Eccentricity of Orbit indicates its Total Energy

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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:

\(\displaystyle K\) \(=\) \(\displaystyle \dfrac {m \mathbf v \cdot \mathbf v} 2\) $\quad$ Dot Product of Vector with Itself $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 2 m \left({r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r}\right) \cdot \left({r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r}\right)\) $\quad$ Velocity Vector in Polar Coordinates $\quad$
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 2 m \left({r^2 \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2 + \left({\dfrac {\mathrm d r} {\mathrm d t} }\right)^2}\right)\) $\quad$ Definition of Dot Product $\quad$


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



\(\displaystyle P\) \(=\) \(\displaystyle - \int_r^\infty \dfrac {G M m} {r^2} \, \mathrm d r\) $\quad$ Newton's Law of Universal Gravitation $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left.{\dfrac {G M m} r}\right\vert_r^\infty\) $\quad$ Primitive of Power $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {G M m} r\) $\quad$ $\quad$
\((2):\quad\) \(\displaystyle \) \(=\) \(\displaystyle -\dfrac {k m} r\) $\quad$ defining $k = G M$: constant for this system $\quad$

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

So:

$E = \dfrac 1 2 m \left({r^2 \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2 + \left({\dfrac {\mathrm d r} {\mathrm d t} }\right)^2}\right) - \dfrac {k m} r$


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:

$e = \sqrt {1 + E \left({\dfrac {2 h^2} {m k^2} }\right)}$

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

$r = \dfrac {h^2 / k} {1 + \sqrt {1 + E \left({2 h^2 / m k^2}\right) \cos \theta} }$

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

an ellipse when $E < 0$
a parabola when $E = 0$
a hyperbola when $E > 0$.

$\blacksquare$


Sources