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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.
An object moving in a circular orbit of radius r with a tangential velocity of v is undergoing a centripetal acceleration of
In terms of the orbital angular velocity ω, where v=ωr, the acceleration is
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
where G is the universal gravitational constant.
From the above it follows that
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 L_{1}, L_{2} and L_{3}. L_{1} lies between the planet and the Sun.
Consider a point at a distance r from the Sun. Let r_{1} be the distance of a planet, say Jupiter, from the Sun. The distance of the point to the planet is then (r_{1}−r). If M_{0} is the mass of the Sun and M_{1} is the mass of the planet then the net inward gravitational attraction per unit mass at the point is
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 ω_{1}²r. But ω_{1}²r is equal to GM_{0}/r_{1}³
Therefore
The factor G can be cancelled out. Division by M_{0} then gives
Multiplying both sides of the above equation by r²(r_{1}−r)² gives
Letting r/r_{1}=z and M_{1}/M_{0}=γ the above equation reduces to
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 point L_{2} 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
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 ω_{1}²r. But ω_{1}²r is equal to GM_{0}/r_{1}³
Therefore
The factor G can be cancelled out. Division by M_{0} then gives
Multiplying both sides of the above equation by r²(r − r_{1})² gives
Letting r/r_{1}=z and M_{1}/M_{0}=γ the above equation reduces to
For the value of γ for Jupiter the root is about z=1.07
The point L_{3} 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
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 ω_{1}²r. But ω_{1}²r is equal to GM_{0}/r_{1}³
Therefore
The factor G can be cancelled out. Division by M_{0} then gives
Multiplying both sides of the above equation by r²(r + r_{1})² gives
Letting r/r_{1}=z and M_{1}/M_{0}=γ the above equation reduces to
For the value of γ for Jupiter the root is about z=1.00008.
These points are much more difficult to analyze but the end result is that they are 60° ahead (L_{4}) and 60° behind (L_{5}) 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 L_{4}. Only 34 of these were numbered and catalogued.
(To be continued.)
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