# Correlation between orbit eccentricity and mechanical energy

1. Dec 16, 2015

1. The problem statement, all variables and given/known data
I have been tasked with showing "how the mechanical energy of a planet determines the shape of its orbit", and I cannot for the life of me make sense of it. I've run into a formula, see below, but I'm not sure how to interpret it nor if it even applies in my case at all (as E is apparently the total mechanic energy, not just the planet's (or something): https://en.wikipedia.org/wiki/Orbital_eccentricity)

2. Relevant equations

3. The attempt at a solution
None. I'm dumbfounded.

Last edited: Dec 16, 2015
2. Dec 16, 2015

### Staff: Mentor

The total mechanical energy for a body in orbit includes both its kinetic energy (due to its velocity) and its potential energy (due to gravity). That is,

$E = m\frac{v^2}{2} - m\frac{\mu}{r}$

More commonly used in astrodynamics is the specific mechanical energy, which is the same thing only the mass m is left out:

$ξ = \frac{v^2}{2} - \frac{\mu}{r}$

That gives you the energy per unit mass (i.e. Joules per kg) for an object with that velocity and distance from the primary. It's very useful because for the most part an orbit is pretty much independent of the mass of the planet, the Sun's mass being so much greater.

But back to your query about orbit shapes. What do you know about orbits shapes and eccentricity? What orbit shapes are there and what ranges or values of eccentricity are associated with them?

3. Dec 16, 2015

I should have been more clear, my apologies (truly!): I've been asked specifically for planetary orbits; I know that for E>0 and E=0 the curve becomes a hyperbola and parabola, respectively, but that is hardly relevant for planets.

4. Dec 16, 2015

### Staff: Mentor

Ah. Well then, you have a bit of a problem. All bound orbits have total energy < 0. So, all circular and elliptical orbits have ξ < 0. When the energy is zero or positive the orbit is unbound (parabolic, hyperbolic). That's the best you can do to use energy to distinguish the eccentricities.

For bound orbits in particular the total energy does not determine the eccentricity. You can have circular or elliptical orbits of any negative energy value (well, disregarding practical details like the Sun not being a point mass so the minimum orbit size is limited by its radius). The magnitude of the energy only gives you the orbit size (semimajor axis for ellipses, radius for circles). Any eccentricity between 0 and 1 is fair game for any negative energy value.

5. Dec 16, 2015

Hm, alright, I began to suspect as much. Thanks!

Now that I've got you, in case you know your Lagrangian points.... -- I'm having trouble understanding the stability of the stable Lagrangian points (L_4 and L_5); Wikipedia explains that if an object in the L_4 or L_5 of a planet is pushed closer towards the common center of gravity of the Sun and the planet, the increased speed that comes from a reduced distance to the center of gravity (due to conserved angular momentum) compensates for this (and the other way around (i.e. if it's pushed farther from the center of gravity)). I don't understand it intuitively nor can I explain it physically, and Google doesn't seem to yield any satisfying explanation either. Grr.

6. Dec 17, 2015

### Staff: Mentor

Working out the stability of the L4 and L5 Lagrange points is not trivial. It involves finding the potential function around the point and showing that there's a relative minimum there. Objects perturbed slightly from such a point tend to orbit the minimum. I've seen the analysis done in a rotating frame of reference so that the two masses (Sun and planet) are fixed. In that non-inertial frame there's an inherent acceleration due to position (centrifugal acceleration) and Coriolis effects.

You might try a web search including "stability", and look for some scholarly articles.