Please support my efforts. I just finished a conic sections and orbital mechanics coloring book. I need help with printing costs. Through this Kickstarter you can pre-order a signed coloring book. I look at conic sections, Kepler's laws, Hohmann transfer orbits, the Oberth effect, space tethers, Tsiolkovsky's rocket equation and lots of other space stuff. The coloring book is $5 plus $5 shipping and handling ($10 shipping and handling if you're outside the U.S.).

Kickstarter for this coloring book ends 4:30 a.m. April 13, 2020.

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**Gravity doesn't cancel at the Lagrange points**

"There are places in the Solar System where the forces of gravity balance out perfectly. Places we can use to position satellites, space telescopes and even colonies to establish our exploration of the Solar System. These are the Lagrange Points."

From Fraser Cain's video on Lagrange points. A lot of pop sci Lagrange articles repeat and spread this bad meme. It just ain't so.

The 5 Lagrange points can be found in many two body systems. They can be Sun-Jupiter, Earth-Moon, Jupiter, Europa -- Any pair of dancers has this retinue of 5 Lagrange regions moving along with them. Above are the 5 Pluto-Charon Lagrange points. Also pictured are the gravity vectors these bodies exert. Pluto's gravity is indicated with purple vectors and these point towards Pluto's center. Charon's gravity is indicated with orange vectors and these point towards Charon's center.

For the gravity vectors to cancel each other, they need to be equal and pointing in opposite directions.

**L1**

The only L-Point where the gravity vectors pull in opposite directions is L1. And here the central body (Pluto) pulls harder than Charon. These two gravities don't balance out.

**L3 and L2**

Zooming in on the L3 and L2 points, we can see both bodies pull the same direction. These don't balance.

**L4 and L5**

Zooming in on the L4 and L5 points. Pluto pulls much harder. The angle between these vectors is 60º

**The So-Called Centrifugal Force**

There is a third player in these Lagrange tug of wars. What we used to call centrifugal force. This is not truly a force but rather inertia in a rotating frame. Here is an XKCD cartoon on this so called force:

Indeed, in a rotating frame, inertia sure feels like a force. The pseudo acceleration can be described as ω

^{2}r where ω is angular velocity in radians per time and r is distance from center of rotation. The vector points away from the center of rotation.

**Putting Gravity and Centrifugal Force Together**

Here's the same diagram but with centrifugal force thrown in (the blue vectors). Also the foot of the Charon gravity vectors are placed on the head of the Pluto gravity vectors -- this is a visual way to carry out vector addition.

For L1, Charon and Centrifugal Force are on the same team and they perfectly balance Pluto's gravity.

For both L2 and L3, Pluto and Charon are on the same team and they neutralize their opponent Centrifugal Force.

But what about L4 and L5? An observant reader may notice that the centrifugal force vector doesn't point away from Pluto's center. Adding Charon's tug to Pluto's tug moves the direction to the side a little bit.

Now the centrifugal force vector points from the barycenter. This is the common point of rotation around which both Pluto and Charon rotate. The same applies to L5.

L4, Charon's center and Pluto's center form an equilateral triangle.

The barycenter lies on the corner of a non-equilateral triangle.

And so it is with all the orbiting systems in our neighborhood. It is a 3 way way tug-of-war between centrifugal force, gravity of the orbiting body and gravity of the central body. Sometimes two players are on the same team, other places they switch. In L4 and L5 everyone pulls in a different direction. But in all 5 Lagrange points, the sum of the three accelerations is zero.

## 14 comments:

Something has gone wrong. For the force vectors for L4/L5, you say "The angle between these vectors is 60º", yet the angle through the barycenter needs to be 60º, and doing both is geometrically impossible.

Ian, it is the line through

Pluto's centerthat that crosses at 60º. Pluto's center, Charon's center and L4 or L5 form an equilateral triangle. But the barycenter does not lie on the corner of an equilateral triangle. It sits on an angle slightly greater than 60º. The top angle becomes a little less than 60º. I will try to clarify this by adding an illustration to this blog post.Ah, it appears you are correct. That is a very good diagram, too!

Thanks for the debunk! I'll be more careful next time I delve into this topic to get my analogies right.

Very interesting explanation.

I hope you posted more often...

Fraser Cain, going to your Lagrange vid and article I see they are still wrong. Your article on the Interplanetary Transit Network is still misleading.

Universe Today remains a source of misinformation.

Great explanation Hop.

Please do cut Fraser a bit of slack though - he and the rest of the UT crew go to great lengths to inform people, misinformation does happen by accident - nobody is perfect! I'm sure Fraser will get around to making a correction, perhaps via a later video.

Lionel W., thanks.

Re: Fraser -- he and his crew do drum up interest for space and space exploration. And we all make mistakes, myself included.

But sources of misinformation can prevent understanding of a topic. So long as Fraser's articles/vids on Lagrange points and Interplanetary Transit System remain uncorrected, I'll be annoyed.

This is a particularly good explanation and illustration.

The 'short text' explanation might be that 'the Lagrange points are those points in a two-body system where the sum of forces acting on a third body balance out in a rotating frame of reference'. That's a simplification that obscures at least as much as it reveals, but might be useful as a reference note.

@ Matter Beam:

As far as posting more often, I know I've had trouble finding suitable topics. My existing posts cover pretty much all the ground I intended to cover, so now it's just incidental or forward-looking posts and occasional 'paper rocket' math adventures.

Hop's body of work is so thorough that it is difficult to find an orbital-mechanics topic that isn't already covered in a clear and informative way. From this outsider's perspective it seems many recent posts here are prompted by current events and are intended to clear up misperceptions.

Hi Hop, sure, and I respect your determination for correctness.

Changing the subject, I recently came across a youtube channel with some really nicely done explanations of mathematical concepts via artwork, and it reminded me of some of the things on your website. Have you come across 3Blue1Brown? Very helpful for learning and explaining math

Lionel, nice link. Thanks. Hope I'll have some time to explore that website. Yeah, presenting math and physic ideas visually is what I try to do.

Hello, Hop! Thank you for an excellent article. I appreciate that your crusade took you to galacticjourney.org, and I've amended the offending sentence. I also responded to your comment.

Hope to see you around in the future.

I find the L1 position at 326000 km along the Terra-Luna line, and the

gravitational balance point at 346000 km, but I am not able to calculate

the single impulse delta-V to go from circular low earth orbit to either

position. Is there a straightforward way to do this, and to find the

position in LEO for the burn? -MBMelcon

Melon, it's a 3 body problem, therefore messy. I don't know of a nice clean simple equations for figuring departure and arrival burns in this case. The best way I know of is Runge-Kutta approximations and chopping the trajectory in many small time intervals.

I have made approximations based on more straightforward 2-body approximations. A LEO to EML1 orbit would be a 300 x 320351 km ellipse (I'm using earth altitudes here, not distance from earth's center). This transfer orbit's perigee speed is 10.81 km/s and apogee speed is .22 km/s. I used the Vis Viva equation to get these numbers.

The LEO burn is pretty much a 2 body scenario. So I go with 10.8 - 7.7 to get 3.1 km/s TLI burn.

But the apogee burn to park at EML1 is trickier. A circular 2 body orbit at that altitude would be moving 1.1 km/s. However EML1 is moving at the same angular velocity as the moon. So EML1 is moving more like .85 km/s. So for a long time I went with .85 -.22 or about a .63 km/s circularization burn at apogee.

Then I grabbed some of Bob Jenkins' orbit sim JAVA code (with his permission) and did my own earth moon sims. The moon exerts an appreciable influence over a fairly large neighorbhood of the apogee. The moon lends a hand. So it's a little less than .63 km/s. Maybe around .55 to .6 km/s.

So adding a 3.1 perigee burn to a .6 apogee burn, I usually say LEO to EML1 is about 3.7 km/s.

And the orbit is time reversible so the same numbers going from EML1 to LEO. Although with LEO you could have the spacecraft pass through the upper atmosphere and using aerobraking to shed velocity. So the LEO circularization burn could be less than 3.1 km/s.

I know that's not a satisfactory answer but it's the best I can give you right now.

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