Does throwing a watch into the air cause it to gain or lose time? Suppose I'm on a non rotating planet. I have two identical, perfect watches. I synchronize them. Then I throw one of them into the air and catch it. Does the one I throw into the air gain or lose time with respect to the one I was holding?
 A: Lawrence has the details, but this question can actually be determined exactly without assuming anything about mass distribution, etc. A geodesic (which is what the watch in free-fall follows) in GR has maximal proper time along it. The "stationary" watch, which is actually accelerated, is following some other path and so must experience a shorter proper time. 
A: What astounds me is there is considerable quibbling over the nature of the question, but nobody answers it!  This is comparatively simple to address.  Let us consider the Schwarzschild metric in a weak gravity field
$$
ds^2~=~-(1~-~2\phi/c^2)dt^2~+~dr^2~+~r^2d\Omega^2
$$
for $\phi~=~GM/r$ the Newtonian gravity potential.  The unit velocity is then
$$
1~=~-(1~-~2\phi/c^2)u_t^2~+~u_r^2~-~\dots
$$
where we can consider the motion in the radial direction for simplicity.  The derivative of this with respect to the proper time $s$ is then
$$
0~=~-(1~-~2\phi/c^2)u_ta_t~+~u_ra_r.
$$
If the gravity potential is zero the solutions are $t~=~g^{-1}\sinh(gs)$ $r~=~g^{-1}\cosh(gs)$, for $g$ the acceleration parameter.  Here $g$ counters the gravitation of the Earth.  If the gravity potential is turned on we can the write the time solution solution as $t~=~g^{-1}\sinh(gs~+~\gamma)$, which we input into the third equation
$$
0~=~-(1~-~2\phi/c^2)g~\cosh(gs)\sinh(gs)~+~g~\cosh(gs~+~\gamma) \sinh(gs~+~\gamma)
$$
$$
=~(1~-~2\phi/c^2)\frac{g}{2}\sinh(2gs~+~2\gamma) ~+~\sinh(2gs)
$$
If we consider weak fields, small accelerations and small proper time $s$ we have
$$
0~\simeq~-(1~-~2\phi/c^2)(g^2s~+~g\gamma)~+~g^2s,
$$
where $\gamma~\simeq~-2\phi gs/c^2$.
The coordinate time is reduced with the turning on of the acceleration.  This implies that the watch on the accelerated frame will mark off a shorter interval of time than the watch which is placed on a geodesic motion in the local gravity field with acceleration $g$.  This is a gravitational version of the twin paradox.  The twin which travels outwards and back is on an accelerated frame, which is a path in spacetime that is non-extremal, or maximal.  As a result the proper time marked off is shorter.
A: I would just point out, there is a twist or two here that would change the result.  For example, if you threw the watch all the way around an airless world and caught it, that thrown watch would record less time than the one you are holding if the angle of launch is less than 60 degrees, more time if the angle is greater than 60 degrees, and exactly the same time if the angle is exactly 60 degrees.
Or, if you cut a tunnel through the planet to the far side, and dropped the watch in, it would fall to the far side, stop, and then fall back to you.  That would record the least amount of time when compared to the standing-still watch.
It is dangerous to use adages like "a geodesic is the path of greatest time between points."  That is not always true.  Plus it is a consequence, not a cause.
It is also not true that the standing watch is being accelerated!  When forces cancel out, the acceleration is zero.  Throw the watch horizontally (at orbital velocity) and both watches have the same gravitational potential energy.  It doesn't matter that one is in free fall and the other is being held up by you.
