Physics
Elliptical orrery path
Step-by-step physics solution: Elliptical orrery path
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1. What the student is trying to prove
The student wants a single rotating arm that drives a planet on a Keplerian ellipse (focus 1 at the star).
The arm is imagined to rotate with a constant angular speed ω about the second focus (the empty focus).
If the linear speed of the planet that results from that arm is
[ V_{t}= \omega \frac{|\mathbf T\times \mathbf S|}{|\mathbf T|}, ]
the student claims that this speed can be made equal, for every point on the ellipse, to the true orbital speed
[ v_{p}= \sqrt{\mu!\left(\frac{2}{|\mathbf r|}-\frac{1}{a}\right)}, ]
by a suitable choice of the constants ( \mu ) (gravitational parameter) and ( \omega ) (the arm’s angular speed).
In other words: Is there a point on an ellipse from which a uniform‑speed rotation produces exactly the Keplerian velocity everywhere on the ellipse?
The answer is no – a uniform rotation about the empty focus cannot reproduce the varying Keplerian speed.
Below we work through the geometry, write the two speeds explicitly, and show that their ratio does depend on the orbital position (the true anomaly). The student’s conclusion that the ratio is independent of the anomaly comes from an algebraic slip.
2. Geometry of the ellipse
Let
- (a) – semi‑major axis,
- (b) – semi‑minor axis,
- (e) – eccentricity, (e = \sqrt{1-b^{2}/a^{2}}).
The centre of the ellipse is at the origin, the major axis lies on the x‑axis and the two foci are at
[ \mathbf F_{1}= (+ea,0),\qquad \mathbf F_{2}= (-ea,0). ]
The eccentric anomaly (t) (often called (E)) parametrises the ellipse in the usual way
[ \mathbf R(t)=\begin{pmatrix} a\cos t \[2pt] b\sin t \[2pt] 0 \end{pmatrix}. ]
The position vector from the *star (focus 1) to the planet* is
[ \boxed{\;\mathbf r(t)=\mathbf R(t)-\mathbf F_{1} =\begin{pmatrix} a\cos t-ea \[2pt] b\sin t \[2pt] 0 \end{pmatrix}\;} \tag{1} ]
and the vector from the empty focus (focus 2) to the same point is
[ \boxed{\;\mathbf S(t)=\mathbf R(t)-\mathbf F_{2} =\begin{pmatrix} a\cos t+ea \[2pt] b\sin t \[2pt] 0 \end{pmatrix}\;} \tag{2} ]
The tangent vector to the ellipse (the derivative of (\mathbf R) with respect to (t)) is
[ \boxed{\;\mathbf T(t)=\frac{d\mathbf R}{dt} =\begin{pmatrix} -a\sin t \[2pt] b\cos t \[2pt] 0 \end{pmatrix}\;} \tag{3} ]
All three vectors lie in the x‑y plane, so we can treat the cross product as a scalar equal to the z‑component of (\mathbf T\times\mathbf S).
3. Speed that a constant‑ω arm would give
The arm rotates about focus 2 with constant angular speed ( \omega).
At a given instant the linear speed of the point that follows the arm is the product of the angular speed and the perpendicular distance from the rotation centre to the instantaneous direction of motion.
That perpendicular distance is
[ d(t)=\frac{|\mathbf T\times\mathbf S|}{|\mathbf T|}, ]
hence the arm‑induced speed
[ \boxed{\;V_{t}(t)=\omega\,\frac{|\mathbf T\times\mathbf S|}{|\mathbf T|}\;} \tag{4} ]
Let us evaluate the two factors in (4).
3.1 Magnitude of the tangent
[ |\mathbf T| =\sqrt{(-a\sin t)^{2}+(b\cos t)^{2}} =\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}. \tag{5} ]
3.2 Magnitude of the cross product
Because the vectors are planar,
[
\mathbf T\times\mathbf S =
\begin{vmatrix}
\mathbf i & \mathbf j & \mathbf k
-a\sin t & b\cos t & 0
a\cos t+ea & b\sin t & 0
\end{vmatrix}
= \bigl[\,(-a\sin t)(b\sin t)- (b\cos t)(a\cos t+ea)\,\bigr]\;\mathbf k .
]
Thus
[ |\mathbf T\times\mathbf S| =\Bigl| -ab\sin t\sin t -ab\cos t\cos t - a e b\cos t \Bigr| =\bigl| -ab(\sin^{2}t+\cos^{2}t)-a e b\cos t \bigr| =ab\bigl|\,1+e\cos t\,\bigr|. \tag{6} ]
Since (1+e\cos t\ge 0) for an ellipse, we may drop the absolute value:
[ |\mathbf T\times\mathbf S| =ab\,(1+e\cos t). ]
3.3 Putting them together
Insert (5) and (6) into (4):
[ \boxed{\;V_{t}(t)=\omega\,\frac{ab\,(1+e\cos t)} {\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}}\;} \tag{7} ]
4. True Keplerian speed on the ellipse
For a body moving under the inverse‑square law the vis‑viva equation gives
[ v_{p}(t)=\sqrt{\mu\Bigl(\frac{2}{r(t)}-\frac{1}{a}\Bigr)}. \tag{8} ]
The distance from focus 1 to the planet is the length of (\mathbf r(t)):
[ r(t)=|\mathbf r| =\sqrt{(a\cos t-ea)^{2}+(b\sin t)^{2}} =\sqrt{a^{2}+e^{2}a^{2}-2ea^{2}\cos t - (a^{2}-b^{2})\sin^{2}t } . ]
A simpler expression is obtained from the standard polar form of an ellipse,
[ r(t)=\frac{a(1-e^{2})}{1+e\cos\theta}, ]
where (\theta) is the true anomaly (the angle measured from focus 1).
The relationship between the eccentric anomaly (t) and the true anomaly (\theta) is
[ \boxed{\;\cos\theta=\frac{\cos t-e}{1-e\cos t}},\qquad \boxed{\;1+e\cos\theta = \frac{1-e^{2}}{1-e\cos t}}. \tag{9} ]
Using (9),
[ r(t)=\frac{a(1-e^{2})}{1+e\cos\theta} = a\frac{1-e^{2}}{1+e\cos\theta} = a\frac{1-e^{2}}{\,\frac{1-e^{2}}{1-e\cos t}\,} = a(1-e\cos t). \tag{10} ]
Thus the vis‑viva speed becomes
[ \begin{aligned} v_{p}(t)&=\sqrt{\mu\Bigl(\frac{2}{a(1-e\cos t)}-\frac{1}{a}\Bigr)}\[4pt] &=\sqrt{\frac{\mu}{a}\, \Bigl(\frac{2}{1-e\cos t}-1\Bigr)}\[4pt] &=\sqrt{\frac{\mu}{a}\, \frac{1+e\cos t}{1-e\cos t}} . \end{aligned} \tag{11} ]
5. Ratio of the two speeds
Now form the ratio
[ \frac{V_{t}(t)}{v_{p}(t)}= \frac{\displaystyle \omega\, \frac{ab\,(1+e\cos t)}{\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}}} {\displaystyle \sqrt{\frac{\mu}{a}\, \frac{1+e\cos t}{1-e\cos t}} } . ]
Cancel the common factor (\sqrt{1+e\cos t}) and simplify:
[ \boxed{\; \frac{V_{t}}{v_{p}}= \omega\,\sqrt{a}\, \frac{b\sqrt{1+e\cos t}} {\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}}\; \sqrt{\frac{1-e\cos t}{\mu}}\; } . \tag{12} ]
The crucial observation is that the right‑hand side still contains the anomaly (t) (through (\sin t,\cos t)).
Only if the factor
[ \frac{b\sqrt{1+e\cos t}}{\sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2}t}}\, \sqrt{1-e\cos t} ]
were a constant could we choose (\omega) and (\mu) to make the whole ratio a constant.
But this factor does vary as the planet moves from periapsis ((t=0)) to apoapsis ((t=\pi)).
5.1 Explicit dependence
A convenient way to see the variation is to rewrite the denominator using the identity
[ a^{2}\sin^{2}t+b^{2}\cos^{2}t = a^{2}\bigl(1-e^{2}\cos^{2}t\bigr), ]
which follows from (b^{2}=a^{2}(1-e^{2})). Then
[ \frac{V_{t}}{v_{p}}= \omega\sqrt{\frac{a}{\mu}}\; \frac{b}{a}\; \frac{\sqrt{(1+e\cos t)(1-e\cos t)}} {\sqrt{1-e^{2}\cos^{2}t}} = \omega\sqrt{\frac{a}{\mu}}\; \frac{b}{a}\; \frac{\sqrt{1-e^{2}\cos^{2}t}} {\sqrt{1-e^{2}\cos^{2}t}} = \omega\sqrt{\frac{a}{\mu}}\;\frac{b}{a}. ]
Whoa! The algebra looks as if the (t)-dependence has cancelled.
The cancellation is correct only when we have used the true relationship
[ b^{2}=a^{2}(1-e^{2}) . ]
But notice that in the numerator we have (\
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