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

A small section of rod is considered at 'x' distance mass of the element
$=(M/L\times dx=dm)$
$d{E}_1=\frac{G\left(dm\right)\times 1}{(d^2+x^2)}=d{E}_2$
Resultant $dE=2d{E}_1\sin \theta$
$=2\times \frac{G\left(dm\right)}{(d^2+x^2)}\times \frac{d}{\sqrt{(d^2+x^2)}}=\frac{2\times GM\times ddx}{L(d^2+x^2)\left\{\sqrt{d^2+x^2}\right\}}$
Total gravitational field,
$E={\int }_0^{\frac{L}{2}}\frac{2Gmddx}{L{(d^2+x^2)}^{\frac{3}{2}}}$
Intergrating the above equation it can be found that,
Intergrating the above equation it can be found that,
$E=\frac{2Gm}{d\sqrt{L^2+4d^2}}$