Multi-Wavelength Observations of Cluster

  • Thread starter Thread starter rstein66
  • Start date Start date
AI Thread Summary
Astronomers studied a galaxy cluster with observations across multiple wavelengths, revealing a redshift of z=0.3 and a distance of approximately 1286 Mpc. The cluster's radius was calculated to be about 3.09 Mpc, and the mass of the hot gas was estimated at 2.16 x 10^45 kg. The total luminosity was determined to be around 9.088 x 10^15 Lsun, with the stellar mass calculated at approximately 9731 Msun. The analysis concluded that dark matter constitutes about 98% of the cluster's total mass, highlighting the significant presence of dark matter in galaxy clusters.
rstein66
Messages
6
Reaction score
0

Homework Statement



Hi, I am having difficulties with a first-year astronomy question that was issued on a final a few years back.
A group of astronomers make various observations of a cluster spanning 10arcmin (600arcsec) on the sky. X-ray astronomers found the cluster to be full of hot gas, with λmax=0.0377nm.
Optical astronomers found the bolometric apparent magnitude of the whole cluster to be, m=10.4.
The mass to light ration of the stellar populations in the galaxies in the cluster was equal to 2. Meanwhile, radio astronomers found the neutral hydrogen line associated with the cluster (λHI)=27.3cm. Finally, for this problem use the Hubble constant of 70km/s/Mpc and the density of the intercluster medium to be equal to 10^-27g/cm^3

Find the following:
1. The redshift, z and distance to the cluster d, in pc.
2. The radius of the cluster r, in pc.
3. The mass of the hot gas in the cluster, Mgas.
4. Luminosity and mass of the stellar content, in solar units.
5. Cluster's total mass Mtot
6. Percentage of cluster's mass in dark matter.

Homework Equations


Stated in the question it says use basic formula and do not use modified inverse square laws for distant galaxies for brightness or size.
This question applies many formulas, I am not sure which to use for every part.

The Attempt at a Solution


1. z(of hydrogen line, using normal atomic hydrogen radius of 21cm found in textbook):
z=(λs-λr)/λr = (27.3cm-21.0cm)/21.0cm = 0.3

d=1/p = 1/600" = 0.00167pc.

2. Not sure what to do here.

3. Mgas= (r(Vgas)^2)/G
Vgas(X-ray)=140(sqrt(T)), where T is found by 2.9e6/λmax to be 7.69e7K. Thus, Vgas is equal to 1.23e6m/s, and therefore Mgas = r(1.23e6)^2/6.67e-11

4. L/Lsun=100^(4.75-M)/5
M=m+5-5log(d)
M=10.4+5-5log(0.00167) = 29.3, thus L=1.23e-5Lsun?
I don't know if this was done correctly?

5. Not sure what to do here.
6. I believe that Mtot-Mstars=Mdark matter then I just find its percentage but still don't have either Mtotal or Mstars.

Thanks so much!
 
Physics news on Phys.org
1. the reason to find z first is so you can get the distance from that red-shift (with Hubble)
2. now that you know the distance (from redshift), and the angular size, find the real size.
 
ah so for 1.
z=0.3
v=Ho*d, thus d=v/Ho, where v=cz

so 3e5km/s(0.3)/70km/s/MPC = 1286Mpc

Then for 2, using the formula: 2.5e5"(r/d)=theta(in arcsec)

r=1286e6pc(600") / 2.5e5" = 3.09Mpc

Did I do this correctly, I edited because I used a wrong formula

Continuing with this data i plugged in the radius for the mass of the gas formula and found it to be 2.2e45kg which seems to high.

And I think I know how to get the total cluster mass using its density stated and the formula density, n= m/v which I converted the radius in Mpc to m and used the volume of a sphere and made sure to change the n to kg/m^3 where the Mtot ended up being 3.62e41kgs which is somehow less then Mgas?
 
Last edited:
Ok, disregard my previous post, I think I got it. If anyone could verify my answers and such I would really appreciate it.

1. z=(λshift,hydrogen - λrest,hydrogen)/λrest, hydrogen
z=27.3cm-21cm/21cm , z= 0.3

v=Ho*d, d=v/Ho where v is the recession velocity which is equal to c*z.

v=cz=3e5km/s(0.3)=90000km/s, thus d=(90000km/s)/70km/s/Mpc = 1286Mpc

2. 2.5e5"(R/d)=θ(in arcsec")
600"/2.5e5" = R/1286Mpc ... R=3.09Mpc (Stated in question it says this is "the gigantic Pandora cluster")

3. Mgas=R(Vgas)^2 / G , where Vgas is found with the X-ray data, using the formula Vgas=140(sqrt(T)) where T=2.9e6nm/0.0377nm = 7.69e7K
Thus Vgas = 123e4m/s and Mgas is equal to (9.53e22m<---[3.09e6pc])(1.52e12m/s)/6.67e-11
...Mgas=2.16e45kg=1.08e15Msun

4. L/Lsun=100^(4.75-M)/5
M=m+5-5log(d) = 10.4+5-5log(1286e6pc)
M=-30
...L=9.088e15Lsun

Thus, L/Lsun=(M/Msun)^3.5, L^-3.5=Mass of stars = 9731Msun

5.Mgalaxy,total=rv^2/G , have radius from before and use v as the recession velocity
...Mgalaxy,total=5.787e16Msun

6. (Mgalaxy,total)-Mstars-Mgas=Mdark matter, which ends up being ~98% of cluster's mass which according to my lecture notes seems accurate of clusters.

I am new to this site so I am not too sure how to write the formulas neatly and such but if you want me to expand on anything really just ask, I kinda just summed up my work.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top