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Motion in Orbital Potential Field

Motion under inverse square force
The force experienced by a particle moving under the invrse force is
\(f(r)=\frac{-k}{r^2}\Rightarrow v(r)=\frac{-k}{r}\;\;\;\;\;\;[f(r)=\frac{-dv}{dr}]\)
and the equation of orbit is
\(\frac{d^2u}{d\theta^2}+u=\frac{-m}{l^2u^2}f(\frac 1u)\)
\(=\frac{-m}{l^2u^2}\times(\frac{-k}{r^2})=\frac{mk}{l^2}\)
\(=\frac{d^2u}{d\theta^2}+\biggl(u_\frac{Mk}{l^2}\biggr)=0\
Let \(u-\frac{mk}{l^2}=y\)
\(\Rightarrow\frac{du}{d\theta}=\frac{dy}{d\theta} and \frac{d^2y}{d\theta^2}=\frac{d^2u}{d\theta^2}\)
\(\therefore\) The equation of orbit reduces to
\(\frac{d^2y}{d\theta^2}+y=0\dotsm(1)\)
The solution of equation (1) can be expressed as
\(y=u'cos(\theta-\theta'\))
Here u' and \(\theta\)' are constants.
Now,
\(u=u'cos(\theta-\theta')+\frac{mk}{l^2}\)
or, \(u=\frac{mk}{l^2}+u'cos\theta\) here\(( \theta'=0\))
\(\Rightarrow \frac1r=\frac{mk}{l^2}+u'cos\theta\)
\(\Rightarrow r=\frac{1}{\frac{mk}{l^2}+u'cos\theta}\)
\(\Rightarrow r=\frac{\frac{l^2}{mk}}{1+\frac{u_1 l^2}{mk} cos\theta}\dotsm(2)\)
The equation (2) will give a conic (conic is the locus in which the ratio of distance from fixed point focus to the distance from fixed straight line (directrix) remain constant.

Fig:
Fig:

From the figure,
\(p=rcos\theta+d\dotsm(1)\)
Here\(\frac rd\)=constant \((\epsilon\))=eccentricity
Let, P=\(\epsilon\) p
\(\therefore \; \frac{p}{\epsilon}+rcos\theta\)
\(\Rightarrow r=\frac{P}{1+\epsilon cos\theta}\dotsm(3)\)
Comparing the equation (2) and (3)
\(p=\frac{l^2}{mk}\) and\( \epsilon=\frac{u'l^2}{mk}\)
From the figure
\(x=rcos\theta,y=rsin\theta\)
\(\therefore\) \(r=\sqrt{x^2+y^2}\)
=\(p-\epsilon x\)
\(\Rightarrow x^2+y^2=p^2-2p\epsilon x+\epsilon^2x^2\dotsm(4)\)
Special cases:
1. If \(\epsilon\)=0
The equation 4 reduces to \(x^2+y^2=p^2\dotsm(5)\)
It refers to a circle
2. If \(\epsilon=1\) then, \(y^2=p^2-2px\dotsm(6)\)
It refers to parabola.
3. If \(\epsilon\)<1, equation (4) reduces to
\(x^2+\frac{2p\epsilon}{1-\epsilon^2}+\frac{y^2}{1-\epsilon^2}=\frac{p^2}{1-\epsilon^2}\)
or,\(\biggl(x+\frac{p\epsilon}{(1-\epsilon^2)}\biggr)^2-\frac{p^2\epsilon^2}{(1-\epsilon^2)^2}+\frac{y^2}{(1-\epsilon^2)}=\frac{p^2}{(1-\epsilon^2)}\)
or, \(\bigg[x+\frac{p\epsilon}{(1-\epsilon^2}\biggr]^2\times\frac{(1-\epsilon^2)^2}{p^2\epsilon^2}-1+\frac{y^2}{1-\epsilon^2}\times\frac{(1-\epsilon^2)^2}{p^2\epsilon^2}=\frac{p^2}{(1-\epsilon^2)}\frac{(1-\epsilon^2)^2}{p^2\epsilon^2}\)
or, \(\frac{\biggl[x+\frac{p\epsilon}{(1-\epsilon^2)}\bigg]^2}{\frac{p^2}{(1-\epsilon^2)^2}}+\frac{\frac{y^2}{p^2}}{(1-\epsilon^2}=1\)
Let, \(\frac{p\epsilon}{1-\epsilon^2}=x_\circ\), \(\frac{p}{1-\epsilon}=a\) and \(\frac{p}{\sqrt{1-\epsilon^2}}=b\)
The equation becomes
\(\frac{(x+x_\circ)^2}{a^2}+\frac{y^2}{b^2}=1\dotsm(7)\)
The equation (7) refers to ellipse
4. If \(\epsilon\)>1, equation (4) reduces to
\(\frac{\biggl[x+\frac{p\epsilon}{(\epsilon^2-1)}\bigg]^2}{\frac{p^2}{(\epsilon^2-1)^2}}+\frac{\frac{y^2}{p^2}}{(\epsilon^2-1)}=1\dotsm(8)\)
The equation (8) refers to hyperbola.
The orbiting nature of particle can be explained with range of eccentricity \((\epsilon\)). The value of eccentricity obtained from \(\epsilon=\frac{u'l^2}{mk}\)
here,
l= Angular momentum
u'= Constant
m= mass
k= force constant
It is difficult to calculate the value of u' under our consideration. Therefore let us try to calculate eccentricity in terms of total energy of the system. We know, the total energy of particle moving under the inverse square force is ,
\(E=\frac12 M(\dot r^2+r^2\dot\theta^2)+vr\)
\(=\frac12 m(\dot r^2+r^2\dot\theta^2)-\frac kr\)
At minimum distance, \(r=r_{min}\Rightarrow \dot r=0\)
\(E=\frac12 m( r_{min})^2\times \frac{l^2}{m^2(r_{min})^4}-\frac{k}{r_{min}}\dotsm(9)\)
Also \(r=\frac{p}{1+ecos\theta}\) and at minimum distance.
\(r_{min}=\frac{p}{1+e}=\frac{l^2}{mk(1+e)}\dotsm(10)\)
Solving the equation (9) and (10)
\(E=\frac12 \frac{l^2}{m\frac{l^4}{m^2k^2(1+\epsilon)^2}}-\frac{k}{\frac{l^2}{mk(1+\epsilon)}}\)
=\(\frac{mk^2(1+\epsilon)^2}{2l^2}-\frac{mk^2}{l^2}(1+\epsilon)\)
\(=\frac{mk^2}{2l^2}[(1+\epsilon)^2-2(1+\epsilon)]\)
\(=\frac{mk^2}{2l^2}[\epsilon^2-1]\)
\(\Rightarrow \epsilon=\sqrt{1+\frac{2El^2}{mk^2}}\dotsm(11)\)
The equation (11) gives eccentricity in terms of total energy
if E=0, then \(\epsilon\)=1, it refers to a parabola
if E>0, then\(\epsilon\)>1, it refers to a hyperbola.
if E<0, then\(\epsilon\)<1, it refers to a ellipse.

Reference
Goldstein, Herbert.Classical Mechanics. Cambridge, MA: Addison-Wesley, 1950. Print.
Morin, David J.Introduction to Classical Mechanics: With Problems and Solutions. Cambridge, UK: Cambridge UP, 2008.
Kibble, T. W. B., and F. H. Berkshire.Classical Mechanics. London: Imperial College, 2004.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


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