Two masses `m_(1)` and `m_(2)` at an infinite distance from each other are initially at rest, start interacting gravitationally. Find their velocity of approach when they are at a distance `r` apart.

Explanation

To find the velocity of approach of two masses \( m_1 \) and \( m_2 \) when they are at a distance \( r \) apart, we can follow these steps:

### Step 1: Understand the gravitational force
When two masses \( m_1 \) and \( m_2 \) are at a distance \( r \) apart, the gravitational force \( F \) acting between them can be expressed using Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant.

### Step 2: Determine the acceleration of each mass
The acceleration \( a_1 \) of mass \( m_1 \) due to the gravitational force exerted by mass \( m_2 \) is given by:
\[ a_1 = \frac{F}{m_1} = \frac{G m_2}{r^2} \]
Similarly, the acceleration \( a_2 \) of mass \( m_2 \) due to the gravitational force exerted by mass \( m_1 \) is:
\[ a_2 = \frac{F}{m_2} = \frac{G m_1}{r^2} \]

### Step 3: Find the relative acceleration
The relative acceleration \( a \) of the two masses towards each other is the sum of their accelerations:
\[ a = a_1 + a_2 = \frac{G m_2}{r^2} + \frac{G m_1}{r^2} = \frac{G (m_1 + m_2)}{r^2} \]

### Step 4: Relate acceleration to velocity
The relative acceleration can also be expressed in terms of the relative velocity \( v \) and the distance \( r \):
\[ a = \frac{dv}{dt} = \frac{dv}{dr} \cdot \frac{dr}{dt} \]
Since \( \frac{dr}{dt} \) is negative (as \( r \) decreases), we can write:
\[ a = -v \frac{dv}{dr} \]

### Step 5: Set up the equation
Substituting the expression for acceleration:
\[ -v \frac{dv}{dr} = \frac{G (m_1 + m_2)}{r^2} \]

### Step 6: Rearrange and integrate
Rearranging gives:
\[ v dv = -\frac{G (m_1 + m_2)}{r^2} dr \]
Now, we can integrate both sides. The left side integrates from \( 0 \) to \( v \) and the right side from \( \infty \) to \( r \):
\[ \int_0^v v \, dv = -G (m_1 + m_2) \int_{\infty}^{r} \frac{1}{r^2} \, dr \]

### Step 7: Perform the integration
The left side becomes:
\[ \frac{v^2}{2} \]
The right side evaluates to:
\[ -G (m_1 + m_2) \left[-\frac{1}{r}\right]_{\infty}^{r} = G (m_1 + m_2) \left(0 - \left(-\frac{1}{r}\right)\right) = \frac{G (m_1 + m_2)}{r} \]

### Step 8: Combine results
Thus, we have:
\[ \frac{v^2}{2} = \frac{G (m_1 + m_2)}{r} \]
Multiplying both sides by 2 gives:
\[ v^2 = \frac{2G (m_1 + m_2)}{r} \]

### Step 9: Solve for \( v \)
Taking the square root of both sides, we find the velocity of approach:
\[ v = \sqrt{\frac{2G (m_1 + m_2)}{r}} \]

### Final Answer
The velocity of approach of the two masses when they are at a distance \( r \) apart is:
\[ v = \sqrt{\frac{2G (m_1 + m_2)}{r}} \]