spidey said:
From equivalence principle, being on Earth is equivalent to being on lift accelerating at 9.8 m/s2...so if we are in a lift and that accelerates at 9.8 m/s2 then at some time we will cross the velocity of light.. my questions are
1) is Earth accelerating ?
2) what happens if the velocity crosses velocity of light?
3) what is the equivalent velocity related to Earth ,when we say "velocity of lift"...is it escape velocity of earth?
In relativity, velocities cannot simply be added to each other. Instead we have to use the formula
\frac{u + v}{1 + uv/c^2}
So, if you are experiencing a constant acceleration of 9.8 m/s
2 (technically "proper acceleration"), after one second you are moving at 9.8 m/s, but after two seconds, your velocity is not 9.8 + 9.8 = 19.6 m/s, it is instead
\frac{9.8 + 9.8}{1 + 9.8^2/c^2} = 19.59999999999997905565337 \mbox{m/s}
Of course, that difference is tiny, but as time goes by, these tiny differences accumulate and become highly significant. The net effect is, you never go faster than the speed of light.
If you do the calculus (and with an infinitesimal time change instead of the one second approximation of above), you get the accurate formula
\left(x+\frac{c^2}{9.8}\right)^2 - c^2t^2 = \frac{c^4}{9.8^2}
as Fredrik indicated. That answers question 2.
The answer to 1 is "yes, relative to any endless stream of falling apples". These apples define a "local inertial frame" (subject to some technicalities). Any point on the surface of the Earth accelerates relative to any of its local inertial frames ("undergoes proper acceleration").
The answer to 3 depends on what height you released the apples from. But it never exceeds the speed of light. Remember there are lots of different inertial frames (all moving at constant velocities relative to each other) so there is no unique speed of anything -- it all depends which frame you choose.