Ahmed Samra said:
You mean if I am away from the center of the Earth time will go slower? But I read in Wikipedia that time will move faster when I am away from the center of the earth.
There are two effects, which work in opposite directions. What you read in Wikipedia is talking about gravitational time dilation, which A.T. referred to; clocks run faster at higher altitudes above a gravitating body.
However, if the gravitating body is rotating, as the Earth is, then you have to take that motion into account as well; moving clocks run slower than stationary clocks because of ordinary, kinematic time dilation. If you are at rest relative to the rotating Earth, then the higher up you are the faster you are moving (because you have to cover a larger distance around a circle in the same time, 24 hours), and the faster you are moving, the slower your clock will run. This is what CrazyMechanic referred to.
An approximate formula that includes both of these effects is:
\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^2 r} - \frac{v^2}{c^2}}
where ##\frac{d\tau}{dt}## is the clock rate relative to a clock that is at rest "at infinity" (i.e., very, very far away from the rotating Earth and not moving with it), ##G## is Newton's gravitational constant, ##M## is the mass of the Earth, ##r## is the distance of the clock from the Earth's center, and ##v## is the clock's velocity, again relative to an observer that is at rest at infinity. Increasing altitude means increasing ##r##, which decreases the second term under the square root and makes the clock rate faster; increasing ##v## obviously increases the last term under the square root and makes the clock rate slower.
For example, a clock on Earth's surface at sea level at the equator, and at rest relative to the (rotating) surface, would have an ##r## of about 6.4 million meters and a ##v## of about 450 meters/second. This gives a clock rate of about 1 - 6.96 x 10^-10, i.e., about 7 parts in 10 billion slower than a clock at rest at infinity.
A clock at the top of Mount Everest would have an ##r## about 9000 meters larger and a ##v## about 1 meter/second larger; if you run the numbers, you get a clock rate of about 1 - 6.95 x 10^-10, i.e., about 1 part in a trillion faster (faster because we're subtracting a slightly smaller number from 1) than the clock at sea level at the equator. (I've cheated a bit by assuming Mount Everest is on the equator.)
So CrazyMechanic was actually not correct; there is a slowing effect due to increased velocity for a clock on top of a mountain, but it's more than compensated for by the speeding up effect of the increased altitude.
- one of the SAs worked it out and posted the derivation fairly recently.