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No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).

In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer, or $M/r=3c^2/(2G)$ (that is, never, because that is smaller than the Schwarzschild radius).

Note: this is an answer about gravitational versus motion time dilation, because I misunderstood the original question.

No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).

In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer, or $M/r=3c^2/(2G)$ (that is, never, because that is smaller than the Schwarzschild radius).

No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).

In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer, or $M/r=3c^2/(2G)$ (assuming the parameters make the approximations accurate enough).

No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).

In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer, or $M/r=3c^2/(2G)$ (that is, never, because that is smaller than the Schwarzschild radius).

No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).

In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer.

No, in general they will not cancel, but they will for some parameters. Imagine the limit of a large and small mass, so the smaller mass is in a circular orbit an mostly experiences time dilation due to motion, and the heavier one at the center mostly experiences time dilation due to gravity. These as seen from an observer far from both mases (close to infinitely far away).

In such a case, the observer will see that one revolution takes a time T. He will also see that one revolution will show, in a clock located at the smaller mass, a time $T_v=T/\sqrt{ 1-v^2/c^2 }=T/\sqrt{ 1-GM/(rc^2) }$, because $v=\sqrt{GM/r}$, and will see that a clock located at the large mass will show $T_g=T\sqrt{ 1-2GM/(rc^2) }$, which are different functions of r and M. The two effects will cancel when both clocks show the same dilation for the far away observer, or $M/r=3c^2/(2G)$ (assuming the parameters make the approximations accurate enough).