Question

Find the gravitational field strength at the centre of arc of linear mass density $\lambda$ subtending an angle $2\alpha$ at the centre

• A. $E=\frac{G\lambda }{R}\sin \alpha$
  
• B. $E=\frac{2G\lambda }{R}\sin \alpha$
  
• C. $E=\frac{2G\lambda }{R}\cos \alpha$
  
• D. $E=\frac{2G\lambda }{R}\sin 2\alpha$

Answer 1

The correct option is B $E=\frac{2G\lambda }{R}\sin \alpha$


Given,
Linear mass density$=\lambda$,

Consider a small element at an angle $\theta$ from the central line as shown in figure.

Mass of the element, $dm=\lambda Rd\theta$

The gravitational field intensity $dE$ due to this element is given by ,

$dE=\frac{Gdm}{R^2}$

$\Rightarrow dE=\frac{G\lambda Rd\theta }{R^2}$

$\Rightarrow dE=\frac{G\lambda }{R}d\theta$

Component of $dE$ in the vertical direction $=dE\cos \theta$

Component of $dE$ in the horizontal direction $=dE\sin \theta$

Similar mass element is considered on the other side of central line. The horizontal components of gravitational field intensity will get cancelled due the symmetry .



Now, the total gravitational field strength due to the arc,

$E={\int }_{-\alpha }^{\alpha }dE\cos \theta$

Substituting the value of $dE$,

$\Rightarrow E=\frac{G\lambda }{R}{\int }_{-\alpha }^{\alpha }\cos \theta$ $d\theta$

$\Rightarrow E=\frac{G\lambda }{R}{\left[\sin \alpha \right]}_{-\alpha }^{\alpha }$

$\Rightarrow E=\frac{2G\lambda }{R}\sin \alpha$

Hence, option (b) is the correct answer.

why this question : To make students practice field calculation due to continuous mass distribution.