It is possible to detect Doppler shifts in the light emitted from each side of a rapidly spinning star. Suppose the light from a nearby star of radius $2.5 \times 10^{8} \mathrm{~km}$ is blue-shifted on one side of the star and red-shifted on the other side. Assume the center of mass of the star is stationary relative to Earth. If the star emits light of wavelength $525 \mathrm{~nm}$ that is shifted, in the received light, to $523 \mathrm{~nm}$ on one side of the star and to $527 \mathrm{~nm}$ on the other side, what is the rotational period of the star?
A) 34 days
B) 24 days
C) 32 days
D) 16 days

Question

Quizplus · Accepted Answer

The change in wavelength due to the Doppler effect is related to the velocity of the source. The velocity $v$ of the star's surface can be calculated using the formula for the Doppler shift in light: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$, where $\Delta \lambda$ is the change in wavelength, $\lambda$ is the original wavelength, and $c$ is the speed of light. The rotational period $T$ can then be found using the circumference of the star and the velocity: $T = \frac{2\pi R}{v}$, where $R$ is the radius of the star. Given $\Delta \lambda = 2 \mathrm{~nm}$ (from 525 nm to 527 nm or from 525 nm to 523 nm), $\lambda = 525 \mathrm{~nm}$, and $R = 2.5 \times 10^{8} \mathrm{~km}$, we can calculate the rotational period, resulting in approximately 16 days.

It Is Possible to Detect Doppler Shifts in the Light 2.5×108 km2.5 \times 10^{8} \mathrm{~km}2.5×108 km

It Is Possible to Detect Doppler Shifts in the Light $2.5 \times 10^{8} \mathrm{~km}$