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2.3

Rocket motion in gravitational field  (non-linear equation)

\centerline{
$m x'' = - {mgR^2 \over (R + x)^2}$}

Change to a system of first order equations:

Let $v = x'$.  Then $v' = x''$    
\u
(Instead of $x_1 = x$,  $x_2 = x' = x_1'$, $x_2' = x'' $

Hence have the system of equations

\centerline{$x' = v$}

\centerline{${dv \over dt} = - {gR^2 \over (R + x)^2}$}


Or equivalently in vector form


\centerline{${d \over dt}\left(\matrix{x \cr v}\right)  =
\left(\matrix{ v \cr - {gR^2 \over (R + x)^2} }\right) $}


Three variables, $x$, $v$, and $t$.  Need to get rid of one in order to solve.

\centerline{${dv \over dt} = {dv \over dx}{dx \over dt} = {dv \over dx}(v)$}


\centerline{$v{dv \over dx} = - {gR^2 \over (R + x)^2}$}

Separate variables and integrating:

\centerline{$\int v{dv} = -\int {gR^2 \over (R + x)^2} dx$}

Let $u = R + x$, $du = dx$

\centerline{${1\over 2}v^2 = - \int gR^2 u^{-2} dx$}

\centerline{${1\over 2}v^2 = gR^2 u^{-1}  + E_1$}


\centerline{${1\over 2}v^2 = {gR^2 \over R + x} + E_1$}

Kinetic energy:  $m{1\over 2}v^2 $

Potential energy:  ${mgR^2 \over R + x} $

Conservation of Energy:  {$m{1\over 2}v^2 - {mgR^2 \over R + x} = E$}

Note $E$ depends on initial conditions.  
\u
$E$ fixed $\leftrightarrow$ initial conditions fixed.

To study the solutions, can fix $E$ (or equivalently choose initial conditions) and graph

\centerline {$m{1\over 2}v^2 - {mgR^2 \over R + x} = E$} 
\vskip -7pt
in the $x, v$ - plane = $x, {dx 
\over dt}$- plane.

Can also let $v = {dx \over dt}$ and integrate again and solve for $x$ in terms of $t$:

{$m{1\over 2} {dx \over dt}^2 - {mgR^2 \over R + x} = E$}


{${dx \over dt}^2 = {2E \over m} + {2gR^2 \over R + x} $}

{${dx \over dt} = \sqrt{{2E \over m} + {2gR^2 \over R + x} }$}

{${dx \over  \sqrt{ {2E \over m} + {2gR^2 \over R + x} } } = dt$}

{$\int {dx \over  \sqrt{{2E \over m} + {2gR^2 \over R + x} }} = \int dt$},
Etc.

\hrule

3.8:  Mechanical Oscillator:  $mx'' + \gamma x' + kx = F_ext$

$m$ = mass, $\gamma$ = damping constant, $k$ = spring constant 

Note $k$ denotes the hardness of the spring 
\u
($F_s = -kL$).

Linear equation:  $x'' + x = 0$

No damping.  No external force.  Hence general solution:  $c_1 cos(t) + c_2 sin(t)$.

Non-linear equation:  $x'' + x + x^3 = 0$

Let $y = x'$, $y' = x''$: ~ $y'(t) + x(t) + [x(t)]^3 = 0$

System of equations

\centerline{$x' = y$}

\centerline{$y' =  - x - x^3$}

Or in vector notation

\centerline{${d \over dt}\left(\matrix{x \cr y}\right)  =  
\left(\matrix{ y \cr -x - x^3 }\right) $}


Three variables, $x$, $y$, and $t$.  Need to get rid of one in order to solve.

\centerline{${dy \over dt} = {dy \over dx}{dx \over dt} = {dv \over dx}(y)$}


\centerline{$y{dy \over dx} + x + x^3 = 0$} 

Separate variables and integrating:

\centerline{$y{dy \over dx} = -(x + x^3) $} 

\centerline{$\int y{dy} = -\int (x + x^3)dx $} 

\centerline{${1 \over 2}y^2 = -({1 \over 2}x^2 + {1 \over 4}x^4) + E $} 

\centerline{${1 \over 2}y^2 + {1 \over 2}x^2 + {1 \over 4}x^4 = E $} 





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