Two bodies of mass `m_(1) and m_(2)` are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Show that their relative velocity of approach at separation r between them is:

Options

A. `sqrt((2G(m_1+m_2))/r)`
B. 2`sqrt((2Gm_1m_2)/((m_1+m_2)r))`
C. `sqrt((G(m_1+m_2))/r)`
D. `sqrt((Gm_1 m_2)/((m_1+m_2)r))`

Answer

To solve the problem of finding the relative velocity of approach of two bodies of mass \( m_1 \) and \( m_2 \) under mutual gravitational attraction at a separation \( r \), we can follow these steps:

### Step 1: Understand the Initial Conditions
Initially, both bodies are at rest and placed at an infinite distance apart. Therefore, the initial potential energy (PE) and kinetic energy (KE) of the system are both zero:
\[ PE_{\text{initial}} = 0, \quad KE_{\text{initial}} = 0 \]

### Step 2: Use Conservation of Energy
When the two bodies are at a separation \( r \), the potential energy due to their gravitational attraction is given by:
\[ PE = -\frac{G m_1 m_2}{r} \]
where \( G \) is the gravitational constant. The total mechanical energy of the system must remain constant, so we can write:
\[ 0 = -\frac{G m_1 m_2}{r} + \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

### Step 3: Express Kinetic Energy in Terms of Potential Energy
Rearranging the equation gives us:
\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{G m_1 m_2}{r} \]

### Step 4: Apply Conservation of Momentum
Since there are no external forces acting on the system, the momentum is conserved. Initially, the momentum is zero, so:
\[ 0 = m_1 v_1 + m_2 v_2 \]
This implies:
\[ m_1 v_1 = -m_2 v_2 \quad \Rightarrow \quad v_1 = -\frac{m_2}{m_1} v_2 \]

### Step 5: Substitute \( v_1 \) into the Energy Equation
Substituting \( v_1 \) into the energy equation:
\[ \frac{1}{2} m_1 \left(-\frac{m_2}{m_1} v_2\right)^2 + \frac{1}{2} m_2 v_2^2 = \frac{G m_1 m_2}{r} \]
This simplifies to:
\[ \frac{1}{2} \frac{m_2^2}{m_1} v_2^2 + \frac{1}{2} m_2 v_2^2 = \frac{G m_1 m_2}{r} \]
Factoring out \( \frac{1}{2} v_2^2 \):
\[ \frac{1}{2} v_2^2 \left(\frac{m_2}{m_1} + m_2\right) = \frac{G m_1 m_2}{r} \]
This leads to:
\[ v_2^2 \left(\frac{1}{m_1} + 1\right) = \frac{2G}{r} \]

### Step 6: Solve for \( v_2 \)
Rearranging gives:
\[ v_2^2 = \frac{2G}{r} \cdot \frac{m_1}{m_1 + m_2} \]
Taking the square root:
\[ v_2 = \sqrt{\frac{2G m_1}{r(m_1 + m_2)}} \]

### Step 7: Find \( v_1 \)
Using the relationship \( v_1 = -\frac{m_2}{m_1} v_2 \):
\[ v_1 = -\frac{m_2}{m_1} \sqrt{\frac{2G m_1}{r(m_1 + m_2)}} \]

### Step 8: Calculate the Relative Velocity of Approach
The relative velocity of approach \( V_{\text{approach}} \) is given by:
\[ V_{\text{approach}} = |v_1| + |v_2| = \frac{m_2}{m_1} \sqrt{\frac{2G m_1}{r(m_1 + m_2)}} + \sqrt{\frac{2G m_1}{r(m_1 + m_2)}} \]
Factoring out the common term:
\[ V_{\text{approach}} = \left(\frac{m_2}{m_1} + 1\right) \sqrt{\frac{2G m_1}{r(m_1 + m_2)}} \]

### Final Result
Thus, the relative velocity of approach at separation \( r \) is:
\[ V_{\text{approach}} = \sqrt{\frac{2G(m_1 + m_2)}{r}} \]