Re: Time Difference

Leigh Palmer (palmer@sfu.ca)
Tue, 1 Oct 1996 17:47:13 -0700

>This is off the beaten track for this group, so I
>apologize beforehand for this posting. But, since
>you guys can calculate orbital elements from a simple
>visual observation this math ought to be easy enough
>for you.
>
>I'm wondering, since Shannon Lucid spent 6 months or
>so traveling at 5 miles per second, how much did time
>slow down for her as compared to her earthbound neighbors?
>
>I'm not asking this from a "teacher" point of view where
>I already know the answer. I'm truely curious.

I'm slow on the calculator, but this is the result I get
taking special and general relativistic effects into account.
An astronaut in a circular orbit around the Earth will have
a watch which ticks at a rate A times as fast as a clock on
Earth's surface, where A is given by the expression

                     GM  /  1     3   \
            A = 1 + ---- | --- - ---- |
                    c^2  \  R     2r  /

Where G = universal gravitational constant, M = mass of Earth,
c^2 = speed of light squared, R = radius of Earth, and
r = radius of orbit.

The amount of time Shannon Lucid's watch would have differed
from an earthbound watch if both were perfect timekeepers
would be given by the expression

                                 GMT  /  1     3   \
           delta t = (A - 1)T = ----- | --- - ---- |
                                 c^2  \  R     2r  /

where T is the time elapsed on Earth (six months). Note that
if the astronaut is in an orbit with a radius of 1.5 R she
will neither gain nor loose time on her watch! In this case
she was in a  considerably smaller orbit and her watch would
indeed have lost time.

I'll leave it to the industrious students to plug in the
numbers and post the result here. I will note that in the
derivation of the formula above I have assumed the effects
are small. That is why the result is simple.

It's nice to be able to contribute something to this group.

Leigh