Doppler effect: A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative velocity of 0.40c. The spaceship passes the beacon and departs from it with a DIFFERENT relative velocity. The beacon signal is now detected at a frequency of 40 MHz. What is the new velocity of the spaceship relative to the beacon?
A) 0.20c
B) 0.22c
C) 0.24c
D) 0.26c
E) 0.28c

Question

Quizplus · Accepted Answer

Using the formula for Doppler effect:

$f_o = f_s \frac{c + v}{c - v}$

where:
$f_o$ = observed frequency
$f_s$ = source frequency
$c$ = speed of light
$v$ = relative velocity between source and observer

When the spaceship approaches the beacon:

$f_o = 40	ext{ MHz}$
$f_s = 50	ext{ MHz}$
$c = 3	imes10^8 	ext{ m/s}$
$v = 0.4c$

Solving for $v$:

$40	ext{ MHz} = 50	ext{ MHz} \frac{3	imes10^8 	ext{ m/s} + v}{3	imes10^8 	ext{ m/s} - v}$
$0.8 = \frac{1.4c}{1.6c - v}$
$1.12c - 0.8v = 1.4c$
$0.6v = 0.28c$
$v = 0.4667c$

When the spaceship departs the beacon, the observed frequency is still 40 MHz. Solving for the new velocity:

$f_o = 40	ext{ MHz}$
$f_s = 50	ext{ MHz}$
$c = 3	imes10^8 	ext{ m/s}$
$v = ?$

$40	ext{ MHz} = 50	ext{ MHz} \frac{3	imes10^8 	ext{ m/s} + v}{3	imes10^8 	ext{ m/s} - v}$
$0.8 = \frac{1.4c}{1.6c + v}$
$1.12c + 0.8v = 1.4c$
$0.8v = 0.28c$
$v = 0.35c$

Therefore, the new velocity of the spaceship relative to the beacon is:

$v_{new} = 0.35c$

The best choice is B because it matches the calculated value.

Doppler Effect: a Navigational Beacon in Deep Space Broadcasts at a Radio