Doppler effect: calculate speed of a moving star

AI Thread Summary
The discussion focuses on calculating the speed of a moving star using the Doppler effect. The initial calculations indicate that the star is moving away from Earth, with a wavelength shift from 6250 Å to 6500 Å. The derived speed of the star is initially calculated as 11760 km/s, but it is clarified that this value represents velocity, which can be negative depending on the defined direction. The distinction between speed as a scalar quantity and velocity as a vector is emphasized, confirming that speed should always be expressed as a positive value. Ultimately, the correct interpretation is that the star's velocity is negative, indicating its movement away from Earth.
Pushoam
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Homework Statement



upload_2017-12-23_12-33-26.png

Homework Equations

The Attempt at a Solution



When the source is not moving, ## \lambda = 6250 A ## ...(1)

When the source is moving, ## \lambda ' = 6500 A ## ...(2)

From (1) and (2),

## \lambda ' > \lambda ## ...(3)

This means that the source is moving away from the Earth.

## \nu ' = \nu \sqrt{ \frac { c+v}{c-v} ##

## {\frac { \lambda }{\lambda '} }^2 = { \frac { c+v}{c-v} ##

0.9245c –c = v(1+0.9245)

V =11760 km/s

Is this correct?
 

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Pushoam said:
0.9245c –c = v(1+0.9245)
I hope you meant to write c - 0.9245c in the left-hand side. Otherwise the argumentation and the result seem fine.
 
Orodruin said:
I hope you meant to write c - 0.9245c in the left-hand side.
I meant to write the following:
Pushoam said:
0.9245c –c = v(1+0.9245)
The following is wrong:
Pushoam said:
V =11760 km/s
The speed v is negative implying that the star is moving away from the Earth.
v = - 11760 km/s
 
Pushoam said:
## \nu ' = \nu \sqrt{ \frac { c+v}{c-v}##
##

{\frac { \lambda }{\lambda '} }^2 = { \frac { c+v}{c-v} ##
## \nu ' = \nu \sqrt{ \frac { c+v}{c-v}} ##
##

{\frac { \lambda }{\lambda '} }^2 = { \frac { c+v}{c-v} }##
 
Speed can never be negative as it is the absolute value of velocity. Whether the velocity is negative or not depends on which direction you defined as positive.
 
Orodruin said:
Speed can never be negative as it is the absolute value of velocity. Whether the velocity is negative or not depends on which direction you defined as positive.
I had the impression that speed is a scalar quantity, which can be either positive or negative.
Now, I got to know that speed is defined as magnitude of velocity.

Then ## \vec v = v \hat v ## where ##\hat v ## is a positive constant unit vector and v could be either positive or negative and it has no specific name.
So, the following is wrong.
Pushoam said:
The speed v is negative implying that the star is moving away from the Earth.
The correct one is :
v is negative implying that the star is moving away from the Earth.
v = - 11760 km/s
 
If you defined the positive direction to be moving towards the star, yes. Rounding the speed gives you one of the available answers.
 
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