Answer
$-2.37KHz$
Work Step by Step
We know that
$f^{\prime}-f=\frac{fu}{c}$
We plug in the known values to obtain:
$f^{\prime}-f=\frac{(8.00\times 10^9)(44.5)}{3.00\times 10^8}$
$f^{\prime}-f=-1190Hz$
Now $f^{\prime \prime}-f=2(f^{\prime -f})$
$f^{\prime \prime -f}=2(-1190)=-2.37KHz$
