The New Calculation Of Lagrangian Point 1, 2, 3
Regarding the n-body problem, by choosing the center of mass or another fixed point as the reference point, we can list the following equations of motion: \ddot{\vec{r}}_k = \sum_{i=1, i \neq k}^{n} G m_i \frac{\vec{r}_i - \vec{r}_k}{|\vec{r}_i - \vec{r}_k|^3} \tag{19} Now we only consider the three-body problem. Astronomers have always hoped to find a concise solution to the three-body problem, but unfortunately, Poincaré has proved that the solutions to the three-body problem are chaotic. This means that any tiny perturbation can lead to unpredictable consequences (vividly metaphorized as: the flapping of a butterfly’s wings in Brazil might cause a tornado in the United States).
Three-Body Problem Animation:
[Flash Animation:
Three-Body Problem]
However, seeking "harmony" within "disharmony" has always been the goal of scientists, and thus they aimed at certain particular solutions. When we find it difficult to obtain a general solution to a problem, we can always try to find particular solutions; perhaps they can guide us on the path to finding the general solution. For the three-body problem, Lagrange found five particular solutions, which are the famous Lagrangian points.
What are Lagrangian points? Lagrangian points originate from the "Circular Restricted Three-Body Problem" (CR3BP). This problem assumes a situation where two finite-mass bodies move in circular orbits around their common center of mass under their mutual gravitational attraction, and a third infinitesimal-mass body does not affect the motion of the first two bodies but moves only under their combined gravitational pull. Lagrangian points are the positions where the third body remains stationary relative to the first two bodies. This characteristic can be easily seen from the following animation:
In other words, a small-mass celestial body located at a Lagrangian point of the Earth has the same orbital period as the Earth. Based on this characteristic, we can use high school knowledge of centripetal force to simply find the positions of the Lagrangian points. Since the small-mass body is also in circular motion, its centripetal force comes from the projection of the resultant force of the Sun’s gravity and the Earth’s gravity onto the position vector. By setting up the equation according to the centripetal force formula, the answer can be obtained. Interested readers can derive it themselves; it will not be detailed here.
Currently, some of the space probes and artificial satellites we have built have been launched to Lagrangian points. Limited by current technology, the mass of the spacecraft can be ignored. However, the circular restricted three-body problem always has certain limitations. When our technology develops to the point where we can build probes as massive as the Moon, we will no longer be able to ignore our own mass. We hope to obtain these "five particular solutions" in the complete three-body problem. In this case, the distance from the Lagrangian point to the Sun or the Earth is no longer constant but changes continuously, while the ratio remains constant. This calculation requires the use of Equation (19) given at the beginning of the article. We attempt to find the Lagrangian points that are collinear with the Sun-Earth position vector.
Since the three are collinear, we can set \vec{r}_2 - \vec{r}_1 = \vec{r}, \vec{r}_3 - \vec{r}_1 = k\vec{r}, and |\vec{r}| = r. Let the masses of the Sun, Earth, and the third body be M, m, m' respectively. We can derive: \begin{aligned} \ddot{\vec{r}}_1 &= G m \frac{\vec{r}_2 - \vec{r}_1}{|\vec{r}_2 - \vec{r}_1|^3} + G m' \frac{\vec{r}_3 - \vec{r}_1}{|\vec{r}_3 - \vec{r}_1|^3} = G m \frac{\vec{r}}{r^3} + G m' \frac{k\vec{r}}{|k|^3 r^3} \\ \ddot{\vec{r}}_2 &= G M \frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|^3} + G m' \frac{\vec{r}_3 - \vec{r}_2}{|\vec{r}_3 - \vec{r}_2|^3} = -G M \frac{\vec{r}}{r^3} - G m' \frac{(1-k)\vec{r}}{|1-k|^3 r^3} \\ \ddot{\vec{r}}_3 &= G M \frac{\vec{r}_1 - \vec{r}_3}{|\vec{r}_1 - \vec{r}_3|^3} + G m \frac{\vec{r}_2 - \vec{r}_3}{|\vec{r}_2 - \vec{r}_3|^3} = -G M \frac{k\vec{r}}{|k|^3 r^3} + G m \frac{(1-k)\vec{r}}{|1-k|^3 r^3} \end{aligned}
To simplify the problem, we transform to the Sun as the reference point, then: \ddot{\vec{r}} = \ddot{\vec{r}}_2 - \ddot{\vec{r}}_1 = -G(M+m)\frac{\vec{r}}{r^3} - G m' \left( \frac{1-k}{|1-k|^3} + \frac{k}{|k|^3} \right) \frac{\vec{r}}{r^3} \tag{20} k\ddot{\vec{r}} = \ddot{\vec{r}}_3 - \ddot{\vec{r}}_1 = -G(M+m')\frac{k\vec{r}}{|k|^3 r^3} + G m \left( \frac{1-k}{|1-k|^3} - 1 \right) \frac{\vec{r}}{r^3} \tag{21}
For the Lagrangian point to hold, the \ddot{\vec{r}} expressed by Equations (20) and (21) must be equal, therefore: (M+m) + m' \left( \frac{1-k}{|1-k|^3} + \frac{k}{|k|^3} \right) = \frac{M+m'}{|k|^3} - m \left( \frac{1-k}{k|1-k|^3} - \frac{1}{k} \right) \tag{22}
Equation (22) is a quintic equation in one variable regarding k. The Lagrangian points are given by this equation, which depends only on the masses. Since it contains absolute values, Equation (22) must be discussed in three cases, so it actually contains three quintic equations. If m' is small enough to be ignored, Equation (22) can be simplified to: M+m = \frac{M}{|k|^3} - m \left( \frac{1-k}{k|1-k|^3} - \frac{1}{k} \right) \tag{23}
Substituting M = 332918.215 and m = 1 (the mass ratio of the Sun to the Earth): when 0 < k < 1, we can find a real root k = 0.9900293204354188\dots, which is the First Lagrangian Point (L1); when k > 1, we can find a real root k = 1.0100374005377246\dots, which is the Second Lagrangian Point (L2); when k < 0, we can find a real root k = -0.9999982478231744\dots, which is the Third Lagrangian Point (L3). (Note: here the distance refers to the distance between the Lagrangian point and the Sun, not the distance to the Earth-Sun barycenter.)
Thus, we have derived the three Lagrangian points for the complete
three-body problem, no longer limited to the "restricted three-body
problem". Of course, L4 and L5 still remain. Due to the limited
scope of BoJone’s knowledge, I have not been able to use the method
discussed in this article to explain them, and I hope for guidance from
experts... During the derivation process, BoJone also realized one
thing: complex numbers can be used to study geometry conveniently!!
The remaining L4 and L5 will be discussed in the following text!
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