= Equal-time sectors of a Kepler ellipse
Take an illustrative <elliptic Kepler orbit> with <semi-major axis> $a=1$, <orbital eccentricity> $e=0.6$ and <gravitational parameter> $\mu=1$, in consistent dimensionless units. Its <semi-minor axis> is $b=a\sqrt{1-e^2}=0.8$, <mean motion> is $n=\sqrt{\mu/a^3}=1$, and <specific angular momentum> magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,
$$
\dot A=0.4,\qquad \Delta A=0.4\,\Delta t.
$$
The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near <periapsis> the short radius sweeps a wide <angle>; near <apoapsis> the long radius sweeps a narrow <angle>. Constant <areal velocity> therefore does not require constant <angular velocity> or constant <speed>.
\Image[/kepler-ellipse-equal-time-swept-sectors.png]
{title=Equal <time> intervals sweep equal <areas> about the focus of an illustrative <elliptic Kepler orbit>}
To compute the sectors without mistaking equal <angles> for equal <times>, parametrize the <ellipse> by <eccentric anomaly> $E$:
$$
x=a(\cos E-e),\qquad y=b\sin E,\qquad
A(E_2)-A(E_1)=\frac{ab}{2}\bigl(E_2-E_1-e(\sin E_2-\sin E_1)\bigr).
$$
This follows by integrating $\tfrac12(x\,dy-y\,dx)$. <Kepler's equation> gives $E-e\sin E=M=n(t-t_{\mathrm p})$, with <mean anomaly> $M$ measured from <periapsis> <time> $t_{\mathrm p}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use <mean anomaly> intervals $[-0.25,0.25]$ and $[\pi-0.25,\pi+0.25]$, rather than equal <eccentric anomaly> increments.
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