We are just starting to learn about relativity and special relativity in class and I feel like I am having a really hard time wrapping my brain around it, especially when it comes to spaceship problems (which, lucky me, are 90% of the questions). This isn't an exact question but it's the same type of framework that keeps popping up and I'm finding it hard to get guidance.

So example:

Let's say a spaceship leaves earth and is traveling to a planet that is 10 light years away (as measured from earth). However, according to the observers on the spaceship, they arrive after 8 years of travel. What would be the speed of the ship relative to Earth, and would this differ if you were measuring the speed as an observer on the ship?

My instinct says to just do v=d/t, and so the speed would be 1.25c (which I know isn't possible). But I really feel like I'm missing some sort of consideration for length contraction/time dilation? I also tried to do L=8v for contracted/observed distance to the planet, and then setting that equal to 10*sqrt(1-v2) to solve for v but I keep getting really wonky answers that don't fit when plugged back in to relevant equations. This feels more in the right direction, and maybe I'm just miscalculating because of squares and square roots. But I keep seeing these types of questions (and questions that build off this concept) so if someone could give me insight on how to solve this type of problem, that would be really helpful.

  • $\begingroup$ The best thing you can do with these types of problems is Draw a Spacetime Diagram. You can look up what that is if needed $\endgroup$
    – RC_23
    May 23, 2023 at 23:03

1 Answer 1


I think your approach makes sense. First of all, what velocity does an observer on the ship measure? The velocity of the ship as measured on the ship is obviously going to be zero, so what you are measuring is the velocity of the planet relative to the ship.

So the travel time as measured by the ship's clock is going to be $t$=8 years. The distance to the planet as measured from earth is $L_0=10$ light years. As you correctly stated, the distance to the planet as measured from the ship is going to be $$L = \frac{L_0}{\gamma} = L_0 \sqrt{1-v^2/c^2}.$$ Thus, the velocity of the planet relative to the ship will be $$v = \frac{L}{t} = \frac{L_0 \sqrt{1-v^2/c^2}}{t}$$ we can solve this for the velocity. First, let's define $\beta := v/c$, which is just the velocity in units of lightspeed. With that, we have, $$\beta c = \frac{L_0\sqrt{1-\beta^2}}{t}$$ Now, bring the $\beta$s to the right-hand side: $$\frac{\sqrt{1-\beta^2}}{\beta} = \frac{tc}{L_0}$$ Take the square: $$\frac{1-\beta^2}{\beta^2} = \frac{1}{\beta^2}-1 = (tc/L_0)^2$$ and now, solve for $\beta$: $$\frac{1}{\beta^2} = 1 + (tc/L_0)^2$$ $$\beta^2 = \frac{1}{1+(tc/L_0)^2}$$ $$\beta = \frac{1}{\sqrt{1+(tc/L_0)^2}}$$ plugging in $tc/L_0$ = 8/10 gives $$\beta = 0.78$$


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