# Gravitational Forces

1. Nov 20, 2008

### NivekOh

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Fg= GMm/r^2

3. 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

2. Nov 20, 2008

### naresh

Is the acceleration uniform?

3. Nov 21, 2008

### NivekOh

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...

4. Nov 21, 2008

### naresh

That is what you need to do. I wouldn't say it is impossible, just needs a little more thought.

5. Nov 21, 2008

### aerospaceut10

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.