Two objects of masses `m` and `4m` are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If `G` is the universal gravitaitonal constant, then at separation `r`

Options

A. the total mechanical energy of the two objects is zero
B. their relative velocity is `sqrt((10Gm)/(r))`
C. the total kinetic energy of the object is `(4Gm^(2))/(r)`
D. their relative velocity is zero.

Explanation

To solve the problem, we need to analyze the motion of two objects of masses `m` and `4m` that are initially at rest at an infinite separation and move towards each other due to their mutual gravitational attraction. We will use the principles of conservation of mechanical energy and conservation of momentum.

### Step-by-Step Solution:

1. **Initial Conditions**:
- The two objects are at rest at an infinite separation.
- Initial kinetic energy (KE_initial) = 0 (since they are at rest).
- Initial gravitational potential energy (PE_initial) = 0 (at infinite separation).
- Therefore, the total mechanical energy (E_initial) = KE_initial + PE_initial = 0 + 0 = 0.

2. **At Separation `r`**:
- Let the masses be `m` and `4m`.
- Let the velocities of the masses `m` and `4m` at separation `r` be `v1` and `v2`, respectively.

3. **Conservation of Momentum**:
- Since there are no external forces acting on the system, the momentum is conserved.
- Initially, the momentum is 0 (as both objects are at rest).
- At separation `r`, the momentum can be expressed as:
\[ 4m \cdot v2 - m \cdot v1 = 0 \]
- Rearranging gives:
\[ 4m \cdot v2 = m \cdot v1 \implies v1 = 4v2 \]

4. **Conservation of Mechanical Energy**:
- The total mechanical energy must remain constant. Therefore:
\[ E_initial = E_final \]
- We know that:
\[ E_final = KE_final + PE_final \]
- The kinetic energy at separation `r` is:
\[ KE_final = \frac{1}{2} m v1^2 + \frac{1}{2} (4m) v2^2 = \frac{1}{2} m (4v2)^2 + \frac{1}{2} (4m) v2^2 \]
\[ = \frac{1}{2} m (16v2^2) + \frac{1}{2} (4m) v2^2 = 8mv2^2 + 2mv2^2 = 10mv2^2 \]

5. **Potential Energy at Separation `r`**:
- The gravitational potential energy (PE) between the two masses at separation `r` is given by:
\[ PE = -\frac{G \cdot m \cdot 4m}{r} = -\frac{4Gm^2}{r} \]

6. **Setting Up the Energy Equation**:
- Since total mechanical energy is conserved:
\[ 0 = KE_final + PE_final \]
- Therefore:
\[ 0 = 10mv2^2 - \frac{4Gm^2}{r} \]
- Rearranging gives:
\[ 10mv2^2 = \frac{4Gm^2}{r} \]
- Dividing both sides by `m` (assuming `m ≠ 0`):
\[ 10v2^2 = \frac{4Gm}{r} \]
- Solving for `v2^2`:
\[ v2^2 = \frac{4Gm}{10r} = \frac{2Gm}{5r} \]
- Therefore:
\[ v2 = \sqrt{\frac{2Gm}{5r}} \]

7. **Finding the Relative Velocity**:
- From the momentum conservation, we have:
\[ v1 = 4v2 = 4\sqrt{\frac{2Gm}{5r}} = \sqrt{\frac{16 \cdot 2Gm}{5r}} = \sqrt{\frac{32Gm}{5r}} \]
- The relative velocity of approach (the sum of the speeds towards each other) is:
\[ v_{relative} = v1 + v2 = 4v2 + v2 = 5v2 = 5\sqrt{\frac{2Gm}{5r}} = \sqrt{25 \cdot \frac{2Gm}{5r}} = \sqrt{\frac{50Gm}{r}} \]

### Final Results:
- The velocities of the masses at separation `r` are:
- \( v2 = \sqrt{\frac{2Gm}{5r}} \)
- \( v1 = 4\sqrt{\frac{2Gm}{5r}} \)
- The relative velocity of approach is:
- \( v_{relative} = \sqrt{\frac{50Gm}{r}} \)