Deflecting an Asteroid with a Low-Thrust Tangential Engine to the Orbit

Abstract

In order to solve the problem of deflecting a dangerous asteroid from a collision orbit with the Earth, using a low-thrust engine directed tangentially to the trajectory is considered. The engine can be mounted on the asteroid or on a “gravity tractor.” The purpose of this study is to establish the fundamental possibility of steering away an asteroid to a safe distance over times of approximately a month and a year. This is acceptable since an asteroid with about a 100-m diameter is unlikely to strike immediately after its discovery. We limited ourselves to a model statement of the problem: the engine provides constant tangential acceleration. Previously, we transformed the respective Euler equations using the averaging method. Here, we solve them by the method of series in powers of “slow time” and demonstrate the adequacy of the solution on the time intervals of decades. It turns out that asteroids up to 55 m in diameter can be deflected in a year with an engine thrust of 1 N. With a thrust of 20 N, asteroids up to 50 m in diameter can be deflected in a month, and asteroids with a diameter of up to 150 m, in a year. Diverting larger asteroids requires more time or more powerful engines. The results are compared with the previously obtained similar data for the case of the transversal perturbing acceleration. The tangential traction leads to better results in all cases; however, both variants nearly coincide for orbits with eccentricities up to 0.4. The difference becomes significant at $e > 0.5$.

ESTIMATES OF SOME COMPLEX VARIABLE FUNCTIONS

Let $0 < {{e}_{0}} < 1$, $0 \leqslant \alpha \leqslant 1$, $0 \leqslant \psi < 2\pi$, $z \in \mathbb{C}$, ${\text{|}}z{\text{|}} \leqslant 1$,

$$\begin{gathered} e(\psi ) = {{e}_{0}} + (1 - {{e}_{0}})\text{Exp} \psi , \\ {{{\text{g}}}_{1}}(\alpha ,{\text{z}}) = \frac{{1 - {\text{z}}}}{{1 - \alpha {\text{z}}}}, \\ {{{\text{g}}}_{2}}({\text{e}}) = {\text{e}}(1 - {{{\text{e}}}^{2}}){\mathbf{D}}({\text{e}}). \\ \end{gathered} $$

(37)

Let us estimate the absolute values of functions (37). Obviously, ${\text{|}}e(\psi ){\text{|}} \leqslant 1$,

$$\begin{gathered} 1 - {{e}^{2}}(\psi ) = (1 - {{e}_{0}})\{ 1 + {{e}_{0}} - 2{{e}_{0}}\cos\psi \\ - \;(1 - {{e}_{0}})\cos2\psi - \mathfrak{i}{\kern 1pt} \text{[}2{{e}_{0}}\sin\psi + (1 - {{e}_{0}})\sin2\psi ]\} , \\ {\text{|}}1 - {{e}^{2}}{{{\text{|}}}^{2}} = 2{{(1 - {{e}_{0}})}^{2}}{{g}_{3}}({{e}_{0}},\psi ), \\ {{g}_{3}} = 1 + 3e_{0}^{2} - 4e_{0}^{2}\cos\psi - (1 - e_{0}^{2})\cos2\psi . \\ \end{gathered} $$

The derivative

$$\frac{{\partial {{g}_{3}}}}{{\partial \psi }} = 4\sin\psi [e_{0}^{2} + (1 - e_{0}^{2})\cos\psi ]$$

vanishes at

$$\psi = 0,\quad \psi = \pi ,$$

and when ${{e}_{0}} < 1{\text{/}}\sqrt 2$ also at

$$\cos\psi = - \frac{{e_{0}^{2}}}{{1 - e_{0}^{2}}},\quad \cos2\psi = - \frac{{1 - 2e_{0}^{2} - e_{0}^{4}}}{{{{{(1 - e_{0}^{2})}}^{2}}}}.$$

From here, we easily derive

$$\mathop {\max}\limits_\psi {\text{|}}1 - {{e}^{2}}(\psi ){\text{|}} = {{g}_{4}}({{e}_{0}}),$$

(38)

where

$${{g}_{4}}({{e}_{0}}) = \left\{ \begin{gathered} 2\sqrt {\frac{{1 - {{e}_{0}}}}{{1 + {{e}_{0}}}}} ,\quad {\text{if}}\quad {{e}_{0}} \leqslant 1{\text{/}}\sqrt 2 {\kern 1pt} , \hfill \\ 4{{e}_{0}}(1 - {{e}_{0}}),\quad {\text{if}}\quad {{e}_{0}} \leqslant 1{\text{/}}\sqrt 2 {\kern 1pt} {\kern 1pt} . \hfill \\ \end{gathered} \right.$$

The triangle axiom for points $1$, $z$, $\alpha z$ of the complex plane gives

$${\text{|}}1 - z{\text{|}} \leqslant \left| {1 - \alpha z} \right| + \left| {z - \alpha z} \right|.$$

(39)

Further,

$${\text{|}}z - \alpha z{{{\text{|}}}^{2}} \leqslant {\text{|}}1 - \alpha z{{{\text{|}}}^{2}}.$$

(40)

Indeed, (40) follows from the obvious inequality

$$1 - 2\alpha \leqslant 1 - \alpha (z + \bar {z}).$$

Relations (39) and (40) imply the inequality

$${\text{|}}{{g}_{1}}(\alpha ,z){\text{|}} \leqslant 2.$$

Hence, we obtain the estimate of ${{g}_{2}}(e)$:

$$\begin{gathered} {\text{|}}{\kern 1pt} {{g}_{2}}(e){\kern 1pt} {\text{|}} \leqslant \left| {e\sqrt {1 - {{e}^{2}}} } \right|\int\limits_0^{\pi /2} \,\left| {\sqrt {{{g}_{1}}({{\sin}^{2}}x,{{e}^{2}})} } \right|{{\sin}^{2}}xdx \\ \, \leqslant \sqrt {2{{g}_{4}}({{e}_{0}})} \int\limits_0^{\pi /2} \,{{\sin}^{2}}xdx = \frac{\pi }{{\sqrt 8 }}\sqrt {{{g}_{4}}({{e}_{0}})} . \\ \end{gathered} $$

(41)