Question

A rocket flies between two planets that are one light-year apart. What should the rocket's speed be so that the time elapsed on the captain's watch is one year?

Note: the answer is not the speed of light ..

Answer 1

The easiest way to do this:
The interval
ds^2 = (c*dt)^2 - dx^2
is invariant: it gives the same result, no matter what frame is used to calculate it.

In the Earth frame,
dx = L = T*c, where T = 1 year
dt = L/v , where v = speed of ship, so:
ds^2 = (cL/v)^2 - L^2
= T^2*((c^2/v)^2 - c^2)
= (cT)^2 * ((c/v)^2 - 1)

In the ship frame:
dx' = 0
dt' = T , so:
ds'^2 = (cT)^2 - 0
= (cT)^2

Since:
ds^2 = ds'^2 , we find:
(cT)^2 = (cT)^2 *((c/v)^2 - 1)
1 = (c/v)^2 - 1
(c/v)^2 = 2, so:
v/c = 1/sqrt(2)

Therefore, v = c/sqrt(2)
= c*0.707
= 2.12e8 (m/s)