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Many detailed proofs are available for Kepler's first law, like this one. But all of them use polar co-ordinates. There is one which doesn't even use differential equations. I'm looking for a proof which is done using cartesian co-ordinates.

For simplicity lets assume the Sun is at origin and the planet is located on the x-axis and is having a suitable velocity which is parallel to positive y-axis.

Background: I'm a high school physics teacher. My students are not familiar with polar co-ordinates, however they are comfortable with calculus

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    $\begingroup$ There's a reason these proofs are done in polar coordinates, doing it in Cartesian coordinates would be much much harder to understand! $\endgroup$
    – knzhou
    Mar 27 '20 at 5:40
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    $\begingroup$ If your students don't understand polar coordinates, I think it's almost certain that they won't understand a very complicated derivation in Cartesian coordinates, either... in fact, I don't think a proof of Kepler's first law would be useful in any high school class except for an honors-level calculus-based one, where all students would be expected to know polar coordinates already. $\endgroup$
    – knzhou
    Mar 27 '20 at 5:47
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    $\begingroup$ In my opinion it's more important to teach them polar coordinates than it is to prove Kepler’s First Law. And it doesn’t take more than an hour to do so. $\endgroup$
    – G. Smith
    Mar 27 '20 at 5:48
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    $\begingroup$ If you want to use Cartesian coordinates, have them numerically integrate the equations of motion to get an (approximate) ellipse, using calculators. That will actually teach them much more about what differential equations really mean. $\endgroup$
    – G. Smith
    Mar 27 '20 at 5:51
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    $\begingroup$ Learning that a good choice of coordinate system can hugely simplify a problem (and conversely, that a bad choice can make it practically impossible) is certainly a good lesson in physics. $\endgroup$ Mar 27 '20 at 8:02
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Take the Sun(mass M) and the planet(mass m) to be point masses. Let the Sun be fixed at the origin and the planet be moving in the x-y plane, initial velocity of the planet be $v_o\hat{j}$ and initial position of the planet be $r_o\hat{i}$. At any given instant, let the planet's position, velocity and acceleration be $\vec{r}, \vec{v}, \vec{a}$ respectively. Let $\theta$ be the angle subtended between $\vec{r}$ and positive x-axis.

The gravitation force is always acting towards the origin, hence torque won't be generated on the planet, about the origin. Therefore, angular momentum of the planet should be conserved about the origin. $$\vec{r}\times\vec{p}=r_omv_o\hat{k}$$ $$\vec{r}\times\vec{v}=r_ov_o....(1)$$ This can also be written as $$I\vec{\omega}=r_omv_o\hat{k}$$ $$mr^2{\frac{d\vec{\theta}}{dt}}=r_omv_o\hat{k}$$ $$r^2\frac{d{\theta}}{dt}\hat{k}=r_ov_o\hat{k}$$ $$r^2\frac{d{\theta}}{dt}=r_ov_o....(2)$$ From Newton's law of Gravitation, $$\vec{F}=\frac{-GMm}{r^3}\vec{r}$$ From Newton's second law of motion, $$\vec{F}=m\vec{a}$$ $$m\vec{a}=\frac{-GMm}{r^3}\vec{r}$$ $$\frac{d\vec{v}}{dt}=\frac{-GM}{r^3}\vec{r}$$ Multiplying and diving Left Hand Side by $d\theta$ and substituting $$\vec{r}=r cos\theta\hat{i}+r sin\theta\hat{j}~(where~r^2=x^2+y^2,cos\theta=\frac{x}{r}~and~sin\theta=\frac{y}{r})$$ $$\frac{d\vec{v}}{d\theta}.\frac{d\theta}{dt}= -\frac{-GM(r cos\theta\hat{i}+r sin\theta\hat{j})}{r^3}$$ $$\frac{d\vec{v}}{d\theta}.r^2\frac{d\theta}{dt}=-GM(cos\theta\hat{i}+sin\theta\hat{j})$$ From (2), $$\frac{d\vec{v}}{d\theta}.r_ov_o=-GM(cos\theta\hat{i}+sin\theta\hat{j})$$ $$Let ~~~\alpha=\frac{GM}{r_ov_o}$$ $$\int_{\vec{v_o}}^{\vec{v}}\vec{dv}=-\alpha(\int_{0}^{\theta}cos\theta.d\theta.\hat{i}+\int_{0}^{\theta}sin\theta.d\theta.\hat{j})$$ $$\vec{v}-\vec{v_o}=-\alpha (sin\theta\hat{i}-cos\theta\hat{j})\Big|_0^{\theta}$$ $$\vec{v}-\vec{v_o}=-\alpha (sin\theta\hat{i}-(1-cos\theta\hat{j}))$$ $$\vec{v}=-\alpha sin\theta\hat{i}+( \alpha cos\theta-\alpha+v_o)\hat{j}$$ Now, $$\vec{r}\times\vec{v}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\rcos\theta&rsin\theta&0\\-\alpha sin \theta & \alpha cos\theta+v_o-\alpha&0\end{vmatrix}$$ $$=(rcos\theta)(\alpha cos\theta+v_o-\alpha)-(-\alpha sin\theta)(sin\theta)\hat{k}$$ $$=(r)(\alpha cos^2\theta+\alpha sin^2\theta-\alpha cos\theta+v_ocos\theta)\hat{k}$$ $$=(r)(\alpha+v_ocos\theta-\alpha cos\theta)$$ From (1), $$r_ov_o\hat{k}=(r)(\alpha+(v_o-\alpha)cos\theta)$$ $$r=\frac{r_ov_o}{\alpha(1+(\frac{v_o-\alpha}{\alpha})cos\theta)}$$ $$Let ~\frac{r_0v_0}{\alpha}=h~~~and ~~~\frac{v_o-\alpha}{\alpha}=p$$ $$Then~~~r=\frac{h}{1+pcos\theta}$$ $$Substituting~~~cos\theta=\frac{x}{r},$$ $$r(1+p\frac{x}{r})=h$$ $$r+px=h$${\tiny } $$r^2=(h-px)^2$$ $$x^2+y^2=h^2+p^2x^2-2hpx$$ $$x^2(1-p^2)+2hpx+y^2=h^2$$ $$When~~~~~(1-p^2)\neq0,$$ $$x^2+\frac{y^2}{1-p^2} +\frac{2hpx}{1-p^2}=\frac{h^2}{1-p^2}$$ $$Adding ~~~\frac{h^2}{(1-p^2)^2} ~~~on ~both~ sides,$$ $$x^2+\frac{h^2p^2}{(1-p^2)^2}+\frac{y^2}{1-p^2} +\frac{2hpx}{1-p^2}=\frac{h^2}{1-p^2}+\frac{h^2p^2}{(1-p^2)^2}$$ $$(x+\frac{hp}{1-p^2})^2+\frac{y^2}{1-p^2}=\frac{h^2}{(1-p^2)^2}$$ This takes the form $$\frac{(x+\frac{hp}{1-p^2})^2}{\frac{h^2}{(1-p^2)^2}}+\frac{y^2}{\frac{h^2}{(1-p^2)}}=1$$ If$(1-p^2)>0$, then the equation will take the form of a shifted ellipse. $\frac{(x+x_o)^2}{a^2}+\frac{y^2}{b^2}=1$ $$1-p^2>0$$ $$p^2-1<0$$ $$(p-1)(p+1)<0$$ $$-1<p<1$$ $$-1<\frac{v_o^2}{(\frac{GM}{r_o})}-1<1$$ $$0<\frac{v_o^2}{(\frac{GM}{r_o})}<2$$ $$0<v_o^2<\frac{2GM}{r_o}$$ $$0<v_o<\sqrt{\frac{2GM}{r_o}}$$ Hence for a suitable velocity, the planet will be orbiting Sun in an elliptical path.

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  • $\begingroup$ Here I have used the basic idea of polar co-ordinates; at the same time, have kept the analysis at high school level. $\endgroup$ Mar 28 '20 at 7:22
  • $\begingroup$ The only idea that I've borrowed from polar co-ordinate system is $$\vec{r}=r cos\theta\hat{i}+r sin\theta\hat{j}$$ $\endgroup$ Mar 28 '20 at 7:26
  • $\begingroup$ Nice. But I think that your students will get a message of mathematical complexity, missing a lot of much more interesting physics. For example, why not spending the same amount of time to discuss the more interesting point of the passage from the true two-body problem to the effective one-body description, instead of introducing the simplifying hypothesis of a fixed Earth which here appears like an approximation? $\endgroup$
    – GiorgioP
    Mar 29 '20 at 18:27
  • $\begingroup$ @Vilvanesh: The equation $\vec{r}=r cos(\theta)\hat{i}+r \sin(\theta)\hat{j}$ is the polar coordinate transformation, namely, $x=\cos\theta$ and $y=\sin\theta$. The best way teach them polar coordinates to shown them how you converted $r=\frac{h}{1+p\cos(\theta)}$ to Cartesian coordinates. But you should write $r$ and trig functions in Cartesian coordinates too otherwise the conversion is half baked. $\endgroup$ Mar 29 '20 at 19:42
  • $\begingroup$ @GiorgioP Thank you. As I told, I understand that it's complex and lengthy one. But my perspective is that this will give them confidence that whatever they know right now(which will not include polar co-ordinates) is good enough to tackle the problem. I just want to increase their confidence by giving them a feel that "I'm well equipped". That's all. $\endgroup$ Mar 30 '20 at 0:10
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Indeed, as already suggested by other answers, it is straightforward to transform the ellipse from Cartesian co-ordinates to polar co-ordinates even for students not familiar with polar co-ordinates.

Once you have expressed the ellipse in terms of r and 𝜃, the attached figure proves, using only the most elementary calculus in Cartesian coordinates, that the ellipse is a solution. The only assumptions are Newtonian (square law) gravity and conservation of angular momentum.

proof of Kepler1

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  • $\begingroup$ In its present form, there is no attached figure to this answer. $\endgroup$
    – GiorgioP
    Jul 9 at 4:13

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