Lagrangian Point (L1) Calculation
Miftahur Rahman, Ph.D.
September 8, 2023
Preamble:
In May 2019, during the 8th Interplanetary CubeSat Workshop (iCubeSat-2019) held at Politecnico di Milano, Milan, Italy, we introduced a white paper titled “Deep Space Communication and Exploration of the Solar System via Inter-Lagrangian Data Relay Satellite Constellation” [1].
Outlines of the paper:
The paper, originating in Milan, Italy, delves into the realm of interplanetary communication within our solar system and the extensive research on deep space exploration carried out by prominent international space agencies. The essence of seamless communication between the solar system’s celestial bodies is crucial for fostering future space missions and the eventual human habitation of both our immediate and more distant cosmic neighbors. At present, our infrastructure lacks the cohesive interplanetary network required to fulfill the envisioned objectives of ambitious deep space exploration endeavors. To address this exigency, we proposed the development of a comprehensive ‘Cluster-Based Data Relay Satellite Network Architecture’ spanning the entirety of our solar system.
To materialize this architectural vision, a strategic divergence from the current Direct-to-Earth (DTE) communication system is necessary, involving a transformation into a multifaceted relay satellite-based approach. This proposed system hinges on the deployment of satellite constellations within the Earth-Moon Lagrangian orbits, as well as the Earth-Sun Lagrangian orbits. It establishes direct connections with Lagrangian orbit satellites situated around other planets, thus extending the network’s reach to ensure continuous interplanetary data connectivity. Consequently, the transmission of signals ‘Direct to Earth’ (DTE) from deep space spacecraft becomes obsolete, permitting the design of considerably more compact spacecraft, thereby enabling CubeSat missions throughout the solar system. Such a transition carries the potential to significantly boost data transfer rates while mitigating the risk of communication disruptions or system failures.
The augmented data capacity directly corresponds to the diversity of forthcoming deep space missions. The proposed interplanetary satellite network stands poised to bolster the development of future deep space ventures, facilitate human colonization of the Moon and other celestial bodies, promote asteroid mining, and enable exploration across both the inner and outer reaches of our solar system. Furthermore, our system envisions the deployment of three primary or core interplanetary data communication hubs in the Earth-Moon Lagrangian orbits L3, L4, and L5, supplemented by two alternative or secondary data relay constellations in the Earth-Sun Lagrangian orbits L1 (Fig.1) and L2. This configuration incorporates cutting-edge free-space optical laser communication, dual-band RF data links (X-band and Ka-band), as well as dedicated Telemetry and Telecommand data relay channels operating within the UHF band.
Newtonian Gravity
Newtonian mechanics initially emerged to elucidate the intricate dance of celestial bodies as they traverse their orbits around the radiant Sun. In this realm, the fundamental agent at play is gravity, an omnipresent force intrinsic to all matter, compelling it to draw neighboring matter towards its core. The intensity of this gravitational force finds its footing in the mass of the entity in question. Notably, the Sun, an astronomical behemoth, dwarfs the Earth in mass by a factor of approximately 328, yielding a gravitational prowess surpassing that of our home planet. To illustrate this, consider the colossal Jupiter, outstripping Earth’s mass by a factor of 318. Consequently, Jupiter wields a gravitational pull registering at a formidable \(22.9\space m/s^2\), a stark contrast to Earth’s surface gravity at a more familiar \(9.8\space m/s^2\).
Nevertheless, objects residing at a substantial remove from the Sun’s vicinity remain largely untouched by its gravitational influence. It is the gravitational force, as elucidated by Sir Isaac Newton in his seminal work “Philosophiae Naturalis Principia Mathematica” (published in 1687), that anchors planets in their elliptical trajectories around the Sun. According to Newton’s doctrine, any two point masses (or spherically symmetric entities with finite extent) partake in a mutual gravitational attraction. This gravitational tether acts along the line that joins the centers of the objects, directly correlating with the product of their masses, yet inversely relating to the square of the separation between them.
To anchor this concept in mathematical rigor, consider the Sun as the primary object, endowed with a mass denoted as \(M\) and occupying the origin of our spatial coordinate system. Now, envision a planet, characterized by its mass \(m\), positioned at a specific location marked by the position vector \({\vec r}\). The gravitational force exerted on this planet by the Sun manifests as follows:
\[\begin{equation} \tag{1-1} \vec F = - \frac{G\,M\,m}{r^3}\,{\vec r}. \end{equation}\]
The constant of proportionality, \(G\), is called the universal gravitational constant, and its value in SI unit is given by:
\[\begin{equation} \tag{1-2} G = 6.67300\times 10^{-11}\space \frac{m^3}{kg-s^2}. \end{equation}\]
Furthermore, given that the force is centripetal in character, Equation (1-1) can be altered in the following manner:
\[\begin{equation} \tag{1-3} \frac{G\,M\,m}{r^2}=\frac{mv^2}{r} \end{equation}\]
By employing Equation (1-3), one can readily compute Earth’s constant orbital velocity of 30 km/sec as it orbits the Sun.
\[\begin{equation} \tag{1-4} {v} = \sqrt{\frac{GM}{r}}= \sqrt{\frac{(6.67 \times 10^{-11}\frac{N-m^2}{kg^2}(1.99 \times 10^{30}kg)}{1.5 \times 10^{11} m}} = 29.747\times 10^3 \frac{m}{s} \approx 30 \frac{km}{s} \end{equation}\]
Also using,
\[\begin{equation} \tag{1-5} v =\frac{2\pi r}{T} \end{equation}\]
\[\begin{equation} \tag{1-6} v =\frac{2\pi r}{T} \end{equation}\]
\[\begin{equation} \tag{1-7} \frac{GM}{r} =\frac{4\pi^2r^2}{T^2} \end{equation}\]
\[\begin{equation} \tag{1-8} T^2 =\frac{4\pi^2r^3}{GM} \end{equation}\]
Equation (1-8) is recognized as Kepler’s third law, applicable to circular motion.
Imagine a spacecraft or satellite with mass \(m_s\), positioned along the line connecting Earth and the Sun. This spacecraft is situated at a distance of R from Earth and (r-R) from the Sun, as depicted in Figure 2. The force F, which draws it towards the Sun, is counteracted by Earth’s gravitational attraction, thus diminishing it in the opposite direction.
\[\begin{equation} \tag{1-9} F =\frac{GMm_s}{(r-R)^2}-\frac{Gmm_s}{R^2} \end{equation}\]
Assume that the satellite also moves in a circle around the sun, with velocity \(v_s\).
\[\begin{equation} \tag{1-10} \frac{GMm_s}{(r-R)^2}-\frac{Gmm_s}{R^2}=\frac{m_sv_s^2}{r-R} \end{equation}\]
\[\begin{equation} \tag{1-11} \frac{GM}{(r-R)}-\frac{Gm(r-R)}{R^2}=v_s^2 \end{equation}\]
\[\begin{equation} \tag{1-12} v_s^2=\frac{4\pi^2(r-R)^2}{T_s^2} \end{equation}\]
\[\begin{equation} \tag{1-13} \frac{GM}{(r-R)^3}-\frac{Gm}{R^2(r-R)}=\frac{4\pi^2}{T_s^2} \end{equation}\]
Will the Earth be always located where its pull on the satellile is exactly opposite to the Sun? No, unless the two orbital period are the same.
\[\begin{equation} \tag{1-14} T_s = T \end{equation}\]
Then satellite’s motion match that of the Earth and the distance between the two stays constant. That happens exactly at one point or one value or R and is unknown that needs to be found.
If \(T_s = T\), then from the Kepler’s third law,
\[\begin{equation} \tag{1-15} \frac{4\pi^2}{T^2}=\frac{GM}{r^3} \end{equation}\]
Substituting Eq.(1-15) in Eq.(1-14)
\[\begin{equation} \tag{1-16} \frac{GM}{(r-R)^3}-\frac{Gm}{R^2(r-R)}=\frac{GM}{r^3} \end{equation}\]
Divide both sides by GM
\[\begin{equation} \tag{1-17} \frac{1}{(r-R)^3}-\frac{m}{MR^2(r-R)}=\frac{1}{r^3} \end{equation}\]
The Earth to Sun mass ratio:
\(y = \frac{m}{M}= \frac{5.98\times 10^{24}kg}{1.99 \times 10^{30}kg} \approx 3 \times 10^{-6}\)
Hence,
\[\begin{equation} \tag{1-18} \frac{r^3}{(r-R)^3}-\frac{yr^3}{R^2(r-R)}=1 \end{equation}\]
\[\begin{equation} \tag{1-19} \frac{r^3}{(1-\frac{R}{r})^3}-\frac{yr^2}{R^2(1-\frac{R}{r})}= 1 \end{equation}\]
Let’s take \(z = \frac{R}{r}\)
\[\begin{equation} \tag{1-20} \frac{1}{(1-z)^3}-\frac{y}{z^2(1-z)}= 1 \end{equation}\]
It is a complicated cubic equation to find the unknown z. James Webb space telescope revealed that
\[r = 1.5 \times 10^{11}m\] \[R = 1 \space million \space miles = 1.6 \times 10^9 m\] But \[z = \frac{R}{r}=\frac{1.6 \times 10^9 m}{1.5 \times 10^{11} m} \approx 10^{-2}\]
Using Binomial theorem
\[(1+z)^3 = 1 + 3z + \frac{3(3-1)z^2}{2!}+.....\approx 1+3z\] Also, \[(1-z)^{-1} \approx (1+z)\] Therefore, \[\frac{1}{(1-z)^3}-y\frac{1}{z^2(1-z)}=1\] \[(1+3z)-y\frac{(1+z)}{z^2}\approx 1\] \[3z \approx \frac{y(1+z)}{z^2}\] \[3z^3 \approx \frac y(1+z)\] \[3z^3 \approx y = 3 \times 10^{-6}\] \[z^3 \approx 10^{-6}\] \[\frac{R}{r}=z \approx (10^{-6})^{1/3}=10^{-2}=0.01\] Therefore, \(R = \frac{r}{100}\)
Thus, the distance to L1 is one hundreadth of the Sun-Earth distance. That is,
\[R = \frac{r}{100}=\frac{1.5 \times 10^{11}m}{100}=1.5 \times 10^6 \space km = 1.5 \space million \space km\]
The L1 point, located 1.5 million kilometers within Earth’s orbit, equivalent to just one percent of the Earth-Sun distance, represents a prime example of a Lagrangian point. Lagrangian points denote regions where the gravitational forces between two celestial bodies nullify each other, rendering them advantageous locales for spacecraft to effectively ‘hover.’
One notable occupant of the L1 point is the European Space Agency
(ESA) and NASA’s solar sentry, the Solar and Heliospheric Observatory
(SOHO). SOHO maintains its position at L1, matching the Earth’s orbital
speed around the Sun. Positioned here, SOHO perpetually observes the
Sun, conducting vital scientific investigations while also serving as an
early-warning sentinel. It is equipped to alert us to potentially
disruptive solar storms that could imperil communication and navigation
satellites orbiting our planet.
where M1 and M2 are the masses of Sun and Earth respectively
hence \(\alpha = \frac{M2}{M1+M2}=\frac{5.98
\times 10^{24}kg}{1.99 \times 10^{30}kg + 5.98 \times 10^{24}kg} \approx
3 \times 10^{-6}\)
