Celestial mechanics


2 - Body Problem

If vectors are written without vector notation it represents the magnitude of the vector.


Time Derivatives are represented by a “.” above them
r˙=drdt
r¨=d2rdt2



Now, Equation of Motion
F1=m1r1˙̇(1)
F1=Gm1m2|r21|2r21^=Gm1m2|r21|3r21=(Gm1m2|r12|3)r12
Putting this value in (1) gives
r1¨=(Gm2r123)r12(2)
Similarly
r2¨=(Gm1r213)r21(3)
From (2) and (3)
m1r1¨+m2r2¨=0
Double Integrating
m1r1+m2r2=ct+d

If we divide by

m1+m2
we get
rCM
Since it is a linear function of t we can say that C.O.M. travels in a straight line.


Subtracting (3) from (2). We get position vector of first body w.r.t. second body.
r12¨=r1¨r2¨=(Gr123)(m2r12m1r21)
Represent r12=r21=r , |r12|=r and μ=G(m1+m2)
r¨+μr3r=0




Take cross product with r .
Gives
r¨×r=0
Also ddr(r˙×r)=r¨×r .
Therefore Angular momentum per unit mass is constant i.e. r˙×r=h
Where h is a constant. Also Motion is planar since orbital plane is fixed. Polar coordinates can be used.
Some Points
r=rr^ r˙=r˙r^+rr^˙ r^˙=θ˙θ^ θ^˙=θ˙r^
This gives r˙=r˙r^+rθ˙θ^
r¨=(r¨rθ˙2)r^+(2r˙θ˙+rθ¨)θ^
Now putting in above derived equation
r¨+μr^r2=0
r¨=(μr2)r^
(r¨rθ˙2)r^+(2r˙θ˙+rθ¨)θ^=(μr2)r^
Comparing
r¨rθ˙2=(μr2)......(4)
2r˙θ˙+rθ¨=0......(5)
in (5), Multiply both sides by r
2rr˙θ˙+r2θ¨=0
d(r2θ˙)dt=0
r2θ˙=h
h is same constant that we got earlier. Just writing it in different way.
Now for (4)
Substitute r=1/u
r˙=1u2θ˙(dudθ)
r˙=r2θ˙(dudθ)
from earlier r2θ˙=h
r˙=h(dudθ)
r¨=hθ˙(d2udθ2)
Now (4) becomes
h\dot{\theta}\left(\frac{d^2u}{d{\theta}^2}\right)-\frac^2}{u}=-\mu u^2
Simplyfying and using θ˙=u2h. We get
d2udθ2+u=μh2
Solving This gives
u=μh2(1+cos(θω¯))
r=h2μ1+cos(θω¯)
r=p1+ecos(θω)=p1+ecosf
Here e is eccentricity, f is true anamoly or angle measured from major axis from closest distance of approach (Pericentre).
p=h2μ,
p represents Semi-Latus Rectum of the orbit.
Centre is position of sun/star.

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Planets have elliptical orbits because all other paths lead planets to be thrown away.

Eccentricity Shape Semi-Latus Rectum
0 Circle h2μ
0<e<1 Ellipse a(1e2)
0<e<1 Parabola 2q
e>1 Hyperbola a(e21)
     

Energy Relations determining orbit

T.E.>0    Hyperbola    Unbound >
T.E.=0    Parabola    Unbound
T.E.<0    Circle/Ellipse    Bound

Analysis for Ellipse

rp=a(1e)ra=a(1+e)

Parametric Equation

r=a(1e2)1+ecosfx=rcosfy=rsinf
Area of ellipse % <![CDATA[A=\pi ab$ &amp $\frac{d}{dt}(A)=\frac{1}{2}h %]]>. if T is orbital period.
πabT=12h
Also
a(1e2)=h2μh2=aμ(1e2)
From above equations. We get
4π2a2b2T2=aμ(1e2)T2=(4π2μ)a3
This proves Kepler’s 3rd Law in elliptical orbit and hence can be extended to other orbital shapes.
Let ms and mp denote mass of sun and planet respectively.T2=(4π2G(ms+mp))a3
Also $m_s+m_p\approx m_s$. So for mass of sun.
ms4π2a3GT2



Starting again with our earlier derived equation
r¨+μr3r=0 Take dot product with r˙
r¨r˙+μrr˙r3=0
Substituting r=rr^$and$r˙=r˙r^+rθ˙θ^
r¨r˙+μ(rr^)(r˙r^+rθ˙θ^)r3=0
r¨r˙+μr˙r2=0 This v can be written as
ddt(12|r˙|2μr)=0

Now
r˙=v

ddt(v22μr)=0
v22μr=c

Total Energy is constant

Now finding that constant
We know $h$ (Angular Momentum) and Energy are constant at apicentre and pericentre
h=rava=rpvp. This gives vp=rarpva
va22μra=vp22μrp

Energy Conservation

va22(ra2rp21)=μ(1rp1ra)
va22=μrp+ra(rpra)

Now
2a=rp+ra

va22=μ2a(2arara)=μra(1ra2a)
Total Energy=va22+μra=μ2a

Velocity in elliptical orbit

v22+μr=μ2a
v2=μ(2r1a)
v=μ(2r1a)
Mean motion (n)=2πT. We get
μ=a3n2. Putting this value we get
vp=na1+e1e
va=na1e1+e

Energy for Ellipse

T.E.=μ2a

Energy for Parabola

T.E.=0

Energy for Hyperbola

T.E.=μ2a



3 - Body Problem

Analysis of 3 Body problem is not very simple due to addition of various variables due to 3rd body.
3 body problem can’t be solved completely.

Restricted 3 Body problem

Let one of the masses is infinitesimal small. m30
Now we have to solve equations only for m3. Because for m1 & m2, it is like 2 body problem
m3r3¨=Gm1m3r31r313Gm2m3r32r323
r3¨=Gm1r31r313Gm2r32r323
Using Cartesian Coordinates
x¨=Gm1(xx1)(xx1)2+(yy1)2+(zz1)2
y¨=Gm1(yy1)(xx1)2+(yy1)2+(zz1)2
z¨=Gm1(zz1)(xx1)2+(yy1)2+(zz1)2


Circular Restricted 3 body problem

Suppose that the first two masses, m1 and m2, execute circular orbits about their common center of mass.

Let us define a Cartesian coordinate system (ξ,η,ζ) in an inertial reference frame whose origin coincides with the center of mass, C, of the two orbiting masses. Let the orbital plane of these masses coincide with the ξ$$η plane, and let them both lie on the ξ-axis at time t=0–see Figure. Suppose that R is the constant distance between the two orbiting masses, r1 the constant distance between mass m1 and the origin, and r2 the constant distance between mass m2 and the origin. Moreover, let ω be the constant orbital angular velocity.

(xyz)=(cosωtsinωt0sinωtcosωt0001)(ξηζ)
Differentiate it twice put it in cartesian equations of motion of $m_3$. We get
ξ¨2ωη=Uξ(7)
η¨+2ωξ=Uη(8)
ζ¨=Uζ(9)

(7)×ξ˙+(8)×η˙+(9)×ζ˙

ξ˙ξ¨+η˙η¨+ζ˙ζ¨=ξ˙Uξ+η˙Uη+ζ˙Uζ(10)
if U=U(ξ,η,ζ)
dUdt=Uξξt+Uηηt+Uζζt
It is R.H.S. of (10). Also
ddtξ˙2=2ξ˙ξ¨
eqn $(10)$ becomes
12ddt(ξ˙2+η˙2+ζ˙2)=dUdt
We know ${\dot{\xi}}^2+{\dot{\eta}}^2+{\dot{\zeta}}^2=v^2$.
By Jacobi Integral (External Link) - For Extra Information
v2=2UC
For v20$,$2UC. This gives
U=ω22(ξ2+η2)+μ1r1+μ2r2
If I term is large
ξ2+η22Cω2
Path would be enclosing m1 and m2. On changing C we get different Lagrangian Points



LAGRANGIAN POINTS

Wikipedia Article

In celestial mechanics, the Lagrangian points; also Lagrange points, L-points, or libration points) are positions in an orbital configuration of two large bodies where a small object affected only by gravity can maintain a stable position relative to the two large bodies. The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to orbit with them. There are five such points, labeled L1 to L5, all in the orbital plane of the two large bodies. The first three are on the line connecting the two large bodies and the last two, L4 and L5, each form an equilateral triangle with the two large bodies. The two latter points are stable, which implies that objects can orbit around them in a rotating coordinate system tied to the two large bodies.L1, L2, L3 are unstable

We have considered

m2m1