Two bodies attracting via gravitation

[Rats_N_Cats]

Homework Statement

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.

Homework Equations

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

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.

 

[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...

 

[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?

 

[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.

 

[Redbelly98]

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.

 

[ehild]

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?

Yes.

ehild