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
A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm$\text{dE}_1=\frac{\text{G}(\text{dm})\times1}{(\text{d}^2+\text{x}^2)}=\text{dE}_2$
Resultant $\text{dE}=2\text{dE}_1\sin\theta$$=2\times\frac{\text{G(dm)}}{(\text{d}^2+\text{x}^2)}\times\frac{\text{d}}{\sqrt{(\text{d}^2+\text{x}^2})}=\frac{2\times\text{GM}\times\text{d dx}}{\text{L}(\text{d}^2+\text{x}^2)\big(\sqrt{\text{d}^2+\text{x}^2}\big)}$
Total gravitational field$\text{E}=\int\limits_{0}^{\frac{\text{L}}{2}}\frac{2\text{Gmd dx}}{\text{L}\big(\sqrt{\text{d}^2+\text{x}^2}\big)^{\frac{3}{2}}}$
Integrating the above equation it can be found that,$\text{E}=\frac{2\text{GM}}{\text{d}\sqrt{\text{L}^2+4\text{d}^2}}$
Resultant $\text{dE}=2\text{dE}_1\sin\theta$$=2\times\frac{\text{G(dm)}}{(\text{d}^2+\text{x}^2)}\times\frac{\text{d}}{\sqrt{(\text{d}^2+\text{x}^2})}=\frac{2\times\text{GM}\times\text{d dx}}{\text{L}(\text{d}^2+\text{x}^2)\big(\sqrt{\text{d}^2+\text{x}^2}\big)}$
Total gravitational field$\text{E}=\int\limits_{0}^{\frac{\text{L}}{2}}\frac{2\text{Gmd dx}}{\text{L}\big(\sqrt{\text{d}^2+\text{x}^2}\big)^{\frac{3}{2}}}$
Integrating the above equation it can be found that,$\text{E}=\frac{2\text{GM}}{\text{d}\sqrt{\text{L}^2+4\text{d}^2}}$
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