Binary star system ratio of force

AI Thread Summary
In a binary star system with masses m and 3m, the gravitational force exerted by each star at the common center of mass is being analyzed. The user calculated the force ratio as 1:9 using Newton's law of gravitation, but expressed uncertainty about the correctness of this result. Other participants questioned the calculation of the center of mass and emphasized that the force depends on both mass and distance from the center. They noted the ambiguity in asking for the force "at the center of mass," as no force exists at that point without a mass present. Clarification on these points is essential for accurate problem-solving.
Ljungberg92
Messages
7
Reaction score
0

Homework Statement


2 stars rotating around the common centre of mass have masses m and 3m. What is the ratio of the force due to each star at the common centre of mass.


Homework Equations


None


The Attempt at a Solution


I tried to find the centre of mass and applied Newtons law of gravitation F=(GMm)/R^2 to each star and solve i got the ratio 1:9. Is this correct?Thanks
 
Physics news on Phys.org
It doesn't look right to me, but it is hard to tell without seeing the work.
How did you calculate the center of mass and how close to it are each of the masses? Did you remember that the force depends on the mass as well as the distance? Kind of odd that it asks for the force "at the center of mass" - there won't be a force unless there is a mass located there.
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Question about gravitational force(Binary star)
Replies
3
Views
2K
Centre of mass of binary system calculation
Replies
8
Views
2K
Binary system of stars (##\alpha## - centauri)
Replies
5
Views
2K
Newton's shell theorem - all mass is concentrated at its centre
Replies
16
Views
1K
Combined magnitudes of two solar-type stars
Replies
6
Views
2K
Properties of a Star Homework help
Replies
3
Views
2K
What is the distance and mass of a binary star system with given parameters?
Replies
5
Views
2K
I Why Do Stars in Binary Systems Assume a Teardrop Shape?
Replies
14
Views
2K
Finding the mass of a planet is a binary system
Replies
2
Views
5K
Challenging High School Planetary Mechanics Problem
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top