Let `r` be the distance of a particle from a fixed point to which it is attracted by an inverse square law force given by `F=k//r^2` (k=constant). Let m be the mass of the particle and L be its angular momentum with respect to the fixed point. Which of the following formulae is correct about the total energy of the system?

Options

A. (a) `1/2m((dr)/(dt))^2-k/r+(L)/(2mr^2)=Constant`
B. (b) `1/2m((dr)/(dt))^2-k/r=Const ant`
C. (c) `1/2m((dr)/(dt))^2+k/r+(L^2)/(2mr^2)=Const ant`
D. (d) None

Explanation

To solve the problem, we need to derive the total energy of a particle under the influence of an inverse square law force, given by \( F = \frac{k}{r^2} \), where \( k \) is a constant, \( r \) is the distance from a fixed point, \( m \) is the mass of the particle, and \( L \) is its angular momentum.

### Step-by-Step Solution:

1. **Identify the Total Energy**:
The total energy \( E \) of the system consists of kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]

2. **Calculate the Potential Energy**:
The potential energy \( U \) associated with a conservative force can be found by integrating the force:
\[ U(r) = -\int F \, dr = -\int \frac{k}{r^2} \, dr \]
The integral of \( \frac{k}{r^2} \) is:
\[ U(r) = -\left[-\frac{k}{r}\right] + C \]
Assuming the potential energy at infinity is zero (\( U(\infty) = 0 \)), we have:
\[ U(r) = -\frac{k}{r} \]

3. **Calculate the Kinetic Energy**:
The kinetic energy \( K \) can be expressed in terms of the radial and angular components of velocity. The total velocity \( v \) can be expressed as:
\[ v^2 = v_r^2 + v_\theta^2 \]
where \( v_r \) is the radial component and \( v_\theta \) is the tangential component. The tangential component can be related to angular momentum \( L \):
\[ v_\theta = \frac{L}{mr} \]
Thus, the kinetic energy becomes:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} m\left(v_r^2 + \left(\frac{L}{mr}\right)^2\right) \]
Simplifying this, we get:
\[ K = \frac{1}{2} mv_r^2 + \frac{L^2}{2mr^2} \]

4. **Combine Kinetic and Potential Energy**:
Now, substituting the expressions for \( K \) and \( U \) into the total energy equation:
\[ E = \frac{1}{2} mv_r^2 + \frac{L^2}{2mr^2} - \frac{k}{r} \]

5. **Final Expression for Total Energy**:
The total energy can be expressed as:
\[ E = \frac{L^2}{2mr^2} + \frac{1}{2} mv_r^2 - \frac{k}{r} \]

### Conclusion:
The total energy of the system is given by:
\[ E = \frac{L^2}{2mr^2} + \frac{1}{2} mv_r^2 - \frac{k}{r} \]