How long would it take a rocket to reach .999c if it was accelerating at 1g

To find out how long it would take for a rocket to reach a speed of \( 0.999c \) (where \( c \) is the speed of light) while accelerating at a constant acceleration of \( 1g \), we can use the principles of special relativity. 1. **Acceleration**: We define the proper acceleration as \( a = 1g \approx 9.81 \, \text{m/s}^2 \). 2. **Time Dilation**: The relativistic acceleration means that the relationship between the time experienced by an observer on the rocket (proper time, \( \tau \)) and the time experienced by an observer at rest (coordinate time, \( t \)) must consider relativistic effects. The relativistic equations of motion give the relationship between the velocity \( v \), acceleration \( a \), and time. The relationship used in this context is derived based on the hyperbolic functions of spacetime and can be expressed as: \[ v = c \tanh\left(\frac{a t}{c}\right) \] 3. We want to find the time \( t \) when \( v = 0.999c \): \[ 0.999c = c \tanh\left(\frac{a t}{c}\right) \] To solve for \( t \): \[ 0.999 = \tanh\left(\frac{a t}{c}\right) \] 4. To find \( \frac{a t}{c} \), we can use the inverse hyperbolic tangent function: \[ \frac{a t}{c} = \tanh^{-1}(0.999) \] Using a calculator or a computational tool, we find: \[ \tanh^{-1}(0.999) \approx 5.013 \] So: \[ \frac{a t}{c} \approx 5.013 \] 5. Now, solving for \( t \): \[ t = \frac{c}{a} \cdot 5.013 \] Using \( c \approx 3 \times 10^8 \, \text{m/s} \) and \( a \approx 9.81 \, \text{m/s}^2 \): \[ t \approx \frac{3 \times 10^8 \, \text{m/s}}{9.81 \, \text{m/s}^2} \cdot 5.013 \approx 1.53 \times 10^9 \, \text{s} \] This corresponds to: \[ t \approx \frac{1.53 \times 10^9 \, \text{s}}{3600 \, \text{s/h}} \approx 425,000 \, \text{hours} \approx 49 \, \text{years} \] However, that's the coordinate time experienced by a stationary observer. ### Proper Time: The proper time \( \tau \), which is the time experienced by passengers on the rocket, can be found from: \[ \tau = t \sqrt{1 - \frac{v^2}{c^2}} = t \sqrt{1 - 0.999^2} \] Calculating gives: \[ \sqrt{1 - 0.999^2} \approx \sqrt{0.001998} \approx 0.0447 \] So, \[ \tau \approx (1.53 \times 10^9 \, \text{s}) \cdot 0.0447 \approx 68.3 \times 10^6 \, \text{s} \approx 2.16 \, \text{years} \] ### Conclusion: - **Coordinate Time (time for a stationary observer)**: Approximately **49 years**. - **Proper Time (time as experienced by the passengers)**: Approximately **2.16 years** to reach a speed of \( 0.999c \) with a constant acceleration of \( 1g \).