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- Did the Apollo astronauts age more or less during their trip than those who stayed behind on Earth?
I recently read Michael Collins' excellent book Carrying the Fire in which, among other things, he describes the Apollo 11 mission to the Moon and back. After the astronauts returned and were quarantined inside an isolation facility, they spoke to President Nixon, and Collins remarks in describing this conversation that Nixon referred to the astronauts having aged a bit less during their trip because of relativity. I have seen other references to similar statements made at the time, and it seems to have been a fairly common belief (and from what Collins says in the book it seems like he shared that belief).
My question is, is that correct? If we just take into account special relativity (which is what I think Nixon and others were thinking of), it would be, because of the much greater speeds the astronauts were traveling relative to the Earth. However, that is not the only factor involved; general relativity says that clocks at higher altitudes in a gravity well tick faster, and the Apollo astronauts spent most of their trip much higher in Earth's gravity well than those who stayed behind on Earth. In fact, for an object moving at Earth escape velocity, at any altitude higher than one Earth radius (i.e., distance from Earth's center greater than twice Earth's radius), the astronaut's clock will be ticking faster than Earth-bound clocks.
Here are some quick rough order of magnitude computations to try to quantify the effects. I invite comment from anyone who is interested.
Under the given conditions, we can use the weak field, slow motion approximation to the time dilation factor, which, in an Earth-centered inertial frame, will be (in units where ##G = c = 1##):
$$
\gamma = \sqrt{ 1- \frac{2M_E}{r} - \frac{2M_M}{R_M - r} - v^2}
$$
where ##M_E## is the mass of the Earth, ##M_M## is the mass of the Moon, ##r## is the distance from the center of the Earth to the astronaut, ##R_M## is the distance from the center of the Earth to the center of the Moon, and ##v## is the astronaut's speed relative to Earth. (Note that we are idealizing all motion as taking place along a radial line between the Earth and the Moon, which is not correct for the actual Apollo trajectories but is a good enough approximation for this discussion. Note also that, for the portions of the mission where the astronauts are in orbit around the Earth or the Moon, we will fudge somewhat on how we handle the distance ##r##, but the fudging won't affect the results significantly.)
The rate of Earth-bound clocks is now simple to express: we take ##r## to be the radius ##r_E## of the Earth, and ##v## to be the speed ##v_R## of an object on the Earth's equator and rotating with the Earth (about 450 meters/second), and obtain:
$$
\gamma_0 =\sqrt{ 1 - \frac{2 M_E}{r_E} - \frac{2 M_M}{R_M - r_E} - v_R^2 }
$$
For purposes of this discussion, we can split up the mission into six phases. In each phase we will make assumptions about ##r## and ##v## that will allow us to simplify the above general equation considerably.
(1) Earth orbit. Here we take ##r## to be the radius of the Earth, and ##v## to be the free-fall orbital velocity, which gives ##v^2 = M_E / r##. So the time dilation factor becomes:
$$
\gamma_1 = \sqrt{1 - \frac{3 M_E}{r_E} - \frac{2 M_M}{R_M - r_E}}
$$
(2) Trans-lunar coast, up to point where speed stops decreasing (because the Moon's gravity starts overcoming Earth's and the spacecraft starts speeding up towards the Moon). Here we take ##v## to be Earth escape velocity, which gives ##v^2 = 2 M_E / r##, and ##r## gradually increases to the distance ##R_S## where speed stops decreasing. Rather than do the messy integral over ##r## that would be required to take the changing ##r## into account, we adopt an average value for ##r## of ##R_S / 2## (which is actually an underestimate since the ship is decelerating away from Earth). This gives for the time dilation factor:
$$
\gamma_2 = \sqrt{1 - \frac{8 M_E}{R_S} - \frac{2 M_M}{R_M - R_S / 2}}
$$
(3) Approach to Moon. Here we take ##v## to be Moon escape velocity, which gives ##v^2 = 2 M_M / (R_M - r)##, and we use an average value for ##r## of ##(R_M - R_S) / 2##. This gives for the time dilation factor:
$$
\gamma_3 = \sqrt{1 - \frac{4 M_E}{R_M - R_S} - \frac{6 M_M}{R_M + R_S}}
$$
(4) Lunar orbit. Here we take ##r## to be ##R_M## minus the radius of the Moon, and ##v## to be the lunar free-fall orbital velocity such that ##v^2 = M_M / r_M##, so we have for the time dilation factor:
$$
\gamma_4 = \sqrt{1 - \frac{2 M_E}{R_M - r_M} - \frac{3 M_M}{r_M}}
$$
(5) and (6) are the same as (3) and (2), since the trip back to Earth is just the reverse of the trip to the Moon. Since the spacecraft re-enters and splashes down without going back into Earth orbit, there is no repeat of (1).
The total elapsed times to compare are thus as follows: using ##T_1##, ##T_2##, ##T_3##, and ##T_4## as the times for trip phases 1 through 4, we have
$$
T_{\text{Earth}} = \left( T_1 + 2 T_2 + 2 T_3 + T_4 \right) \gamma_0
$$
$$
T_{\text{Astronaut}} = T_1 \gamma_1 + 2 \left( T_2 \gamma_2 + T_3 \gamma_3 \right) + T_4 \gamma_4
$$
This gives
$$
T_{\text{Astronaut}} - T_{\text{Earth}} = T_1 \left( \gamma_1 - \gamma_0 \right) + 2 \left[ T_2 \left( \gamma_2 - \gamma_0 \right) + T_3 \left( \gamma_3 - \gamma_0 \right) \right] + T_4 \left( \gamma_4 - \gamma_0 \right)
$$
The only term on the RHS of the above that will be negative is the first, since ##\gamma_1 < \gamma_0## (as is well known for low Earth orbit), but all the other ##\gamma##s are greater than ##\gamma_0## (since for all of those values of ##r## the altitude effect well outweighs the speed effect, and the Moon's gravity well is too shallow to make much difference even in ##\gamma_4##). Since ##T_1## is much shorter than all of the other times (a couple of hours vs. days -- about 2 days for ##T_2## and about a day for ##T_3## and ##T_4##), without even doing a detailed calculation it seems to me that the difference above will be positive by a fair margin, i.e., the astronaut's elapsed time will end up being larger. However, I will fill in some actual numbers in a follow-up post.
My question is, is that correct? If we just take into account special relativity (which is what I think Nixon and others were thinking of), it would be, because of the much greater speeds the astronauts were traveling relative to the Earth. However, that is not the only factor involved; general relativity says that clocks at higher altitudes in a gravity well tick faster, and the Apollo astronauts spent most of their trip much higher in Earth's gravity well than those who stayed behind on Earth. In fact, for an object moving at Earth escape velocity, at any altitude higher than one Earth radius (i.e., distance from Earth's center greater than twice Earth's radius), the astronaut's clock will be ticking faster than Earth-bound clocks.
Here are some quick rough order of magnitude computations to try to quantify the effects. I invite comment from anyone who is interested.
Under the given conditions, we can use the weak field, slow motion approximation to the time dilation factor, which, in an Earth-centered inertial frame, will be (in units where ##G = c = 1##):
$$
\gamma = \sqrt{ 1- \frac{2M_E}{r} - \frac{2M_M}{R_M - r} - v^2}
$$
where ##M_E## is the mass of the Earth, ##M_M## is the mass of the Moon, ##r## is the distance from the center of the Earth to the astronaut, ##R_M## is the distance from the center of the Earth to the center of the Moon, and ##v## is the astronaut's speed relative to Earth. (Note that we are idealizing all motion as taking place along a radial line between the Earth and the Moon, which is not correct for the actual Apollo trajectories but is a good enough approximation for this discussion. Note also that, for the portions of the mission where the astronauts are in orbit around the Earth or the Moon, we will fudge somewhat on how we handle the distance ##r##, but the fudging won't affect the results significantly.)
The rate of Earth-bound clocks is now simple to express: we take ##r## to be the radius ##r_E## of the Earth, and ##v## to be the speed ##v_R## of an object on the Earth's equator and rotating with the Earth (about 450 meters/second), and obtain:
$$
\gamma_0 =\sqrt{ 1 - \frac{2 M_E}{r_E} - \frac{2 M_M}{R_M - r_E} - v_R^2 }
$$
For purposes of this discussion, we can split up the mission into six phases. In each phase we will make assumptions about ##r## and ##v## that will allow us to simplify the above general equation considerably.
(1) Earth orbit. Here we take ##r## to be the radius of the Earth, and ##v## to be the free-fall orbital velocity, which gives ##v^2 = M_E / r##. So the time dilation factor becomes:
$$
\gamma_1 = \sqrt{1 - \frac{3 M_E}{r_E} - \frac{2 M_M}{R_M - r_E}}
$$
(2) Trans-lunar coast, up to point where speed stops decreasing (because the Moon's gravity starts overcoming Earth's and the spacecraft starts speeding up towards the Moon). Here we take ##v## to be Earth escape velocity, which gives ##v^2 = 2 M_E / r##, and ##r## gradually increases to the distance ##R_S## where speed stops decreasing. Rather than do the messy integral over ##r## that would be required to take the changing ##r## into account, we adopt an average value for ##r## of ##R_S / 2## (which is actually an underestimate since the ship is decelerating away from Earth). This gives for the time dilation factor:
$$
\gamma_2 = \sqrt{1 - \frac{8 M_E}{R_S} - \frac{2 M_M}{R_M - R_S / 2}}
$$
(3) Approach to Moon. Here we take ##v## to be Moon escape velocity, which gives ##v^2 = 2 M_M / (R_M - r)##, and we use an average value for ##r## of ##(R_M - R_S) / 2##. This gives for the time dilation factor:
$$
\gamma_3 = \sqrt{1 - \frac{4 M_E}{R_M - R_S} - \frac{6 M_M}{R_M + R_S}}
$$
(4) Lunar orbit. Here we take ##r## to be ##R_M## minus the radius of the Moon, and ##v## to be the lunar free-fall orbital velocity such that ##v^2 = M_M / r_M##, so we have for the time dilation factor:
$$
\gamma_4 = \sqrt{1 - \frac{2 M_E}{R_M - r_M} - \frac{3 M_M}{r_M}}
$$
(5) and (6) are the same as (3) and (2), since the trip back to Earth is just the reverse of the trip to the Moon. Since the spacecraft re-enters and splashes down without going back into Earth orbit, there is no repeat of (1).
The total elapsed times to compare are thus as follows: using ##T_1##, ##T_2##, ##T_3##, and ##T_4## as the times for trip phases 1 through 4, we have
$$
T_{\text{Earth}} = \left( T_1 + 2 T_2 + 2 T_3 + T_4 \right) \gamma_0
$$
$$
T_{\text{Astronaut}} = T_1 \gamma_1 + 2 \left( T_2 \gamma_2 + T_3 \gamma_3 \right) + T_4 \gamma_4
$$
This gives
$$
T_{\text{Astronaut}} - T_{\text{Earth}} = T_1 \left( \gamma_1 - \gamma_0 \right) + 2 \left[ T_2 \left( \gamma_2 - \gamma_0 \right) + T_3 \left( \gamma_3 - \gamma_0 \right) \right] + T_4 \left( \gamma_4 - \gamma_0 \right)
$$
The only term on the RHS of the above that will be negative is the first, since ##\gamma_1 < \gamma_0## (as is well known for low Earth orbit), but all the other ##\gamma##s are greater than ##\gamma_0## (since for all of those values of ##r## the altitude effect well outweighs the speed effect, and the Moon's gravity well is too shallow to make much difference even in ##\gamma_4##). Since ##T_1## is much shorter than all of the other times (a couple of hours vs. days -- about 2 days for ##T_2## and about a day for ##T_3## and ##T_4##), without even doing a detailed calculation it seems to me that the difference above will be positive by a fair margin, i.e., the astronaut's elapsed time will end up being larger. However, I will fill in some actual numbers in a follow-up post.
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