If you use the semimajor axis of an elliptical orbit, isn't it the same as using the radius of a circular orbit? I don't know why, but I've heard that it has something to do with the energy of an orbit staying constant. Not sure though, but thanks a ton!
Oops - neglected to check this thread, sorry! (Astronomy usually isn't my thing, but orbital dynamics is.)
The answer to your question is no - it isn't the same. If an elliptical orbit is circular (think of a circle as an ellipse with zero eccentricity), then the semimajor axis is equal to the radius of the circle. Moreover, however, the distance between the two masses is always constant and equal to the radius. For an elliptic orbit with nonzero eccentricity, however, the radius is not defined, but that distance (between the two masses) is changing continuously with time.
When you formulate the equation for energy in an orbit, it's the distance between the two bodies that you're concerned with, which happens to be constant and equal to the radius in the circular case - but not in the eccentric case. Let's look at the formulation:
Define
to be the distance between the two bodies. Note that this is also equivalent to the radius of a circular orbit, but this is
not the formal definition.
If we assume that mass
is stationary (inertially fixed), then the total kinetic energy of the system is:
or just due to the motion of
. The potential energy is:
Note that here, I've used
, which is known as the "gravitational parameter." This is common in astrodynamics use, as for any given body,
can be determined much more precisely than G. If
, as is the case the orbit of a planet around a star, etc, then we can say
. I'll proceed with this definition, but analytically you could proceed forward with either. (Note:
is more commonly used in astrodynamics than G, because for a given body, it can be determined to much greater precision than the universal constant G. I've never seen
used in high school physics courses or the like - they usually stick to the formulation with G, but it's what orbital dynamics people use.)
Now, we can combine the two together to get the total energy:
Great, right? Well yes, but we still have
, or the velocity in the equation. We know that for an elliptic orbit, this is related to
by other orbital parameters.
Let's now look at the Vis-Viva equation. Note that this is actually derived from the fact that energy
and angular momentum must be constant throughout an orbit - so the energy at apogee is equal to the energy at perigee, and the angular momentum at apogee is equal to the angular momentum at perigee. I'll skip the derivation here, but Wikipedia's derivation is clear if you're interested. Anyway, the Vis-Viva equation:
where
is the semimajor axis.
Now, let's substitute the Vis-Viva equation into the total energy equation:
This simplifies to:
So total energy in an orbit is not a function of the current position
, but of the semimajor axis
! This makes sense, as we know the total energy cannot change in an orbit.
Now, if your orbit is circular, then this becomes
as
is constant (
only for the circular orbit). Then, we know the potential energy at any point (since
is constant!) is
- or total energy is half the potential energy. So we've proven this for the circular case! But what happens in the elliptical case?
The answer is, since
, we can't say anything about the kinetic or potential energies for an arbitrary point, except that they must equal the total mechanical energy,
unless we know . If we're concerned about actually propagating the orbit, there are a few ways to determine
, using a choice of anomaly (true, eccentric, or mean) to determine the position vector, and then taking the magnitude of the position vector. For periapsis and apoapsis, however, we have the relations:
We can use these to determine the kinetic and potential energy components for periapsis and apoapsis. At periapsis, potential energy is
and at apoapsis, potential energy is
So yes, it has to do with the energy (and momentum) of an orbit being constant, which will give you the Vis-Viva equation. Substitute this into the definition of kinetic and potential energies, and the result follows.
Hope this makes sense!