TUG OF WAR

How does the balance between gravitational force and forces opposing it shape the lives of stars? What happens when this balance is disturbed? We will try to answer the following questions in this article. We will follow the journey of a star from its youth to its old age and we won't stop there! We will also visit the graves and have a look at the dead bodies-the stellar remnants, and in our journey, we will witness spectacular powers of the universe at play-All of them the consequence of a tug of war!


THE DAY THE SUN STOPPED SHINING

The historic solar eclipse of 1919 that Eddington used to finally prove Einstein's general theory of relativity, that established the end of the Newtonian era and went on to lay the foundation of cosmology and the understanding of the modern day Universe.


THE APOLLO MISSION

On the 50th anniversary of the Apollo Mission, this article takes us back in history to the 1960s when humans were struggling to grasp with the idea of a world outside our own.


TWO BODY PROBLEM

This article is a concise and mathematical read on the famous 2 body problem.


Three Body Problem

Analysis of 3 Body problem is not very simple due to addition of various variables due to 3rd body.
3 body problem can’t be solved completely.

Restricted 3 Body problem

Let one of the masses is infinitesimal small. m30
Now we have to solve equations only for m3. Because for m1 & m2, it is like 2 body problem
m3r3¨=Gm1m3r31r313Gm2m3r32r323
r3¨=Gm1r31r313Gm2r32r323
Using Cartesian Coordinates
x¨=Gm1(xx1)(xx1)2+(yy1)2+(zz1)2
y¨=Gm1(yy1)(xx1)2+(yy1)2+(zz1)2
z¨=Gm1(zz1)(xx1)2+(yy1)2+(zz1)2


Circular Restricted 3 body problem

Suppose that the first two masses, m1 and m2, execute circular orbits about their common center of mass.

Let us define a Cartesian coordinate system (ξ,η,ζ) in an inertial reference frame whose origin coincides with the center of mass, C, of the two orbiting masses. Let the orbital plane of these masses coincide with the ξ$$η plane, and let them both lie on the ξ-axis at time t=0–see Figure. Suppose that R is the constant distance between the two orbiting masses, r1 the constant distance between mass m1 and the origin, and r2 the constant distance between mass m2 and the origin. Moreover, let ω be the constant orbital angular velocity.

(xyz)=(cosωtsinωt0sinωtcosωt0001)(ξηζ)
Differentiate it twice put it in cartesian equations of motion of $m_3$. We get
ξ¨2ωη=Uξ(7)
η¨+2ωξ=Uη(8)
ζ¨=Uζ(9)

(7)×ξ˙+(8)×η˙+(9)×ζ˙

ξ˙ξ¨+η˙η¨+ζ˙ζ¨=ξ˙Uξ+η˙Uη+ζ˙Uζ(10)
if U=U(ξ,η,ζ)
dUdt=Uξξt+Uηηt+Uζζt
It is R.H.S. of (10). Also
ddtξ˙2=2ξ˙ξ¨
eqn $(10)$ becomes
12ddt(ξ˙2+η˙2+ζ˙2)=dUdt
We know ${\dot{\xi}}^2+{\dot{\eta}}^2+{\dot{\zeta}}^2=v^2$.
By Jacobi Integral (External Link) - For Extra Information
v2=2UC
For v20$,$2UC. This gives
U=ω22(ξ2+η2)+μ1r1+μ2r2
If I term is large
ξ2+η22Cω2
Path would be enclosing m1 and m2. On changing C we get different Lagrangian Points