## College Physics (4th Edition)

$L =6.5\times 10^{10}~m$
Let $L_0 = 0.75\times 10^{11}~m$ We can find the distance $L$ according to the astronaut: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $L = (0.75\times 10^{11}~m)~\sqrt{1-\frac{(0.50~c)^2}{c^2}}$ $L =6.5\times 10^{10}~m$