# Eccentricity of Orbit indicates its Total Energy

## Theorem

Consider a planet $p$ of mass $m$ orbiting a star $S$ of mass $M$ under the influence of the gravitational field which the two bodies give rise to.

Then the total energy of the system determines the eccentricity of the orbit of $p$ around $S$.

## Proof

Let:

- $\mathbf u_r$ be the unit vector in the direction of the radial coordinate of $p$
- $\mathbf u_\theta$ be the unit vector in the direction of the angular coordinate of $p$.

By Kinetic Energy of Motion, the kinetic energy of $p$ is:

- $K = \dfrac {m v^2} 2$

where $v$ is the magnitude of the velocity of $p$.

Thus:

\(\displaystyle K\) | \(=\) | \(\displaystyle \dfrac {m \mathbf v \cdot \mathbf v} 2\) | Dot Product of Vector with Itself | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac 1 2 m \paren {r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r} \cdot \paren {r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r}\) | Velocity Vector in Polar Coordinates | ||||||||||

\((1):\quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2}\) | Definition of Dot Product |

The potential energy $P$ of the system is the negative of the work required to move $p$ to infinity:

\(\displaystyle P\) | \(=\) | \(\displaystyle -\int_r^\infty \dfrac {G M m} {r^2} \rd r\) | Newton's Law of Universal Gravitation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \intlimits {\dfrac {G M m} r} r \infty\) | Primitive of Power | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\dfrac {G M m} r\) | |||||||||||

\((2):\quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle -\dfrac {k m} r\) | defining $k = G M$: constant for this system |

By the Principle of Conservation of Energy, the total energy in the system remains constant: $E$, say.

So:

- $E = \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2} - \dfrac {k m} r$

Without loss of generality, let us arrange the polar axis so as to make $r$ a minimum when $\theta = 0$.

By Kepler's First Law of Planetary Motion, the position of $p$ in polar coordinates is:

- $(3): \quad r = \dfrac {h^2 / k} {1 + e \cos \theta}$

At the instant when $\theta = 0$, we therefore have:

- $r = \dfrac {h^2 / k} {1 + e}$

At this point, $r$ is at a local minimum.

Hence:

- $\dfrac {m r^2} 2 \dfrac {h^2} {r^4} - \dfrac {k m} r = E$

Eliminating $r$ from these gives:

- $e = \sqrt {1 + E \paren {\dfrac {2 h^2} {m k^2} } }$

Thus equation $(3)$ for the orbit of $p$ can be written as:

- $r = \dfrac {h^2 / k} {1 + \sqrt {1 + E \paren {2 h^2 / m k^2} \cos \theta} }$

Thus from Equation of Conic Section in Polar Form, it can be seen that the orbit is:

$\blacksquare$

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 3.21$: Newton's Law of Gravitation