(1) When observations were made of a spectral line with a rest wavelength of 500
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(1) When observations were made of a spectral line with a rest wavelength of 500.7 nm of interstellar gas on opposite sides of the galaxy M87, it was found that one side peaked at 499.8 nm and the other at 501.6 nm. The gas that was observed is located at a distance of 30.7 pc. (a) How can you explain this observation in terms of rotation of the galaxy and its gas, and what is the observed speed of the rotation (assume u ? c)? (b) Estimate the mass inside the radius where this gas is located in solar masses.Explanation / Answer
1. given spectral line wavelength lambdao = 500.7 nm
peaks are found at wavelengths lambdar = 499.8 nm, lambdab = 501.6 nm
from dopplers effect, for an object moving away (+ve velocity) or coming towards (-ve velocity) the observer, the observed wavelength is given by
v/c = (lambda - lambdao)/lambdao, where v is velocityof the object
now for the galaxy
the two velocities we find are
v1/c = (0.9nm)/500.7 nm
v1 = -0.001797483523c
v2/c = 0.9 nm/500.7 nm
v2 = 0.001797483523 c
a. hence
we can see that one side of the galaxy is moving away from us while the other comes towards us, both at same speed
hence, the center of the galaxy is at relative rest to the frame of reference
also, the galaxy is inour plane of sight and hence we can see its two tips at equal coming ang gpiong speeds, hecne its rotating with aangular speed w such that
w*r = v2
where r is the distance of this point onthe galaxy from the center of the galaxy
hence
rotation speed = 0.001797483523 c
where c = 3*10^8 m/s
b. mass inside the ring = M
then
GM/r^2 = v^2/r
GM/r = v^2
M = v^2*r/G
r = 30.7/2 = 15.35 Pc = 47.3701*10^16 m
hence
M = 1.0325731251*10^9 M. ( where M. is mass of sun)
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