For this "tiny" acceleration towards the Sun, there's an easy way to figure the gravity growth equivalent, taking inertial mass as equivalent to gravity mass, so if G is growing, then inertial mass is growing by like proportion: Conceptually, if distant G is growing, then dividing it by AU distance would give us what is that rate of growth per AU; and dividing that by the known Newton's G constant would give us the rate of growth of G per Newton's G, which would account for the "acceleration towards the Sun". Now taking it the other way, starting with that acceleration towards the Sun, we reverse the process for the Pioneers accelerating at ~8e-8 cm/s^2 towards the Sun:

a (towards Sun) = 8e-8 cm/s^2 = 8e-10 m/s^2

multiply by distance in meters of one astronomical unit (AU), distance of Earth from Sun:

a*d per AU = 8e-10 m/s^2 times 150e9 m = 1200e^-1 m^2/s^2 = 1.20e^2 m^2/s^2

multiply this by (Earth's) G, as in F = GMm/r^2:

a*d per AU*G = 1.20e2 m^2/s^2 times 6.67e-11 m^3 kg^-1 s^-2 = 8.00e-9 m^5 kg^-1 s^-4,

which represent G growth (per m^2 per s^2) per AU, or per acceleration and distance (m^2/s^2),

so that G/AU = ~8.0e-9 m^3 kg^-1 s^-2 (m^2/s^2).

However, this number is actually high (by a power of two), which means "IF" the Pioneers are slowing due to increasing G, the real G/AU number is probably lower (more like ~8e-11 Nm^2kg^-2), with the difference of 10e-2 accounted for by space dust, heat vents, signal redshift, Kuiper belt tugs, and other...etc., so the probes are slowing faster than gravity differential can account for alone.

But, as you can see, this is NOT a very small acceleration towards the Sun, but a rather LARGE number!

(for those who want to play around with more math)

Separately, I figured out by how much this distant G is growing per AU by first working out the total orbital Energy for each of the planets: Sun's irradiance (Watts/m^2) times distance (meters) times kinetic energy (KE = 1/2 mv^2) for each planet, and found that E grows parabolically TOWARDS the Sun. (Note, the final result is adjusted by 1000 to adjust for kilogram mass, viz. m = 1 gram, net for any orbital body). In using another equation to figure for proton mass, the mass drops off towards the Sun as E increases, by using: E = hc/ (em lambda)* (proton mass) where em lambda remains constant at about 1.32e-15 m). Conversely, away from the Sun, E decreases and proton mass increases. By taking this ratio and applying it to proton-to-proton gravitational coupling constant (~5.9e-39) and coverting this to Newton's G shows that G grows LINEARLY away from the Sun by an amount not too different from what was figured above, and confirmed by Pioneers. But this is becoming a very long story, so I stop here. But if anyone likes to play with numbers, try figuring total orbital Energy for the planets, and you'll discover that Earth's orbit sits at ~9e16 Watts, which is oddly similar to E=mc^2 = 90 petajoules (per second). How about that! #-o

[Edited 11-10-04, for mass m = 1 g in KE = 1/2mv^2]

__________________
Caveat Lector. Experimentum summus judex...