Find the potential energy of the gravitational interation of a point mass of mass m and a thin uniform rod of mass `M` and length l, if they are located along a straight line such that the point mass is at a distance 'a' from one end. Also find the force of their interaction.

Explanation

To solve the problem of finding the gravitational potential energy of a point mass \( m \) and a thin uniform rod of mass \( M \) and length \( l \), we will follow these steps:

### Step 1: Set up the problem
We have a thin uniform rod of mass \( M \) and length \( l \). The point mass \( m \) is located at a distance \( a \) from one end of the rod. We can denote the position of the point mass and the rod in a coordinate system.

### Step 2: Define the small mass element
Consider a small element of the rod of length \( dx \) at a distance \( x \) from the point mass \( m \). The mass of this small element can be expressed as:
\[ dm = \frac{M}{l} \, dx \]
where \( \frac{M}{l} \) is the mass per unit length of the rod.

### Step 3: Write the expression for potential energy
The gravitational potential energy \( dU \) between the point mass \( m \) and the small mass element \( dm \) is given by:
\[ dU = -\frac{G m \, dm}{x} \]
Substituting for \( dm \):
\[ dU = -\frac{G m}{x} \cdot \frac{M}{l} \, dx \]

### Step 4: Determine the limits of integration
The limits of integration for \( x \) will be from \( a \) (the position of the point mass) to \( a + l \) (the far end of the rod).

### Step 5: Integrate to find total potential energy
Now we can integrate \( dU \) to find the total potential energy \( U \):
\[ U = -\frac{G m M}{l} \int_{a}^{a+l} \frac{1}{x} \, dx \]
The integral of \( \frac{1}{x} \) is \( \ln x \):
\[ U = -\frac{G m M}{l} \left[ \ln x \right]_{a}^{a+l} \]
Evaluating the limits:
\[ U = -\frac{G m M}{l} \left( \ln(a+l) - \ln(a) \right) \]
Using the property of logarithms:
\[ U = -\frac{G m M}{l} \ln\left(\frac{a+l}{a}\right) \]

### Step 6: Find the force of interaction
The force \( F \) of their interaction can be found by taking the derivative of the potential energy \( U \) with respect to \( a \):
\[ F = -\frac{dU}{da} \]
Calculating the derivative:
\[ F = -\frac{G m M}{l} \cdot \frac{1}{\frac{a+l}{a}} \cdot \frac{d}{da}\left(\frac{a+l}{a}\right) \]
Using the quotient rule:
\[ F = -\frac{G m M}{l} \cdot \left( \frac{1}{a} - \frac{1}{a+l} \right) \]
Simplifying:
\[ F = -\frac{G m M}{l} \cdot \left( \frac{(a+l) - a}{a(a+l)} \right) = -\frac{G m M}{l} \cdot \frac{l}{a(a+l)} \]
Thus, the force of interaction is:
\[ F = -\frac{G m M}{a(a+l)} \]

### Final Results
1. The potential energy \( U \) of the gravitational interaction is:
\[ U = -\frac{G m M}{l} \ln\left(\frac{a+l}{a}\right) \]
2. The force of interaction \( F \) is:
\[ F = -\frac{G m M}{a(a+l)} \]