A small objected of mas m moves in a circular orbit under an attractive central force `kr^(3) ( i.e., vec( F) = - kr^(3) hat( r ))`. The radius of the orbit is `a_(0)`. Take the potential energy to be zero at the origin i.e., r = 0. The total mechanical energy of the object is

Options

A. `ka_(0)^(4)`
B. `(3)/(4) ka_(0)^(4)`
C. `(1)/(2) ka_(0)^(4)`
D. `(1)/(4) ka_(0)^(4)`

Answer

To find the total mechanical energy of a small object of mass \( m \) moving in a circular orbit under the influence of a central force \( \vec{F} = -k r^3 \hat{r} \), we will calculate both the kinetic energy and potential energy of the object.

### Step-by-Step Solution:

1. **Identify the Central Force**:
The central force acting on the object is given by:
\[ \vec{F} = -k r^3 \hat{r} \]
where \( k \) is a constant, \( r \) is the distance from the center, and \( \hat{r} \) is the unit vector in the radial direction.

2. **Equate Forces for Circular Motion**:
For an object moving in a circular orbit of radius \( a_0 \), the centripetal force required to keep the object in circular motion is provided by the central force:
\[ F_{\text{centripetal}} = \frac{mv^2}{a_0} \]
Setting the magnitudes of the forces equal gives:
\[ k a_0^3 = \frac{mv^2}{a_0} \]

3. **Solve for Velocity**:
Rearranging the equation, we find:
\[ k a_0^4 = mv^2 \quad \Rightarrow \quad v^2 = \frac{k a_0^4}{m} \]

4. **Calculate Kinetic Energy**:
The kinetic energy \( K \) of the object is given by:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} m \left( \frac{k a_0^4}{m} \right) = \frac{k a_0^4}{2} \]

5. **Calculate Potential Energy**:
The potential energy \( U \) associated with the force can be found by integrating the force from \( 0 \) to \( a_0 \):
\[ U(r) = -\int_0^{a_0} F \, dr = -\int_0^{a_0} (-k r^3) \, dr = \int_0^{a_0} k r^3 \, dr \]
Evaluating the integral:
\[ U(a_0) = k \left[ \frac{r^4}{4} \right]_0^{a_0} = k \frac{a_0^4}{4} \]

6. **Total Mechanical Energy**:
The total mechanical energy \( E \) is the sum of kinetic and potential energy:
\[ E = K + U = \frac{k a_0^4}{2} + \frac{k a_0^4}{4} \]
Combining these terms:
\[ E = \frac{2k a_0^4}{4} + \frac{k a_0^4}{4} = \frac{3k a_0^4}{4} \]

### Final Answer:
The total mechanical energy of the object is:
\[ E = \frac{3k a_0^4}{4} \]