SRT - Distance to the moon for a rocket

Question

I shall use units where time and distance are in seconds and speed is dimensionless between zero and one. Suppose a rocket $B$ passes by the earth $A$ on its way to the moon at speed $v$. At that event $(1)$ both clocks are synchronized to zero. According to $A$, the moon is $d$ seconds away. How far it is according to $B$ at event $(1)$? $B$ sends a photon to the moon and it comes back after $2t$ seconds. According to $B$, the photon travelled $t$ seconds to the moon and $t$ seconds back. But during that $t$ seconds the moon moved toward $B$ a distance of $tv$. So the distance to the moon according to $B$ at event $(1)$ is $t(1+v)$. What is $t$? According to $A$, when $2t$ seoconds passed on $B$'s clock, on $A$'s clock passed $2t/\sqrt {1-v^2}$. During that time the photon travelled distance $d$ to the moon and distance $d(1-v)$ back to $B$ and $d+d(1-v)=2t/\sqrt{1-v^2}$. So $t=d \sqrt{1-v^2}(2-v)/2$. And $t(1+v)=d \sqrt{1-v^2}(2-v)(1+v)/2$. This result is strange, since for small $v$ this is actually greater than $d$, maximizes to almost $1.06d$ when $v=0.232$. Did I make a mistake?

Answer 1

and distance $d(1-v)$ back to $B$

This is wrong. The correct expression is $d(1-v)/(1+v).$ At the moment when the photon hits Moon and is reflected, the distance between the rocket and Moon is $d(1-v),$ but then the rocket continues moving towards Moon at the same time as the photon moves back to the rocket. So the actual distance the photon travels during the return is shorter than $d(1-v).$

If $\tau$ is the time the return takes, then the distance the photon travels is start distance minus the distance the rocket moves during that time, $d(1-v) - v\tau,$ and since it's a light pulse, this distance is also $\tau.$ Thus, $$d(1-v) - v\tau = \tau,$$ which gives $$\tau = d\frac{1-v}{1+v}.$$