How long is a second in zero gravity? General relatively predicts that gravity creates time dilation and a relatively slower passage of time for the observer experiencing larger acceleration. For an observer experiencing zero acceleration, including being far enough from any mass that the gravitational forces are approaching zero, how long is one of their seconds compared to a second on the surface of the Earth? 
It seems like Schwarzschild Coordinates as a means for calculating time dilation outside a non-rotating sphere is the right approach, but I’m not a physicist, just an engineer who saw Intersellar last night and needs some expert help to wrap my mind around this.
 A: For the record, your opening statement:

General relatively predicts that gravity creates time dilation and a relatively slower passage of time for the observer experiencing larger acceleration

is incorrect, or at least misleading. For example an observer moving in a circle at speed $v$ in flat spacetime is experiencing acceleration towards the centre of the circle (the centripetal force). However the time dilation is just dependant on the speed $v$ and is given by the usual equation for motion in a straight line:
$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} $$
Obviously the acceleration depends on the velocity so I suppose you could write the time dilation as a function of acceleration, but this would be misleading as it's the displacement in space (i.e. the velocity) that causes the time dilation.
Anyhow, the time measured on a clock carried by the astronaut in the gravitational field of a non-rotating sphere is given by the Schwarzschild metric:
$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 - \frac{dr^2}{\left(1-\frac{2GM}{c^2r}\right)} - r^2 d\Omega^2 $$
If the atronaut is hovering at a fixed position then $dr = d\Omega = 0$ and the expression simplifies to:
$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 $$
and the time dilation (relative to an observer far from the black hole) is therefore:
$$ \frac{d\tau}{dt} = \sqrt{1-\frac{2GM}{c^2r}} $$
But, this will not correctly describe the time dilation in Interstellar because that was around a rotating black hole. In fact for a static black hole there is a minimum distance $6GM/c^2r$ below which nothing can maintain a stable orbit, and even at this distance the time dilation is not as great as in Interstellar. To correctly describe the very high time dilation in the film you have to use the Kerr metric that describes a rotating black hole. You also need to take into account the time dilation from both the spacetime curvature and the orbital velocity of the astronaut.
The trouble is that while the Schwarzschild metric is simple enough that I can describe the calculation to you in a few lines this is not true of the Kerr metric.
A: Gravitational time dilation relates to the gravitational potential $\Phi$, not the gravitational field $g$, so comparing earth's $g$ with $g=0$ doesn't give the relevant information.
An example of a question that does have an answer would be how long is a second aboard a GPS satellite, compared to a second at the earth's surface? The answer is that gravitational time dilation makes the clock on the satellite run faster by a factor of $1+\Delta\Phi/c^2=1.00000000045$.
