- #1
Daaavde
- 29
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As the title says, I'm interested in the precision a Doppler shift can be revealed.
That's why I read that it's possible to detect exo-planets by measuring the Doppler shift of the star's spectra. Works like this: the stars moves, since it moves back and forth is spectrum suffer from Doppler shift, by measuring this shift you can deduce at which speed the star is moving.
Now, my professor said once that state of the art spectrographs can measure the star moving at 10 cm/s.
But, using Doppler's law: V = c \frac{\lambda -\lambda_0}{\lambda_0} \qquad \Rightarrow \qquad \Delta \lambda = \frac{\lambda_0 V}{c} and setting, for now, V = 1 m/s, c = 3 \cdot 10^{8} m/s that means that we can reveal a \Delta \lambda = \frac{\lambda_0}{10^8}.
My professer intended that they use a special tecnique to achieve this precision and that he would probably ask me this at my exam (which is tomorrow) so please, if someone could answer very fast i would be very grateful.
That's why I read that it's possible to detect exo-planets by measuring the Doppler shift of the star's spectra. Works like this: the stars moves, since it moves back and forth is spectrum suffer from Doppler shift, by measuring this shift you can deduce at which speed the star is moving.
Now, my professor said once that state of the art spectrographs can measure the star moving at 10 cm/s.
But, using Doppler's law: V = c \frac{\lambda -\lambda_0}{\lambda_0} \qquad \Rightarrow \qquad \Delta \lambda = \frac{\lambda_0 V}{c} and setting, for now, V = 1 m/s, c = 3 \cdot 10^{8} m/s that means that we can reveal a \Delta \lambda = \frac{\lambda_0}{10^8}.
My professer intended that they use a special tecnique to achieve this precision and that he would probably ask me this at my exam (which is tomorrow) so please, if someone could answer very fast i would be very grateful.