Calculating Main Sequence Lifetime.

  • Thread starter Thread starter adwodon
  • Start date Start date
  • Tags Tags
    Lifetime Sequence
AI Thread Summary
To calculate the main sequence lifetime of a 10-M☉ star with a luminosity of 10^4 L☉ and 10% mass conversion from hydrogen to helium, the formula tms ≈ M/L is suggested. The user attempted to apply this formula but found the resulting lifetime of approximately 5 years to be implausibly short. They also calculated the total energy from the mass conversion using E=mc^2, arriving at a similar conclusion. The discussion highlights confusion over the application of the formula and the interpretation of the results. Clarification on the expected lifetime for such a star is sought, indicating a need for further understanding of stellar evolution.
adwodon
Messages
10
Reaction score
0
1. Calculate the main sequence lifetime in years of a 10-M\odot star if it has luminosity of 104L\odot and 10% of its mass will be converted from hydrogen to helium in the core. What will the end state of this star be?

Ok so I wasn't really sure what to put for working here as I have no idea where to begin, I've been searching around for a forumla or something to calculate the lifetime but my lecturers notes are very vague and not helpful at all and numerous different google searches haven't helped in giving me a straight answer.

I found this formula somewhere but it doesn't make sense to me:

tms\approx\frac{M}{L}

Its part of a larger question involving the Hertzsprung Russell diagram which I've done however this bit I am totally stumped, if any of you guys could point me in the right direction or explain it to me i'd really appreciate it.

Oh and I should point out this is from a past paper, the only relevant information I could imagine is given to help on the front of the paper is (excluding the standards like Planck / speed of light etc):

Solar mass: 2x1030kg
Solar Bolometric Luminosity: 3.9x1026W
 
Physics news on Phys.org
Ok so I used E=mc^2 and got the total energy of the converted material as 6x10^38J
Which would sustain that luminosity for 5 years...

Thats roughly the same answer I got from the equation in my original post but I thought it was wrong because 5 years seems waaaaay too short.

Is that a plausible answer?
 
Last edited:
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