The Many-body Problem:
Many-body problem may be explained as follows: given at any time positions and velocities of three or more massive particles, moving under mutual gravitational fields, masses also being known, compute their positions and velocities for any other time. First precise formulation of many-body problem was provided by Sir Isaac Newton.
Many-body problem is very complicated particularly when shapes and internal constitutions of bodies have to be taken into consideration. Point-mass many-body problem has inspired several scientists. Even three-body problem is much more complicated than two-body problem. In three-body problem, every body is subjected to complex variable gravitational fields because of its attraction by other two. Consequently, close encounters with each of the other two bodies may be brought about. Each near-collision gives rise to entirely new kind of orbit.
It would need a general formulation of inconceivable complexity to explain all the results of all possible close encounters. Though, many general and helpful statements can easily be made concerning many-body problem. Such statements are embodied in ten known integrals of motion. Additionally, particular solutions were found by Lagrange. Such solutions exist only when certain relationships hold among initial conditions. The earth, the moon, and a space vehicle in earth-moon space comprise approximate example of the three-body problem.
Equations of Motion in Many-body Problem:
Consider n massive particles of masses mi , i = 1,2, ,n. Assume that their radius vectors from unaccelerated reference point O are, R→i whereas their mutual radius vectors are provided by r→ij where
r→ij = Rj→ - Ri→
Then, Newton's law of motion becomes
F = miR→i..
While Newton's law of gravitation becomes
F = G j=1Σn(mimj/r3ij)r→ij, j ≠ i, i = 1,2,...,n
Here G is the universal constant of gravitation. When these two laws are combined, you obtain
miRi.. = G j=1Σn(mimj/r3ij)r→ij, j ≠ i, i = 1,2,...,n
r→ij implies that vector between mi and mj is directed from mi to mj so that
r→ij = -r→ji
Equation represents required set of equations of motion in many-body problem.
The Ten known Integrals of the Motion:
Ten known integrals of motion comprise six integrals of centre-of-mass, three integrals of area and one integral of energy.
Integrals of Centre-of-Mass:
By summing equations, you get
j=1ΣnmiRi→.. = 0
Integrating, we get
j=1ΣnmiRi→. = a→
Where a→ is constant vector. Integrating once more, we get
j=1ΣnmiRi→. = a→t + b→
By definition, centre-of-mass of system has radius vector R provided by
MR→ = i=1ΣnmiRi→
Here M = i=1Σnmi
Therefore, equation becomes
MR→.i = a→ and equation becomes
MR→ = a→t + b→
Therefore, radius vector R→ of center-of-mass is provided by:
R→ = (a→t + b→)/M
As the velocity of center of mass is provided by above Relation illustrates that center-of-mass of system moves through space in straight line. Relation illustrates that center-of-mass of system moves through space with the stable speed. This brings you to conclusion that center-of-mass of the system moves through space with the stable velocity (that is with stable speed in straight line).
Constant of integration in equation is b→, while constant of integration in equation is a→. If equations are resolved with respect to set of three un accelerated rectangular axes through O, we get six constants of integration, namely ax, ay, az, bx, by and bz.
Three Integrals of Area:
Taking the vector product of Ri→ and R→..i for each of the set and summing we get
i=1ΣnmiRi→X R→i.. = GΣi=1nΣj=1(mimj/rij3)Ri→ x r→ij, j ≠ ii. But rij→ = Rj→ - Ri→
So that Ri→ x r→ij = R→I x (R→j - R→i) = R→I x R→j (as R→i x R→i = 0)
Therefore, right-hand side of equation reduces in pairs to zero, providing
j=1ΣnmiRi→xR→i.. = 0
Integrating the preceding equation, you obtain
j=1ΣnmiRi→xR→.i = C→
Equation defines that sum of angular momenta (i.e. moments of momenta) of masses in system is a constant. This constant vector C states a plane defined as invariable plane of Laplace. In planetary system, the invariable plane of Laplace is inclined at about 1.50 to plane of ecliptic. This plane lies between orbital planes of Jupiter and Saturn. If relation given above is resolved with respect to set of un accelerated rectangular axes through O, three integrals of area are attained, i.e.,
i=1Σnmi(xiy.i - yixi.) = C1,
i=1Σnmi(yiz. - ziyi.) = C2,
i=1Σnmi(zix.i - xizi.) = C3
Here C2 = C12 + C22 + C32,
Relation given above gives three more constants of integration, C1, C2, C3. Therefore, sums of angular momenta of n masses about each of the reference axes are constants.
The related constants of integration are nine, i.e., ax, ay, az, bx, by, bz, C1, C2 and C3. Next, observe tenth integral of motion, i.e., energy integral.
The Integral of Energy:
To get tenth and final constant, take scalar product of R.→i with equation in i and sum over all i. This leads to expression
i=1ΣnmiR.→i.R..i = GΣi=1nΣj=1n(mimj/r3ij).(R.i→.r→ij,) j≠i
Now R.→ir→ij = R.→i.(R→j - R→i),
While R.→j.r→ji = R.→ji(R→i - R→j).
By adding we get.
=-(R.→j - R.→i).(R→j - R→i)
Now, velocity of i th mass is vi, where
vi2 = R.→i.R.→i
Also u = 1/2G i=1Σn j=1Σn(mimj/rij) equation becomes
T - u = E Where T = 1/2 i=1Σnmivi2
Here, T is the kinetic energy of the system while -u is its potential energy.
Therefore, equation defines that total energy of system of n particles is the constant. This is tenth (and final) constant of integration. Systems of constant total energy are known as conservative systems
Consider more closely the function u stated by
u = 1/2G i=1Σn j=1Σn(mimj/rij), i≠j
This is a symmetrical function of all masses and mutual distances apart. Neither time nor particle's radius vectors from origin enter u explicitly. This is important as, if u didn't have these properties, it would have been practically impossible to derive ten integrals of motion. The first nine integrals result from property that u is invariant with respect to rotations of axes or transformations of the origin. The energy integral arises as u doesn't have time explicitly (though it is, of course, function of time through the r→ij).
If we introduce unit vectors iˆ, ˆj, and kˆ along the axes Ox, Oy and Oz, gradient of u is provided by
∇u = grad u=i^∂u/∂x + j^∂u/∂y + k^∂u/∂z
Then, for particle of mass mi, it is seen that, as
R‾..i = i^x..i + j^y..i + k^z..i,
Following relation should hold: miR‾..i = grad u. Therefore, equating coefficients of unit vectors, we get
mix..i =∂u/∂xi, miy..i = ∂u/∂yi, miz..i = ∂u/∂zi
Relations are equations of motion of particle of mass mi in rectangular coordinates. Consequently, u is known as force function as partial derivatives of u with respect to coordinates provide components of force acting on the particle.
Moment of Inertia and Force Function:
Assume I is a function defined by
I = Σi=1nmiRi→2
If you differentiate I twice with respect to time, you obtain
I.. = 2Σi=1nmiR‾..2i + 2 i=1ΣnmiRi‾.R‾i..
or I.. = 4T + 2Σi=1nR→i.∇u,
Where ∇u = miR→..i
Equation associates second time derivative of moment of inertia I to kinetic energy through the force function.
Transfer between Orbits:
It is frequently desirable for the satellite to transfer from one orbit to another. Consider the satellite in orbit about massive spherical body. In absence of perturbations by other masses, satellite moves in a central force field. Orbit of satellite remains conic section as long as motors are not fired. Though, if motors are fired, there will be changes in satellite orbit. These changes will, generally, influence all six elements of conic section. Consequently, satellite goes into completely new orbit (with six new elements of the conic section).
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