Special relativity; spaceship moving towards a planet [closed]

Question

A spacecraft starts travelling from Earth, moving at constant speed, towards a yet-to-be-discovered planet, which is $20$ light hours away from Earth. It takes $25$ hours (according to an Earth-based observer) for the spacecraft to reach this planet. Assuming that the clocks are synchronized at the beginning of the journey, compare the time elapsed for an Earth-based clock and for a clock in the spacecraft.

If the unprimed frame is the Earth-based observer's reference frame, then $\triangle t=25 h$ and $\triangle x=c*20h$ (I think this is what the 20 light hours means). Although, it's a little unclear from which frame the distance is measured, it seems reasonable to assume it was measured by an Earthling.

Then using the spacetime interval $(\triangle S)^2=(c \triangle t)^2-(\triangle x)^2=(\triangle S')=(c \triangle t')^2-(\triangle x')^2$. But $\triangle x'$ should be $0$, since - in my mind, anyway - the two events are the front of the ship leaving Earth, and the front of the ship arriving at the new planet. Both happen at the front of the ship.

Then $(c \triangle t')^2=(c \triangle t)^2-(\triangle x)^2$

$\implies c \triangle t'=\sqrt{(c \triangle t)^2-(\triangle x)^2}=\sqrt{c^2(25h)^2-c^2(20h)^2}=c \sqrt{(25h)^2-(20h)^2}$

$\implies\triangle t'=\sqrt{(25h)^2-(20h)^2}=15h=0.6 \triangle t$

Points of subtlety for me are where the distance to the new planet is measured from, and whether $\triangle x'=0$.

Answer 1

According to my understanding your reasoning is correct.

Concerning your subtle points:

First, when you say that the distance between points $x$ and $y$ is equal to $d$ meters, you (if otherwise not specified) mean that it is the distance in your reference frame, measured by a ruler at rest. And when you say that the distance is $t$ light seconds, this means that the distance is $d=t\times c$ meters. In other words, this is really the value of the space-like interval between the two endpoints of the route taken at the same moment in your reference frame. This is just what you have used in your computation.

Second, the $x'$-coordinate system is associated with the spaceship, so if two objects are at the same relative position to the spaceship the do have they same value of coordinates in $x'$-coordinate system. This is just a reformulation of what you have said and justifies your assumption of $\Delta x'=0$

Actually, I find your explanation to the problem almost exhaustive and, once again, correct.