Cepheid Variables and the Small Magellanic Cud

  • Thread starter Thread starter Berdi
  • Start date Start date
  • Tags Tags
    Variables
AI Thread Summary
The discussion revolves around calculating the distance to a Cepheid variable in the Small Magellanic Cloud (SMC) based on its apparent brightness compared to δ Cephei. The parallax of δ Cephei is given as 0.0033 arcseconds, establishing its distance at 303 parsecs. The SMC Cepheid is noted to be 11.5 magnitudes fainter, indicating it is further away. The relationship between magnitude difference and distance is explored using the equation m-M=5log(d)-5, emphasizing that the same period implies equal luminosity. Ultimately, understanding these relationships allows for the calculation of the SMC's distance based on the observed brightness difference.
Berdi
Messages
12
Reaction score
0

Homework Statement



The Galactic Cepheid variable δ Cephei has a parallax of 0.0033 arcseconds. An astronomer locates another Cepheid, with the same period as δ Cephei, in the Small Magellanic Cloud (SMC). If this Cepheid appears 11.5 magnitudes fainter than δ Cephei, calculate the distance to the SMC.

Homework Equations



m-M=5\log(d)-5

L=4\pi R^2F

m_{1}-m_{2}=2.5\log( L_{2}/L_{1})

The Attempt at a Solution



Well, I know the distance to the first Cepheid, as it is 1/Parallax = 303pc. I understand that the same period means the same luminosity, and that 11.5 fainter will mean that the m will be actually 11.5 larger, But I'm unsure how this will help me. Unless you say that m_{1}-m_{2} is 11.5 and change the L's for 1/d^2 ?
 
Physics news on Phys.org
The absolute magnitudes of the two variables are the same (same period)
If one appears 11.5 mag fainter then it's obvisouly further away.

The first equation tells you have far away an object would have to be to be less bright by (m-M) magnitudes compared to an object 1pc away.

Another way to think of it. Work out the relative apparent luminosity of the two stars, then if the second one was 1/4 as bright it woudl have to be twice as far away.
 
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 '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...
Back
Top