Answer
$9\times10^9$ light-years.
Work Step by Step
Use equation 33–5b, and then use equation 33–3 to express the ratio of wavelengths.
$$z=\sqrt{\frac{1+v/c}{1-v/c}}-1$$
$$z+1=\sqrt{\frac{1+v/c}{1-v/c}}=2$$
$$ \frac{1+v/c}{1-v/c}=4$$
$$4(1-v/c)=1+v/c$$
$$4-4v/c=1+v/c$$
$$5v/c=3$$
$$v=0.6c$$
Find the distance from Hubble’s law, equation 33–4.
$$d=\frac{v}{H_0}$$
$$d=\frac{0.6(3.00\times10^8m/s)}{ (21000m/s/Mly)}=8571\;Mly\approx9\times10^9 ly$$
