Lagrange points

kowloon

I think I'm clear how the first three Lagrange points are a thing, but can anyone explain what forces are in play for 4 and 5? Pretend you're explaining it to a moron .

kowloon

Just popped into my head as I was making dinner: Are things at 4 and 5 orbiting not under enough G to be pulled in by the blue mass and so orbit at the same rate? Is there more to it than that?

ps200306

I know your picture is from Wikipedia but it's actually misleading as it implies objects at L4 and L5 on the same orbital track as the secondary, orbiting a perfectly central primary. If such a setup were possible, any point on the orbit would be stable relative to the secondary, not just L4 and L5.

Let's digress for a moment. Picture the International Space Station. It's 400 km up, and travelling at a little under 8 km per second. Why that speed? Because that's the speed you have to be travelling to be in orbit at that height.

The maths isn't all that important, but we can work out the orbital speed of the ISS using Newton's Law of Universal Gravitation. We set the centripetal acceleration equal to the gravitational acceleration which causes it:



where:

• and are the centripetal and gravitational accelerations,
• is the linear velocity of the ISS,
• is the distance of the ISS from the centre of the Earth,
• is the mass of the Earth,
• is the gravitational constant.

Notice that the mass of the ISS doesn't enter into the equation. The orbital speed for any object at a given height is the same regardless. So if you were in the space shuttle alongside the ISS, doing the same speed in the same direction, the two of you would happily orbit in the same relative position without drifting apart.

So what's the point of the L4 and L5 Lagrange points? We shouldn't need any special stable position in an orbit as every position ought to be stable. Just match your height, speed and direction and Bob's yer uncle.

Not so! What we're ignoring is that the orbital velocity formula above is an approximation which only works where the mass of the orbiting body is negligible compared to the primary body. In reality a gravitational orbit involves two bodies co-orbiting around their common centre of gravity (or barycentre):



Let's suppose the larger of those two bodies (the primary) is the Sun, and the smaller one (the secondary) is the Earth. And let's suppose further that we have a third body of negligible mass and we want it to orbit the same barycentre in a fixed position relative to the Earth without drifting apart. The problem is that this third body wants to orbit the centre of the Sun, not the Sun-Earth barycentre. It cannot maintain a fixed relationship to Earth.

Except ... there is a way we can make this happen. We draw a line between the centres of the primary and secondary, and make it the baseline of an equilateral triangle. From the apex of this triangle, the sum of the gravitational force vectors to the primary and secondary points toward the barycentre. What's more, a third object at this location will have the same orbital angular speed around the barycentre as the secondary and thus maintains its position relative to the secondary. (Note: not the same linear speed -- the third object is on a bigger orbit so must travel faster).



Obviously by symmetry there are two equivalent vertices, above and below the baseline, and these are the L4 and L5 Lagrange points. Note, the picture above is an exaggeration. A two-body system with such a large secondary wouldn't have stable L4 and L5 points. In reality, the Sun-Earth barycentre is only 450 km from the centre of the Sun.

kowloon

Thanks for the reply! I get now that I was ignoring the centre of the system and only thinking in terms of the centre of the largest body. I dun lurned a ting.