What is "perceived gravity"?
In the sense of the gravitic force calculation,
F=G(m1*m2)/r^2
converted from the result(in kg*m/s^2) to m/s^2 based on the mass of the smaller body (usually m2).
Effectively, the Grav engines work fine as long as you're in a gravity well (stellar or planetary), but they lose effectiveness when you're outside one much faster than the base rules.
Assume 100dT craft, with a 2G M-Drive, and a density of 6T/dT, for a mass of 6x10^5kg. At surface level on Earth (5.972x10^24kg), it has a perceived gravity of 9.8m/s^2, and with a 2G drive, would have an acceleration of 19.6m/s^2. Based on the above, the two objects would cease to have a gravitic relationship at:
(Setting F to 0, reconfiguring)
r=SQRT(G*(m1*m2)
r= SQRT(6.67x10^-11*6x10^5*5.972x10^24) = 1.549x10^10m, or about 15 million km. It's a quadratic relationship, so you can eyeball it along a standard square of x slope for how the thrust falls off.
The thing is, for a given craft, you can have the G*m2 calculation already completed, and just do a quick calculation of the relative weight of the planet. In our case, the Gm2 number is 4.002x10^-5, so if we want to calculate the slope out from Mars (6.39x10^23kg) we just plug that into the equation in place of the mass of Earth for
r=SQRT(6.39x10^23*4.002x10^-5) = 5.056x10^9m, or about 5 million km.
the same would hold true for the Sun (once you leave the planetary wells, you may still be within the Solar gravity well) and the calculation becomes:
r=SQRT(1.989x10^30*4.002x10^-5 = 8.921x10^12m, or 8.9 billion km, somewhere in the neighborhood of 60AU is where the drives would fall off to nothing.
Notably, the drive would be more efficient the closer you get to heavy bodies (such as the sun) so as long as you keep out of the corona, you could potentially get a huge boost while travelling in-system by passing within a few million km of the sun. It's horribly slow (in Traveller terms) otherwise.
For instance, at about the orbit of Mercury, assuming you're not in the influence of other bodies, that 2G M-Drive would have an effective acceleration of .08m/s^2, but moving to half the distance (~30 million km) would increase the effective thrust to 408 km/s^2. Solar slingshots become a serious THING.
So.. all that is pretty much to say I like doing math. d= .5at^2 is the same as calculating the stat mods in MgT to me, so I like to complicate things sometimes.
Edit: I left out the units in a lot of the calculations. G cancels a lot (because it's in N) and most of the rest are in meters.
Edit 2: Corona, not Heliosphere. Heliosphere extends out to 120AU, so not what I meant...