Energy conservation problem (two star system)

AI Thread Summary
In a binary star system with two sun-like stars, a comet approaches the midpoint between them due to gravitational attraction. The gravitational potential energy (PE) between the stars is calculated using the formula Ug = -G(m1m2)/r, yielding approximately 2.65 X 10^8 J. To find the comet's velocity at the midpoint, it's suggested to focus on the PE difference between infinity and the comet's closest approach, considering symmetry for both stars. An alternative approach involves adjusting for the center of mass of the two stars, as their separation affects the gravitational force exerted on the comet. Understanding these concepts will help in determining the comet's velocity as it reaches the midpoint.
Jordanosaur
Messages
10
Reaction score
0
Hi guys -

Here's the problem I am chewing on:

A binary star system consists of two stars, each equal to the sun in mass. The distance between the two stars is 1.0 X 10^12m. A comet which is essentially at rest, begins to make its journey toward the binary star system as a result of gravity acting upon the comet. If the comet begins a straight line approach that will result in it passing through the midpoint of the distance between the stars, what will the velocity of the comet be at the midpoint?

I believe this to be a problem in which we need to identify the gravitational PE between the two stars, and use the answer to calculate the gravitational force acting on the comet being pulled into the star system.

I can find Ug by using the PE formula for two bodies: Ug = -G(m1m2)/r

I am coming up with roughly 2.65 X 10^8 J, but I don't understand how I can turn this around to apply to the comet's approach to the system. Any suggestions as to where I can start?

Thanks

Jordan
 
Physics news on Phys.org
Jordanosaur said:
Hi guys -

Here's the problem I am chewing on:

AI believe this to be a problem in which we need to identify the gravitational PE between the two stars, and use the answer to calculate the gravitational force acting on the comet being pulled into the star system.

I can find Ug by using the PE formula for two bodies: Ug = -G(m1m2)/r

I am coming up with roughly 2.65 X 10^8 J, but I don't understand how I can turn this around to apply to the comet's approach to the system. Any suggestions as to where I can start?

Thanks

Jordan

I'm not sure that you need to worry about the PE between the stars themselves.

It may be more fruitful to consider the PE difference between infinity and the comets closest approach to one of the stars, ie the 1/2 way point between the two. Realize that you have twice this quantity from symmetry, and convert to kinetic energy. The other approach I might consider would be to realize that the center of mass of the two stars is midway between them, make an adjustment in terms of distance or mass, as two masses separated by a huge distance will in unison exert less force than two side by side.
 
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