Eccentricity of Orbit indicates its Total Energy

From ProofWiki
Jump to navigation Jump to search

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
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2}\) 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
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {G M m} r\)
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle -\dfrac {k m} r\) 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:

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

$\blacksquare$


Sources