Two objectes of mass m and 4m are at rest at and infinite seperation. They move towards each other under mutual gravitational attraction. If G is the universal gravitational constant. Then at seperation r

Options

A. the total energy of the two objects is zero
B. their relative velocity of approach is `((10GM)/( r))^(1//2)` in magnitude
C. the total kinetic energy of the objects is `(4 Gm^(2))/( r)`
D. net angular momentum of both the particles is zero about any point

Answer

To solve the problem, we need to analyze the gravitational interaction between two masses, \( m \) and \( 4m \), that are initially at rest and then move towards each other under their mutual gravitational attraction. We will determine the total energy, total kinetic energy, and the relative velocity at a separation \( r \).

### Step-by-Step Solution:

**Step 1: Initial Energy Calculation**
- At infinite separation, both objects are at rest, so their initial kinetic energy \( KE_i = 0 \).
- The gravitational potential energy \( PE_i \) at infinite separation is also considered to be \( 0 \).
- Therefore, the total initial energy \( E_{total, initial} = KE_i + PE_i = 0 + 0 = 0 \).

**Step 2: Final Energy Calculation at Separation \( r \)**
- As the two masses move towards each other, they will lose potential energy and gain kinetic energy.
- The gravitational potential energy \( PE \) at a separation \( r \) is given by:
\[ PE = -\frac{G \cdot m \cdot 4m}{r} = -\frac{4Gm^2}{r} \]
- The total energy \( E_{total, final} \) at separation \( r \) will be the sum of the kinetic energy and potential energy:
\[ E_{total, final} = KE + PE \]

**Step 3: Conservation of Energy**
- The total energy must remain constant, so:
\[ E_{total, final} = E_{total, initial} = 0 \]
- Thus, we have:
\[ KE + \left(-\frac{4Gm^2}{r}\right) = 0 \implies KE = \frac{4Gm^2}{r} \]

**Step 4: Total Kinetic Energy Calculation**
- The total kinetic energy of the system when the objects are at separation \( r \) is:
\[ KE = \frac{4Gm^2}{r} \]

**Step 5: Relative Velocity Calculation**
- We can find the relative velocity \( V_r \) using the concept of reduced mass \( \mu \):
\[ \mu = \frac{m \cdot 4m}{m + 4m} = \frac{4m^2}{5m} = \frac{4m}{5} \]
- The kinetic energy in terms of relative velocity is given by:
\[ KE = \frac{1}{2} \mu V_r^2 \]
- Setting the two expressions for kinetic energy equal:
\[ \frac{1}{2} \cdot \frac{4m}{5} V_r^2 = \frac{4Gm^2}{r} \]
- Solving for \( V_r^2 \):
\[ V_r^2 = \frac{2 \cdot \frac{4Gm^2}{r}}{\frac{4m}{5}} = \frac{10Gm}{r} \]
- Thus, the relative velocity \( V_r \) is:
\[ V_r = \sqrt{\frac{10Gm}{r}} \]

### Final Results:
- Total energy at separation \( r \): \( E_{total} = 0 \)
- Total kinetic energy at separation \( r \): \( KE = \frac{4Gm^2}{r} \)
- Relative velocity at separation \( r \): \( V_r = \sqrt{\frac{10Gm}{r}} \)