Proof

Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function [latex]{\bf{r}}\,(t)[/latex] represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as [latex]{\bf{F}}=m{\bf{a}}[/latex], where [latex]{\bf{F}}[/latex] represents the net force acting on the planet. His law of universal gravitation can be written in the form [latex]{\bf{F}}=-\frac{GmM}{\left\Vert{\bf{r}}\right\Vert^{2}}\cdot\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}[/latex], which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude [latex]\frac{GmM}{\left\Vert{\bf{r}}\right\Vert^{2}}[/latex] (Figure 9).

Setting these two forces equal to each other, and using the fact that [latex]{\bf{a}}\,(t)={\bf{v}}'\,(t)[/latex], we obtain

[latex]\large{m{\bf{v}}'\,(t)=-\frac{GmM}{\left\Vert{\bf{r}}\right\Vert^{2}}\cdot\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}}[/latex],

which can be rewritten as

[latex]\large{\frac{d{\bf{v}}}{dt}=-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}{\bf{r}}}.[/latex]

This equation shows that the vectors [latex]d{\bf{v}}/dt[/latex] and [latex]{\bf{r}}[/latex] are parallel to each other, so [latex]d{\bf{v}}/dt\,\times\,{\bf{r}}=0[/latex]. Next, let’s differentiate [latex]{\bf{r}}\,\times\,{\bf{v}}[/latex] with respect to time:

[latex]\large{\frac{d}{dt}({\bf{r}}\,\times\,{\bf{v}})=\frac{d{\bf{r}}}{dt}\,\times\,{\bf{v}}+{\bf{r}}\,\times\frac{d{\bf{v}}}{dt}={\bf{v}}\,\times\,{\bf{v}}+{\bf{0}}={\bf{0}}}.[/latex]

This proves that [latex]{\bf{r}}\,\times\,{\bf{v}}[/latex] is a constant vector, which we call [latex]{\bf{C}}[/latex]. Since [latex]{\bf{r}}[/latex] and [latex]{\bf{v}}[/latex] are both perpendicular to [latex]{\bf{C}}[/latex] for all values of [latex]t[/latex], they must lie in a plane perpendicular to [latex]{\bf{C}}[/latex]. Therefore, the motion of the planet lies in a plane.

Next we calculate the expression [latex]d{\bf{v}}/dt\,\times\,{\bf{C}}[/latex]:

[latex]\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}=-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}{\bf{r}}\,\times\,({\bf{r}}\,\times\,{\bf{v}})=-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}[({\bf{r}}\cdot{\bf{v}})\,{\bf{r}}-({\bf{r}}\cdot{\bf{r}})\,{\bf{v}}].[/latex]

The last equality in the above equation is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for [latex]{\bf{r}}\cdot{\bf{v}}[/latex]. To calculate this, we differentiate [latex]{\bf{r}}\cdot{\bf{r}}[/latex] with respect to time:

[latex]\frac{d}{dt}({\bf{r}}\,\cdot\,{\bf{r}})=\frac{d{\bf{r}}}{dt}\cdot{\bf{r}}+{\bf{r}}\cdot\frac{d{\bf{r}}}{dt}=2{\bf{r}}\cdot\frac{d{\bf{r}}}{dt}=2{\bf{r}}\cdot{\bf{v}}.[/latex]

Since [latex]{\bf{r}}\,\cdot\,{\bf{r}}=\left\Vert{\bf{r}}\right\Vert^{2}[/latex], we also have

[latex]\frac{d}{dt}({\bf{r}}\cdot{\bf{r}})=\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert^{2}=2\left\Vert{\bf{r}}\right\Vert\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert[/latex].

Combining the last two equations, we get

[latex]\begin{array}{ccc}\hfill {2{\bf{r}}\cdot{\bf{v}}} & =\hfill & {2\left\Vert{\bf{r}}\right\Vert\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert} \hfill \\ \hfill {{\bf{r}}\cdot{\bf{v}}}& =\hfill & {\left\Vert{\bf{r}}\right\Vert\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert{.}}\hfill \\ \hfill \end{array}[/latex]

Substituting into our expression for [latex]\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}[/latex] gives us

[latex]\begin{array}{ccc}\hfill {\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}} & =\hfill & {-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}[({\bf{r}}\cdot{\bf{r}})\,{\bf{r}}-({\bf{r}}\cdot{\bf{r}})\,{\bf{v}}]} \hfill \\ \hfill & =\hfill & {-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}\big[\left\Vert{\bf{r}}\right\Vert{\big(}\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert\big)\,{\bf{r}}-\left\Vert{\bf{r}}\right\Vert^{2}{\bf{v}}\big]}\hfill \\ \hfill & =\hfill & {-GM\Big[\frac{1}{\left\Vert{\bf{r}}\right\Vert^{2}}\big(\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert\big)\,{\bf{r}}-\frac{1}{\left\Vert{\bf{r}}\right\Vert}{\bf{v}}\Big]}\hfill \\ \hfill & =\hfill & {GM\Big[\frac{\bf{v}}{\left\Vert{\bf{r}}\right\Vert}-\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert^{2}}\big(\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert\big)\Big].}\hfill \\ \hfill \end{array}[/latex]

However,

[latex]\begin{array}{ccc}\hfill {\frac{d}{dt}\,\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}} & =\hfill & {\frac{\frac{d}{dt}({\bf{r}})\left\Vert{\bf{r}}\right\Vert-{\bf{r}}\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert}{\left\Vert{\bf{r}}\right\Vert}} \hfill \\ \hfill & =\hfill & {\frac{\frac{d{\bf{r}}}{dt}}{\left\Vert{\bf{r}}\right\Vert}-\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert^{2}}\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert}\hfill \\ \hfill & =\hfill & {\frac{\bf{v}}{\left\Vert{\bf{r}}\right\Vert}-\frac{{\bf{r}}}{\left\Vert{\bf{v}}\right\Vert^{2}}\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert.}\hfill \\ \hfill \end{array}[/latex]

Therefore, we get

[latex]\large{\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}=GM\Big(\frac{d}{dt}\,\frac{\bf{r}}{\left\Vert{\bf{r}}\right\Vert}\Big)}.[/latex]

Since [latex]\bf{C}[/latex] is a constant vector, we can integrate both sides and obtain

[latex]{\bf{v}}\,\times\,{\bf{C}}=GM\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}+{\bf{D}},[/latex]

where [latex]\bf{D}[/latex] is a constant vector. Our goal is to solve for [latex]\left\Vert{\bf{r}}\right\Vert[/latex]. Let’s start by calculating [latex]{\bf{r}}\cdot({\bf{v}}\,\times\,{\bf{C}})[/latex]:

[latex]{\bf{r}}\cdot({\bf{v}}\,\times\,{\bf{C}})={\bf{r}}\cdot\Big(GM\frac{\bf{r}}{\left\Vert{\bf{r}}\right\Vert}+{\bf{D}}\Big)=GM\frac{\left\Vert{\bf{r}}\right\Vert^{2}}{\left\Vert{\bf{r}}\right\Vert}+{\bf{r}}\cdot{\bf{D}}=GM\left\Vert{\bf{r}}\right\Vert+{\bf{r}}\cdot{\bf{D}}.[/latex]

However, [latex]{\bf{r}}\cdot({\bf{v}}\,\times\,{\bf{C}})=({\bf{r}}\,\times\,{\bf{v}})\,\times\,{\bf{C}}[/latex], so

[latex]({\bf{r}}\,\times\,{\bf{v}})\,\times\,{\bf{C}}=GM\left\Vert{\bf{r}}\right\Vert+{\bf{r}}\cdot{\bf{D}}[/latex].

Since [latex]{\bf{r}}\,\times\,{\bf{v}}={\bf{C}}[/latex], we have

[latex]\left\Vert{\bf{C}}\right\Vert^{2}=GM\left\Vert{\bf{r}}\right\Vert+{\bf{r}}\cdot{\bf{D}}[/latex].

Note that [latex]{\bf{r}}\cdot{\bf{D}}=\left\Vert{\bf{r}}\right\Vert\,\left\Vert{\bf{D}}\right\Vert\cos{\theta}[/latex], where [latex]\theta[/latex] is the angle between [latex]\bf{r}[/latex] and [latex]\bf{D}[/latex]. Therefore,

[latex]\left\Vert{\bf{C}}\right\Vert^{2}=GM\left\Vert{\bf{r}}\right\Vert+\left\Vert{\bf{r}}\right\Vert\,\left\Vert{\bf{D}}\right\Vert\cos{\theta}[/latex].

Solving for [latex]\left\Vert{\bf{r}}\right\Vert[/latex],

[latex]\left\Vert{\bf{r}}\right\Vert=\frac{\left\Vert{\bf{C}}\right\Vert^{2}}{GM+\left\Vert{\bf{D}}\right\Vert\cos{\theta}}=\frac{\left\Vert{\bf{C}}\right\Vert^{2}}{GM}\bigg(\frac{1}{1+e\cos{\theta}}\bigg)[/latex],

where [latex]e=\left\Vert{\bf{D}}\right\Vert/GM[/latex]. This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if [latex]e\,>\,1[/latex], a parabola if [latex]e=1[/latex], or an ellipse if [latex]e\,<\,1[/latex]. Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic. [latex]_\blacksquare[/latex]