Given a and T as average radius and orbit, Kepler's third law says that T2 is proportional to a3. That is, T2= Ka3.
If your graph of log(T) against log(a) is a straight line, then y= mx+ b or
log(T)= m log(a)+ b= log(am)+ b so that T= eb am.
The most important you should be able to derive from that is that your slope, m, should be, approximately, 3/2.
If you are really interested in what that "K" is, here is what I just did:
Since elliptic orbits are a nuisance, let's approximate by a circular orbit: imagine a satellite of mass m orbiting a body of mass M, in a circular orbit of radius R with angular velocity ω. Taking the center of the orbit as the origin of coordinate system, we can write the position vector as x= <R cos(ω t), R sin(ω t)>.
Differentiating gives the speed vector: v= x'= <-Rω sin(ω t), Rω cos(ωt)>.
Differentiating again gives the acceleration vector a= v'= <-Rω2 cos(ωt), -Rω2 sin(ω t) .
That is, the scalar value for acceleration is Rω2.
Since the force of gravity is \frac{-GmM}{R^2} and F= ma
we must have a= R\omega^2= \frac{GM}{R^2} or
\omega^2= \frac{GM}{R^3}.
Since sine and cosine both have period 2\pi, if T is the period of the orbit, \omega T= 2\pi so \omega = \frac{2\pi}{T} and \omega^2= \frac{GM}{R^3} becomes \frac{T^2}{4\pi^2}= \frac{R^3}{GM}. That is the "K" in T2= Ka3 is \frac{4\pi^2}{GM}. Since you graphed log(T) against log(a), your y-intercept is 1/2 the logarithm of that:
\frac{1}{2}log(\frac{4\pi^2}{GM})!
(What I did here, really, was use Newton's gravity formula to derive Kepler's third law. Newton did it the other way around: used Kepler's third law to derive the formula for gravity. Of course, he had to develop calculus in order to be able to do that!)
