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==Mathematical solution of the differential equation ({{EquationNote|1}}) above== |
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For movement under any central force, i.e. a force parallel to '''r''', the [[specific relative angular momentum]] <math> \bold{H} = \bold{r} \times {\dot{\bold{r}}} </math> stays constant:<br/> |
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<math> \dot {\bold{H}} = \frac{d}{dt}\left(\bold{r} \times {\dot{\bold{r}}}\right) = \dot{\bold{r}} \times {\dot{\bold{r}}} + \bold{r} \times {\ddot{\bold{r}}} =\bold{0} + \bold{0} = \bold{0}</math><br/> |
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Since the cross product of the position vector and its velocity stays constant, they must lie in the same plane, orthogonal to <math> \bar{H} </math>. This implies the vector function is a [[plane curve]]. <br/> |
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Because the equation has symmetry around its origin, it is easier to solve in polar coordinates. However, it is important to note that equation ({{EquationNote|1}}) refers to linear acceleration <math> \left (\ddot{\bold{r}} \right )</math>, as opposed to angular <math> \left (\ddot{\theta} \right )</math> or radial <math> \left (\ddot{r} \right )</math> acceleration. Therefore, one must be cautious when transforming the equation. |
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Introducing a cartesian coordinate system <math>(\hat{\bold{x}} \ , \ \hat{\bold{y}})</math> and [[unit vector#Cylindrical coordinates|polar unit vectors]] <math>(\hat{\bold{r}} \ , \ \hat{\boldsymbol\theta})</math> in the plane orthogonal to <math> \bold{H} </math>:<br/> |
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<math> \hat{\bold{r}}=\cos(\theta)\hat{\bold{x}} + \sin(\theta)\hat{\bold{y}} </math><br/> |
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<math> \hat{\boldsymbol\theta}=-\sin(\theta)\hat{\bold{x}} + \cos(\theta)\hat{\bold{y}} </math><br/> |
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We can now rewrite the vector function <math>\bold{r}</math> and its derivatives as:<br/> |
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<math> \bold{r} =r ( \cos\theta \hat{x} + \sin \theta \hat{y}) = r\hat{\mathbf{r}} </math><br/> |
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<math> \dot{\bold{r}} = \dot r \hat {\mathbf r} + r \dot \theta \hat {\boldsymbol{\theta}} </math><br/> |
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<math> \ddot{\bold{r}} = (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r \ddot\theta + 2 \dot r \dot\theta) \hat{\boldsymbol\theta} </math><br/> |
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(see "[[Polar coordinates#Vector calculus]]"). Substituting these into ({{EquationNote|1}}), we find:<br/> |
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<math> (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r \ddot\theta + 2 \dot r \dot\theta) \hat{\boldsymbol\theta} = \left (-\frac{\mu}{r^2}\right )\hat{\mathbf{r}} + (0)\hat{\boldsymbol\theta}</math><br/> |
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This gives the non-ordinary polar differential equation: |
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# {{NumBlk|:|<math>\ddot{r} - r {\dot{\theta}}^2 = - \frac {\mu} {r^2}</math><br/>|{{EquationRef|2}}}} |
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In order to solve this equation, we must first eliminate all time derivatives. We find that: |
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<math> H = |\bold{r} \times {\dot{\bold{r}}}| = |(r\cos(\theta), r\sin(\theta), 0) \times (\dot{r}\cos(\theta)-r\sin(\theta)\dot{\theta}, \dot{r}\sin(\theta)+r\cos(\theta)\dot{\theta}, 0)| = |(0,0,r^2\dot\theta)| = r^2\dot\theta</math><br/> |
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# {{NumBlk|:|<math> \dot\theta = \frac{H}{r^2} </math>|{{EquationRef|3}}}} |
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Taking the time derivative of ({{EquationNote|3}}), we get |
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# {{NumBlk|:|<math>\ddot{\theta} = - \frac {2 \cdot H \cdot \dot{r}} {r^3}</math>|{{EquationRef|4}}}} |
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Equations ({{EquationNote|3}}) and ({{EquationNote|4}}) allow us to eliminate the time derivatives of <math>\theta</math>. In order to eliminate the time derivatives of <math>r</math>, we must use the chain rule to find appropriate substitutions: |
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# {{NumBlk|:|<math>\dot{r} = \frac {dr} {d\theta} \cdot \dot {\theta}</math>|{{EquationRef|5}}}} |
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# {{NumBlk|:|<math>\ddot{r} = \frac {d^2r} {d\theta^2} \cdot {\dot {\theta}}^2 + \frac {dr} {d\theta} \cdot \ddot {\theta}</math>|{{EquationRef|6}}}} |
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Using these four substitutions, all time derivatives in ({{EquationNote|2}}) can be eliminated, yielding an [[ordinary differential equation]] for <math>r</math> as function of <math>\theta\,</math>.<br/> |
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<math>\ddot{r} - r {\dot{\theta}}^2 = - \frac {\mu} {r^2}</math><br/> |
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<math>\frac {d^2r} {d\theta^2} \cdot {\dot {\theta}}^2 + \frac {dr} {d\theta} \cdot \ddot {\theta} - r {\dot{\theta}}^2 = - \frac {\mu} {r^2}</math><br/> |
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<math>\frac {d^2r} {d\theta^2} \cdot \left (\frac{H}{r^2} \right )^2 + \frac {dr} {d\theta} \cdot \left (- \frac {2 \cdot H \cdot \dot{r}} {r^3} \right ) - r\left (\frac{H}{r^2} \right )^2 = - \frac {\mu} {r^2}</math><br/> |
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# {{NumBlk|:|<math>\frac {H^2} {r^4} \cdot \left ( \frac{d^2 r} {d\theta ^2} - 2 \cdot \frac{\left (\frac {dr} {d\theta} \right ) ^2} |
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{r} - r\right )= - \frac {\mu} {r^2}</math>|{{EquationRef|7}}}} |
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The differential equation ({{EquationNote|7}}) can be solved analytically by the variable substitution |
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# {{NumBlk|:|<math>r=\frac{1} {s}</math>|{{EquationRef|8}}}} |
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Using the chain rule for differentiation one gets: |
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# {{NumBlk|:|<math>\frac {dr} {d\theta} = -\frac {1} {s^2} \cdot \frac {ds} {d\theta}</math>|{{EquationRef|9}}}} |
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# {{NumBlk|:|<math>\frac {d^2r} {d\theta^2} = \frac {2} {s^3} \cdot \left (\frac {ds} {d\theta}\right )^2 - |
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\frac {1} {s^2} \cdot \frac {d^2s} {d\theta^2}</math>|{{EquationRef|10}}}} |
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Using the expressions ({{EquationNote|10}}) and ({{EquationNote|9}}) for <math>\frac {d^2r} {d\theta^2}</math> and <math>\frac {dr} {d\theta}</math> |
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one gets |
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# {{NumBlk|:|<math>H^2 \cdot \left ( \frac {d^2s} {d\theta^2} + s \right ) = \mu</math>|{{EquationRef|11}}}} |
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with the general solution |
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# {{NumBlk|:|<math>s = \frac {\mu} {H^2} \cdot \left ( 1 + e \cdot \cos (\theta-\theta_0)\right )</math>|{{EquationRef|12}}}} |
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where ''e'' and <math>\theta_0\,</math> are constants of integration depending on the initial values for ''s'' and <math>\frac {ds} {d\theta}</math>. |
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Instead of using the constant of integration <math>\theta_0\,</math> explicitly one introduces the convention that the unit vectors <math> \hat{x} \ , \ \hat{y}</math> defining the coordinate system in the orbital plane are selected such that <math> \theta_0\,</math> takes the value zero and ''e'' is positive. This then means that <math>\theta\,</math> is zero at the point where <math> s</math> is maximal and therefore <math> r= \frac {1}{s}</math> is minimal. Defining the parameter p as <math> \frac {H^2}{\mu}</math> one has that<br/> |
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<math>r = \frac {1}{s} = \frac {p}{1 + e \cdot \cos \theta}</math><br/><br/> |
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===Alternate derivation=== |
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Another way to solve this equation without the use of polar differential equations is as follows:<br/>Define a unit vector <math>\bold{u}</math> such that <math>\bold{r} = r\bold{u}</math> and <math> \ddot{\bold{r}} = -\frac{\mu}{r^2}\bold{u}</math>. It follows that<br/> |
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<math>\bold{H} = \bold{r} \times \dot{\bold{r}} = r\bold{u} \times \frac{d}{dt}(r\bold{u}) = r\bold{u} \times (r\dot{\bold{u}}+\dot{r}\bold{u}) = r^2(\bold{u} \times \dot{\bold{u}}) + r\dot{r}(\bold{u} \times \bold{u}) = r^2\bold{u} \times \dot{\bold{u}}</math><br/> |
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Now consider<br/> |
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<math>\ddot{\bold{r}} \times \bold{H} = -\frac{\mu}{r^2}\bold{u} \times (r^2\bold{u} \times \dot{\bold{u}}) = -\mu\bold{u} \times (\bold{u} \times \dot{\bold{u}}) = -\mu[(\bold{u}\cdot\dot{\bold{u}})\bold{u}-(\bold{u}\cdot\bold{u})\dot{\bold{u}}]</math><br/> |
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(see [[Triple product#Vector triple product]]). Notice that<br/> |
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<math>\bold{u}\cdot\bold{u} = |\bold{u}|^2 = 1</math><br/> |
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<math>\bold{u}\cdot\dot{\bold{u}} = \frac{1}{2}(\bold{u}\cdot\dot{\bold{u}} + \dot{\bold{u}}\cdot\bold{u}) = \frac{1}{2}\frac{d}{dt}(\bold{u}\cdot\bold{u}) = 0 </math><br/> |
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Therefore<br/> |
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<math>\dot{\bold{r}}\times\bold{H}=\mu\dot{\bold{u}}</math><br/> |
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Integrating both sides:<br/> |
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<math>\dot{\bold{r}}\times\bold{H}=\mu\bold{u} + \bold{c}</math><br/> |
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Where '''c''' is a constant vector. Dotting this with '''r''' yields an interesting result:<br/> |
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<math> \bold{r}\cdot(\dot{\bold{r}}\times\bold{H})=\bold{r}\cdot(\mu\bold{u} + \bold{c}) = \mu\bold{r}\cdot\bold{u} + \bold{r}\cdot\bold{c} = \mu r(\bold{u}\cdot\bold{u})+rc\cos(\theta)=r(\mu + c\cos(\theta))</math><br/> |
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Where <math>\theta</math> is the angle between <math>\bar{r}</math> and <math>\bar{c}</math>. Solving for r: <br/> |
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<math> r = \frac{\bold{r}\cdot(\dot{\bold{r}}\times\bold{H})}{\mu + c\cos(\theta)} = \frac{(\bold{r}\times\dot{\bold{r}})\cdot\bold{H}}{\mu + c\cos(\theta)} = \frac{|\bold{H}|^2}{\mu + c\cos(\theta)}</math><br/> |
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Notice that <math>(r,\theta)</math> are effectively the polar coordinates of the vector function. Making the substitutions <math>p=\frac{|\bold{H}|^2}{\mu}</math> and <math>e=\frac{c}{\mu}</math>, we again arrive at the equation<br/> |
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# {{NumBlk|:|<math>r = \frac {p}{1 + e \cdot \cos \theta}</math>|{{EquationRef|13}}}} |
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This is the equation in polar coordinates for a [[conic section]] with origin in a focal point. The argument <math>\theta\,</math> is called "true anomaly".<br/> |
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===Properties of Trajectory Equation=== |
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For <math>e\ =\ 0\,</math> this is a circle with radius ''p''. |
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For <math>0\ < e\ <\ 1\,</math> this is an [[ellipse]] with |
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# {{NumBlk|:|<math>a = \frac {p}{1-e^2}</math>|{{EquationRef|14}}}} |
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# {{NumBlk|:|<math>b = \frac {p}{\sqrt{1-e^2}} = a \cdot \sqrt{1-e^2}</math>|{{EquationRef|15}}}} |
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For <math>e\ =\ 1\,</math> this is a [[parabola]] with focal length <math>\frac {p}{2}</math> |
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For <math>e\ >\ 1\,</math> this is a [[hyperbola]] with |
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# {{NumBlk|:|<math>a = \frac {p}{e^2-1}</math>|{{EquationRef|16}}}} |
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# {{NumBlk|:|<math>b = \frac {p}{\sqrt{e^2-1}} = a \cdot \sqrt{e^2-1}</math>|{{EquationRef|17}}}} |
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The following image illustrates an ellipse (red), a parabola (green) and a hyperbola (blue) |
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[[File:Kepler orbits.svg|thumb|An elliptic Kepler orbit with an eccentricity of 0.7, a parabolic Kepler orbit and a hyperbolic Kepler orbit with an eccentricity of 1.3. The distance to the focal point is a function of the polar angle relative to the horizontal line as given by the equation ({{EquationNote|13}})]] |
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The point on the horizontal line going out to the right from the focal point is the point with <math>\theta = 0\,</math> for which the distance to the focus takes the minimal value <math>\frac {p}{1 + e}</math>, the pericentre. For the ellipse there is also an apocentre for which the distance to the focus takes the maximal value <math>\frac {p}{1 - e}</math>. For the hyperbola the range for <math>\theta\,</math> is |
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:<math>\left [ -\cos^{-1}\left(-\frac{1}{e}\right) < \theta < \cos^{-1}\left(-\frac{1}{e}\right)\right ]</math> |
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and for a parobola the range is |
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:<math>\left [ -\pi < \theta < \pi \right ]</math> |
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Using the chain rule for differentiation ({{EquationNote|5}}), the equation ({{EquationNote|2}}) and the definition of ''p'' as <math>\frac {H^2}{\mu}</math> one gets that the radial velocity component is |
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# {{NumBlk|:|<math>V_r = \dot{r} = \frac {H}{p} \cdot e \cdot \sin \theta = \sqrt{\frac {\mu}{p}} \cdot e \cdot \sin \theta</math>|{{EquationRef|18}}}} |
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and that the tangential component (velocity component perpendicular to <math>V_r</math>) is |
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# {{NumBlk|:|<math>V_t = r \cdot \dot{\theta} = \frac {H}{r} = \sqrt{\frac {\mu}{p}} \cdot (1 + e \cdot \cos \theta)</math>|{{EquationRef|19}}}} |
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The connection between the polar argument <math>\theta\,</math> and time ''t'' is slightly different for elliptic and hyperbolic orbits. |
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For an elliptic orbit one switches to the "[[eccentric anomaly]]" ''E'' for which |
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# {{NumBlk|:|<math>x = a \cdot (\cos E -e)</math>|{{EquationRef|20}}}} |
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# {{NumBlk|:|<math>y = b \cdot \sin E</math>|{{EquationRef|21}}}} |
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and consequently |
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# {{NumBlk|:|<math>\dot{x} = -a \cdot \sin E \cdot \dot{E}</math>|{{EquationRef|22}}}} |
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# {{NumBlk|:|<math>\dot{y} = b \cdot \cos E \cdot \dot{E}</math>|{{EquationRef|23}}}} |
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and the angular momentum ''H'' is |
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# {{NumBlk|:|<math>H = x \cdot \dot{y} - y \cdot \dot{x}=a \cdot b \cdot ( 1 - e \cdot \cos E) \cdot \dot{E}</math>|{{EquationRef|24}}}} |
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Integrating with respect to time ''t'' one gets |
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# {{NumBlk|:|<math>H \cdot t = a \cdot b \cdot ( E - e \cdot \sin E)</math>|{{EquationRef|25}}}} |
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under the assumption that time <math>t=0</math> is selected such that the integration constant is zero. |
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As by definition of ''p'' one has |
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# {{NumBlk|:|<math>H = \sqrt{\mu \cdot p}</math>|{{EquationRef|26}}}} |
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this can be written |
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# {{NumBlk|:|<math>t = a \cdot \sqrt{\frac{a} {\mu}} ( E - e \cdot \sin E)</math>|{{EquationRef|27}}}} |
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For a hyperbolic orbit one uses the [[hyperbolic functions]] for the parameterisation |
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# {{NumBlk|:|<math>x = a \cdot (e - \cosh E)</math>|{{EquationRef|28}}}} |
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# {{NumBlk|:|<math>y = b \cdot \sinh E</math>|{{EquationRef|29}}}} |
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for which one has |
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# {{NumBlk|:|<math>\dot{x} = -a \cdot \sinh E \cdot \dot{E}</math>|{{EquationRef|30}}}} |
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# {{NumBlk|:|<math>\dot{y} = b \cdot \cosh E \cdot \dot{E}</math>|{{EquationRef|31}}}} |
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and the angular momentum ''H'' is |
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# {{NumBlk|:|<math>H = x \cdot \dot{y} - y \cdot \dot{x}=a \cdot b \cdot ( e \cdot \cosh E-1) \cdot \dot{E}</math>|{{EquationRef|32}}}} |
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Integrating with respect to time ''t'' one gets |
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# {{NumBlk|:|<math>H \cdot t= a \cdot b \cdot ( e \cdot \sinh E-E)</math>|{{EquationRef|33}}}} |
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i.e. |
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# {{NumBlk|:|<math>t = a \cdot \sqrt{\frac{a} {\mu}} (e \cdot \sinh E-E)</math>|{{EquationRef|34}}}} |
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To find what time t that corresponds to a certain true anomaly <math>\theta\,</math> one computes corresponding parameter ''E'' connected to time with relation ({{EquationNote|27}}) for an elliptic and with relation ({{EquationNote|34}}) for a hyperbolic orbit. |
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Note that the relations ({{EquationNote|27}}) and ({{EquationNote|34}}) define a mapping between the ranges |
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:<math>\left [ -\infin < t < \infin\right ] \longleftrightarrow \left [-\infin < E < \infin \right ] </math> |
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=Planets= |
=Planets= |
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Revision as of 16:10, 26 July 2013
 | This article possibly contains original research. |
Planets
Mercury
I feel that there needs to be a table of elongations of Mercury. I have occasionally frustrated by the fact that there is no page for this data available on wikipedia, for some reason the page Aspects of Mercury was deleted some time ago. Also noteworthy is that Aspects of Venus does not have any sources listed. I suspect that the page on Mercury had a similar problem.
The page on Elongation (astronomy) has links to fourmilab's Mercury Chaser app as well as the Heavens Above site, which provide data relating to elongations and conjunctions. Now I'm questioning if its even worth it.
Some of this info could probably be obtained from http://ssd.jpl.nasa.gov HORIZONS system. Unfortunately, they don't produce tables of elongations and conjunctions. For some reason, published lists seem to be hard to come by. Perhaps an astronomical almanac?
Because Earth and Venus both have very low eccentricities, the elongations of Venus are generally around the same angular separation. Mercury however has the largest eccentricity of the terrestrial planets. Thus the angle of elongation varies significantly between occurrences.
Western Elongations
Mercury is west of the sun so it is visible in the morning, immediately before sunrise. In the northern hemisphere at sunrise, during the months of August through October, the ecliptic is at its steepest with regards to the local horizon; this means that Mercury will rise significantly earlier than the sun, allowing easier observations. In February through April, the ecliptic is much closer to the horizon; the time between Mercury rise and sunrise will be much shorter. In the southern hemisphere, the effect is the opposite: favorable observations occur in February through April, while unfavorable elongations occur in August through October.
(source for that besides "I looked at Stellarium (computer program)")
| Date | Elongation Angle | Apparent Magnitude |
|---|---|---|
| 2012 Aug 16 | 18.7°W | +0.2 |
| 2012 Dec 4 | 20.6°W | -0.3 |
| 2013 Mar 31 | 27.8°W | +0.5 |
| 2013 Jul 30 | 19.6°W | +0.4 |
| 2013 Nov 18 | 19.5°W | -0.3 |
| 2014 Mar 14 | 27.6°W | +0.4 |
| 2014 Jul 12 | 20.9°W | +0.6 |
| 2014 Nov 1 | 18.7°W | -0.3 |
| 2015 Feb 24 | 26.7°W | +0.3 |
| 2015 Jun 24 | 22.5°W | +0.7 |
| 2015 Oct 16 | 18.1°W | -0.3 |
Eastern Elongations
Eastern elongation occurs when a body is farthest East from the sun as viewed from Earth, meaning it will set after the sun. As with western elongation, there are favorable and unfavorable observing seasons. For the northern hemisphere, August through October are favorable, February through April are unfavorable. Again, the opposite conditions apply for the southern hemisphere.
| Date of Elongation | Elongation Angle | Apparent Magnitude |
|---|---|---|
| 2012 Oct 26 | 24.1°E | +0.1 |
| 2013 Feb 16 | 18.1°E | -0.2 |
| 2013 Jun 12 | 24.3°E | +0.7 |
| 2013 Oct 9 | 25.3°E | +0.2 |
| 2014 Jan 31 | 18.4°E | -0.3 |
| 2014 May 25 | 22.7°E | +0.7 |
| 2014 Sep 21 | 26.4°E | +0.3 |
| 2015 Jan 14 | 18.9°E | -0.4 |
| 2015 May 7 | 21.2°E | +0.5 |
| 2015 Sep 4 | 27.1°E | +0.5 |
| 2015 Dec 29 | 19.7°E | -0.3 |
Venus
This is from the page on aspects of Venus as a template for Aspects of Mercury
The table contains special positions of Venus until 2021.
| Greatest eastern elongation | Greatest brilliancy | Stationary, then retrograde | Inferior conjunction | Stationary, then prograde | Greatest brilliancy | Greatest western elongation | Superior conjunction |
|---|---|---|---|---|---|---|---|
| March 29, 2004 - 46° | May 3, 2004 | May 18, 2004 | June 8, 2004 | June 29, 2004 | July 13, 2004 | August 17, 2004 - 45.8° | March 31, 2005 |
| November 3, 2005 - 47.1° | December 12, 2005 | December 23, 2005 | January 13, 2006 | February 3, 2006 | February 14, 2006 | March 25, 2006 - 46.5° | October 27, 2006 |
| June 9, 2007 - 45.4° | July 14, 2007 | July 25, 2007 | August 18, 2007 | September 7, 2007 | September 23, 2007 | October 28, 2007 - 46.5° | June 9, 2008 |
| January 14, 2009 - 47.1° | February 20, 2009 | March 5, 2009 | March 27, 2009 | April 15, 2009 | April 29, 2009 | June 5, 2009 - 45.9° | January 11, 2010 |
| August 20, 2010 - 46° | September 27, 2010 | October 7, 2010 | October 29, 2010 | November 16, 2010 | December 2, 2010 | January 8, 2011 - 47° | August 16, 2011 |
| March 27, 2012 - 46° | April 30, 2012 | May 15, 2012 | June 6, 2012 | June 27, 2012 | July 10, 2012 | August 15, 2012 - 45.8° | March 28, 2013 |
| November 1, 2013 - 47.1° | December 10, 2013 | December 20, 2013 | January 11, 2014 | January 31, 2014 | February 11, 2014 | March 22, 2014 - 46.6° | October 25, 2014 |
| June 6, 2015 - 45.4° | July 12, 2015 | July 23, 2015 | August 15, 2015 | September 5, 2015 | September 20, 2015 | October 26, 2015 - 46.4° | June 6, 2016 |
| January 12, 2017 - 47.1° | February 18, 2017 | March 2, 2017 | March 25, 2017 | April 12, 2017 | April 26, 2017 | June 3, 2017 - 45.9° | January 9, 2018 |
| August 17, 2018 - 45.9° | September 25, 2018 | October 5, 2018 | October 26, 2018 | November 24, 2018 | November 30, 2018 | January 6, 2019 - 47° | August 14, 2019 |
| March 24, 2020 - 46.1° | April 28, 2020 | May 13, 2020 | June 3, 2020 | June 24, 2020 | July 8, 2020 | August 13, 2020 - 45.8° | March 26, 2021 |
Note: Greatest brilliancy is often confused with "maximum brightness". Although they are related, they are not quite the same thing. The "greatest brilliancy" is really a geometric maximum: it occurs when the apparent area of the sunlit part of Venus is greatest. Only if the luminance of Venus' apparent surface would be constant (i.e. the same at every point and at every phase) would the "greatest brilliancy" of Venus coincide with its maximum brightness. However, the reflection of sunlight on Venus more closely follows Lambert's law, which means that the maximum brightness occurs at a somewhat larger phase of Venus than its greatest brilliancy.