update lagrange points

theory/celestial mechanics/lagrange points.md

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 ## Lagrange Points
+
+<br />
+The point near two large bodies in orbit such that if a third body (of negligible mass) is placed at that point, it will remain stationary with respect to the two large bodies is called a [Lagrange point](#lagrange-points).
+
+### Derivation
+
+Assume two bodies of masses $$m_1$$ and $$m_2$$ are in circular orbits around their center of mass (placed at the origin), with the same angular velocity, $$\omega$$. Without loss of generality, let's further assume that the two bodies are in the XY-plane.
+
+Using the law of [gravitation](./newtonian%20gravity.html), we can find out the distances of these two bodies from their center of mass (i.e. the origin). Let's assume that the distance of the first body from the origin is $$r_1$$ (i.e. placed at $$(-r_1, 0, 0)$$) and the distance of the second body from the origin is $$r_2$$ (i.e. placed at $$(r_2, 0, 0)$$). Hence, we get:
+
+$$
+\begin{equation}
+  m_1r_1 = m_2r_2
+\end{equation}
+$$
+
+We can very easily calculate the orbital speed $$\omega$$ from this:
+
+$$
+\begin{equation}
+  \begin{split}
+    \frac{Gm_1m_2}{(r_1+r_2)^2} &= m_1 \omega^2 r_1 \\
+    \frac{Gm_1m_2}{(r_1+r_2)^2} &= m_2 \omega^2 r_2
+  \end{split}
+\end{equation}
+$$
+
+Hence, we get,
+
+$$
+\begin{equation}
+  \begin{split}
+    Gm_1 &= r_2 (r_1+r_2)^2 \omega^2 \\
+    Gm_1 &= r_2 (r_1+r_2)^2 \omega^2 \\
+    \Rightarrow G(m_1+m_2) &= (r_1+r_2)^3 \omega^2 \\
+    \Rightarrow \omega &= \sqrt{\frac{G(m_1+m_2)}{(r_1+r_2)^3}}
+  \end{split}
+\end{equation}
+$$
+
+Now, let's introduce a third body in this two body system of mass $$m_3 (\ll m_1, m_2)$$ at $$(x, y, z)$$. For this body to be at a [lagrange point](#lagrange-points), it must, by definition, remain stationary w.r.t. $$m_1$$ and $$m_2$$. That is, it must also orbit the origin with only an angular speed $$\omega$$ along the Z-axis.
+
+Before we start calculating the [gravitational](./newtonian%20gravity.html) forces on $$m_3$$, we need to keep in mind that if it is located at $$z \neq 0$$, it will always experience a force towards the XY-plane due to the presence of $$m_1$$ and $$m_2$$ in the XY-plane. Hence, $$a_z$$ (acceleration in the z-direction) will never be equal to 0, thereby failing to be at a [lagrange point](#lagrange-points). So, there exist no [lagrange points](#lagrange-points) having $$z \neq 0$$. Consequently, we shall restrict ourselves to finding point(s) in the XY-plane only and will be of the form $$(x, y, 0)$$.
+
+<br />
+![Derivation](../../assets/images/theory/celestial%20mechanics/lagrange%20points/derivation.png)
+<br />
+
+Coming back to calculating the [gravitational](./newtonian%20gravity.html) forces acting on $$m_3$$, the [gravitational](./newtonian%20gravity.html) force acting on $$m_3$$ due to $$m_1$$ can be given as:
+
+$$
+\begin{equation}
+  \begin{split}
+    \vec{F_1} &= -\frac{Gm_1m_3}{r_{31}^2}\hat{r_{31}} \\
+    &= -\frac{Gm_1m_3}{((x+r_1)^2 + y^2)^{3/2}} ((x+r_1)\hat{i} + y\hat{j})
+  \end{split}
+\end{equation}
+$$
+
+Similarly, the [gravitational](./newtonian%20gravity.html) force acting on $$m_3$$ due to $$m_2$$ can be written as:
+
+$$
+\begin{equation}
+  \begin{split}
+    \vec{F_2} &= -\frac{Gm_2m_3}{r_{32}^2}\hat{r_{32}} \\
+    &= -\frac{Gm_2m_3}{((x-r_2)^2 + y^2)^{3/2}} ((x-r_2)\hat{i} + y\hat{j})
+  \end{split}
+\end{equation}
+$$
+
+The sum of these two forces should be equal to the centripetal force required by $$m_3$$ to orbit the origin with an angular velocity of $$\omega$$ about the Z-axis. Mathematically,
+
+$$
+\begin{equation}
+  \vec{F_1} + \vec{F_2} = -m_3 \omega^2 (x\hat{i} + y\hat{j})
+\end{equation}
+$$
+
+Separating the force balance equation into its components along the X- and Y-axis, we get,
+
+$$
+\begin{equation}
+  \begin{split}
+    m_3 \omega^2 x &= \frac{Gm_1m_3}{((x+r_1)^2 + y^2)^{3/2}} (x+r_1) + \frac{Gm_2m_3}{((x-r_2)^2 + y^2)^{3/2}} (x-r_2) \\
+    m_3 \omega^2 y &= \frac{Gm_1m_3}{((x+r_1)^2 + y^2)^{3/2}} y + \frac{Gm_2m_3}{((x-r_2)^2 + y^2)^{3/2}} y
+  \end{split}
+\end{equation}
+$$
+
+Easy solution(s) of this can be obtained by finding [lagrange point](#lagrange-points)(s) on the X-axis, i.e., by taking $$y = 0$$. This reduces the above equations to:
+
+$$
+\begin{equation}
+  \begin{split}
+    m_3 \omega^2 x &= \frac{Gm_1m_3}{(x+r_1)^2} + \frac{Gm_2m_3}{(x-r_2)^2} \\
+    0 &= 0 \\
+  \end{split}
+\end{equation}
+$$
+
+Substituting the value of $$\omega$$, we end up with a third order polynomial in $$x$$, i.e., a cubic equation. It will have three roots, denoting three [lagrange points](#lagrange-points), namely, $$L1$$, $$L2$$ and $$L3$$. The calculation of the exact values of these are left as an exercise to the you :p
+
+{: .fun }
+
+> These three [lagrange points](#lagrange-points) will always be found in the following three regions:
+>
+> 1. Before $$x = -r_1$$
+> 2. Between $$x = -r_1$$ and $$x = r_2$$
+> 3. Beyond $$x = r_2$$
+>
+> And will, as proven above, have $$y = 0$$ and $$z = 0$$
+
+Now, if we take $$y \neq 0$$, we get 2 solutions, giving us two more [lagrange points](#lagrange-points), namely, $$L4$$ and $$L5$$.
+
+{: .fun }
+
+> Connect a line segment from $$m_1$$ to $$m_2$$ and using this length, draw two equilateral triangles on the XY-plane with this line segment as one of the edges. The third vertex of these triangles will be the $$L4$$ and $$L5$$ [lagrange points](#lagrange-points).
+>
+> Hence, they are also known as equilateral [lagrange points](#lagrange-points)
+
+### Stability
+
+The [stability](#stability) of a [lagrange point](#lagrange-points) is defined by its ability to retain the object at that point if slight perturbation is introduced in the position of $$m_3$$. If $$m_3$$ returns back to the same [lagrange point](#lagrange-points), that [lagrange point](#lagrange-points) is considered to be stable. It is declared unstable otherwise.
+
+In orbiting two-body systems, the [lagrange points](#lagrange-points) $$L_1$$, $$L_2$$ and $$L_3$$ are unstable and the [lagrange points](#lagrange-points) $$L_4$$ and $$L_5$$ are stable as can be seen in the image below (Credits: [wordpress.mrreid.org](http://wordpress.mrreid.org/))
+
+![Sun-Earth lagrange points stability](../../assets/images/theory/celestial%20mechanics/lagrange%20points/stability.png)
+
+### Utility
+
+[Lagrange points](#lagrange-points) offer a lot of advantages. A few of them include:
+
+- Placement of telescopes near $$L1$$ and $$L2$$. $$L1$$ is an absolutely brilliant point for placement of solar observatories as they'll be able to get un-blocked view of the Sun 24x7. Examples include the [Aditya L1](https://www.isro.gov.in/Aditya_L1.html) mission. Similarly, $$L2$$ is an ideal location for an infrared observatory, for example, the [James Webb Space Telescope](https://webb.nasa.gov/). (Image credits: [ESA Webb](esawebb.org))
+
+![JWST L2](../../assets/images/theory/celestial%20mechanics/lagrange%20points/webb.jpg)
+
+- Protection from rouge asteroids and comets. A massive planet, such as Jupiter, captures and holds potentially devastating asteroids and comets that may come towards the inner planets (such as the Earth), near its [lagrange points](#lagrange-points), particulartly the stable $$L4$$ and $$L5$$ [lagrange points](#lagrange-points). (Image credits: [Mdf](https://en.wikipedia.org/wiki/User:Mdf))
+
+{: .fun }
+
+> These are called Trojan asteroids, which are named after characters from Homer's lliad (the legendary siege of Troy).
+>
+> Asteroids at the $$L4$$ points, which leads Jupiter, are referred to as the "Greek camp", while at the $$L5$$ points are referred to as the "Trojan camp". These asteroids are largely named after characters from the respective sides of the Trojan war.
+
+![Trojan Asteroids](../../assets/images/theory/celestial%20mechanics/lagrange%20points/trojan%20asteroids.png)