Doppler effect: calculate speed of a moving star

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SUMMARY

The discussion focuses on calculating the speed of a moving star using the Doppler effect. The initial calculations indicated a speed of 11760 km/s, but it was clarified that this value represents velocity, which can be negative depending on the defined direction. The correct interpretation is that the star is moving away from Earth, resulting in a negative velocity of -11760 km/s. The distinction between speed as a scalar quantity and velocity as a vector quantity was emphasized, highlighting that speed cannot be negative.

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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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