Is orbital eccentricity determined by specific mechanical energy and angular momentum?

[Contadoren11]

Hi

I'm really at a loss: How should this formula be interpreted? Is e simply dependent on the specific mechanical energy of, say, a planet in orbit around the Sun as well as its angular momentum?

 

[Khashishi]

You need to provide some context. Where it's from and what the symbols mean.

 

[marcus]

Here's some context indicating what the symbols mean:
==quote Wikip. ==
...or in the case of a gravitational force:

where

is the specific orbital energy (total energy divided by the reduced mass),

the standard gravitational parameter based on the total mass, and

the specific relative angular momentum (angular momentum divided by the reduced mass).
==endquote==

In a two body problem the "reduced mass" is Mm/(M+m)

 

[tfr000]

Hi

I'm really at a loss: How should this formula be interpreted? Is e simply dependent on the specific mechanical energy of, say, a planet in orbit around the Sun as well as its angular momentum?

I'd say the answer to your question is "yes", but I think you're looking for more than that. Maybe you should elaborate.