The time period of rotation of the sun is 25 | vedclass

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(A) The Doppler shift $\Delta \lambda$ is given by the formula $\Delta \lambda = \lambda \frac{v}{c}$,where $v$ is the tangential velocity of the sun'

Vedclass · Accepted Answer

$0.04$

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(A) The Doppler shift $\Delta \lambda$ is given by the formula $\Delta \lambda = \lambda \frac{v}{c}$,where $v$ is the tangential velocity of the sun's surface and $c$ is the speed of light.First,calculate the angular velocity $\omega = \frac{2\pi}{T}$,where $T = 25 \ 	ext{days} = 25 	imes 24 	imes 3600 \ 	ext{s}$.$\omega = \frac{2\pi}{25 	imes 86400} \ 	ext{rad/s}$.The tangential velocity $v = r\omega = (7 	imes 10^8 \ 	ext{m}) 	imes \frac{2\pi}{2160000} \ 	ext{s} \approx 2036 \ 	ext{m/s}$.Using $c = 3 	imes 10^8 \ 	ext{m/s}$ and $\lambda = 6000 \ \mathring{A}$:$\Delta \lambda = 6000 	imes \frac{2036}{3 	imes 10^8} \ \mathring{A} \approx 0.04 \ \mathring{A}$.

The time period of rotation of the sun is $25$ days and its radius is $7 \times 10^8 \ m$. The Doppler shift for the light of wavelength $6000 \ \mathring{A}$ emitted from the surface of the sun will be ...... $\mathring{A}$.

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