With what speed relative to the Earth did the rocket move if time in it went 3 times slower than on Earth?

To calculate the speed of the rocket relative to the Earth, we will use the formula: tz = tp / √ (1 – V ^ 2 / C ^ 2), whence 1 – V ^ 2 / C ^ 2 = √ (tp / tz); V ^ 2 / C ^ 2 = 1 – (t0 / t) ^ 2 and V = √ (1 – (tr / t3) ^ 2) * C.

Constants and variables: tр – time on the rocket; tz – time on Earth (tz = 3tр); C is the speed of light (C = 3 * 10 ^ 8 m / s).

Let’s perform the calculation: V = √ (1 – (tr / tz) ^ 2) * C = √ ((1 – (1/3) ^ 2) * 3 * 10 ^ 8 ≈ 2.828 * 10 ^ 8 m / s.

Answer: The rocket was supposed to move relative to the Earth at a speed of 2.828 * 10 ^ 8 m / s.