Binary-star system, gravity + centripetal force question?

AI Thread Summary
In a binary-star system with each star having a mass of 1.36e30 kg and a distance of 1.1e8 km between them, the gravitational force equals the centripetal force for their revolution around the center of mass. The user initially struggled with the calculations, particularly with the conversion of kilometers to meters and the correct application of the formulas. After some confusion regarding the values for radius and diameter, the user realized the mistake was simply in unit conversion. Once corrected, the calculations became straightforward. This highlights the importance of accurate unit conversion in astrophysical calculations.
basenne
Messages
20
Reaction score
0
1. In a certain binary-star system, each star has the a mass of 1.36e30 kg, and they revolve about their center of mass. The distance between them is 1.1e8 km. What is their period of revolution in Earth years?
2.
gravity = centripetal force
(g*m*m)/d^2 = (mv^2)/r

3. I've tried plugging in numbers, tried 1/2 of 1.1e8 for R and 1.1e8 for D(diameter), didn't work. Not quite sure what I'm doing wrong. I know that v^2 becomes (diameter/time)^2 and to solve from there, however, I'm just not sure where to start. (
 
Last edited:
Physics news on Phys.org
Nevermind, got it. It's actually really easy, I just forgot to change kilometers into meters... fail!
 
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