I'll assume that says we're traveling at
[tex]v= 3 \times 10^4 \textrm{ meters / second}[/tex]
Speed of light in the ether frame
[tex]v= 3 \times 10^8 \textrm{ meters / second}[/tex]
Let's say initially we're travelling toward the light,so we expected to measure a speed of [tex]v+c.[/tex]
Six months later we're travelling away from the light, so we expected its speed to be [tex]v-c.[/tex]
That's [tex]v= 3 \times 10^8 \textrm{ m/s } - 3 \times 10^4 \textrm{ m/s} = 2.9997 \times 10^8 \textrm{ m/s}[/tex]
