Question

A point mass $m$ is placed at the centre of the square $A B C D$ of side $a$ units as shown below.

The resultant gravitational force on mass $m$ due to masses $m_{1}$ and $m_{2}$ placed on the vertices of square is

• A. $\frac{G m_{1} m_{2}}{(a \sqrt{2})^{2}}$
• B. $\frac{2 G m\left(m_{1}+m_{2}\right)}{a^{2}}$
• C. zero
• D. $\frac{G m\left(m_{1}+m_{2}\right)}{(a \sqrt{2})^{2}}$

Answer 1

Answer: (C)

A point mass $m$ is located at the centre of the square $A B C D$. Let gravitational force $m$ and $m_{1}$ is
$\left| F _{1}\right|=\left[\frac{G m m_{1}}{(a / \sqrt{2})^{2}}\right]$
and between $m$ and $m_{2}$ is $\left| F _{2}\right|=\left[\frac{G m m_{2}}{(a / \sqrt{2})^{2}}\right]$ as shown below
$\because$ Along $B D$ there are two forces of same magnitude $F_{2}$ acting between $m$ and $m_{2}$ (on $B$ or $\left.D\right)$
Net gravitational force along $B D$,
$F _{\text {net }(B D)}= F _{2}- F _{2}=0$
Similarly, net gravitational force along $A C$,
$F _{\text {net }(A C)}= F _{1}- F _{1}=0$
$\therefore$ Net force on $m$ is $\left| F _{\text {net }}\right|=$ zero