IAO 2014, Теория, Група β — Задача 6. Sombrero Galaxy

Solution

6.1. All the lines in the spectra are shifted somewhat to longer wavelengths (redshifted). Obviously, M104 is receding from us due to the expansion of the Universe.

6.2. The overall receding velocity of M104 may be found from the redshift of the spectrum of the central region. Taking the average redshift of the four emission lines, we find that

and since $z\ll1$, the overall receding velocity of M104 is given by:

6.3. According to the Hubble law the distance to Sombrero Galaxy is

Note: In reality the distance to Sombrero Galaxy is 9 Mpc. But we continue solution according to the calculated value, 15 Mpc.

The first and second spectra have different shifts than the third one. It indicates that the gas which is generating the emission lines has a motion relative to the center of M104. The amounts of these redshifts are not too clear but evidently less than the third one. To find the mass of the smaller region near the center, we will use the first spectrum as obtained closer to the center region. The distance from the center of the galaxy to this near-center region can be found by measuring sizes at the lower picture. The angular distance is approximately 0.05 of the width of the lower image, $\alpha=0.05\times10.8''=0.54''$. To get the actual distance in a.u., it is easy to use the formula

Note: In reality $R(\mathrm{a.u.})=0.54\times9\cdot10^6\approx5\cdot10^6$.

The average difference in redshifts of the first and third spectra may be estimated as about $\Delta z\approx0.0013$. Thus the velocity of the gas relative to the galactic center is about

Then, let us assume that the gas in the region of the third spectrum revolves around the galactic center in a circular orbit with the period T. Its revolving velocity is

By comparing with the average velocity of the Earth revolving around the Sun (A=1 au, P=1 year),

we find that

Assuming that the mass of the gas is much less than that of the galactic center and applying the Kepler’s III law (in units solar mass–au–year), we obtain the mass of the galactic center

Note: In reality, we should write $M\approx200\times5\cdot10^6m_{\odot}\approx10^9m_{\odot}$.

It is an approximation for the mass of the supermassive black hole in the center of M104.

Source note: the key switches from z=0.0035 to 0.0036 while retaining 1050 km/s; the latter multiplication would give 1080. It also changes the moving gas from the first spectrum to the third in its prose. Both inconsistencies are retained. The “in reality” values above are statements of the 2014 source.

Official grading criteria

• 6.1. Spectra redshifted, receding from us — 1 pt.
• 6.2. z_av=(Δλ/λ)_average≈0.0035 — 1 pt; V=cz=0.0036×300000 km/s=1050 km/s — 1 pt.
• 6.3. L=V/H=1050 km/s /71 km/s/Mpc≈15 Mpc — 1 pt; R(au)=0.54×15·10⁶≈8·10⁶ — 1 pt; other correct calculations and obtaining mass of galactic center — 3 pt.

Jury note: evaluate the solution in essence rather than formal presence of sentences/formulae; award points when subsequent steps de facto use the required positions.