The absorption spectrum of light galaxy is measured and the wavelength of one of the lines identified as the calcium line is found to be 478 nm. The same line has a wavelength of 397 nm when measured in a laboratory. (a) Is the galaxy moving towards or away from the earth? (b) Calculate the speed of the galaxy relative to Earth (Speed of light 3 ｘ 10⁸ m s⁻¹). (Ans: (a) away from the Earth (b) 6.1 ｘ10⁷ m s⁻¹)


Data Given:


Apparent Wavelength = λ' = 478 nm = 478 ｘ10⁻⁹ m = 4.78 ｘ10⁻⁷ m

Measured (Actual) Wavelength = λ = 397 nm = 397 ｘ10⁻⁹ m = 3.97ｘ10⁻⁷ m

Speed of light = r = 3ｘ10⁸ m s⁻¹

To Find:


(a) Direction of motion of a Galaxy (toward or away)  =  ?

(b) Relative speed of the galaxy = `\v_s` = ?


Solution:

(a) Direction of motion of a Galaxy (toward or away)  =  ?


We know that when

f ' > f then the two objects are approaching toward each other and when
f ' < f then the two object are moving away each other.

So, we will find first the actual frequency f and apparent frequency f ' from the given data

As we know that for actual frequency f

v = f λ

or

f = `\frac {v}{λ}`

putting values

f = `\frac {3ｘ10⁸ m s⁻¹}{3.97ｘ10⁻⁷ m}`

f = 0.756 ｘ10¹⁵ Hz

or

f = 7.56 ｘ10¹⁴ Hz ----------------(1)



for apparent frequency f '

f ' = `\frac {v}{λ'}`

putting values

f '= `\frac {3ｘ10⁸ m s⁻¹}{4.78ｘ10⁻⁷ m}`

f ' = 0.628 ｘ10¹⁵ Hz

or

f ' = 6.28 ｘ10¹⁴ Hz ----------------(2)
 
comparing equation (1) an (2) we get

f ' < f 

Thus, galaxy is moving away from the earth ----Ans. 1


(b) Relative speed of the galaxy = `\v_s` = ?


The formula for apparent frequency f ' when the Source is moving away from the observer  is


f ' = `\frac {v}{v + v_s}` f -------------(1)

or (by simplifying) 

`\v_s` = `\frac {v f}{f '}` - v

by putting the corresponding values

`\v_s` = `\frac {3ｘ10⁸ m s⁻¹ ｘ 7.56 ｘ10¹⁴ Hz }{6.28 ｘ10¹⁴ Hz}` - 3ｘ10⁸ m s⁻¹

`\v_s` = 3.611ｘ10⁸ m s⁻¹ - 3ｘ10⁸ m s⁻¹ 

`\v_s` = 0.611 ï½˜10⁸ m s⁻¹ 

or

`\v_s` = 6.11 ï½˜10⁷ m s⁻¹   ------------Ans.(2)



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