$A$ star producing light of wavelength $6000 \ \mathring{A}$ moves away from the Earth with a speed of $5 \ km/s$. Due to the Doppler effect, the shift in wavelength will be $...... \ \mathring{A}$ $(c = 3 \times 10^8 \ m/s)$.

$A$ heavenly body is receding from Earth such that the fractional change in wavelength $\lambda$ is $1$. What is its velocity?

$A$ rocket is moving away from the Earth at a speed of $0.2c$,where $c$ is the speed of light. It emits a signal of frequency $4 \times 10^7 \text{ Hz}$. What will be the frequency observed by an observer on the Earth?

Write the formula for the Doppler shift for light.

$A$ star moves away from the Earth at a speed of $0.8\,c$ while emitting light of frequency $6 \times 10^{14} \,Hz$. What frequency will be observed on the Earth? ($c =$ speed of light)

In the spectrum of light of a luminous heavenly body,the wavelength of a spectral line is measured to be $4747 \ \mathring{A}$,while the actual wavelength of the line is $4700 \ \mathring{A}$. The relative velocity of the heavenly body with respect to the Earth will be (velocity of light is $3 \times 10^8 \ m/s$).