>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