Doppler effect: calculate speed of a moving star

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Homework Help Overview

The discussion revolves around calculating the speed of a moving star using the Doppler effect. Participants are examining the relationship between observed wavelengths and the motion of the star relative to Earth.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are analyzing the equations related to the Doppler effect and questioning the correctness of the calculations presented. There is a focus on the interpretation of negative velocity and its implications for the direction of motion.

Discussion Status

The discussion is active, with participants providing feedback on the calculations and clarifying concepts related to speed and velocity. There is an exploration of different interpretations regarding the sign of velocity and its physical meaning.

Contextual Notes

Participants are navigating the definitions of speed and velocity, particularly in the context of directionality and the implications of negative values in their calculations. There is an emphasis on the importance of defining positive direction in the context of the problem.

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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