# Gravitation between two spheres

1. Dec 3, 2013

### dzidziaud

Hello, I'm new here :) I'm stuck on a homework problem, even though it's probably really easy and I'm just being stupid.

1. The problem statement, all variables and given/known data
In outer space, two solid spheres of radius 10.0 cm are held 30.0 cm apart center to center. One has a mass of 20.0 kg while the other has a mass of 50.0 kg. They are then released. Ignoring all forces except the mutual gravitational attraction of the two spheres, what is the speed of each sphere when they make contact?

2. Relevant equations
U = -Gm1m2/r

3. The attempt at a solution
So I found the potential energy before the spheres are released by using the equation above. I got -2.22e-7 J. When the spheres make contact, the centers are 20 cm apart, so I found the potential energy at r=20cm, and got -3.34e-7 J.
Then I used conservation of energy: KE1 + U1 = KE2 + U2. The initial KE was 0, so the KE when the spheres make contact is U1-U2 = 1.12e-7 J. So that is the total kinetic energy when the spheres make contact.
What I don't know is how to find the kinetic energy of EACH sphere individually. I thought that it would be proportional to their masses, but then I end up with a redundant answer where the masses cancel out and the velocities are equal, which I don't think is right.

2. Dec 3, 2013

### voko

What about the momentum of the entire system?

3. Dec 3, 2013

### dzidziaud

I don't know :( Are you saying that momentum will be the same before and after collision? But using that turns into an ugly system of equations that requires me to use the quadratic formula. Is there no simpler way?

4. Dec 3, 2013

### voko

What is the total momentum at the very beginning, not at the collision?

5. Dec 3, 2013

### dzidziaud

WHOA. It's zero. So m1v1=-m2v2. So v1 = -m2/m1v2. Okay. That makes things a lot easier :) So now KE of m1 = 1/2(m1)(m2^2/m1)(v2)^2 and KE of m2 is just 1/2(m2)(v2)^2. Putting that together, I get that the final KE, which I got above to be 1.12e-7, = 1/2((m2)^2/m1)(v2)^2 + 1/2(m2)(v2)^2. Solving for v2, I got 3.58e-5 m/s. Finding v1 from that will be simple. Is that right? :D

6. Dec 3, 2013

### voko

I have not checked the numbers, but the method is good.

7. Dec 3, 2013

### dzidziaud

Ok. Thanks so much!