Exercises — True anomaly from eccentric anomaly

• $\nu$ (true anomaly) is measured at the focus — it is the real angle you'd see from the Sun.
• $E$ (eccentric anomaly) is measured at the center of the ellipse, on the auxiliary circle.
• $M$ (the mean anomaly) is not a geometric angle; it grows linearly with time (see Mean anomaly and time).

With $e=0$: $r=a(1-0)=a,\cos \nu =\frac{\cos E}{1}=\cos E,\sin \nu =\frac{1\cdot \sin E}{1}=\sin E,\tan \frac{\nu }{2}=\sqrt{\frac{1}{1}}\tan \frac{E}{2}=\tan \frac{E}{2}.$ Every equation says $\nu =E$. A circle has its focus at the center, so the two angles collapse into one. This is your sanity anchor for all later work.

• $E=0$: $\cos 0=1$, so $r=a(1-e)=a-ae=a-c$ — the smallest distance (perihelion).
• $E=\pi$: $\cos \pi =-1$, so $r=a(1+e)=a+ae=a+c$ — the largest distance (aphelion). So perihelion is $c$ inside $a$ and aphelion is $c$ outside $a$: the focus offset $c=ae$ is literally the half-difference $(r_{\text{apo}}-r_{\text{peri}})/2=ae=c$. The body is closest at perihelion, farthest at aphelion.

Figure s01 (below): an $e=0.6$ ellipse drawn in black with the center $O$ (black dot) and the focus $F$ (the single red dot — $F$ is the key object, so it is the only thing drawn in accent red). Two red arrows run from $F$ to the perihelion point and to the aphelion point, labelled $r=a(1-e)$ and $r=a(1+e)$; these red arrows are the "accent-red features" the text points to. The gap between the black center dot and the red focus dot is the focus offset $c=ae$.

$\tan (E/2)=\tan 45^{\circ }=1$. The stretch factor: $\sqrt{\frac{1+0.6}{1-0.6}}=\sqrt{\frac{1.6}{0.4}}=\sqrt{4}=2.$ So $\tan (\nu /2)=2\cdot 1=2$, giving $\nu /2=\arctan 2=63.43^{\circ }$, hence $\nu =126.87^{\circ }.$ Note $\nu >E$: seen from the offset focus the body has already swept past the halfway mark — the perihelion-side stretching in action.

$\cos \nu =\frac{\cos 90^{\circ }-0.6}{1-0.6\cos 90^{\circ }}=\frac{-0.6}{1}=-\mathrm{0.6.}$ $\sin \nu =\frac{\sqrt{1-0.36}\sin 90^{\circ }}{1}=\sqrt{0.64}=+\mathrm{0.8.}$ $\cos \nu <0$ and $\sin \nu >0$ ⟹ second quadrant. $\arctan \frac{0.8}{-0.6}$ resolved into Q2 gives $\nu =126.87^{\circ }$ ✓ — same as L2.1.

$r=a(1-e\cos E)=2(1-0.6\cos 90^{\circ })=2(1-0)=2\text{ AU}.$ At $E=90^{\circ }$ the $\cos E$ term vanishes, so $r=a$ regardless of $e$ — a handy checkpoint.

Invert the half-angle formula for $E$: $\tan \frac{E}{2}=\sqrt{\frac{1-e}{1+e}}\tan \frac{\nu }{2}.$ $\tan (\nu /2)=\tan 60^{\circ }=1.7320508$. Factor $\sqrt{0.7/1.3}=\sqrt{0.5384615}=0.7337994$. $\tan \frac{E}{2}=0.7337994\times 1.7320508=\mathrm{1.2709803.}$ $E/2=\arctan 1.2709803=51.798^{\circ }$, so $E=103.597^{\circ }.$ Sanity: $E<\nu$ here (the reverse stretch), consistent with a body on the perihelion-approaching side.

By symmetry with L2.1, this is the mirror image below the axis. Use signed components: $\cos \nu =\frac{\cos 270^{\circ }-0.6}{1-0.6\cos 270^{\circ }}=\frac{0-0.6}{1-0}=-0.6,$ $\sin \nu =\frac{\sqrt{0.64}\sin 270^{\circ }}{1}=\frac{0.8\cdot (-1)}{1}=-\mathrm{0.8.}$ $\cos \nu <0$, $\sin \nu <0$ ⟹ third quadrant: $\nu =233.13^{\circ }$ (equivalently $-126.87^{\circ }$). Now the half-angle check. We take $E=270^{\circ }$, so $E/2=135^{\circ }$ and $\tan (E/2)=\tan 135^{\circ }=-1$, giving $\tan (\nu /2)=2\times (-1)=-2$. The equation $\tan (\nu /2)=-2$ has two candidate half-angles: the principal $\arctan (-2)=-63.43^{\circ }$ and the shifted $-63.43^{\circ }+180^{\circ }=116.57^{\circ }$. Why we must pick $116.57^{\circ }$ and not $-63.43^{\circ }$: the half-angle $\nu /2$ must live in the same half-turn as $E/2$. Here $E/2=135^{\circ }$ lies in $(90^{\circ },180^{\circ })$, so $\nu /2$ must also lie in $(90^{\circ },180^{\circ })$ — that forces the $116.57^{\circ }$ branch, not the negative one. Doubling gives $\nu =233.13^{\circ }$ ✓, matching the signed-component answer. This is exactly where naive $\arccos$ fails — it would return $126.87^{\circ }$, losing the sign.

At $E=10^{\circ }$: $\tan (E/2)=\tan 5^{\circ }=0.0874886$. $\tan (\nu /2)=2\times 0.0874886=0.1749773$, so $\nu /2=9.923^{\circ }$, $\nu =19.846^{\circ }$. Gap $\nu -E=+9.85^{\circ }$. At $E=170^{\circ }$: $\tan (E/2)=\tan 85^{\circ }=11.4300523$. $\tan (\nu /2)=2\times 11.4300523=22.8601$, so $\nu /2=87.494^{\circ }$, $\nu =174.988^{\circ }$. Gap $\nu -E=+4.99^{\circ }$. Near perihelion (small $E$) the body is close to the focus, so a small $E$-step sweeps a large $\nu$ — the stretch is strong, big gap. Near aphelion (large $E$) the body is far, the same $E$-step sweeps a smaller $\nu$ — small gap. The $\sqrt{(1+e)/(1-e)}$ factor encodes this focus-offset distortion.

Figure s02 (below): the horizontal axis is $E$ from $0^{\circ }$ to $180^{\circ }$; the vertical axis is the gap $\nu -E$ in degrees. The single red curve (the key object, so it is the only red element) is the gap for $e=0.6$; it rises steeply out of perihelion at the left edge and returns gently toward aphelion at the right edge. Two black dots mark the $E=10^{\circ }$ and $E=170^{\circ }$ points you just computed — these are the "black dots" the text refers to.

$\nu -E$ is $0$ at $E=0$ and $0$ again at $E=\pi$ (both angles hit $0$ and $\pi$ together). Being $0$ at both ends and positive between, it peaks somewhere in the middle — this is the crest of the red curve in figure s02. Check $E=90^{\circ }$: $\nu =126.87^{\circ }$ (L2.1), gap $=36.87^{\circ }$. Check $E=60^{\circ }$: $\tan (30^{\circ })=0.5773503$, $\tan (\nu /2)=1.1547005$, $\nu /2=49.107^{\circ }$, $\nu =98.21^{\circ }$, gap $=38.21^{\circ }$. So the maximum sits a bit before $90^{\circ }$ (around $E\approx 60^{\circ }$ for this $e$), on the perihelion-leaning side — consistent with the strongest stretching happening while the body is still fairly close in.

This chains Mean anomaly and time → Kepler's Equation → this note. With $E=73.90^{\circ }$: $\tan (E/2)=\tan 36.95^{\circ }=0.7522$. Stretch factor $\sqrt{1.5/0.5}=\sqrt{3}=1.7320508$. $\tan \frac{\nu }{2}=1.7320508\times 0.7522=\mathrm{1.3028.}$ $\nu /2=\arctan 1.3028=52.49^{\circ }$, so $\nu =104.98^{\circ }.$ As expected $M(45^{\circ })<E(73.9^{\circ })<\nu (105^{\circ })$: each successive angle is "stretched" toward the perihelion side. This ordering $M<E<\nu$ on the outgoing perihelion side is a great final sanity check.

Direct: $r=a(1-e\cos E)=4(1-0.5\cos 73.90^{\circ })=4(1-0.5\times 0.27731)=4\times 0.86134=3.4454\text{ AU}$. Cross-check via true anomaly: $\cos \nu =\cos 104.98^{\circ }=-0.25834$. $r=\frac{4(1-0.25)}{1+0.5\times (-0.25834)}=\frac{3.0}{0.87083}=3.4454\text{ AU}.✓$ Both routes agree — the two radius formulas are the same physics wearing different clothes.

Step 1 — square and rewrite with the half-angle identity. Square the given relation: ${\tan }^2\frac{E}{2}=\frac{1-e}{1+e}{\tan }^2\frac{\nu }{2}.$ Now apply Half-angle trigonometric identities, ${\tan }^2\frac{\theta }{2}=\frac{1-\cos \theta }{1+\cos \theta }$, to both sides: $\frac{1-\cos E}{1+\cos E}=\frac{1-e}{1+e}\cdot \frac{1-\cos \nu }{1+\cos \nu }.$ Step 2 — name the unknowns to declutter. Let $u=\cos E$ (what we want) and $w=\cos \nu$ (known). The equation is $\frac{1-u}{1+u}=\frac{(1-e)(1-w)}{(1+e)(1+w)}.$ Step 3 — cross-multiply. Clear both denominators: $(1-u)(1+e)(1+w)=(1+u)(1-e)(1-w).$ Step 4 — expand each side. Multiply out the $e,w$ brackets first: $(1+e)(1+w)=1+e+w+ew,(1-e)(1-w)=1-e-w+ew.$ So the equation reads $(1-u)(1+e+w+ew)=(1+u)(1-e-w+ew).$ Step 5 — collect the $u$ terms on one side, constants on the other. Expand: $\underset{\text{const}}{\underbrace{(1+e+w+ew)}}-u(1+e+w+ew)=\underset{\text{const}}{\underbrace{(1-e-w+ew)}}+u(1-e-w+ew).$ Move all $u$-terms right, all constants left: $(1+e+w+ew)-(1-e-w+ew)=u[(1-e-w+ew)+(1+e+w+ew)].$ Step 6 — simplify each bracket. On the left the $1$'s and $ew$'s cancel, leaving $2e+2w$. On the right the $e,w$ terms cancel, leaving $2+2ew$: $2e+2w=u(2+2ew).$ Step 7 — divide by 2 and solve for $u$. Cancel the factor $2$: $e+w=u(1+ew)\Rightarrow u=\frac{w+e}{1+ew}.$ Restoring names: $\boxed{\cos E=\frac{\cos \nu +e}{1+e\cos \nu }}.$ The "why" of the sign flip: compared to the forward $\cos \nu =\frac{\cos E-e}{1-e\cos E}$, the inverse has $+e$ instead of $-e$. Going from center-view to focus-view subtracts the perihelion kick; going back adds it — the offset reverses direction, and that is exactly what the algebra in Steps 5–6 produced when the $ew$ terms cancelled.

As $e\to 1^{-}$, $1-e\to 0^{+}$ so the factor $\sqrt{\frac{1+e}{1-e}}\to +\infty$. Then $\tan \frac{\nu }{2}\to \infty$ for any fixed $E>0$, i.e. $\nu \to 180^{\circ }$. Physically: a near-parabolic orbit is enormously elongated, the focus is extremely offset, so almost the entire perihelion passage sweeps $\nu$ rapidly toward $180^{\circ }$ while $E$ barely moves. The eccentric-anomaly picture (auxiliary circle) degenerates as the ellipse stretches unbounded — which is exactly why parabolic orbits use a different parameter (a whole other note).

From $\sin \nu =\frac{\sqrt{1-e^2}\sin E}{1-e\cos E}$: the prefactor $\frac{\sqrt{1-e^2}}{1-e\cos E}$ is strictly positive (numerator $>0$ for $e<1$; denominator $1-e\cos E\ge 1-e>0$). So $\sin \nu$ carries exactly the sign of $\sin E$ — they never disagree. This is why the upper/lower half is preserved and the half-angle formula stays consistent. Numeric at $E=120^{\circ }$, $e=0.4$: ratio $=\frac{\sqrt{1-0.16}}{1-0.4\cos 120^{\circ }}=\frac{\sqrt{0.84}}{1-0.4\times (-0.5)}=\frac{0.916515}{1.2}=0.763763$.

• I can state $r=a(1-e\cos E)$ and both $\cos \nu ,\sin \nu$ formulas from memory.
• I use atan2 / half-angle, never bare arccos, for the full orbit.
• I flip the stretch factor when inverting $\nu \to E$.
• I flip the sign of $e$ in the $\cos E(\nu )$ inverse.
• I can chain $M\to E\to \nu \to r$ without early rounding, keeping 5+ significant figures until the last step.