Derive an expression fro the gravitational field due to a unifrom rod of length L and M at a point on its perpendicular bisector at a distance d from the centre.

Explanation

To derive the expression for the gravitational field due to a uniform rod of length \( L \) and mass \( M \) at a point on its perpendicular bisector at a distance \( d \) from the center, we can follow these steps:

### Step 1: Define the system
Consider a uniform rod of length \( L \) and mass \( M \). We want to find the gravitational field at point \( P \), which is located on the perpendicular bisector of the rod at a distance \( d \) from the center of the rod.

### Step 2: Divide the rod into small elements
We can divide the rod into infinitesimally small elements of length \( dx \). The mass of each small element \( dm \) can be expressed as:
\[ dm = \frac{M}{L} dx \]
where \( \frac{M}{L} \) is the mass per unit length of the rod.

### Step 3: Determine the distance from the element to point P
Let \( x \) be the distance from the center of the rod to the small element \( dx \). The distance \( r \) from the element \( dm \) to point \( P \) is given by:
\[ r = \sqrt{d^2 + x^2} \]

### Step 4: Calculate the gravitational force due to the element
The gravitational force \( dF \) exerted by the small mass \( dm \) on a test mass \( m \) placed at point \( P \) is given by Newton's law of gravitation:
\[ dF = \frac{G M dm}{r^2} = \frac{G M \left(\frac{M}{L} dx\right)}{d^2 + x^2} \]

### Step 5: Resolve the gravitational force into components
The gravitational force \( dF \) can be resolved into two components: one along the vertical direction (towards the rod) and one along the horizontal direction. The vertical component \( dF_y \) is given by:
\[ dF_y = dF \cos \theta \]
where \( \cos \theta = \frac{d}{\sqrt{d^2 + x^2}} \). Thus,
\[ dF_y = \frac{G M \left(\frac{M}{L} dx\right)}{d^2 + x^2} \cdot \frac{d}{\sqrt{d^2 + x^2}} = \frac{G M^2 d}{L (d^2 + x^2)^{3/2}} dx \]

### Step 6: Integrate to find the total gravitational force
To find the total gravitational force \( F_y \) acting on the mass \( m \) at point \( P \), we need to integrate \( dF_y \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ F_y = \int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{G M^2 d}{L (d^2 + x^2)^{3/2}} dx \]

### Step 7: Evaluate the integral
The integral can be evaluated using the substitution \( x = d \tan \theta \):
\[ dx = d \sec^2 \theta d\theta \]
Changing the limits accordingly, we can solve the integral:
\[ F_y = \frac{G M^2 d}{L} \int_{-\tan^{-1}(\frac{L/2}{d})}^{\tan^{-1}(\frac{L/2}{d})} \sec^2 \theta \cdot \frac{1}{(d^2(1 + \tan^2 \theta))^{3/2}} d\theta \]

### Step 8: Final expression for gravitational field
After evaluating the integral, we find the total gravitational field \( E \) at point \( P \) by dividing the total gravitational force \( F_y \) by the mass \( m \):
\[ E = \frac{F_y}{m} = \frac{2 G M}{d \sqrt{L^2 + 4d^2}} \]

### Final Result
Thus, the expression for the gravitational field due to a uniform rod of length \( L \) and mass \( M \) at a point on its perpendicular bisector at a distance \( d \) from the center is:
\[ E = \frac{2 G M}{d \sqrt{L^2 + 4d^2}} \]