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 The Lagrangian Points for a Planetary Orbit

A planet's orbit is determined by a balance between the gravitation attraction and the centrifugal force from its movement in an approximately circular orbit. For a particular planet there are five points that involve a balance involving the attraction of the Sun and the planet.

For illustration, consider the planet Jupiter.

There is a point between the Sun and Jupiter where a mass would be equally attracted to the Sun and Jupiter and therefore in balance. However, considering that a mass would be moving around the Sun with Jupiter there would be some degree of centrifugal force involved.

## Centripetal Acceleration and Centrifugal Force

An object moving in a circular orbit of radius r with a tangential velocity of v is undergoing a centripetal acceleration of

#### aC = v²/r

In terms of the orbital angular velocity ω, where v=ωr, the acceleration is

#### aC = ω²r

The force per unit mass required to keep the object in that circular orbit is called the centrifugal force and is equal to that the above centripetal acceleration.

If the radial force is due to the gravitation of a central body of mass M then

#### v²/r = ω²r = GM/r²

where G is the universal gravitational constant.

From the above it follows that

#### ω² = GM/r³

This is a formula which will be useful later.

There are three equilibrium points which lie on the line between the Sun and the planet. These are labeled L1, L2 and L3. L1 lies between the planet and the Sun.

## The Lagrangian Point L1

Consider a point at a distance r from the Sun. Let r1 be the distance of a planet, say Jupiter, from the Sun. The distance of the point to the planet is then (r1−r). If M0 is the mass of the Sun and M1 is the mass of the planet then the net inward gravitational attraction per unit mass at the point is

#### GM0/r² − GM1/(r1−r)²

At an equilibrium point this net attraction must balance the centrifugal force per unit mass. The equilibrium point must move with the same angular velocity as the planet so this force is ω1²r. But ω1²r is equal to GM0/r1³

Therefore

#### GM0/r² − GM1/(r1−r)² = GM0/r1³

The factor G can be cancelled out. Division by M0 then gives

#### 1/r² − (M1/M0)/(r1−r)² = r/r1³

Multiplying both sides of the above equation by r²(r1−r)² gives

#### (r1−r)² − r²(M1/M0) = (r³/r1³)(r1−r)² or with division by r1² (1−r/r1)² − (r/r1)² = (r/r1)³(1−r/r1)²

Letting r/r1=z and M1/M0=γ the above equation reduces to

#### (1−z)² − z²γ = z³((1−z)²) or, upon expansion of the binomials 1 − 2z + (1−γ)z2 = z3 −2z4 + z5and upon rearrangement z5 −2z4 + z3 − (1−γ)z2 + 2z − 1 = 0

This equation has five roots, all of them real. The relevant root is positive and less than one. When γ=0, z=1 satisfies the equation.

The value of γ for Jupiter and the Sun is 9.55×10−4. The solution for this value of γ is z=0.93332.

## The Lagrangian Point L2

The point L2 is on the line between the Sun and the planet, but beyond the planet. The net inward gravitational attraction per unit mass at the point is

#### GM0/r² + GM1/(r − r1)²

At an equilibrium point this net attraction must balance the centrifugal force per unit mass. The equilibrium point must move with the same angular velocity as the planet so this force is ω1²r. But ω1²r is equal to GM0/r1³

Therefore

#### GM0/r² + GM1/(r − r1)² = GM0/r1³

The factor G can be cancelled out. Division by M0 then gives

#### 1/r² + (M1/M0)/(r − r1)² = r/r1³

Multiplying both sides of the above equation by r²(r − r1)² gives

#### (r − r1)² + r²(M1/M0) = (r³/r1³)(r − r1)² or with division by r1² (r/r1 −1)² + (r/r1)² = (r/r1)³(r/r1 − 1)²

Letting r/r1=z and M1/M0=γ the above equation reduces to

#### (z−1)² + z²γ = z³(z−1)² or, upon expansion of the binomials (1+γ)z2 − 2z + 1 = z5 −2z4 + z3and upon rearrangement z5 −2z4 + z3 − (1+γ)z2 + 2z − 1 = 0

For the value of γ for Jupiter the root is about z=1.07

## The Lagrangian Point L3

The point L3 is on the line between the Sun and the planet, but on the other side of the Sun from the planet. The net inward gravitational attraction per unit mass at the point is

#### GM0/r² + GM1/(r + r1)²

At an equilibrium point this net attraction must balance the centrifugal force per unit mass. The equilibrium point must move with the same angular velocity as the planet so this force is ω1²r. But ω1²r is equal to GM0/r1³

Therefore

#### GM0/r² + GM1/(r + r1)² = GM0/r1³

The factor G can be cancelled out. Division by M0 then gives

#### 1/r² + (M1/M0)/(r + r1)² = r/r1³

Multiplying both sides of the above equation by r²(r + r1)² gives

#### (r + r1)² + r²(M1/M0) = (r³/r1³)(r + r1)² or with division by r1² (r/r1 +1)² + (r/r1)² = (r/r1)³(r/r1 + 1)²

Letting r/r1=z and M1/M0=γ the above equation reduces to

#### (z+1)² + z²γ = z³(z+1)² or, upon expansion of the binomials (1+γ)z2 + 2z + 1 = z5 +2z4 + z3and upon rearrangement z5 +2z4 + z3 − (1+γ)z2 − 2z − 1 = 0

For the value of γ for Jupiter the root is about z=1.00008.

## The Lagrangian Points L4 and L5

These points are much more difficult to analyze but the end result is that they are 60° ahead (L4) and 60° behind (L5) of the planet in its orbit. For Jupiter the Trojan (or Trojan and Greek) asteroids are found clustered around these points. A 1996 investigation found four hundred Trojan asteroids clustered around L4. Only 34 of these were numbered and catalogued.

(To be continued.)