This question is brought up due to confusion over that cool video of an asteroid cricling the Earth and not understanding the explanation of what the L1 represents. Thank you.
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If you have a two body system, like the Sun and the Earth, most orbital points are unstable. This means if you put something there and you don’t use rockets to keep it there it will fall into the Sun or the Earth in a few years. There are 5 points where this doesn’t happen, the Legrange points.
Its a point of equilibrium for an object with a small mass between 2 other objects with much larger masses. L1 is smack in-between 2 large orbiting bodies, in our case between the Earth and the Sun. Because of Earth’s gravitational influence, at L1 the object’s orbital period matches Earth’s exactly.
https://en.wikipedia.org/wiki/Lagrange_point
> At the Lagrange points, the gravitational forces of the two large bodies and the centrifugal force balance each other.
> Small objects placed in orbit at Lagrange points are in equilibrium in at least two directions relative to the center of mass of the large bodies.
Basically, when you have two objects orbiting each other (sun and earth, earth and moon, etc.), there are 5 points where gravity and centrifugal force cancel each other out. You can plop something at a lagrange point and it’ll just stay there (relative to the two objects). Which is why we put the new James Webb telescope at L2.
In other words, if you look at the sun and earth from above and rotate to keep the earth “stationary”, the lagrange points are also stationary. If you instead don’t rotate and watch the earth orbit the sun, it’ll look like the lagrange points “follow” the earth.
Very cool 😎 I feel smarter now 😃
So, when two bodies are orbiting each other (like earth orbiting the sun), it turns out there are five places where their mutual gravity and orbits mean that an object can stay in the same place relative both of them. These places are called “legrange points”
L1 – the first Legrange point – is the place in between them where their gravity is equal. The easiest way to imagine this is to imagine a really big trampoline, and two people sitting on it. If you look at the trampoline between the two people, there’s going to be a point at which the trampoline is flat – it’s going to be closer to the lighter person; but there’s going to be that flat point.
Basically, if a light object (compared to Earth) is sitting at L1 it will have the same orbit time as the Earth as it tries to orbit the Sun and Earth at the same time, can’t figure out which one to orbit, and so stays right in between them.
However, it’s an “unstable” point, meaning that if the object moves just a little, it will orbit faster or slower and tend to end up orbiting one body or the other. You can solve this with a spacecraft using power to stay in that location – the James Webb Space Telescope is at L2 (a point past Earth in a direct line with the sun, which is also unstable; but has the advantage of being mostly hidden from the Sun by Earth), because it will mostly stay in place.
3 body gravitational systems are chaotic unstable. Chaotic meaning that a small, almost insignificant change to the initial conditions will completely change the outcome of the system, and unstable meaning that almost always one of the bodies will be ejected (launched away with escape velocity by gravitational forces) resulting in a much more stable 2 body system.
Lagrange points are points that will be stable in a 3 body system where M1 >> M2 >> m. M1 being a star, M2 being a planet, and m being a satellite, and >> means much greater than.
L1, L2, and L3 were the first ones discovered because they are easy to figure out with Newtonian mechanics, that’s just where the forces of M1 and M2 on m balance out. (They originally had different names, but were renamed after L4 and L5 were discovered)
L1 is between M1 and M2, so if you place something there perfectly, the forces allow it to maintain its orbit at a slower speed than it would if M2 didn’t exist. That’s because for circular motion, a = v^2 /r, and a (acceleration) is reduced because the force from M1 is being very slightly canceled out by M2. Now this would apply at every point between M1 and M2, but it needs to be at L1 specifically so it can also have the same angular speed (have the same year length) as M2 around M1, otherwise, it would drift away from being between M1 and M2.
Since M2 >> m, we can ignore the effect m has on M2.
ω(angular velocity) = 2π/T(period) = v(linear velocity)/r
We know the periods, and therefore angular velocities of m and M2 must be the same, so we will be using ω=v/r, and remember a = v^2 /r. Combine the two, and we have a = ω^2 r. And then acceleration due to gravity is a = GM/r^2 (formula for force of gravity, but mass of the object canceled out)
For M2 we get GM1/R1^2 = ω^2 R1 and for m we get GM1/R2^2 – GM2/(R1 – R2)^2 = ω^2 R2. (R1 is M2’s distance to M1 and R2 is m’s distance to M1)
Do some math, we get GM1/R1 = ω^2 and G(M1/R2 – M2R2/(R1 – R2)^2) = ω^2, since angular velocity has to be equal, then so does angular velocity squared. GM1/R1 = G(M1/R2 – M2R2/(R1 – R2)^2)
Do some more math, M1/R1 = M1/R2 – M2R2/(R1 – R2)^2, and then M1(1/R2 – 1/R1) = M2R2/(R1-R2)^2 and then M1/M2 (R1-R2/R1R2) = R2/(R1-R2)^2
M1/M2 = R1R2^2 / (R1-R2)^3 the ratio of the masses determines the ratio of the radii
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