Question

The net gravitational field at point $P$ is

• A. $\frac{\sqrt{65}}{72}\frac{GM}{R^2}$
• B. $\frac{1}{9}\frac{GM}{R^2}$
• C. $\frac{\sqrt{3}}{2}\frac{GM}{R^2}$
• D. $\frac{\sqrt{65}}{36}\frac{GM}{R^2}$
  

Answer 1

The correct option is A $\frac{\sqrt{65}}{72}\frac{GM}{R^2}$

Gravitational field due to spherical shell,
$E_1=\frac{GM}{{(2R+R)}^2}=\frac{GM}{9R^2},\text{towards left}$

Gravitational field due to solid sphere $1$ is,
$E_2=\frac{G\left(\frac{M}{2}\right)}{{(R+R)}^2}=\frac{GM}{8R^2},\text{towards right}$

Gravitational field due to solid sphere $2$ is,
$E_3=\frac{GM}{{(R+2R)}^2}=\frac{GM}{9R^2},\text{downwards}$

Net gravitational field due to $E_1\text{and}{E}_2$ will be,
$E^{\prime }=E_1+E_2$

Considering the rightwards direction as $+ve$
$E^{\prime }=\frac{-GM}{9R^2}+\frac{GM}{8R^2}=\frac{GM}{72R^2},\text{towards right}$


Net gravitational field at $P$ is the vector sume of $E^{\prime }$ and $E_3$,

$E_{\text{net}}=\sqrt{E^{\prime }+E_3}$

$E_{\text{net}}=\sqrt{{\left(\frac{GM}{9R^2}\right)}^2+{\left(\frac{GM}{72R^2}\right)}^2}=\frac{GM}{9R^2}\sqrt{1+\frac{1}{64}}$

$\therefore E_{\text{net}}=\frac{\sqrt{65}}{72}\frac{GM}{R^2}$