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 an attractive central force \( \vec{F} = -k r^3 \hat{r} \), we can follow these steps:

### Step 1: Understand the relationship between force and potential energy
The force \( F \) is related to the potential energy \( U \) by the equation:
\[ F = -\frac{dU}{dr} \]
Given that \( F = -k r^3 \), we can express this relationship as:
\[ -k r^3 = -\frac{dU}{dr} \]

### Step 2: Integrate to find the potential energy
To find the potential energy \( U \), we can integrate the force with respect to \( r \):
\[ dU = k r^3 dr \]
Integrating both sides:
\[ U = \int k r^3 dr = \frac{k r^4}{4} + C \]
where \( C \) is the constant of integration.

### Step 3: Determine the constant of integration
We are given that the potential energy is zero at the origin, \( r = 0 \):
\[ U(0) = 0 \Rightarrow \frac{k(0)^4}{4} + C = 0 \Rightarrow C = 0 \]
Thus, the potential energy becomes:
\[ U = \frac{k r^4}{4} \]

### Step 4: Evaluate potential energy at the radius \( a_0 \)
At the radius of the orbit \( r = a_0 \):
\[ U(a_0) = \frac{k a_0^4}{4} \]

### Step 5: Find the kinetic energy
The centripetal force required for circular motion is provided by the attractive force:
\[ F = m \frac{v^2}{r} \]
Substituting \( F = k r^3 \):
\[ k r^3 = m \frac{v^2}{r} \]
Rearranging gives:
\[ m v^2 = k r^4 \]
Thus, the kinetic energy \( K \) is:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} \left( \frac{k r^4}{m} \right) = \frac{1}{2} k r^4 \]
At \( r = a_0 \):
\[ K(a_0) = \frac{1}{2} k a_0^4 \]

### Step 6: Calculate total mechanical energy
The total mechanical energy \( E \) is the sum of kinetic and potential energy:
\[ E = K + U = \frac{1}{2} k a_0^4 + \frac{k a_0^4}{4} \]
Combining these terms:
\[ E = \frac{1}{2} k a_0^4 + \frac{1}{4} k a_0^4 = \frac{2}{4} k a_0^4 + \frac{1}{4} k a_0^4 = \frac{3}{4} k a_0^4 \]

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

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