Hi Sigblips

Because the observation is at high frequency (~ 8 GHz), a drift rate of 0.5 Hz/s is quite reasonable. The Doppler effect is related to velocity (for velocities << c) by

f = fo * (1- v/c)                               (http://en.wikipedia.org/wiki/Doppler_effect)

where f is the observed frequency, fo is the emitted frequency, c is the speed of light, and v is the radial velocity between of emitter in the rest frame of the receiver. Taking the derivative w.r.t. time, we get the drift rate R

R = df / dt = - fo * a / c.

where a is the acceleration. So for R = 0.5 and fo = 8 GHz, the acceleration is -0.019 m / s^2. Did I do that right?

This is compared with the centripetal acceleration on the equator caused by Earth rotation (given that earth radius (r_e) is 6000 km and earth rotates once per day)

a_earthmax = v_e^2 / r_e                    (http://en.wikipedia.org/wiki/Centripetal_force)

where v_e = 2 * pi * r_e / (24 * 3600 seconds per day)  = 436 m/s.

Hence a_earthmax = -0.032 m / s^2. This is less by about a factor of 2 from the observed drift rate. Notice that since we can't see through the earth, the acceleration and drift rate are always negative (but that doesn't include acceleration at the transmitter, which could be present but not for Voyager).

The observed drift rate should be smaller depending on the angle of the arriving signal. For example, the rotation velocity decreases with increasing latitude as cos(latitude). Also, when Voyager appears overhead (that is, on the meridian), you get the maximum drift rate. On the horizon the drift rate is zero (acceleration is zero) where the velocity is at maximum. Also a sinusoidal variation, but complicated by the direction of arrival of the radio waves.

Gerry