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We were taught that Lagrange points are where the resultant gravitational intensity is zero. Does that mean the resultant gravitational attraction of an object kept there is zero? If that's the case how can it orbit, the resultant should be equal to the centripetal force right?

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  • $\begingroup$ There is no force on the object in the rotating reference frame of the barycenter. That only means it is in a stable orbit. $\endgroup$ Dec 31 '20 at 7:38
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We were taught that lagrangian points are where the resultant gravitational intensity is zero.

No this is not the case. Lagrange points are the points where the net gravitational force of Sun and Earth provides the necessary centripetal force to make an object go around their common center of mass.

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  • $\begingroup$ missing the part that says, that the third body does not therefore, move in relation with the first two bodies. Further, Lagrangian points are stable, in the same way that a perfectly vertically balanced pencil is stable. $\endgroup$
    – Aron
    Dec 31 '20 at 4:10
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    $\begingroup$ @Aron An object in orbit at close distance to L4/L5 will continue to orbit at close distance to L4/L5 indefinitely. (With the standard caveats that gravitational influence from yet another planet may disturb things. In the case of a three-body system, with the mass of the third body negligable compared to the Primary and the Secondary, the Tertiary will remain at close distance to L4/L5 forever. $\endgroup$
    – Cleonis
    Dec 31 '20 at 7:27
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In the three-body problem, the Lagrange points are those points in space where two bodies with large mass (Earth and Sun), through the interaction of the respective gravitational force, allow a third body with a much lower mass to maintain a stable position relative to them.

In a planetary system it implies that a small object, such as a satellite or an asteroid, which shares the same orbit of a planet and positioned in a Lagrange point, will keep constant the distances between the major celestial bodies, the star and the planet with which shares the orbit.

For this to happen, the resultant of the gravitational accelerations imparted by the celestial bodies to the object must be exactly the centripetal acceleration necessary to keep the object in orbit at that particular distance from the largest celestial body, with the same angular velocity as the planet.

Out of curiosity: you can also verify using Lagrangian mechanics these five Lagrangian points.
Let be $(P-O)=\rho \vec{e}_\rho$ the distance between our third point and the mass centre of the system; $(T-O)=x_E \vec{e}_x$ the distance between the Earth and the CM and $(S-O)=x_S \vec{e}_x$ the distance between the Sun and the CM (very near to the centre of the Sun).
You will find that the system potential energy is $$U_{tot}=U_{grav}+U_{centr}=-G\frac{M_E\cdot m}{||P-T||}-G\frac{M_S\cdot m}{||P-S||}-\frac{1}{2}m\Omega^2\rho^2$$ If you derivate this energy potential respect to $\rho$ and $\theta$ you will find these points. And also, if you make the Hessian matrix you will find that 3 of them are points of instability $(L_1, L_2, L_3)$ and the other 2 are points of stability $(L_4, L_5)$.
I do not pretend that you understand perfectly these calculations but only the principal idea that is behind.
I hope that the first part of the answer could be helpful for you and maybe also the second one.

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    $\begingroup$ So not where the gravitational intensity is zero? $\endgroup$ Dec 30 '20 at 14:21
  • $\begingroup$ Exactly, there is the centripetal force that is essential $\endgroup$ Dec 30 '20 at 14:24
  • $\begingroup$ Is null point and lagrangian points the same or not? $\endgroup$ Dec 30 '20 at 14:26
  • $\begingroup$ I do not really know what null point is for you. Lagrangian points are those points that solve the derivates of this system energy potential. You can see them like null points of the derivates of the system energy potential. Using Lagrangian Mechanics it becomes easier to understand. $\endgroup$ Dec 30 '20 at 14:29
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Decades ago I learned there were five Lagrangian points around the orbit of a planet correction: with or without a moon but I understood that these were just simple solutions for the three body problem (digression - 'Three Body Problem' is an excellent novel by Liu Cixin). At these points gravity acts on the third body in a manner that maintains stable orbits for all three bodies. I suggest you return to your tutor and ask them to clarify because the statement posted is wrong. A short explanation is that the gravitational force provides the centripetal force exactly balancing the object's inertia at L1, L2 & L3; at L4 & L5 there is stable oscillation (digression, old novel 'Integral Trees' by Larry Niven).

(Further digression: the two more famous points, L4 and L5, are occupied by the Greek and Trojan asteroid clusters either side of Jupiter in its orbit. Do not confuse fictional Antichthon in L3 with fictional Nemesis from Sailor Moon. Gaia, the Milky Way telescope, occupies L2 and the second correction: SOHO telescope is in L1.)

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    $\begingroup$ L1, L2, and L3, are not long time stable points. There are spacecrafts that have been positioned at a Lagrange point. The solar system is very big, so it doesn't take much thruster corrections to maintain the position of a spacecraft at L1, L2, or L3. However, if you would not execute any correction at all then over time the spacecraft can end up anywhere at roughly its orbital radius. Oribiting at close distance to L4/L5 on the other hand, will remain close distance indefinitely $\endgroup$
    – Cleonis
    Dec 30 '20 at 15:25
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    $\begingroup$ Do you mean "with a moon?" What are your 3 bodies: planet, object, and ?? $\endgroup$
    – Bill N
    Dec 30 '20 at 15:25
  • $\begingroup$ Both telescopes are at L2, I think. $\endgroup$
    – TonyK
    Dec 31 '20 at 11:51
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In this answer I will name the celestial bodies as follows: the Primary, the Secondary, and the Tertiary.

For example, in the case of the Sun, Jupiter, and the Trojans the Sun is the Primary, Jupiter is the Secondary, and any member of the Trojans is the Tertiary.

For a Tertiary orbiting at a Lagrange point the following applies: at the position of the Lagrange point the resultant of the gravitational attractions of the Primary and the Secondary is such that the Tertiary is orbiting with the same orbital period as the Secondary.

I assume Wikipedia is already covering the L1, L3, and L3 adequately.


L4/L5

The following applies for L4 and L5.

For an object located at L4/L5:
The gravitational attraction on the Tertiary is the resultant of the gravitational attractions from the Primary and the Secondary.

This means that when positioned at L4/L5 the Tertiary is orbiting the common center of mass of the Primary and Secondary. From here on I will refer to that as the 'COM' (The mass of the Tertiary is treated as negligable).

At L4/L5 The distance of the Tertiary to the COM is slightly larger than the distance of the Secondary to the COM, since the tertiary is attracted by both the Primary and Secondary, and the Secondary is attracted only by the Primary.

That is, at L4/L5 the resultant attraction on the Tertiary gives the amount of centripetal acceleration that is required to orbit the COM with the same period as the Secondary.

Next I define a vector $\vec{r}$ for a radial vector from the COM to the Tertiary.

I will refer to 'tangential direction' as the direction perpendicular to the radial vector.

For an object at L4/L5: In the tangential direction:
At the precise location of L4/L5 the gravitational attractions from the primary and the secondary are balanced.

What if that object strays from the Lagrange point in tangential direction? Then the fact that gravity is an inverse square force comes into play. When straying towards the Primary, hence away from the Secondary (in tangential direction), the resultant attraction is unbalanced, now the resultant attraction tends to pull the Tertiary further away from the Lagrange point. Straying towards the Secondary: same thing; the tendency is to pull the Tertiary further away from the Lagrange point.

Combining effects in radial direction and tangential direction: in both directions, when the Tertiary strays from the Lagrange point the tendency is for the Tertiary to move further away from the Lagrange point. In effect L4 and L5 act as locations of gravitational repulsion. Since that is counterintuitive I repeat: repulsion.


Orbital mechanics

Essential to the mechanics of the Lagrange points L4 and L5 is orbital mechanics in general. When you are in orbit you always return to where you came from. This is obvious in the case of a circular orbit, perhaps a tad less obvious in the case of ellipse-shaped orbit. Circular orbit or ellipse-shaped orbit: after a full orbital period you arrive back at where you started from.

A Tertiary, in orbit close to L4/L5, is continuously repelled by the nearby Lagrange point. But no matter what happens, after a full period of revolution the orbital mechanics delivers the Tertiary back to the Lagrange point.

(Some time ago I created an orbital mechanics simulation specifically for the case of L4/L5 mechanics. Watching the simulation proceed was hilarious. I would place the Tertiary precisely on the Lagrange point, and it would orbit there for a while. But eventually the integration error of the simulation accumulates to a point where the Tertiary starts to stray. So then the Tertiary starts to slide down the gravitatonal potential. But the Tertiary doesn't get far; the orbital mechanics keeps bringing it back.)

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