## College Physics (4th Edition)

Note that $v = 1.2\times 10^8~m/s = 0.40~c$ We can find the frequency received $f_r$ by the astronauts: $f_r = f_s~\sqrt{\frac{1-v/c}{1+v/c}}$ $f_r = (55~kHz)~\sqrt{\frac{1-(0.40~c)/c}{1+(0.40~c)/c}}$ $f_r = (55~kHz)~\sqrt{\frac{1-0.40}{1+0.40}}$ $f_r = (55~kHz)~\sqrt{\frac{0.60}{1.40}}$ $f_r = 36~kHz$ The astronauts should tune to a frequency of 36 kHz.