Question

Calculate the gravitational field at $P$ at a distance $a$ on perpendicular bisector if the mass of rod is $M$ and distributed uniformly over length $L$ .

• A. $\frac{2 G M}{a \left(L^{2} + a^{2}\right)}$
• B. $\frac{2 G M}{a \left(L^{2} + a^{2}\right)^{1 / 2}}$
• C. $\frac{G M}{2 a \left(L^{2} + 4 a^{2}\right)}$
• D. None of these

Answer 1

Answer: (B)

$dm=\frac{M}{L}dx$
$dE=\frac{G \left(M / L\right) d x}{\left(x^{2} + a^{2}\right)}dE_{n e t}=dEcos\theta =\frac{G M a d x}{L \left(x^{2} + a^{2}\right)^{3 / 2}}$
$E=∫dEc\text{os\theta }=\frac{G M a}{L}\displaystyle \int _{- 1 / 2}^{1 / 2}\frac{d x}{\left(x^{2} + a^{2}\right)^{3 / 2}}$
Put, $x=a\text{tan\theta }$ then $dx=a\text{sec}^{2}\text{\theta }d\text{\theta }$
$E=\frac{G M a}{L}\displaystyle \int \frac{a sec^{2} \theta d \text{\theta }}{a^{3} \text{sec}^{3} \text{\theta }}=\frac{G M}{L a}\displaystyle \int cos\theta d\theta $
$=\frac{G M}{L a}\text{sin\theta }=\begin{bmatrix} \frac{G M x}{L a \sqrt{x^{2} + a^{2}}} \\ \end{bmatrix}_{- 1 / 2}^{1 / 2}$
$\text{=}\frac{G M}{a L \left(L^{2} + 4 a^{2}\right)^{1 / 2}}=\frac{2 G M}{a \left(L^{2} + 4 a\right)^{1 / 2}}$