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Is there a way to show that motion of Earth around the Sun is elliptical (Kepler's 1st law) from Newton's laws without resorting to the use of differential equations of motion?

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  • $\begingroup$ Observation? $\quad$ $\endgroup$ – Emilio Pisanty Nov 13 '13 at 18:17
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    $\begingroup$ (Sorry, couldn't resist.) It's not clear what your precise question is. As I understand it, you'd like a proof that the orbit of the solution of the Kepler ODE problem is an ellipse, which doesn't depend on a direct, full solution of the ODE problem - much in the way that the orbit can be shown to be planar. Is this correct? $\endgroup$ – Emilio Pisanty Nov 13 '13 at 18:21
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    $\begingroup$ I see no reason why the Earth's orbit couldn't be perfectly circular around the center of mass of the Earth + Sun system. Can you clarify what you're asking? $\endgroup$ – Brandon Enright Nov 13 '13 at 18:27
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    $\begingroup$ If so, I must point out that the standard derivations of Kepler's laws from Newton's, e.g. this, do not use a full solution of the ODE. The equations in $d^2r/dt^2$ are reduced to an equation in $d^2r/d\theta^2$, which is solved to get the orbit. The equation for $\theta(t)$, on the other hand (and $r(t)$ with it) is not actually solvable in terms of elementary functions. $\endgroup$ – Emilio Pisanty Nov 13 '13 at 18:34
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    $\begingroup$ I would agree with the first comment from Emilio Pisanty, since Kepler also made observations from which he derived his laws. $\endgroup$ – fibonatic Nov 18 '13 at 17:10
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Newton's original proof was in fact based on geometry (he hadn't invented calculus yet). Richard Feynman devised his own, simpler geometric proof for one of his famous lectures. You can find it in Feynman's Lost Lecture, by Goodstein & Goodstein, and in this article: Paths of the Planets from Hall & Higson. But since it's so much fun, I'll describe it here as well.

Let's start with a lesser-known way to construct an ellipse, the so-called circle construction. Draw a circle with centre $O$, and fix a point $A$ inside the circle. Pick a point $B$ on the circle, and draw the perpendicular bisector of $\overline{AB}$ (blue line). It intersects $\overline{OB}$ in a point $P$, and as $B$ moves around the circle, these intersection points form an ellipse. Also, the blue biscector lines are tangent lines to the ellipse, and $O$ and $A$ are the foci.

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Why is it an ellipse? Because $\overline{AP}$ has the same length as $\overline{BP}$, so that the sum of the lengths of $\overline{AP}$ and $\overline{OP}$ is constant, i.e. the radius of the circle. In other words, we get the classic tack-and-string definition of an ellipse. It is also straightforward to see that the angles $a$ and $b$ are equal. Since $a$ and $c$ are also equal, this means that $b$ and $c$ are equal, so that the blue line is indeed a tangent line.

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The geometric proof of Kepler's Second Law (planets sweep out equal areas in equal times) from Newton's first two laws is straightforward and can be found in the Hall & Higson article. Now, if a planet traverses an angle $\Delta\theta$ in a small time interval $\Delta t$, it sweeps out an area $$ \text{area}\approx \frac{1}{2}\Delta\theta\, r^2. $$ enter image description here

At this point, Feyman's argument deviates from Newton: while Newton breaks up the orbit it equal-time pieces, Feyman considers equal-angle pieces. In other words, Feynman breaks up the orbit in subsequent pieces with areas $$ \text{area}\approx \text{constant}\cdot r^2. $$ Newton's inverse-square law (which can be derived from Kepler's Third Law) states that the acceleration of a planet is proportional to the inverse square of its distance $r$: $$ \left\|\frac{\Delta\boldsymbol{v}}{\Delta t}\right\| = \frac{\text{constant}}{r^2}. $$ Eliminating $r^2$, we get $$ \left\|\Delta\boldsymbol{v}\right\| \approx \text{constant}\cdot\frac{\Delta t}{\text{area swept out in $\Delta t$}}. $$ But Kepler's Second Law states that the area swept out in $\Delta t$ is a constant multiple of $\Delta t$. Therefore, $$ \left\|\Delta\boldsymbol{v}\right\| \approx \text{constant}, $$ that is, intervals of constant $\Delta\theta$ also have a constant change in velocity. We can use this fact to construct a so-called velocity diagram. Break up the orbit into equal-angle pieces, draw the velocity vectors, and translate these vectors to the same point.

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Since $\left\|\Delta\boldsymbol{v}\right\|$ is constant, the resulting figure is a polygon with $\dfrac{360^\circ}{\Delta\theta\,}$ sides. The smaller the angles, the more it approaches a circle.

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Now, let's draw the velocity diagram of an orbiting planet. If $l$ is the tangent line to the orbit at point $P$ (parallel to the velocity vector in $P$), then $l'$ in the corresponding velocity diagram is also parallel to $l$. Also note that $\theta$ in both diagrams is the same.

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Rotate the velocity diagram clockwise by $90^\circ$, so that $l'$ becomes perpendicular to $l$. Construct the perpendicular bisector $p$ to the line $\overline{AB}$, and the intersection $P'$ with $\overline{OB}$. It turns out that we are in the exact same situation as the circle construction for the ellipse: as $B$ moves on the velocity diagram, the points $P'$ form an ellipse.

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The lines $p$ are the tangent lines to the ellipse. However, these lines are also parallel to the lines $l$, which are the tangent lines to the planet's orbit. Because of the tangent principle, if two curves have the same tangent lines at every point, then those curves are the same. In other words, the lines $l$ are also the tangent lines of an ellipse. This proves that the orbit of a planet is indeed an ellipse.

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  • $\begingroup$ Thank you for the answer, but also for the Article of Hall end Higson. It's amazing and the exercises there really help to understand what's going on! $\endgroup$ – Bman72 Jun 26 '19 at 9:36

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