Sunday, 8 February 2015

newtonian mechanics - Elliptic Orbit Solution based on initial conditions



¨r=r|r|k|r|2


k here is a constant dependent on the gravitational constant, and the masses of the two objects. If I transform it into cartesian coordinates:


¨X(t)=kX(t)(X(t)2+Y(t)2)3/2


¨Y(t)=kY(t)(X(t)2+Y(t)2)3/2


I can not solve this system of equations. Perhaps it would require some special functions like the elliptic function etc. Maybe I should get rid of the time dependency and represent it as an implicit function but I do not know how. I realize that solving an elliptic orbit is know, I am curious about how one would solve it in the fundamental f=ma kind of way, without any other assumption.



Answer



First of all, it is best to solve the equation in polar coordinates rather than Cartesian coordinates: this works well with the fact that the force vector - and therefore the acceleration vector - is always pointing in a radial direction.


While it is possible to express the equations of motions as differential equations in time, it is not possible to obtain closed form solutions of the orbital position as a function of time, and so these time differential equations cannot be solved as they as. The strategy is to reformulate the time differential equations into "positional" differential equations, i.e. the derivative terms are differentiated with respect to the orbital angular position. Details for this are provided below.


Formulation of polar equation of motion in time


r(t) and θ(t) are the polar coordinates. It is useful to be able to relate the acceleration vector to the polar coordinates and their derivatives. The position vector can be represented as:



r(t)=[rcosθrsinθ]


The acceleration is obtain by double differentiation in time:


¨r(t)=[¨rcosθ2˙r˙θsinθr˙θ2cosθr¨θsinθ¨rsinθ+2˙r˙θcosθr˙θ2sinθ+r¨θcosθ]=(¨rr˙θ2)[rcosθrsinθ]+(2˙r˙θ+r¨θ)[rsinθrcosθ]


Note how ¨rr˙θ2 is the radial component of acceleration and 2˙r˙θ+r¨θ is the tangential component. However, we know that gravity only imparts a radial acceleration on the body, and so two simultaneous equations of motion are obtained:


¨rr˙θ2=kr2


2˙r˙θ+r¨θ=0


These equations cannot be immediately solved as they are, since both equations contain r and θ terms together. However, it can be shown that the second equation amounts to the conservation of angular momentum. Multiply the second equation by r to yield


2r˙r˙θ+r2¨θ=0


Noting that ddtr2=2r˙r, the equation can be re-expressed as


ddt(r2)˙θ+r2ddt(˙θ)=0ddt(r2˙θ)=0



This means that


r2˙θ=constant


We can represent this constant quantity with h=r2˙θ, and h is the specific angular momentum. Now we can substituted ˙θ=h/r2 into the first equation of motion, yielding


¨rh2r3+kr2=0


This is the polar equation of motion in time, but this form is still not sufficient in determining the shape of an orbit.


Obtaining the "positional" equation of motion


A useful step to perform is to defined a new quantity q=1r, and substitute this into the equation of motion. In the process, we can convert time derivatives of r(t) into positional derivatives of q(θ)


Note how ˙r can be expanded using the chain rule:


˙r=drdt=drdθdθdt=ddθ(1q)˙θ=1q2dqdθhq2=hdqdθ


Likewise, ¨r becomes



¨r=d˙rdt=d˙rdθdθdt=ddθ(hdqdθ)˙θ=h2q2d2qdθ2


Therefore, the equation of motion can be re-expressed as:


h2q2d2qdθ2h2q3+kq2=0


or


d2qdθ2+q=kh2


This equation can be solved, and is in fact a linear differential equation with a general solution:


q(θ)=Acosθ+Bsinθ+kh2


Solving the equation of motion


A and B are yet to be known, but can be determined with two boundary conditions. h is also unknown, and can be determined with a third boundary condition.


Let's say that r0 is the initial distance of the body from the orbited planet, u0 is the initial (outward) radial velocity, and v0 is the initial (anticlockwise) tangential velocity. With no loss of generality, we can assume the orbiting body is initially at θ=0.



The first boundary condition is that r=r0 at t=0. In terms of q, this becomes q=1r0 at θ=0.


The second boundary condition involves the radial velocity: ˙r=u0 at t=0. This becomes dq/dθ=u0/h at θ=0.


As for the final boundary condition, note that h=r2˙θ. Then h can be determined from the initial conditions, such that h=r0v0.


With these conditions, the expression for q becomes:


q(θ)=(1r0kr20v20)cosθu0r0v0sinθ+kr20v20


Then, since r=1/q:


r(θ)=r0(1kr0v20)cosθu0v0sinθ+kr0v20


This equation is the polar equation of the orbit, and is the equation of an ellipse for certain choices of initial conditions r0, u0 and v0. This equation will also account for parabolic and hyperbolic "orbits".


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