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Central Forces

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Motion Under A Central Force

Consider the motion of a particle at P ( position vector #~r ) moving under the influence of a force #~F always directed towards the same point O (which we will adopt as the origin). The equation of motion is

~#F _ = _ ~m deriv2{#~r}

Then deriv2{#~r} is also parallel to ~#r, so _ ~#r # deriv2{#~r} _ = _ 0 . _ So

fract{d ( ~#r # deriv{#~r} ), d~t} _ = _ deriv{#~r} # deriv{#~r} + ~#r # deriv2{#~r} _ = _ 0 + 0 _ = _ 0

So _ _ _ _ _ _ ~#r # deriv{#~r} _ = _ ~{constant} _ =#: _ #~h

This means that ~#r and deriv{#~r} must always be in the plane perpendicular to #~h.

 

Motion In A Plane

We set up a system of rectangular coordinates such that the center of attraction O is the origin, and the ~x~y-plane coincides with the plane of motion (which contains ~#r and deriv{#~r}). The ~x-axis is the line from the origin O to the initial position .

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Let &theta. be the angle between #~r and the ~x-axis. &theta. changes with time of course.

We now introduce two unit vectors, #~p which is parallel to #~r, and #~q, which is perpendicular to #~r in the plane of motion ( i.e. the planeperpendicular to #~h ).

#~p _ = _ cos &theta. #~i + sin &theta. #~j

#~q _ = _ - sin &theta. #~i + cos &theta. #~j

 

It is easy to see that

deriv{#~p} _ = _ deriv{&theta.} #~q _ _ _ and _ _ _ deriv{#~q} _ = _ - deriv{&theta.} #~p

We have:

#~r _ = _ | ~#r | #~p _ = _ ~r #~p

deriv{#~r} _ = _ deriv{~r} #~p + ~r deriv{#~p} _ = _ deriv{~r} #~p + ~r deriv{&theta.} #~q

deriv2{#~r} _ = _ deriv2{~r} #~p + deriv{~r} deriv{#~p} + deriv{~r} deriv{&theta.} #~q + ~r deriv2{&theta.} #~q + ~r deriv{&theta.} deriv{#~q}

_ _ _ _ = _ ( deriv2{~r} - ~r ( deriv{&theta.} )^2 ) #~p _ + _ ( 2 deriv{~r} deriv{&theta.} + ~r deriv2{&theta.} ) #~q

Area Swept

We saw that

#~r # deriv{#~r} _ = _ ~{constant vector} _ =#: _ ~#h

In terms of #~p and #~q

~r #~p # ( deriv{~r} #~p + ~r deriv{&theta.} #~q ) _ = _ #~h _ => _ ~r^2 deriv{&theta.} ( #~p # #~q ) _ = _ #~h

and since #~p and #~q are unit vectors, and putting _ ~h _ #:= _ | #~h |

~r^2 deriv{&theta.} _ = _ ~h _ = _ ~{constant}

Let &delta.~A represent the area swept out in a small time interval &delta.~t. There will be a small change &delta.&theta. in the angle, so the area is approximately the area of the segment (see diagram right):

&delta.~A _ ~~ _ fract{~r^2&delta.&theta.,2} _ _ => _ _ fract{&delta.~A,&delta.~t} _ ~~ _ fract{~r^2,2} fract{&delta.&theta.,&delta.~t}

In the limit we have

fract{d~A,d~t} _ = _ fract{~r^2,2} fract{d&theta.,d~t}

deriv{~A} _ = _ ~h ./ 2 _ = _ ~{constant}

~A _ = _ ~~h ~t ./ 2 + ~{integration constant}

 

 

Orbital Equation

Now _ deriv2{#~r} is parallel to ~#p , i.e. so

deriv2{#~r} _ = _ - ~a ~#p + 0 #~q

where _ ~a _ #:= _ | #~F | ./ ~m _ = _ | deriv2{#~r} | _ (i.e. the acceleration or force per unit mass). _ But

deriv2{#~r} _ = _ ( deriv2{~r} - ~r ( deriv{&theta.} )^2 ) #~p _ + _ ( 2 deriv{~r} deriv{&theta.} + ~r deriv2{&theta.} ) #~q

so by equating coefficients we have:

deriv2{~r} - ~r ( deriv{&theta.} )^2 _ = _ - ~a _ _ _ _ _ (#A)

2 deriv{~r} deriv{&theta.} + ~r deriv2{&theta.} _ = _ 0 _ _ _ _ _ (#B)

So far the equations of motion involve ~r and &theta. and their derivatives with respect to time. We will now eliminate the time factor to derive an equation purely in polar coordinates.

Put _ ~u _ #:= _ ~r^{-1} , then _ deriv{&theta.} _ = _ ~u^2 ~h , _ and

deriv{~r} _ = _ fract{d ~u^{-1},d ~t} _ = _ - ~u^{-2} fract{d ~u,d ~t} _ = _ - ~u^{-2} fract{d ~u,d &theta.} fract{d &theta.,d ~t} _ = _ - ~u^{-2} fract{d ~u,d &theta.} ~u^2 ~h

so _ deriv{~r} _ = _ - ~h ( d ~u ./ d &theta. ) , _ and

deriv2{~r} _ = _ - ~h fract{d,d ~t}rndb{fract{d ~u,d &theta.}} _ = _ - ~h fract{d,d &theta.}rndb{fract{d ~u,d &theta.}} # fract{d &theta.,d ~t} _ = _ - ~u^2 ~h ^2 fract{d ^2 ~u,d &theta. ^2}

So substituting for deriv2{~r}, ~r, and deriv{&theta.} in (#A) above

- ~u^2 ~h ^2 fract{d ^2 ~u,d &theta. ^2} _ - _ ~u^{-1} ~u^4 ~h^2 _ = _ - ~a

fract{d ^2 ~u,d &theta. ^2} _ + _ ~u _ = _ fract{ _ ~a _ ,~u^2 ~h ^2}

This second degree differential equation links ~u ( = ~r^{-1} ) to the angle &theta., dependant on ~a = ~a(~t). It describes the ~#{orbit} taken by the paritcle and we will refer to it as the #~{orbital equation}.