Ceres motion from frame 2-600:

dx = 11 pixels * 1.117"/pixel = 12.2870"
dy = 2 pixels *1.074"/pixel = 2.1480"

dr = 12.4733"

dt = 2:52:28am - 2:29:26am = 1382 sec

--> 4.3757e-8 rad/sec

From this, determine how far from Sun Ceres orbits.

acceleration = (G*M_Sun)/r^2 = v^2/r

At opposition, the difference between Earth's, Ceres's velocity determines the angular motion:

d(theta)/dt = (v_Earth - v_Ceres)/(r_Ceres - r_Earth)

let r_Ceres = a*r_Earth

d(theta)/dt = (v_Earth - sqrt(G*M_Sun/r_Ceres)) / (r_Earth*(a-1))

            = (v_Earth - sqrt(G*M_Sun/(a*r_Earth)) / (r_Earth*(a-1))

            = (v_Earth - v_Earth/sqrt(a)) / (r_Earth*(a-1))

            = (v_Earth / r_Earth) * (1-1/sqrt(a)) / (a-1)

let b = d(theta)/dt * r_Earth / v_Earth

      = d(theta)/dt * r_Earth / (2*pi*r_Earth / T_Earth)

      = d(theta)/dt * T_Earth / (2*pi)  (Eq. 1)

Measured value: b = 0.21962

(1-1/sqrt(a)) / (a-1) = b   (Eq. 2)

(sqrt(a) - 1) / (sqrt(a) (sqrt(a)+1) (sqrt(a)-1)) = b

drop a=1 solution,

a + sqrt(a) - 1/b = 0

keeping only sqrt(a)>0 solution (assuming b>0),

sqrt(a) = -1/2 + (1/2)sqrt(1 + 4/b)

a = (1/4) (sqrt(1 + 4/b) - 1)^2

---> a = 2.862

Actual value for Ceres:

a = 2.765349 (Observer's Handbook 2010)

:)