Answer
$4.1\ ly$
Work Step by Step
We can find the required distance in light years as follows:
$L=L_{\circ}\sqrt{1-v^2/c^2}$
$\implies L_{\circ}=\frac{L}{\sqrt{1-v^2/c^2}}$
$\implies L_{\circ}=\frac{L_1}{\sqrt{1-v_1^2/c^2}}=\frac{L_2}{\sqrt{1-v_2^2/c^2}}$
This simplifies to:
$L_2=L_1\frac{\sqrt{1-v_2^2}}{\sqrt{1-v_1^2}}$
We plug in the known values obtain:
$L_2=7.5\ ly\frac{1-\sqrt{(0.89)^2}}{1-(0.55)^2}$
$L_2=4.1\ ly$
