Question

Derive an 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 centre.

Answer 1

Consider a small mass element of length dx at a distance x from the centre of the rod.
Mass of the mass element, dm = (M/L) × dx

Gravitational field due to this element at point P is given by $$dE=\frac{G\left(dm\right)\times 1}{(d^2+x^2)}$$

The components of the gravitational field due to the symmetrical mass element along the length of the rod cancel each other.

Now, resultant gravitational field = 2dE sin θ

$$=2\times \frac{G\left(dm\right)}{(d^2+x^2)}\times \frac{d}{\sqrt{(d^2+x^2)}}$$

$$=\frac{2\times GM\times ddx}{L(d^2+x^2)\left\{\left(\sqrt{d^2+x^2}\right)\right\}}$$

Total gravitational field due to the rod at point P is given by $$E={\int }_0^{L/2}\frac{2Gmddx}{L{(d^2+x^2)}^{3/2}}$$

On integrating the above equation, we get :

 

$$E=\frac{2Gm}{d\sqrt{L^2+4d^2}}$$