Gravitational Forces between three stars

AI Thread Summary
The problem involves three stars, each with the mass and radius of the sun, forming an equilateral triangle and released from rest. The gravitational force equation is applied, but confusion arises regarding the uniformity of acceleration as the stars move toward each other. The discussion suggests using conservation of energy to relate gravitational potential energy to kinetic energy for accurate calculations. The gravitational potential energy is calculated using the distance between the stars, and the goal is to find the final speed as they collide at the center. The expected answer is 3.71*10^5 m/s, highlighting the need for careful consideration of gravitational interactions during the stars' acceleration.
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Homework Statement


Three stars, each with the mass and radius of our sun, form an equilateral triangle 5.0*10^9m on a side. If all three are simultaneously released from rest, what are their speeds as they crash together in the center?

Homework Equations


Fg= GMm/r^2


The Attempt at a Solution


It says the three stars are released from rest and start accelerating towards each other, so I'm assuming there is no centripetal acceleration (?). So I set 2Fgcos30 equal to ma and solved for a, which equals 9.2. Then I plugged it into the v = (2as)^(1/2) formula with s=2.89*10^9m being the midpoint of the circle using trig. I can't figure out what I'm doing wrong.

The given answer in the book is 3.71*10^5m/s
 
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Is the acceleration uniform?
 
I'm not sure, but if it wasn't.. you would have to account for stuff I feel like is impossible, like accounting for the increase in gravitational forces as the stars accelerate towards each other...
 
That is what you need to do. I wouldn't say it is impossible, just needs a little more thought.
 
To solve this problem, you need to start out with the basic equation of conservation of energy, as this is an energy problem.So the gravitational potential energy equation is given by Ue = G M1*M2/R

Kinetic energy is K = 1/2 M V^2

We need to calculate the gravitational potential energy for one of the masses, since the equations of motion for one of them is the same for the other two.

So since G is just the constant, leave it be...M1 can be equated as the other two masses, so 2*mass of sun, then M2 is the mass of the current sun at hand, so M1 = mass of sun. R is going to be the Rcos30 value as we are only concerned with that particular component.

So once you calculate your gravitational potential energy...I got something around 1.22*10^41 or so...there is a negative sign as it indicates direction. Then set that equal to the kinetic energy, 1/2mV^2, and solve for V.
 
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