Proof

The proof starts with the two body equation of motion, derived from Newton's law of universal gravitation:

$${\displaystyle {\ddot {\mathbf {r} }}+{\frac {Gm_{1}}{r^{2}}}{\frac {\mathbf {r} }{r}}=0}$$

where:

• ${\displaystyle \mathbf {r} }$ is the position vector from ${\displaystyle m_{1}}$ to ${\displaystyle m_{2}}$ with scalar magnitude ${\displaystyle r}$.
• ${\displaystyle {\ddot {\mathbf {r} }}}$ is the second time derivative of ${\displaystyle \mathbf {r} }$. (the acceleration)
• ${\displaystyle G}$ is the Gravitational constant.

The cross product of the position vector with the equation of motion is:

$${\displaystyle \mathbf {r} \times {\ddot {\mathbf {r} }}+\mathbf {r} \times {\frac {Gm_{1}}{r^{2}}}{\frac {\mathbf {r} }{r}}=0}$$

Because ${\displaystyle \mathbf {r} \times \mathbf {r} =0}$ the second term vanishes:

$${\displaystyle \mathbf {r} \times {\ddot {\mathbf {r} }}=0}$$

It can also be derived that:

$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)={\dot {\mathbf {r} }}\times {\dot {\mathbf {r} }}+\mathbf {r} \times {\ddot {\mathbf {r} }}=\mathbf {r} \times {\ddot {\mathbf {r} }}}$$

Combining these two equations gives:

$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)=0}$$

Since the time derivative is equal to zero, the quantity ${\displaystyle \mathbf {r} \times {\dot {\mathbf {r} }}}$ is constant. Using the velocity vector ${\displaystyle \mathbf {v} }$ in place of the rate of change of position, and ${\displaystyle \mathbf {h} }$ for the specific angular momentum:

$${\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} }$$

is constant.

This is different from the normal construction of momentum, ${\displaystyle \mathbf {r} \times \mathbf {p} }$, because it does not include the mass of the object in question.

First law

The proof starts again with the equation of the two-body problem. This time one multiplies it (cross product) with the specific relative angular momentum

$${\displaystyle {\ddot {\mathbf {r} }}\times \mathbf {h} =-{\frac {\mu }{r^{2}}}{\frac {\mathbf {r} }{r}}\times \mathbf {h} }$$

The left hand side is equal to the derivative ${\textstyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\dot {\mathbf {r} }}\times \mathbf {h} \right)}$ because the angular momentum is constant.

After some steps (which includes using the vector triple product and defining the scalar ${\displaystyle {\dot {r}}}$ to be the radial velocity, as opposed to the norm of the vector ${\displaystyle {\dot {\mathbf {r} }}}$) the right hand side becomes:

$${\displaystyle -{\frac {\mu }{r^{3}}}\left(\mathbf {r} \times \mathbf {h} \right)=-{\frac {\mu }{r^{3}}}\left(\left(\mathbf {r} \cdot \mathbf {v} \right)\mathbf {r} -r^{2}\mathbf {v} \right)=-\left({\frac {\mu }{r^{2}}}{\dot {r}}\mathbf {r} -{\frac {\mu }{r}}\mathbf {v} \right)=\mu {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\mathbf {r} }{r}}\right)}$$

Setting these two expression equal and integrating over time leads to (with the constant of integration ${\displaystyle \mathbf {C} }$)

$${\displaystyle {\dot {\mathbf {r} }}\times \mathbf {h} =\mu {\frac {\mathbf {r} }{r}}+\mathbf {C} }$$

Now this equation is multiplied (dot product) with ${\displaystyle \mathbf {r} }$ and rearranged

$${\displaystyle {\begin{aligned}\mathbf {r} \cdot \left({\dot {\mathbf {r} }}\times \mathbf {h} \right)&=\mathbf {r} \cdot \left(\mu {\frac {\mathbf {r} }{r}}+\mathbf {C} \right)\\\Rightarrow \left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)\cdot \mathbf {h} &=\mu r+rC\cos \theta \\\Rightarrow h^{2}&=\mu r+rC\cos \theta \end{aligned}}}$$

Finally one gets the orbit equation^[1]

$${\displaystyle r={\frac {\frac {h^{2}}{\mu }}{1+{\frac {C}{\mu }}\cos \theta }}}$$

which is the equation of a conic section in polar coordinates with semi-latus rectum ${\textstyle p={\frac {h^{2}}{\mu }}}$ and eccentricity ${\textstyle e={\frac {C}{\mu }}}$.

Second law

The second law follows instantly from the second of the three equations to calculate the absolute value of the specific relative angular momentum.^[1]

If one connects this form of the equation ${\textstyle \mathrm {d} t={\frac {r^{2}}{h}}\,\mathrm {d} \theta }$ with the relationship ${\textstyle \mathrm {d} A={\frac {r^{2}}{2}}\,\mathrm {d} \theta }$ for the area of a sector with an infinitesimal small angle ${\displaystyle \mathrm {d} \theta }$ (triangle with one very small side), the equation

$${\displaystyle \mathrm {d} t={\frac {2}{h}}\,\mathrm {d} A}$$

Third law

Kepler's third is a direct consequence of the second law. Integrating over one revolution gives the orbital period^[1]

$${\displaystyle T={\frac {2\pi ab}{h}}}$$

for the area ${\displaystyle \pi ab}$ of an ellipse. Replacing the semi-minor axis with ${\displaystyle b={\sqrt {ap}}}$ and the specific relative angular momentum with ${\displaystyle h={\sqrt {\mu p}}}$ one gets

$${\displaystyle T=2\pi {\sqrt {\frac {a^{3}}{\mu }}}}$$

There is thus a relationship between the semi-major axis and the orbital period of a satellite that can be reduced to a constant of the central body.