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\begin{large}\[ L_{i} = mv_{i}b \]\end{large}

\begin{large}\[ E_{\rm total} = KE + PE \]\end{large}

\begin{large}\[ E_{\rm initial\;total} = \frac{1}{2}m{v_{i}}^{2} - \frac{mMG}{r} \]\end{large}

\begin{large}\[ L_{f} = mv_{f}R \]\end{large}

\begin{large}\[ E_{\rm final\;|

|!636px-Jupiter_showing_SL9_impact_sites.jpg!|
|Jupiter, showing the impact marks from fragments of Comet Shoemaker-Levy 9 in July 1994
Photo from Wikimedia Commons. Original by Hubbel Space Telescope Comet Team and NASA|


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{excerpt:hidden=true}Calculation of Effective Cross-Section of a planet with Gravity{excerpt}

Because [gravity] will act to pull meteors, comets, and other space debris toward a planet, the effective cross-section for a planet to capture an object is larger than its geometrical cross-section. What is the size of this effective cross-section in terms of the physical qualities of the planet and the situation? What features of the impacting body is it independent of?

h4. Solution

{toggle-cloak:id=sys} *System:*  {cloak:id=sys}[Point particle] subject to [gravity] but moving with constant [angular momentum].{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}[Gravity].{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod} XXXXX.{cloak}

{toggle-cloak:id=app} *Approach:*  

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{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}

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The Force Diagram of the Meteor approaching the Planet

|Force Diagram of Meteor and Planet|

The [single-axis torque] about the center of the planet is zero, because the force of [gravity] acts along the same direction as the radius *r*. About this point, therefore, [angular momentum] is conserved.

|Sketch showing Torque|

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{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

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We can calculate the effective Capture Radius *b* by setting this as the distance the path of the mass is displaced from one running through the center of the planet for the situation where the mass just touches the surface of the planet at its closest approach. The offset between the path of the object and the center of the planet is usually called the _impact parameter_, and we are looking for its critical value. Once we have it, we can turn it into a cross-section for capture by squaring it and multiplying by π . 

We begin by recognizing that both Energy and Angular Momentum (about the planet's center) are conserved. Some of the object's potential Energy_ is transformed into _Kinetic Energy_, but none is lost. And since there is no torque when the Angular Momentum is calculated about the center of the planet, the angular momentum must be conserved as well. (Both of these statements would not be true if some nonconservative, dissipative [Force] was present, but we are assuming motion through empty space, and a fall onto a surface with no atmosphere.)We write the conditions for the _initial_ state (when the mass is very far from the planet) with subscript *i* and for the _final_ state (when the mass comes down and just grazes the planet tangentially) by the subscript *f*.

{latex}\begin{large}\[ \vec{\tau} = \vec{r} X \vec{F} = rF = I_{\rm total} \alpha \]\end{large}{latex}


{latex}\begin{large}\[ I_{\rm total} = I_{\rm small} + I_{\rm large} \]\end{large}{latex}


{latex}\begin{large}\[ I = \frac{1}{2}m r^2 \] \end{large}{latex}



{latex}\begin{large}\[ I_{\rm total} = \frac{1}{2}(m r^2 + M R^2 )m{v_{f}}^{2} - \frac{mMG}{R} \]\end{large}{latex}





{latex}

\begin{large}\[ \omegamv_{\rm i}b = mv_{f}R \]\end{large}

\begin{large}\[ v_{f} = \frac{b}{R}v_{i} \]\end{large}

= \omega_{\rm i} + \alpha (t_{\rm f} - t_{\rm i}) \] \end{large}{latex}



{latex}\begin{large}\[ \theta_{\rm f} b = R\theta_sqrt{\rm1 i} + \omega_{\rm i} ( t_{\rm f} - t_{\rm i} ) + \frac{1}{2} \alpha ( t_{\rm f} - t_{\rm i} )^2  frac{2MG}{{v_{i}}^{2}}\frac{r_{i}-R}{r_{i}R}} \]\end{large}

{latex}


{latex}\begin{large}\[ \omega_{\rm f} b = R\alpha t_{\rm f}\]\end{large}{latex}

{latex}\begin{large}\[ \theta_{\rm f} = \frac{1}{2} \alpha {t_{\rm f}}^2sqrt{1 + \frac{2MG}{R{v_{i}}^{2}}} \]\end{large}{latex}
where
{latex}

\begin{large} \[ Cross \alpha; Section = \fracpi b^{rF}{I_{\rm total}} = \frac{2rF}{mr^2 + MR^2 }2} = \pi R^{2}\left(1 + \frac{2MG}{R{v_{i}}^{2}}\right) \]\end{large}{latex}

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