# Two bodies attracting via gravitation

1. Jun 30, 2010

### Rats_N_Cats

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

Say, there are two bodies, mass m1 and m2, initially at rest at infinite separation. They start accelerating towards each other because of gravity. Express the relative velocity of approach as a function of the distance between the two bodies.

2. Relevant equations

Laws of Motion, Law of Universal Gravitation, Conservation of Momentum and Energy.

3. The attempt at a solution

I thought the loss of gravitational potential energy should equal the gain in kinetic energy of the two bodies, so $\frac{1}{2}(m_1 v_1^2 + m_2 v_2^2) = \frac{Gm_1 m_2}{r}$. Also, because of conservation of momentum, m1v1 = m2v2, because the two objects have zero initial momentum. But I can't figure out the velocities from these two; and I think that's what i need.

2. Jun 30, 2010

### dulrich

You're on the right track. Maybe you just need one more nudge. The second formula implies v2 = m1 v1 / m2. Substitute that into the energy equation and solve for v1...

3. Jul 1, 2010

### Rats_N_Cats

Doing as suggested, I end up with the values for v1 and v2. Then the velocity of approach should be v1 + v2. I get for that $v_1 + v_2 = \sqrt{\frac{2G(m_1 + m_2)}{r}}$. Is that correct?

4. Jul 2, 2010

### inky

When I calculate the problems for collision,
e=(speed of separation)/(speed of approach)

If two bodies are moving opposite directions, relative velocity is v1+v2.

If two bodies are moving same direction, relative velocity is v2-v1.

After collision two bodies need to separate. It means v2>v1.

5. Jul 2, 2010

### Redbelly98

Staff Emeritus
The relative velocity is either v2 - v1 or v1 - v2.

The only time you would use v1 + v2 is if those are the speeds (not velocities) of two objects traveling in opposite directions.

6. Jul 2, 2010

### ehild

Yes.

ehild

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