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Cornelis
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It's very likely that inhabitants of the atoll Makatea in the South Pacific were the closest to Mars !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew could be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one for Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged, what gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between the atoll Mataiva and the island Huahine)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours at the time.
With Table Settings Quantities 19 and averaging for 9.51 UTC, the Sun-Earth distance was calculated to be 151,159,530.3 km.

The island Huahine is the closest to the 150⁰ 16' W, 15⁰ 47'S spot with a distance of 130 km, and by the curvature of the Earth that would make it 1325 m. further away from Mars at 9.51 UTC.
But 8 minutes earlier the closest point to Mars was near the atoll Makatea 213 km to the east, and with the equations in the answer from @SE-stop firing the good guys a distance of 258 m. further away from Mars was calculated.

The closest point on Mars to Earth at that time appeared to be 25.51⁰ E, 9.08⁰ S.

It's very likely that inhabitants of the atoll Makatea in the South Pacific were the closest to Mars !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew could be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one for Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged, what gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between the atoll Mataiva and the island Huahine)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours at the time.
With Table Settings Quantities 19 and averaging for 9.51 UTC, the Sun-Earth distance was calculated to be 151,159,530.3 km.

The island Huahine is the closest to the 150⁰ 16' W, 15⁰ 47'S spot with a distance of 130 km, and by the curvature of the Earth that would make it 1325 m. further away from Mars at 9.51 UTC.
But 8 minutes earlier the closest point to Mars was near the atoll Makatea 213 km to the east, and with the equations in the answer from @SE-stop firing the good guys a distance of 258 m. further away from Mars was calculated.

The closest point on Mars appeared to be 25.51⁰ E, 9.08⁰ S.

It's very likely that inhabitants of the atoll Makatea in the South Pacific were the closest to Mars !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew could be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one for Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged, what gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between the atoll Mataiva and the island Huahine)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours at the time.
With Table Settings Quantities 19 and averaging for 9.51 UTC, the Sun-Earth distance was calculated to be 151,159,530.3 km.

The island Huahine is the closest to the 150⁰ 16' W, 15⁰ 47'S spot with a distance of 130 km, and by the curvature of the Earth that would make it 1325 m. further away from Mars at 9.51 UTC.
But 8 minutes earlier the closest point to Mars was near the atoll Makatea 213 km to the east, and with the equations in the answer from @SE-stop firing the good guys a distance of 258 m. further away from Mars was calculated.

The closest point on Mars to Earth at that time appeared to be 25.51⁰ E, 9.08⁰ S.

added the last 3 sentences
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Cornelis
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It's very likely that inhabitants of the atoll Makatea in the South Pacific were the onesclosest to Mars !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew probably cancould be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one withfor Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged that, what gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between the atoll Mataiva and the island Teti'aroaHuahine)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours at the time.
With Table Settings Quantities 19 and averaging for 9.51 UTC we could conclude, the Sun-Earth distance thenwas calculated to be 151,159,530.3 km.

The island Huahine is the closest to the 150⁰ 16' W, 15⁰ 47'S spot with a distance of 130 km, and by the curvature of the Earth that would make it 1325 m. further away from Mars at 9.51 UTC.
But 8 minutes earlier the closest point to Mars was near the atoll Makatea 213 km to the east, and with the equations in the answer from @SE-stop firing the good guys a distance of 258 m. further away from Mars was calculated.

The closest point on Mars appeared to be 25.51⁰ E, 9.08⁰ S.

It's very likely that inhabitants of Makatea were the ones !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew probably can be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one with Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged that gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between Mataiva and Teti'aroa)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours.
With Table Settings Quantities 19 and averaging for 9.51 UTC we could conclude the Sun-Earth distance then to be 151,159,530.3 km.

It's very likely that inhabitants of the atoll Makatea in the South Pacific were the closest to Mars !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew could be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one for Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged, what gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between the atoll Mataiva and the island Huahine)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours at the time.
With Table Settings Quantities 19 and averaging for 9.51 UTC, the Sun-Earth distance was calculated to be 151,159,530.3 km.

The island Huahine is the closest to the 150⁰ 16' W, 15⁰ 47'S spot with a distance of 130 km, and by the curvature of the Earth that would make it 1325 m. further away from Mars at 9.51 UTC.
But 8 minutes earlier the closest point to Mars was near the atoll Makatea 213 km to the east, and with the equations in the answer from @SE-stop firing the good guys a distance of 258 m. further away from Mars was calculated.

The closest point on Mars appeared to be 25.51⁰ E, 9.08⁰ S.

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Cornelis
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It's very likely that inhabitants of Makatea were the ones !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew probably can be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one with Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged that gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between Mataiva and Teti'aroa)

Using the R.A.(right ascension) on both the tables of Earth and Mars at the timeaugust 27 and averaging for 9.51 UTC28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours.
With Table Settings Quantities 19 and averaging for 9.51 UTC we could conclude the Sun-Earth distance then to be 151,159,530.3 km.

It's very likely that inhabitants of Makatea were the ones !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew probably can be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one with Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged that gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between Mataiva and Teti'aroa)

Using the R.A.(right ascension) on both the tables of Earth and Mars at the time and averaging for 9.51 UTC gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours.
With Table Settings Quantities 19 and averaging for 9.51 UTC we could conclude the Sun-Earth distance then to be 151,159,530.3 km.

It's very likely that inhabitants of Makatea were the ones !

(As was discovered and announced by @SE-stop firing the good guys in one of his comments to this answer.)

Now that one of the other answers has ruled out any Apollo astronaut, and the ISS crew probably can be excluded too, we can focus on determining the exact location on Earth, and this can be done with the aid of the HORIZONS Web-Interface.
With the Table Settings Quantities 1 and 15, the Target Body Earth and the Observer Location Sun(body center) we can see the Sunsub longitude and the Sunsub latitude at the start of August 27, 2003.
Averaging for 9.51 UTC gives longitude 330.26405⁰ and latitude 10.210355⁰ for the place where the Sun would be in the zenith.
Comparing this table with one with Mars as the Target Body we see that the right ascension for Earth and Mars are almost equal at that special time, so the plane that contains the Sun, Earth and Mars was then almost perpendicular to the ecliptic.(and the angles of declination of the two planets would lie in that plane.)
If the Earth and Mars had also the same declination at that time, we just could draw a line from the "Sunsub" spot on Earth through its centre to the surface on the other side where Mars would then be in the zenith.(supposing the Earth would be a perfect sphere).
That "midnight spot" would then have coordinates 150.264⁰ West and 10.210355⁰ South.
But when we look at still another table with Target Body Mars but with this time Observation Location "Geocentric" at August 27, 2003, Mars has a considerable declination of over -15⁾, minus the declination of Earth of -10⁾ 9', and then averaged that gives a total of -5⁰ 35' for 9.51 UTC.

That makes the final coordinates of the closest spot to Mars: 150⁰ 16'W, 15⁰ 47'S.
(somewhere between Mataiva and Teti'aroa)

Using the R.A.(right ascension) on both the tables of Earth and Mars at august 27 and 28, 2003 gives an angular velocity for Earth relative to Mars of 75.08" per 24 hours.
With Table Settings Quantities 19 and averaging for 9.51 UTC we could conclude the Sun-Earth distance then to be 151,159,530.3 km.

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Cornelis
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changed to Makatea and added the last two sentences
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added the angles of declination and the minus Earth declination
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