Lagrange points are equilibrium points of the earth-sun system (there are also those relative to the other planets).

The James Webb Space Telescope has just arrived at Lagrange point L2.

Let's try to calculate where it is, as a first approximation.

It is not difficult at all, just a few simple steps of algebra.

Knowing how to do this helps to have a deep understanding of what Lagrange points are and their importance in astronomy. There are five Lagrange points.

The first three are located on the straight line earth sun, two at the ends and the other in the middle, as shown in the figure.

A body located in a Lagrange point, subject to the attractions of the earth and the sun, will rotate in a circular motion with the same angular velocity as the earth, and therefore will always see the sun and the earth in the same relative position.

We know that the earth revolves around the sun with an elliptical motion which is almost a uniform circular mition, and we will consider it so, with a distance d from the sun:

d = 150 • 10⁶km

Let us put ourselves in a rotating reference frame, at the same angular velocity ω as the earth, which will therefore appear stationary, with the sun in the center.

(The angular velocity ω is given by the angle traveled divided by the time taken, if we take the revolution angle, the time is the period of revolution, for the earth it is one year.)

We know from the theory that in a rotating system, therefore not inertial, a centrifugal acceleration α, equal to ω²r, appears at every point, which tends to make a body move away radially from the center:

(a): α = ω²r

where r is the distance of the point under examination from the center, that is, from the sun.

(Even if we don't know this formula, we can quickly see that it is the size of an acceleration.

The speed ω is given by an angle, dimensionless, divided by a time.

Therefore ω²r has the dimensions of a length, given by r, divided by a time squared, given by ω², precisely the dimensions of an acceleration.)

In this reference system both the earth and the Lagrange points are fixed, their position does not change over time, it is the system that rotates en bloc.

The earth maintains (approximately) the constant distance d from the sun, so it must have zero total acceleration.

The acceleration of gravity g caused by the sun is radially directed towards the center (in the figure it starts from the earth and is directed towards the left, therefore it requires the minus sign), and is given by Newton's formula

(b): g = -GM/d²,

here G is the universal gravitational constant, and M is the mass of the sun.

The centrifugal acceleration is given by the formula (a), with r = d, (in the figure it starts from the earth and is directed to the right).

(a'):. α = ω²d

For equilibrium the sum of the two accelerations must be zero:

(1): g+α = 0,

-GM/d²+ω²d=0,

GM/d²=ω²d

At Lagrange's points there are two accelerations of gravity, caused by the sun and the earth, we will call them g¹ and g², and their sum with the centrifugal acceleration α must be zero.

Let's see how this happens for the Lagrange point L2.

L2 is located along the straight line sun earth, after the earth, with the sun on the opposite side, as in the figure.

We call x the distance of L2 from the earth, its distance r from the sun is then:

r = d+x

The acceleration g¹ caused by the sun at point L2 is:

g¹ = -GM/r²

The acceleration g² caused by the earth at point L2 is

g² = -Gm/x²

where m is the earth mass

The centrifugal acceleration at point L2 is

α = ω²r = ω²(d+x)

the total acceleration in all Lagrange points must be zero, therefore also in L2:

(2): g¹+g²+α = 0,

-GM/r²-Gm/x²+ω²r=0,

GM/(d+x)²+

+Gm/x² = ω²(d+x)

now a brief development of the computation of (d+x)²:

(3): (d+x)² = d²+2dx+x² ~

~ d²+2dx

in the last passage we have neglected the term x² compared to the other two, considering it much smaller, because we think that x is much smaller than d, since the mass of the sun is much larger than that of the earth.

Then this approximation will have to be confirmed by the final result. Furthermore

(3'): d²+2dx = d²(1+2x/d)

we insert the (3') in (2):

(4): GM/[d²(1+2x/d)]+

+Gm/x² = ω²(d+x)

now let's do another approximation.

If ε is a number much less than one, then

1-ε and 1 + ε are almost the inverse of each other.

In fact their product gives 1:

(1-ε)(1+ε) = 1-ε² ~ 1

the term ε² being negligible with respect to one.

In our case, x is very small compared to d, so 2x/d is much less than one, and we can call it ε:

ε = 2x/d:

(1 +2x/d)(1-2x/d) =

= 1-4x²/d² ~ 1

so 1-2x/d is almost the inverse of 1+2x/d, since their product is 1:

(5): 1/(1+2x/d) = 1-2x/d

we insert (5) in (4):

(6): GM/d²•(1-2x/d) +

+ Gm/x² = ω²(d+x),

GM/d² - 2GMx/d³ +

+ Gm/x² = ω²d + ω²x

thanks to (1) the first terms of both members of the last formula are equal, and we can simplify them, obtaining:

(6'): -2GMx/d³ +

+ Gm/x² = ω²x

again from (1) we obtain ω²:

(1'): ω² = GM/d³

we insert ω² of (1 ') into (6'), and developing the calculations and simplifying G we obtain x:

(7): -2GMx/d³ +

+ Gm/x² = GMx/d³,

Gm/x² = 3GMx/d³,

Gm = 3GMx³/d³,

x³ = d³(m/3M),

and finally, taking the cube root of both sides, we obtain the desired final result for the distance x of L2 from the earth:

(8): x = ³√(m/3M)•d = K•d

The distance x of the Lagrange point is proportional to the distance d of the earth from the sun.

Let's examine the proportionality factor K, which depends on the ratio between the masses of the earth m and the sun M, so this factor is small, as we expected, we calculate how much.

We insert in the cube root the values of the two masses:

m = 6 • 10²⁴kg

M = 2 • 10³⁰kg

(9): K = ³√(m/3M) =

= ³√(6•10²⁴/3•2•10³⁰) =

K =³√(1/10⁶) =1/100

We insert (9) into ( remembering the value of the distance d:

d = 150 • 10⁶km

(10): x = d/100 =

= 150•10⁶/100 =

x = 1.5•10⁶km

we found that the Lagrange point L2 is one and a half million km beyond the earth, on the opposite side of the sun.

We have also found that the distance x is much less than the distance d, as assumed before.

If any steps were not clear, feel free to ask in the comments.

The calculation of the Lagrange points L1 and L3 is similar, the distance x must be redefined, and the direction, and therefore the sign, of the gravitational accelerations g¹ and g² must be changed.

The calculation of the points L4 and L5 is more elaborate, it must be done in two dimensions, as these points are outside the straight line earth sun.

The point L1, for example, is to the left of the earth, so the acceleration of gravity g² due to the earth is positive, while g¹ remains negative.

With just this change of sign of g² the calculations are almost identical, and it turns out that x for L1 is identical to that just found for L2, only changed in sign, negative.

That is, in our approximation, the point L1 is also located one and a half million km from the earth, but on the other side of L2, in the figure on the left of the earth, between the earth and the sun.

We now know why the James Webb Space Telescope traveled one and a half million kilometers!

In the Lagrange point the accelerations cancel each other out, as can be seen from the starting formula (2), therefore also the total force applied to a body present there is zero, and the telescope does not have to spend energy to remain in this position, which it is of equilibrium.

It will actually orbit around point L2, with a minimum of energy to supply to the motion.

(This reasoning is valid in the rotating system, in the inertial system the reasoning is a little different.

There is no centrifugal force, and the sum of the two gravitational forces gives rise to the centripetal force that keeps the Lagrange point in rotation around the sun, in resonance with the Earth's rotation.

The final result, however, is the same, obviously, physics cannot depend on the reference system we have chosen, which nature does not know.)

Point L2 has an advantage over point L1.

In L2 the sun is covered by the earth which eclipses it, even if not completely, a circular solar ring appears.

In fact, the earth has a radius of almost 6400km, the sun has a radius of almost 700000km, almost 110 times greater.

Instead the ratio of the distances of the sun and the earth from the Lagrange point is smaller:

(d+x)/x = 101.

So in L2 the earth cannot totally eclipse the sun, however the L2 point is the best that the telescope can reach.

It has a full view of the sky in the opposite direction to the earth.

At point L1, on the other hand, it can see the sun from one direction and the earth from the opposite direction.

In reality, much more precise calculations were needed to send the telescope to L2, it must be considered that the earth actually travels in a slightly elliptical orbit, with an aphelion and a perihelion, and with it the Lagrange points.

The orbit of the moon must also be considered, the center of gravity of the earth-moon system must be considered.

But the moon has a mass 81 times smaller than the terrestrial one, at first it is negligible, and the calculation just made makes a first very clear idea of the importance of L2.

Unfortunately at that distance it will not be possible to intervene to repair or upgrade the telescope, as was done with the Hubble Telescope, much closer, so much care and checks were required before launching it.

Good job, James Webb Telescope, we know that you will make us amazed with your observations, what surprises you will tell us yet we do not know, certainly will not be inferior to the surprises that Hubble Telescope gave us, thanks in advance!

And thanks also to the Lagrange L2 point that is hosting you!

P. S.

Many are asking me if the concepts and calculations in this post on Lagrange points are applicable to other planets.

Of course!

On the masses M and m no one has applied the label with sun and earth written on it, as is done with the names on backpacks and suitcases!

The beauty of theoretical physics is that one can often make general an argument made in a particular case.

Wherever there are two bodies in rotation with each other and sufficiently isolated from the others, there are Lagrange points, and the formulas of this post are valid.

It is only necessary to be careful of the approximation made, of the mass M much greater than m, and therefore of x / d much less than one, that may no longer be valid.

Thus there are Lagrange points also for exoplanets, provided they are not too close to each other to influence each other, and even for two stars in rotation with each other, there are many.

Do we want to exaggerate?

They would also exist for two kids playing spinning in circles, as long as they were in space, far away from celestial bodies!

## Sunday, January 30, 2022

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