# How does gravity change beyond L2?

Why does gravity appear to increase again on the far side of L2 from earth as indicated on the gravitational contour diagram? One would expect gravity to continue to decrease as the distance from sun and earth becomes greater.

From Wikipedia's Lagrange point; Radial acceleration:

The radial acceleration a of an object in orbit at a point along the line passing through both bodies is given by:

$$a = -\frac{GM_1}{r^2} \text{sgn}(r) + \frac{GM_2}{(R-r)^2} \text{sgn}(R-r) + \frac{G((M_1 + M_2)r - M_2R)}{R^3}$$

where r is the distance from the large body M1, R is the distance between the two main objects, and sgn(x) is the sign function of x. The terms in this function represent respectively: force from M1; force from M2; and centripetal force. The points $$L_3, L_1, L_2$$ occur where the acceleration is zero — see chart at right. Positive acceleration is acceleration towards the right of the chart and negative acceleration is towards the left; that is why acceleration has opposite signs on opposite sides of the gravity wells.

Here's a plot of "radial acceleration" as a function of position along the Earth-Moon line, intersecting the three collinear Lagrange points.

A plot along the Sun-Earth line would look qualitatively similar; it would have the same shape and number of zero crossings, and also rise on the right past L2 as it does to the left of L1.

Net radial acceleration of a point orbiting along the Earth-Moon line. A positive value means the object will be forced to the right. Lagrangian points L3,L1,L2 occur where the line crosses the x-axis, but due to the positive slope on crossover, none of them are stable.

Source

• Most likely because you misunderstand the gravity contour diagram. Please provide a reference to the specific image you are using, so that we can assist you in understanding it. Jan 7, 2022 at 18:02
• Maybe begin with the diagram at NASA's What is a Lagrange Point? Or read the paper "The Lagrange Points" by Neil J. Cornish if you're mathematically inclined.
– Wyck
Jan 7, 2022 at 18:03
• Welcome to the Space Exploration community! May I suggest you include said diagram? Also, your question may be better suited for the Physics community... Jan 7, 2022 at 18:12
• Lagrange point diagrams typically use a rotating reference frame, and show the combined effect of gravity & centrifugal force. Jan 7, 2022 at 20:08
• @Sunfish Welcome to Stack Exchange! I've added a block quote, equation and diagram from Wikipedia that seems to show what you are describing. I think it's enough to satisfy the comments, but if you have another that you'd rather use feel free to edit/update your post.
– uhoh
Jan 7, 2022 at 21:30

The key here is that the graph shows acceleration in a rotating frame of reference.

The line shows the sum of gravitational acceleration and centrifugal acceleration, which are opposite in direction. Gravitational acceleration decreases with the square of distance and centrifugal acceleration increases linearly with distance. So, at a great enough distance, centrifugal will inevitably exceed gravitational.

In the sketch below, blue is centrifugal force, green is gravity and red is the sum of the two

In a non-rotating frame, centrifugal acceleration doesn't exist (it's a fictional force used only in a rotating frame) and so the acceleration due to gravity will still decrease with distance.

In Newtonian mechanics, the centrifugal force is an inertial force (also called a "fictitious" or "pseudo" force) that appears to act on all objects when viewed in a rotating frame of reference.

• @uhoh Thanks for the edit. Concisely put! I really like that graph. It's great at illustrating what an L-point really is: the place where centrifugal acceleration is equal to total gravitational acceleration. Isn't it surprising how useful a fictitious force can be? Kind of like imaginary numbers. Jan 8, 2022 at 2:06
• – uhoh
Jan 8, 2022 at 2:27

Lagrange point contour diagrams typically use a rotating reference frame, and show the combined effect of gravity & centrifugal force. Everything on the diagram is actually rotating, as well as being subjected to gravity.

To illustrate the Lagrange points of the Sun-Earth system we use a rotating frame that has an angular velocity of 1 revolution per year. Here's the potential surface for the Sun's gravity (without the Earth) and the centrifugal force at that speed.

Bear in mind that these diagrams only apply to objects that are revolving with the frame. That is, they are static relative to the rotating frame. A body sitting on the very top of that surface is 1 AU from the Sun. It has the correct angular velocity to remain in a circular orbit, so it remains motionless in the rotating frame. A body with radius <1 AU is going too slow, so it moves inwards, due to gravity. A body with radius >1 AU is going too fast, so it moves outwards, due to centrifugal force. In either case, we can't say exactly what path the body will take because we don't know the Coriolis force, so we can only tell what the initial direction of motion will be.

Due to the Coriolis force, a test mass $$m$$ (without propulsion) will drift along (instead of perpendicular to) equipotential lines!
[...]
The Coriolis force explains the stability of Lagrange points $$L_4$$ & $$L_5$$.

Here's a two-body diagram, where the larger body has a mass 8 times that of the smaller body. (If we use a large ratio it's harder to see what's going on, unless the diagram is huge).

The Lagrange points are marked by small purple spheres. L4 and L5 are sitting on top of hills, but L1, L2, and L3 are at saddle points, which are maxima in the radial direction, but minima in the tangential direction.

It's easier to see what's going on in this interactive 3D diagram, created using Sage. You can pan and rotate using the mouse, and zoom using the scroll wheel. On touchscreen devices, use one finger to rotate, two fingers to pan and zoom.

The potential surface function uses the same scheme as Wikipedia:

$$-\Psi(x,y) = \left(x - \frac{q}{1+q}\right)^2 + y^2 + \frac{2}{(1+q)\sqrt{x^2+y^2+z^2}} + \frac{2q}{(1+q)\sqrt{(x-1)^2+y^2+z^2}}$$

where $$q$$ is the mass ratio, $$z=0$$ (because we're looking at the orbital plane), with the bodies on the X axis. The large body is at $$x=0$$ and the small body is at $$x=1$$. The barycentre of the system is at $$x=\frac{q}{1+q}$$.

• +1 but Why shouldn't we illustrate spacecraft trajectories on top of static zero velocity pseudo-potential surfaces? Is it possible to add a mention that this is a zero-velocity surface and it would look different for something moving in a halo orbit for example? It's not necessary of course, the question doesn't ask about non-zero velocity, but just adding a few extra words might serve as an "Aha!" to some future reader. Thanks!
– uhoh
Jan 10, 2022 at 21:13
• @uhoh Well, I did say "we can't say exactly what path the body will take because we don't know the Coriolis force, so we can only tell what the initial direction of motion will be". But I'll try to think of a way to clearly emphasise that these diagrams are only valid for bodies that are static relative to the rotating frame. Jan 10, 2022 at 21:39
• – uhoh
Jan 10, 2022 at 21:44
• One thing you might want to add. Although L4 and L5 are maxima in the potential field, they represent stable points because ifvthe effect of a non-potential gradient, the Coriolis force. When we add the Coriolis force to that due to the potential gradient, we find that maxima are stabilized if the maximum is not too high or sharp, but not saddle points. Jan 13, 2022 at 15:09