T.R | Title | User | Personal Name | Date | Lines |
---|
1452.1 | | SAUTER::SAUTER | John Sauter | Thu May 02 1991 12:26 | 2 |
| What is "local time" on the space shuttle?
John Sauter
|
1452.2 | orbital time | ANGLIN::BRISCOE | | Thu May 02 1991 16:00 | 5 |
| "local time" refers to the space center not the shuttle. The shuttle
is on "elapsed" time. at 17k mph the shuttle changes time zones VERY
quickly - ie every 17 minutes depending on it's trajectory.
tjb7
|
1452.3 | I suspect "local" is EDT, not CDT | SAUTER::SAUTER | John Sauter | Fri May 03 1991 10:48 | 5 |
| re: .2
The mission started with liftoff about 0730 in Florida, and 4 hours and
11 minutes later would be 1145 Florida time, not Houston time.
John Sauter
|
1452.4 | local time is for the public's use | SHRCAL::MORRILL | | Fri May 03 1991 12:56 | 14 |
|
As a former member of the STDN program at NASA, I offer this bit of
info.
The flight of ALL spacecraft is referenced to GMT. The term Local
Time is referenced to either launch point or the local time at the
tracking sight doing the report.
Since all of the STDN stations (NASA Tracking Stations) are
referneced to GMT, local time is only used in reporting information to
the civilian population.
dlm
|
1452.5 | Used three in my day. | NEWVAX::MZARUDZKI | I am my own VAX | Fri May 03 1991 14:26 | 14 |
|
More info to digest...
NASA Ground Segments actually use three time tracks.
Mission or Space Craft time
GMT
and
EST or local time depending on geographical location.
-Back to billable time.
Mike Z.
|
1452.6 | Relativity | SAUTER::SAUTER | John Sauter | Mon May 06 1991 09:08 | 6 |
| re: .5
That's four time tracks. Mission Elapsed Time and Space Craft Time
are different for a craft that's been moving at 17,000 mph for eight
days. Not much different, I admit.
John Sauter
|
1452.7 | Contact GSG for details | DENTON::AMARTIN | Alan H. Martin | Mon May 06 1991 13:06 | 4 |
| Re .6:
Ever heard the phrase, "Close enough for government work"?
/AHM
|
1452.8 | time_in_space = time_@_NASA ? | USRCV1::RHODESJ | | Mon May 06 1991 14:25 | 3 |
| If they set the time on the computer (or a watch for that matter)
will it be the same when they land again? (assuming it will be the
same timezone as where they lifted off from)
|
1452.9 | yes and no | SAUTER::SAUTER | John Sauter | Mon May 06 1991 15:20 | 8 |
| re: .7, yes
re: .8, nope, if the watch is sufficiently accurate and precise. Mary
Payne told me that she had been asked to compute the time loss for the
Apollo astronauts. She said it was less than a second, so I suppose
they didn't worry about it, but I noticed that elapsed time for Apollo
was referred to as something like "Ground Elapsed Time".
John Sauter
|
1452.10 | | JARETH::EDP | Always mount a scratch monkey. | Tue May 07 1991 08:10 | 11 |
| Re .6:
I believe the dominating factor in the difference of time experienced
on current spacecraft is not their speed but the fact that they are not
as deeply in Earth's gravitational field as we are. Also, I think this
makes the sign different; the astronauts experience more time than
people on the ground. It would be the other way around if special
relativity were the dominating effect.
-- edp
|
1452.11 | Probably Gravity Waves at work! | WLDWST::BRODRIGUES | Fiat Lux | Tue May 07 1991 09:37 | 10 |
| Sorry, but relativistic theory is not dependent on the gravitational
forces at work. The decay rate of an nuclear particle is the same in
space as it is on earth. The vibrational frequencies of electromagnetic
radiation are not dependent on gravity. The only thing effecting the
time difference would be the spped of the ship, which is so small when
compared to the speed of light that I would suspect a bigger effect
just do to human errors in time keeping, than on the speed of the ship.
Brian
|
1452.12 | | WLDWST::BRODRIGUES | Fiat Lux | Tue May 07 1991 10:42 | 11 |
| I decided to see what relativistic efect travelling at 17,000mph
would realy have.
17,000 mph = 7,600 meters/sec
spped of light = 3 x 10^8 m/sec
the effect is t = t'/ square root (1 - v*v/c*c)
thie yield a number .9999999998 divided into 1 second.
The effect isn't even measurable with our time system.
Brian
|
1452.13 | however ... | A1VAX::KREFETZ | Reality is the fiction we live by. | Tue May 07 1991 12:35 | 53 |
| re: .11
That is wrong.
The special relativistic Doppler effect (the difference between the
wavelength of electromagnetic radiation emitted by A and the
wavelength measured by the observer B) depends on the factor
[1 + (v/c)cosx]/[1 - (v/c)**2]**.5
where B is taken (for convenience) as stationary, and x is the angle
between the velocity of A and the line of sight of B (where x=0 means
that A is moving away from B along B's line of sight); v is the speed
of A, c is the speed of light.
The denominator (always =< 1) always tends to increase the wavelength
(redshift the light, decrease the frequency). When A is moving away
from B along B's line of sight, the numerator reinforces this -- and
the light is indeed redder to B than to A. When A is moving directly
toward B, the overall result is a blueshift.
The gravitational effect (in the simplest case of a spherically symmetric
gravitational source where the gravitational field is being measured outside
the source) is
1/[1 - 2GM/rc**2]**.5
where G is the universal constant of gravitation, M is the mass of the source
of the gravitational field (e.g. the earth), and r is the distance from A
to the center of the gravitational source.
You can view this as a statement of the fact that when A is stationary in
a gravitational field, the equivalent of the numerator in the Doppler shift
factor becomes 1, and the denominator reflects the substitution of the
gravitational potential energy of A -- in this case -- for the kinetic energy
of A -- in that case. As viewed by B ('outside' of the gravitational field)
the electromagnetic radiation from A is always redshifted.
In terms of astronauts and earthdwellers, the astronauts clocks are (overall)
effected by 1/[1 - (v/c)**2]**.5 and by 1/[1 - 2GM/rc**2]**.5, whereas
the earthdwellers are just affected by 1/[1 - 2GM/rc**2]**.5. Clearly
the gravitational term is more important for the earthdweller than the
astronaut (r being significantly smaller). I leave the calculations of the
various effects to the reader.
Note: Particle decay rates _are_ 'slower' in a gravitational field.
Note: If you have a source whose radius =< 2GM/c**2 something funny happens
at r = 2GM/c**2. The something funny is called a black hole.
Elliott
|
1452.14 | | JARETH::EDP | Always mount a scratch monkey. | Tue May 07 1991 13:16 | 15 |
| Re .11, .12:
The formulae and effects you have discussed are those of special
relativity. General relativity is, obviously, more general and
considers effects omitted from special relativity. Special relativity
uses as axioms that the speed of light is constant and that physical
laws are invariant under transformation of inertial frames of
references. General relativity is more accurate; it goes beyond that
and considers space-time curvature and effects of non-inertial frames.
The surface of the Earth is not an inertial frame of reference; our
time measurement here cannot be completely accounted for in comparison
to time measurement in free fall solely by special relativity.
-- edp
|
1452.15 | | SSBN1::YANKES | | Tue May 07 1991 17:49 | 29 |
|
What is the relativistic effect of travelling 17,000 mph in orbit
about the earth?
This is a trick question, right? :-)
Don't forget that the earth isn't stationary. Thus calculating the
effects of relativity on a body moving 17,000 mph is incorrect. We are
travelling through space at about 1 million mph when you factor in all the
motions involved (the solar system's speed around the galaxy and the galaxy's
speed through the cosmos are the "biggies"). Lets say for simplicity sake
that the spacecrafts's plane of orbit around the earth was exactly the same
plane in which our solar system's direction vector is currently aimed at (lets
call this "A" from here on). When the spacecraft is on the side of the orbit
paralleling earth's path toward A moving in the direction of A, the spacecraft
is travelling at 1,017,000 mph. (Using, for grins, a figure of exactly
1 million mph for the earth's average speed.) At this instant, the astronauts
are travelling 17,000 mph faster than you and me relative to some intergalactic
fixed reference point. At the other side of the orbit (when the spacecraft
is moving towards the point opposite to A), the spacecraft is travelling
at only 983,000 mph -- 17,000 mph *slower* than you and me relative to that
same fixed point. Granted, the "sometime faster, sometime slower" won't
*exactly* even out due to the squaring of various numbers in the equations,
but the difference is even smaller than hinted at in .12.
And, of course, I'm sure this all has something to do with the Digital
way or working somewhere... ;-)
-craig
|
1452.16 | | ROYALT::KOVNER | Everything you know is wrong! | Tue May 07 1991 18:46 | 12 |
| You don't need to include the Earth's motion to compare the time dilation
for someone in orbit compared with someone on the ground; they both experience
the same time dilation due to this motion. You only need to include the RELATIVE
motion of the 2 observers. Now, if we were comparing the time dilation of
an astronaut relative to someone at rest with respect to the center of the local
group of galaxies, that would be different.
And 17000 mph may not give much time dilation, but it is measurable over
a period of time. An atomic clock was taken around the world on commercial
airliners - that's about 550-600 mph - and compared with a clock that stayed
on the ground. They differed by the predicted amount. (It might have been
more than once around the world. I wonder who got the frequent flyer miles?)
|
1452.17 | Gravity still doesn't matter!!! | WLDWST::BRODRIGUES | Fiat Lux | Tue May 07 1991 23:24 | 20 |
| re .13
I never meant to imply that general or special relativity doesn't
have some effect on physical events, just that it has an immeasurable
effect on things that affect our daily lives. That's why it took so
long to discover.
You say that gravity will have some effect and earlier someone
tried to imply that it somehow would cause the 1 second time
difference. You gave us the formula but you failed to do the
calculation which clearly shows that even the gravitational effect of
being on the moon (much farther away than the shuttle is from earth)
is on ly 1 part in 10E11 out of 1. (i.e. the factor you would divide
your delta time value by is 0.99999999998.
Anytime you try to imply relativistic( either special or general)
science to man made events, the resulting correction is so small that
is doesn't even make sense to discuss it.
Brian
|
1452.18 | but ... | A1VAX::KREFETZ | Reality is the fiction we live by. | Tue May 14 1991 12:38 | 19 |
| .14: No, the gravitational effect mentioned in .13 is a general
relativistic effect -- it comes out of the Schwarzschild metric (the
solution to Einstein's gravitational field equations with a spherically
symmetric source).
.15: It's only the relative speed that counts -- that's why it's called
RELATIVITY (there is no 'fixed point').
.16: Right.
.17: As .16 notes, the time dilation effect we've been discussing is
small but CUMULATIVE, and therefore measurable. As for a special
relativistic effect that affects your daily life, it's called
magnetism. [It's not swamped by the potentially much larger electric
field because we live in a world that is (roughly) electrically
neutral.]
Elliott
|
1452.19 | Just to tie this back to "Working at DIGITAL", ... | YUPPIE::COLE | Lead with a discount, close with an allowance! | Tue May 14 1991 13:45 | 3 |
| ... anybody heard from our Italian(?) compatriot who was selected
for Shuttle crew training last year? Wasn't that part of the European Lab
project the Shuttle is supposed to carry? FWIW! :>)
|