If I accelerated continously at 1 gravity, how long would it take before I experienced time dilation of 100:1?

Question

From a "stationary" body at my point of origin?

Answer 1

Depends how far up you are (distance)

Answer 2

time dilation

t' = t/√(1-v²/c²)

100 = 1/√(1-v²/c²)

10000 = 1/(1-v²/c²)

1-v²/c² = 1/10000

v²/c² = 1 - 1/10000 = 0.9999

v/c = 0.99995

But now I get stuck, I don't think the ordinary equations of motion apply.

try anyway

v = at

t = 3e8/10 = 3e7 seconds or 1 year

and this has to be wrong.

One problem is that from an observer on the ship, the gravity increased as you approached the speed of light, and reached 100G when the time compression hit 100.

.

Answer 3

The rapidity is the same thing as the velocity at low speeds, but has the advantage that rapidities add like r = r1 + r2, instead of velocities which add like v = (v1 + v2) / (1 + v1*v2). Thus we can use the standard kinematic equations if we use rapidity instead of velocity (and proper time instead of coordinate time).

For gamma = 100, the rapidity r = arccosh(gamma) = arccosh(100) = 5.29829.

The acceleration is 9.8 m/s^2 but 3*10^8 m = s, so the acceleration is 3.27*10^-8 / s.

Therefore, t = r/a = 5.29829 / (3.27*10^-8 /s) = 1.62*10^8 s

Which is 5.16 years, proper time on your wristwatch.

To find the time required as measured back on Earth, you would do the same thing with celerity instead of rapidity.