DanHelzer said:
So then why can r be replaced with a if the orbits are not circular? mv^2/r, as you stated, applies to a circular orbit. How then is it possible to simply replace "r" with "a" in the law of periods equation of Kepler's third law?
You have to take into consideration a couple of things.
First, the total energy (kinetic energy+ gravitational potential energy) of any orbit is a constant and at any point of the orbit can be found with:
(1)E = \frac{mv^2}{2}-\frac{GMm}{r}
Where m is the mass of the orbiting body (and in this case, much smaller than M)
Thus if you take the above formula and use the perihelion distance (r
P) and aphelion distance (r
A) for r, you get two equations that equal each other, with this, and knowing that r
P+r
A= a(the semi-major axis), you can work out that the total energy of an elliptical orbit can also be found by
(2)E= -\frac{GMm}{a}
And by equating eq(1) and eq(2), we can solve for v:
(3)v = \sqrt{GM \left ( \frac{2}{r}- \frac{1}{a}\right )}
Another aspect of elliptical orbits is that they sweep out equal volumes in equal times(Kepler's 2nd law)
Now the "Areal" velocity(the "area" it sweeps out at any given instant) of the object at any point of its orbit is.
v_A = \frac{rv \cos \phi}{2}
\phi is the angle between the object's velocity vector and the line perpendicular to the line joining the object and the focus of the orbit.
This, due to Kepler's 2nd law is also a constant throughout the orbit, Thus we can get the Areal velocity for all points of the orbit by finding it at any point of the orbit
Two particular points of interest are aphelion and perihelion, where \phi=0 and thus \cos \phi = 1, so it makes sense use one of these points as we continue.
For any orbit of eccentricity e, the perihelion distance is equal to
(4)r_P=a(1-e)
Thus at perhelion:
(5)v_A = \frac{a(1-e)v_P}{2}
Where v
P is the velocity at perihelion
And since this is a constant and the orbit sweeps out the entire area of its orbit in one orbital period and area of an ellipse is:
(6) \pi a^2 \sqrt{1-e^2}
We can find the period of the orbit with by dividing the area of the ellipse by the Areal velocity.
(7) T=\frac{\pi a^2 \sqrt{1-e^2}}{\frac{a(1-e)v_P}{2}}
(8) T=\frac{2\pi a \sqrt{1-e^2} }{(1-e)v_P}
(9)T=\frac{2 \pi a \sqrt{\frac{1+e}{1-e}}}{v_P}
To find v
P we go back to eq(3) and substitute a(1-e) for r and reduce:
(10) v_P = \sqrt{\frac{GM}{a} \frac{1+e}{1-e}}
(11) v_P = \sqrt{\frac{GM}{a}} \sqrt{ \frac{1+e}{1-e}}
Substitute for v
P in eq(9)
(12)T=\frac{2 \pi a \sqrt{\frac{1+e}{1-e}}}{ \sqrt{\frac{GM}{a}} \sqrt{ \frac{1+e}{1-e}}}
(13)T=\frac{2 \pi a}{ \sqrt{\frac{GM}{a}}}
(14)T= 2 \pi \sqrt{\frac{a^3}{GM}}
As you will note, e drops out of the equation completely leaving only a, and the equation is the same as for a circular orbit with a for r. (technically, a circle is an ellipse with an eccentricity of 0, and in this case, the semi-major axis is equal to the radius)