Estimating the Size of Sgr A*'s Emitting Region

  • Thread starter Thread starter albega
  • Start date Start date
  • Tags Tags
    Astrophysics
AI Thread Summary
The discussion focuses on estimating the size of the emitting region of Sgr A*, which flares at X-ray wavelengths. The initial calculation suggests that light travels approximately 5 AU in 40 minutes, providing an upper bound on the radial extent of the emitting region. However, this does not determine the tangential dimensions, and spherical symmetry is assumed for simplicity. The assumptions made, while not entirely accurate, offer a rough estimate that can be refined with future observations. Overall, the calculations serve as a starting point for understanding the dynamics of Sgr A*'s emissions.
albega
Messages
74
Reaction score
0

Homework Statement


Sgr A* is the radio source at the centre of the galaxy. When it flares at X-ray wavelengths the flux increases by x 40 in 40 minutes. Estimate the size of the emitting region.

The Attempt at a Solution


The solution is to consider the distance traveled by the light in 40 minutes, which gives 40*60*3x10^8m or around 5AU.

However I really don't understand how that is relevant in terms of finding the size of the emitting region. Any hints? Thanks.
 
Physics news on Phys.org
Imagine the region being a cloud of emitters of some radial extent. If all of the emitters begin flashing at the same instant, which ones will you notice first, and when will you finally see all of them?
 
Bandersnatch said:
Imagine the region being a cloud of emitters of some radial extent. If all of the emitters begin flashing at the same instant, which ones will you notice first, and when will you finally see all of them?

Ahhhhh thanks :)

Also, does the above calculation then give the diameter of the region?
 
albega said:
Also, does the above calculation then give the diameter of the region?
It only gives you an upper bound on the radial extent. You can't really extract the tangential dimensions just from the time it takes to reach maximum flux.
You can, though, assume shperical symmetry of the region, and use the radial extent as the diametre.
Neither of the assumptions(shperical symmetry and instantenous flashing of the whole region) is likely to be correct(the second one's physically impossible), but they do give you a ballpark result that later can be narrowed down by future observations.
 
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

Replies
10
Views
3K
Replies
7
Views
3K
Replies
6
Views
2K
Replies
1
Views
1K
Replies
1
Views
5K
Back
Top