The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes.

Consider the standard twin 'paradox':

• twin a hangs around in free-fall;
• twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns.

Well, we know that twin a's clock is fast with respect to twin b's, because twin a has followed a geodesic between the two meetings, and this has the maximal proper time of all smooth paths between the two events and both twins have followed smooth paths (in fact we can make stronger statements than this, but I only want to consider smooth paths, since I am making an argument from continuity and smooth paths are continuous (and much more)).

OK, so now let's modify the experiment: instead of being in free-fall, twin a is on a planet. We can make this planet as light as we like (really, we can make the gravitational acceleration experienced by twin a as low as we like).

Well, now it is obvious that for a sufficiently low gravitational acceleration the result of the experiment is unchanged: twin a still experiences more proper time than twin b. Equally it's clear that for a sufficiently high gravitational acceleration the result will go the other way: if twin a is hanging around near the event horizon of a black hole then they will experience less proper time than twin b unless twin b does something extraordinary. By continuity there's a setup where the proper times are the same.

Finally it remains to show that the twins can always observe each others' clocks, and so they will actually see the clocks running slow or fast: it's easy to convince yourself that this is true.

The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes.

Consider the standard twin 'paradox':

• twin a hangs around in free-fall;
• twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns.

Well, we know that twin a's clock is fast with respect to twin b's, because twin a has followed a geodesic between the two meetings, and this has the maximal proper time of all smooth paths between the two events and both twins have followed smooth paths (in fact we can make stronger statements than this, but I only want to consider smooth paths, since I am making an argument from continuity and smooth paths are continuous (and much more)).

OK, so now let's modify the experiment: instead of being in free-fall, twin a is on a planet. We can make this planet as light as we like (really, we can make the gravitational acceleration experienced by twin a as low as we like).

Well, now it is obvious that for a sufficiently low gravitational acceleration the result of the experiment is unchanged: twin a still experiences more proper time than twin b. Equally it's clear that for a sufficiently high gravitational acceleration the result will go the other way: if twin a is hanging around near the event horizon of a black hole then they will experience less proper time than twin b unless twin b does something extraordinary. By continuity there's a setup where the proper times are the same.

Finally it remains to show that the twins can always observe each others' clocks, and so they will actually see the clocks running slow or fast. It's easy to convince yourself that this is true from continuity: consider versions of the experiment where twin b's path differs only a very small amount from a geodesic and/or gravity is ver small, then it's clear that they can always see each other's clocks because they can in the limit where the paths are the same. Now you can deform the path / increase gravity continuously and nothing goes wrong.

Sorry for the slightly informal nature of these arguments: all this can be made precise, just not on a touchscreen keyboard.

The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes.

Consider the standard twin 'paradox':

• twin a hangs around in free-fall;
• twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns.

Well, we know that twin a's clock is fast with respect to twin b's, because twin a has followed a geodesic between the two meetings, and this has the maximal proper time of all smooth paths between the two events and both twins have followed smooth paths (in fact we can make stronger statements than this, but I only want to consider smooth paths).

OK, so now let's modify the experiment: instead of being in free-fall, twin a is on a planet. We can make this planet as light as we like (really, we can make the gravitational acceleration experienced by twin a as low as we like).

Well, now it is obvious that for a sufficiently low gravitational acceleration the result of the experiment is unchanged: twin a still experiences more proper time than twin b. Equally it's clear that for a sufficiently high gravitational acceleration the result will go the other way: if twin a is hanging around near the event horizon of a black hole then they will experience less proper time than twin b unless twin b does something extraordinary.

Finally it remains to show that the twins can always observe each others' clocks, and so they will actually see the clocks running slow or fast: it's easy to convince yourself that this is true.

The brief answer is 'yes’. Here is a thought experiment which I think makes it easy to see that the answer must be yes.

Consider the standard twin 'paradox':

• twin a hangs around in free-fall;
• twin b zooms off on their spaceship at some enormous speed with respect to twin a, turns around (in some smooth way, undergoing acceleration), and returns.

Well, we know that twin a's clock is fast with respect to twin b's, because twin a has followed a geodesic between the two meetings, and this has the maximal proper time of all smooth paths between the two events and both twins have followed smooth paths (in fact we can make stronger statements than this, but I only want to consider smooth paths, since I am making an argument from continuity and smooth paths are continuous (and much more)).

OK, so now let's modify the experiment: instead of being in free-fall, twin a is on a planet. We can make this planet as light as we like (really, we can make the gravitational acceleration experienced by twin a as low as we like).

Well, now it is obvious that for a sufficiently low gravitational acceleration the result of the experiment is unchanged: twin a still experiences more proper time than twin b. Equally it's clear that for a sufficiently high gravitational acceleration the result will go the other way: if twin a is hanging around near the event horizon of a black hole then they will experience less proper time than twin b unless twin b does something extraordinary. By continuity there's a setup where the proper times are the same.

Finally it remains to show that the twins can always observe each others' clocks, and so they will actually see the clocks running slow or fast: it's easy to convince yourself that this is true.