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Mathematical solution of the differential equation (1) above

For movement under any central force, i.e. a force parallel to r, the specific relative angular momentum ${\mathbf {H}}={\mathbf {r}}\times {\dot {\mathbf {r}}}$ stays constant:
${\dot {\mathbf {H}}}={\frac {d}{dt}}\left({\mathbf {r}}\times {\dot {\mathbf {r}}}\right)={\dot {\mathbf {r}}}\times {\dot {\mathbf {r}}}+{\mathbf {r}}\times {\ddot {\mathbf {r}}}={\mathbf {0}}+{\mathbf {0}}={\mathbf {0}}$

Since the cross product of the position vector and its velocity stays constant, they must lie in the same plane, orthogonal to ${\bar {H}}$. This implies the vector function is a plane curve.
Because the equation has symmetry around its origin, it is easier to solve in polar coordinates. However, it is important to note that equation (1) refers to linear acceleration $\left({\ddot {\mathbf {r}}}\right)$, as opposed to angular $\left({\ddot {\theta }}\right)$ or radial $\left({\ddot {r}}\right)$ acceleration. Therefore, one must be cautious when transforming the equation. Introducing a cartesian coordinate system $({\hat {\mathbf {x}}}\ ,\ {\hat {\mathbf {y}}})$ and polar unit vectors $({\hat {\mathbf {r}}}\ ,\ {\hat {\boldsymbol {\theta }}})$ in the plane orthogonal to ${\mathbf {H}}$:

${\hat {\mathbf {r}}}=\cos(\theta ){\hat {\mathbf {x}}}+\sin(\theta ){\hat {\mathbf {y}}}$
${\hat {\boldsymbol {\theta }}}=-\sin(\theta ){\hat {\mathbf {x}}}+\cos(\theta ){\hat {\mathbf {y}}}$

We can now rewrite the vector function ${\mathbf {r}}$ and its derivatives as:

${\mathbf {r}}=r(\cos \theta {\hat {x}}+\sin \theta {\hat {y}})=r{\hat {\mathbf {r} }}$

${\dot {\mathbf {r}}}={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\theta }}{\hat {\boldsymbol {\theta }}}$

${\ddot {\mathbf {r}}}=({\ddot {r}}-r{\dot {\theta }}^{2}){\hat {\mathbf {r} }}+(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}){\hat {\boldsymbol {\theta }}}$

(see "Polar coordinates#Vector calculus"). Substituting these into (1), we find:

$({\ddot {r}}-r{\dot {\theta }}^{2}){\hat {\mathbf {r} }}+(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}){\hat {\boldsymbol {\theta }}}=\left(-{\frac {\mu }{r^{2}}}\right){\hat {\mathbf {r} }}+(0){\hat {\boldsymbol {\theta }}}$

This gives the non-ordinary polar differential equation:

1. ${\ddot {r}}-r{\dot {\theta }}^{2}=-{\frac {\mu }{r^{2}}}$
   (2)

In order to solve this equation, we must first eliminate all time derivatives. We find that:

$H=|{\mathbf {r}}\times {\dot {\mathbf {r}}}|=|(r\cos(\theta ),r\sin(\theta ),0)\times ({\dot {r}}\cos(\theta )-r\sin(\theta ){\dot {\theta }},{\dot {r}}\sin(\theta )+r\cos(\theta ){\dot {\theta }},0)|=|(0,0,r^{2}{\dot {\theta }})|=r^{2}{\dot {\theta }}$

1. ${\dot {\theta }}={\frac {H}{r^{2}}}$ (3)

Taking the time derivative of (3), we get

1. ${\ddot {\theta }}=-{\frac {2\cdot H\cdot {\dot {r}}}{r^{3}}}$ (4)

Equations (3) and (4) allow us to eliminate the time derivatives of $\theta$. In order to eliminate the time derivatives of $r$, we must use the chain rule to find appropriate substitutions:

1. ${\dot {r}}={\frac {dr}{d\theta }}\cdot {\dot {\theta }}$ (5)

1. ${\ddot {r}}={\frac {d^{2}r}{d\theta ^{2}}}\cdot {\dot {\theta }}^{2}+{\frac {dr}{d\theta }}\cdot {\ddot {\theta }}$ (6)

Using these four substitutions, all time derivatives in (2) can be eliminated, yielding an ordinary differential equation for $r$ as function of $\theta \,$.

${\ddot {r}}-r{\dot {\theta }}^{2}=-{\frac {\mu }{r^{2}}}$

${\frac {d^{2}r}{d\theta ^{2}}}\cdot {\dot {\theta }}^{2}+{\frac {dr}{d\theta }}\cdot {\ddot {\theta }}-r{\dot {\theta }}^{2}=-{\frac {\mu }{r^{2}}}$

${\frac {d^{2}r}{d\theta ^{2}}}\cdot \left({\frac {H}{r^{2}}}\right)^{2}+{\frac {dr}{d\theta }}\cdot \left(-{\frac {2\cdot H\cdot {\dot {r}}}{r^{3}}}\right)-r\left({\frac {H}{r^{2}}}\right)^{2}=-{\frac {\mu }{r^{2}}}$

1. ${\frac {H^{2}}{r^{4}}}\cdot \left({\frac {d^{2}r}{d\theta ^{2}}}-2\cdot {\frac {\left({\frac {dr}{d\theta }}\right)^{2}}{r}}-r\right)=-{\frac {\mu }{r^{2}}}$ (7)

The differential equation (7) can be solved analytically by the variable substitution

1. $r={\frac {1}{s}}$ (8)

Using the chain rule for differentiation one gets:

1. ${\frac {dr}{d\theta }}=-{\frac {1}{s^{2}}}\cdot {\frac {ds}{d\theta }}$ (9)
2. ${\frac {d^{2}r}{d\theta ^{2}}}={\frac {2}{s^{3}}}\cdot \left({\frac {ds}{d\theta }}\right)^{2}-{\frac {1}{s^{2}}}\cdot {\frac {d^{2}s}{d\theta ^{2}}}$ (10)

Using the expressions (10) and (9) for ${\frac {d^{2}r}{d\theta ^{2}}}$ and ${\frac {dr}{d\theta }}$ one gets

1. $H^{2}\cdot \left({\frac {d^{2}s}{d\theta ^{2}}}+s\right)=\mu$ (11)

with the general solution

1. $s={\frac {\mu }{H^{2}}}\cdot \left(1+e\cdot \cos(\theta -\theta _{0})\right)$ (12)

where e and $\theta _{0}\,$ are constants of integration depending on the initial values for s and ${\frac {ds}{d\theta }}$.

Instead of using the constant of integration $\theta _{0}\,$ explicitly one introduces the convention that the unit vectors ${\hat {x}}\ ,\ {\hat {y}}$ defining the coordinate system in the orbital plane are selected such that $\theta _{0}\,$ takes the value zero and e is positive. This then means that $\theta \,$ is zero at the point where $s$ is maximal and therefore $r={\frac {1}{s}}$ is minimal. Defining the parameter p as ${\frac {H^{2}}{\mu }}$ one has that
$r={\frac {1}{s}}={\frac {p}{1+e\cdot \cos \theta }}$

Alternate derivation

Another way to solve this equation without the use of polar differential equations is as follows:
Define a unit vector ${\mathbf {u}}$ such that ${\mathbf {r}}=r{\mathbf {u}}$ and ${\ddot {\mathbf {r}}}=-{\frac {\mu }{r^{2}}}{\mathbf {u}}$. It follows that

${\mathbf {H}}={\mathbf {r}}\times {\dot {\mathbf {r}}}=r{\mathbf {u}}\times {\frac {d}{dt}}(r{\mathbf {u}})=r{\mathbf {u}}\times (r{\dot {\mathbf {u}}}+{\dot {r}}{\mathbf {u}})=r^{2}({\mathbf {u}}\times {\dot {\mathbf {u}}})+r{\dot {r}}({\mathbf {u}}\times {\mathbf {u}})=r^{2}{\mathbf {u}}\times {\dot {\mathbf {u}}}$

Now consider

${\ddot {\mathbf {r}}}\times {\mathbf {H}}=-{\frac {\mu }{r^{2}}}{\mathbf {u}}\times (r^{2}{\mathbf {u}}\times {\dot {\mathbf {u}}})=-\mu {\mathbf {u}}\times ({\mathbf {u}}\times {\dot {\mathbf {u}}})=-\mu [({\mathbf {u}}\cdot {\dot {\mathbf {u}}}){\mathbf {u}}-({\mathbf {u}}\cdot {\mathbf {u}}){\dot {\mathbf {u}}}]$

(see Triple product#Vector triple product). Notice that

${\mathbf {u}}\cdot {\mathbf {u}}=|{\mathbf {u}}|^{2}=1$

${\mathbf {u}}\cdot {\dot {\mathbf {u}}}={\frac {1}{2}}({\mathbf {u}}\cdot {\dot {\mathbf {u}}}+{\dot {\mathbf {u}}}\cdot {\mathbf {u}})={\frac {1}{2}}{\frac {d}{dt}}({\mathbf {u}}\cdot {\mathbf {u}})=0$

Therefore

${\dot {\mathbf {r}}}\times {\mathbf {H}}=\mu {\dot {\mathbf {u}}}$

Integrating both sides:

${\dot {\mathbf {r}}}\times {\mathbf {H}}=\mu {\mathbf {u}}+{\mathbf {c}}$

Where c is a constant vector. Dotting this with r yields an interesting result:

${\mathbf {r}}\cdot ({\dot {\mathbf {r}}}\times {\mathbf {H}})={\mathbf {r}}\cdot (\mu {\mathbf {u}}+{\mathbf {c}})=\mu {\mathbf {r}}\cdot {\mathbf {u}}+{\mathbf {r}}\cdot {\mathbf {c}}=\mu r({\mathbf {u}}\cdot {\mathbf {u}})+rc\cos(\theta )=r(\mu +c\cos(\theta ))$

Where $\theta$ is the angle between ${\bar {r}}$ and ${\bar {c}}$. Solving for r:

$r={\frac {{\mathbf {r}}\cdot ({\dot {\mathbf {r}}}\times {\mathbf {H}})}{\mu +c\cos(\theta )}}={\frac {({\mathbf {r}}\times {\dot {\mathbf {r}}})\cdot {\mathbf {H}}}{\mu +c\cos(\theta )}}={\frac {|{\mathbf {H}}|^{2}}{\mu +c\cos(\theta )}}$

Notice that $(r,\theta )$ are effectively the polar coordinates of the vector function. Making the substitutions $p={\frac {|{\mathbf {H}}|^{2}}{\mu }}$ and $e={\frac {c}{\mu }}$, we again arrive at the equation

1. $r={\frac {p}{1+e\cdot \cos \theta }}$ (13)

This is the equation in polar coordinates for a conic section with origin in a focal point. The argument $\theta \,$ is called "true anomaly".

Properties of Trajectory Equation

For $e\ =\ 0\,$ this is a circle with radius p.

For $0\ <e\ <\ 1\,$ this is an ellipse with

1. $a={\frac {p}{1-e^{2}}}$ (14)
2. $b={\frac {p}{\sqrt {1-e^{2}}}}=a\cdot {\sqrt {1-e^{2}}}$ (15)

For $e\ =\ 1\,$ this is a parabola with focal length ${\frac {p}{2}}$

For $e\ >\ 1\,$ this is a hyperbola with

1. $a={\frac {p}{e^{2}-1}}$ (16)
2. $b={\frac {p}{\sqrt {e^{2}-1}}}=a\cdot {\sqrt {e^{2}-1}}$ (17)

The following image illustrates an ellipse (red), a parabola (green) and a hyperbola (blue)

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 (13)

The point on the horizontal line going out to the right from the focal point is the point with $\theta =0\,$ for which the distance to the focus takes the minimal value ${\frac {p}{1+e}}$, the pericentre. For the ellipse there is also an apocentre for which the distance to the focus takes the maximal value ${\frac {p}{1-e}}$. For the hyperbola the range for $\theta \,$ is

$\left[-\cos ^{-1}\left(-{\frac {1}{e}}\right)<\theta <\cos ^{-1}\left(-{\frac {1}{e}}\right)\right]$

and for a parobola the range is

$\left[-\pi <\theta <\pi \right]$

Using the chain rule for differentiation (5), the equation (2) and the definition of p as ${\frac {H^{2}}{\mu }}$ one gets that the radial velocity component is

1. $V_{r}={\dot {r}}={\frac {H}{p}}\cdot e\cdot \sin \theta ={\sqrt {\frac {\mu }{p}}}\cdot e\cdot \sin \theta$ (18)

and that the tangential component (velocity component perpendicular to $V_{r}$) is

1. $V_{t}=r\cdot {\dot {\theta }}={\frac {H}{r}}={\sqrt {\frac {\mu }{p}}}\cdot (1+e\cdot \cos \theta )$ (19)

The connection between the polar argument $\theta \,$ and time t is slightly different for elliptic and hyperbolic orbits.

For an elliptic orbit one switches to the "eccentric anomaly" E for which

1. $x=a\cdot (\cos E-e)$ (20)
2. $y=b\cdot \sin E$ (21)

and consequently

1. ${\dot {x}}=-a\cdot \sin E\cdot {\dot {E}}$ (22)
2. ${\dot {y}}=b\cdot \cos E\cdot {\dot {E}}$ (23)

and the angular momentum H is

1. $H=x\cdot {\dot {y}}-y\cdot {\dot {x}}=a\cdot b\cdot (1-e\cdot \cos E)\cdot {\dot {E}}$ (24)

Integrating with respect to time t one gets

1. $H\cdot t=a\cdot b\cdot (E-e\cdot \sin E)$ (25)

under the assumption that time $t=0$ is selected such that the integration constant is zero.

As by definition of p one has

1. $H={\sqrt {\mu \cdot p}}$ (26)

this can be written

1. $t=a\cdot {\sqrt {\frac {a}{\mu }}}(E-e\cdot \sin E)$ (27)

For a hyperbolic orbit one uses the hyperbolic functions for the parameterisation

1. $x=a\cdot (e-\cosh E)$ (28)
2. $y=b\cdot \sinh E$ (29)

for which one has

1. ${\dot {x}}=-a\cdot \sinh E\cdot {\dot {E}}$ (30)
2. ${\dot {y}}=b\cdot \cosh E\cdot {\dot {E}}$ (31)

and the angular momentum H is

1. $H=x\cdot {\dot {y}}-y\cdot {\dot {x}}=a\cdot b\cdot (e\cdot \cosh E-1)\cdot {\dot {E}}$ (32)

Integrating with respect to time t one gets

1. $H\cdot t=a\cdot b\cdot (e\cdot \sinh E-E)$ (33)

i.e.

1. $t=a\cdot {\sqrt {\frac {a}{\mu }}}(e\cdot \sinh E-E)$ (34)

To find what time t that corresponds to a certain true anomaly $\theta \,$ one computes corresponding parameter E connected to time with relation (27) for an elliptic and with relation (34) for a hyperbolic orbit.

Note that the relations (27) and (34) define a mapping between the ranges

$\left[-\infty <t<\infty \right]\longleftrightarrow \left[-\infty <E<\infty \right]$

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.