^2}{2} - \dfrac{GM_Em}{r_p} = E$ and $mr_av_a = mr_pv_p$. Solving these equations for velocities, one obtains $v_p^2 = gR_E^2\dfrac{2r_a}{r_p(r_a + r_p)}$ and $v_a^2 = gR_E^2\dfrac{2r_p}{r_a(r_a + r_p)}$. This gives for the total mechanical energy: $E = -\dfrac{gR_E^2m}{r_p + r_a}$. Using this result, one finds that $\left|\dfrac{E_0}{E_1}\right| = \dfrac{r + R_E + h}{2(R_E + h)}\dfrac{m_0}{m_1}$, where $m_0$ and $m_1$ are the ship masses in the initial and the intermediate orbits.

Since the radius of initial orbit equals the distance to the perihelion of intermediate orbit, the transition was performed by increasing the ship velocity from $v_0 = \sqrt{\dfrac{gR_E^2}{R_E + h}} \approx 7{,}7247$ km/s to $v_p = \sqrt{\dfrac{2gR_E^2r}{(R_E + h)(r + R_E + h)}} \approx 10{,}152$ km/s for a short period of time (according to the problem statement the maneuver was «short-term»). Therefore, it is reasonable to neglect both the curvature of ship trajectory and a gravity force and to consider the ship acceleration as being completely due to the engine thrust. Hence, one can use the Tsiolkovsky rocket equation: $m_1 = m_0\cdot e^{-\Delta v_1/u}$, where $\Delta v_1 \equiv v_p - v_0 \approx 2{,}4271$ km/s. Thus, $m_1 = 0{,}66730\cdot m_0$ and $\left|\dfrac{E_0}{E_1}\right| = \dfrac{r + R_E + h}{2(R_E + h)}e^{\Delta v_1/u} \approx 5{,}49$. Note that both energies are negative, i.e. actually $E_0 < E_1$.

2. All ship orbits under consideration (with its engine turned off) near Earth are closed. Therefore, the short-term acceleration for the transition from the intermediate orbit into the GSO must be done immediately before the aphelion of intermediate orbit. To do this, it suffices to increase the ship velocity from $v_a$ at the aphelion to the velocity in the GSO (it is important that both at the aphelion of intermediate orbit and in the GSO the velocity is perpendicular to the radius). Note that $v_a = \sqrt{\dfrac{2gR_E^2(R_E + h)}{r(r + R_E + h)}} \approx 1{,}6040$ km/s. Therefore, the velocity increment required to switch into the GSO must be $\Delta v_2 \equiv v - v_a \approx 1{,}4665$ km/s. Again, use the rocket equation: $m_2 = m_1\cdot e^{-\Delta v_2/u} = m_0\cdot e^{-(\Delta v_1 + \Delta v_2)/u} \approx 0{,}52260\cdot m_0$. The mass increment $m_1 - m_2 \approx 0{,}14470\cdot m_0$ is the mass of the fuel and oxidizer which has to be consumed to switch into the GSO from the intermediate orbit. According to the problem, it was one third of the total supply. Hence, the mass of fuel and oxidizer in the intermediate orbit was $m_F = 0{,}43410\cdot m_0$. Since the ship mass in the intermediate orbit was $m_1 = 0{,}66730\cdot m_0$, the «useful» mass (the engine, the fuel tanks, and the hull) was equal to $m_u = 0{,}23321\cdot m_0$.

3. In order to send the ship to the Sun it is necessary to reduce significantly the ship velocity with respect to the Sun. To do this in a most efficient way, the engine must be started when the ship is closest to the point where its velocity in the intermediate orbit is minimal with respect to the Sun. Obviously, it would be the best to choose the point at which the Earth’s velocity $V$ with respect to the Sun is minimum (i.e. the aphelion of the Earth’s orbit) and the ship velocity relative to Earth is maximum and opposite to $\vec{V}_A$. Therefore, the second maneuver must be performed when the ship approaches the perihelion of its orbit during the summer of the northern hemisphere, i.e. when Earth is at the aphelion, and at noon of the Earth’s time at the point below the ship.

In order to reach the Sun, the ship after leaving Earth must follow an orbit around the Sun for which the distance to the aphelion is approximately equal to the distance to aphelion of the Earth’s orbit and the distance to perihelion is less or equal to the Solar radius. Then the ship velocity relative to the Sun after the it has receded from Earth to a distance sufficiently greater than the Earths’ radius (though much less than $r_A \approx 152$ Mkm) must not exceed $v_A = \sqrt{\dfrac{2GM_SR_S}{r_A(r_A + R_S)}}$.

The Solar mass can be estimated by using the orbital period of Earth and the mean radius of the Earth’s orbit: $V = \dfrac{\pi(r_A + r_P)}{T_0} \approx \sqrt{\dfrac{2GM_S}{r_A + r_P}} \Rightarrow 2GM_S \approx \dfrac{\pi^2(r_A + r_P)^3}{T_0^2}$. Thus, the required velocity is $v_A \approx \dfrac{\pi(r_A + r_P)}{T_0}\sqrt{\dfrac{(r_A + r_P)R_S}{r_A(r_A + R_S)}} \approx 2{,}8273$ km/s. The Earth’s velocity at the aphelion is $V_A = \sqrt{\dfrac{2GM_Sr_P}{r_A(r_A + r_P)}} \approx \dfrac{\pi(r_A + r_P)}{T_0}\sqrt{\dfrac{r_P}{r_A}} \approx 29{,}2898$ km/s, so the ship velocity relative to Earth at the moment of its passage into the orbit leading to the Sun must be $v'_A = V_A - v_A \approx 26{,}4625$ km/s. It is important to keep in mind the following. After turning off the engine, the ship must move somewhat faster than $v'_A$ to overcome the gravitational pull of receding Earth. This effect can be accounted for by means of the energy conservation law: the velocity $\tilde{v}_2$ of the part of the ship (of the mass $m_2$) being sent to the Sun relative to the Earth’s center at the moment right after the engine was turned off is determined by the equation $\dfrac{m_2\tilde{v}_2^2}{2} - m_2gR_E = \dfrac{m_2(v'_A)^2}{2} \Rightarrow \tilde{v}_2 = \sqrt{(v'_A)^2 + 2gR_E} \approx 28{,}7267$ km/s.

4. During the maneuver of transition into the orbit leading to the Sun the increment of the ship velocity is $\Delta\tilde{v}_2 \equiv v_2 - v_p \approx 18{,}575$ km/s. Before the acceleration the ship got rid of some of its mass ($\Delta m$), so the mass which will eventually reach the Sun is $\tilde{m}_2 = (m_1 - \Delta m)\cdot e^{-\Delta\tilde{v}_2/u}$. To send the maximum mass to the Sun, the ship had to use up all the fuel and oxidizer supply, therefore $m_F = m_1 - \Delta m - \tilde{m}_2 = (m_1 - \Delta m)\cdot(1 - e^{-\Delta\tilde{v}_2/u}) \Rightarrow \Delta m = m_1 - \dfrac{m_F}{1 - e^{-\Delta\tilde{v}_2/u}} \approx 0{,}21264\cdot m_0$. This calculation shows that the mass of the ship left after the second maneuver is $\tilde{m}_2 = \dfrac{m_F}{1 - e^{-\Delta\tilde{v}_2/u}}e^{-\Delta\tilde{v}_2/u} = \dfrac{m_F}{e^{\Delta\tilde{v}_2/u} - 1} \approx 0{,}02\cdot m_0$. Thus, no more than 2% of the ship mass in the initial orbit can be delivered to the Sun!

5. According to the third Kepler’s law, a ratio of the squares of orbital periods equals to the cube of the ratio of the semi-major axes of the orbits. Therefore, the orbital period of the accelerated part of the ship in the new orbit is $T' = \left(\dfrac{r_A + R_s}{r_A + r_P}\right)^{3/2}T_0 \approx 133{,}3$ days. The trip to the Sun will take a half of this period, i.e. $t = \dfrac{1}{2}\left(\dfrac{r_A + R_s}{r_A + r_P}\right)^{3/2}T_0 \approx 66{,}7$ days.

6. Detailed analysis of the gravity assisted maneuver requires a more general study of trajectories of a body moving in the gravitational field of a massive spherically uniform object. Let us again turn to the laws of conservation of mechanical energy and angular momentum. A ship following the orbit $\rho(\varphi) = \dfrac{p}{1 + \varepsilon\cdot\cos\varphi}$ has the total mechanical energy $E = \dfrac{m\vec{v}^2}{2} - \dfrac{GMm}{r} = \dfrac{m(\dot{\rho}^2 + \rho^2\dot{\varphi}^2)}{2} - \dfrac{GMm}{\rho} = const$ and the angular momentum $L = m\rho^2\dot{\varphi} = const$, where the dot means differentiation with respect to time (the polar coordinates are employed). From the second relation one gets $\dot{\varphi} = \dfrac{L}{m\rho^2}$ and substituting into the first equation obtains: $\dfrac{E}{m} = \dfrac{\dot{\rho}^2}{2} + \dfrac{L^2}{2m^2\rho^2} - \dfrac{GM}{\rho}$. The distances to aphelion and perihelion, $r_a = \dfrac{p}{1 - \varepsilon}$ and $r_p = \dfrac{p}{1 + \varepsilon}$, are the roots of the equation $\dot{\rho} = 0$. Therefore ($M_V$ is the Venus’s mass):

When the part of the ship approaches Venus in the region of the orbits’ closing, in the Venus’s frame it comes to Venus from «a large distance» at non-zero velocity $v_{\infty}$, i.e. its total mechanical energy $E = \dfrac{mv_{\infty}^2}{2} > 0$ and the orbit eccentricity $\varepsilon > 1$. In this case the orbit is hyperbolic, one can see that at $\varphi = \pm\left[\dfrac{\pi}{2} + \arcsin\left(\dfrac{1}{\varepsilon}\right)\right]$ the radius $\rho \to \infty$. Thus, the velocity relative to Venus rotates at the angle $\theta = 2\arcsin\left(\dfrac{1}{\varepsilon}\right)$. In this case, $L = mv_{\infty}b$ where $b$ (see the Figure) is called the «impact parameter».

One can see that the maximum rotation angle is achieved at the minimum eccentricity. According to the formula for eccentricity, $\varepsilon = \sqrt{1 + \dfrac{b^2v_{\infty}^4}{(GM_V)^2}}$, and the minimum value of the impact parameter is determined by the requirement that the distance between the orbit perihelion and the Venus’s center cannot be less than the planet radius $R_V$: $\dfrac{p}{1 + \varepsilon} \ge R_V \Rightarrow b^2 \ge \left(1 + 2\dfrac{GM}{R_Vv_{\infty}^2}\right)R_V^2 = \left(1 + 2\dfrac{v_{1V}^2}{v_{\infty}^2}\right)R_V^2$. Thus, $\varepsilon_{\min} = 1 + \dfrac{v_{\infty}^2}{v_{1V}^2}$, and the maximum rotation angle achieved during the flyby around Venus is $\theta_{\max} = 2\arcsin\left[\dfrac{v_{1V}^2}{v_{1V}^2 + v_{\infty}^2}\right]$. The ship velocity relative to the Sun after the gravity assist maneuver is a vector sum of velocity $\vec{v}'_{\infty}$ rotated by the angle $\theta$ and the velocity $\vec{V}$ of the massive object relative to the Sun. Thus, the maximum angle $\alpha_{\max}$ the ship velocity vector can rotate toward the Sun during the maneuver can be found by means of the law of sines: $\sin\alpha = \dfrac{v_{\infty}}{|\vec{V} + \vec{v}'_{\infty}|}\sin\theta$. Finally,

Let us determine the velocity of the part of ship at the perihelion of its orbit: $v_P = \sqrt{\dfrac{2GM_Sr_A}{r'_P(r_A + r'_P)}}$, or $v_P = \dfrac{\pi(r_A + r_P)}{T_0}\sqrt{\dfrac{r_A(r_A + r_P)}{r'_P(r_A + r'_P)}} \approx 38{,}0332$ km/s (recall that the aphelion of the orbit leading from Earth to Venus coincides with the aphelion of the Earth’s orbit and the perihelion with the perihelion of the Venus’s orbit.). So, $v_{\infty} = v_P - V'_P \approx 2{,}7832$ km/s, $V = V'_P \approx 35{,}25$ km/s, and $\theta_{\max} = 2\arcsin\left[\dfrac{v_{1V}^2}{v_{1V}^2 + v_{\infty}^2}\right] \approx 121{,}9^{\circ}$. Therefore $\alpha_{\max} \approx 4^{\circ}$. However, to hit the Sun at this velocity the angle rotation must be close to 90°. Thus, the gravity assisted maneuver near Venus cannot help in delivering a substantially larger ship mass to the Sun.