Wavelengths of light from a distant galaxy are found to be 0.44% longer than the corresponding wavelengths measured in a terrestrial laboratory. What is the speed that the wave is approaching?

Question

Wavelengths of light from a distant galaxy are found to be 0.44% longer than the corresponding wavelengths measured in a terrestrial laboratory.  What is the speed that the wave is approaching?

Hailey Alexander · Accepted Answer

Light always travels at the speed of light, in a vacuum, #2.9979*10^8m/s# When solving wave problems, the universal wave equation, #v=flamda#, is often used. And if this were a general wave problem an increased wavelength would correspond with an increased speed (or decreased frequency). But the speed of light remains the same in a vacuum, for any observer, the constant known as #c#.

Olivia Aldridge · Answer

undefined To calculate the speed at which the galaxy is approaching or receding, you can use the formula for redshift (z), which is related to the velocity (v) of the galaxy relative to us and the speed of light (c). The observed redshift (z) is given by the fractional increase in wavelength (Δλ/λ), where Δλ is the change in wavelength, and λ is the original wavelength.

Given:
- Δλ/λ = 0.44% = 0.0044 (converting percentage to decimal)

The formula relating redshift to velocity is:

$$v = z \times c$$

where:
- $v$ is the velocity of the galaxy,
- $z$ is the redshift,
- $c$ is the speed of light ($c \approx 3.00 \times 10^8$ m/s).

Substitute the given values into the equation:

$$v = 0.0044 \times 3.00 \times 10^8 \, \text{m/s}$$
$$v = 1.32 \times 10^6 \, \text{m/s}$$

Therefore, the galaxy