Question

A point mass $m$ is placed at the centre of the square $ABCD$ 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. $\frac{2Gm\left(m_1+m_2\right)}{a^2}$
• C. zero
• D. $\frac{Gm\left(m_1+m_2\right)}{(a\sqrt{2}{)}^2}$

Answer 1

Answer: (C)

A point mass $m$ is located at the centre of the square $ABCD$. Let gravitational force $m$ and $m_1$ is
$\left|F_1\right|=\left[\frac{Gm{m}_1}{(a/\sqrt{2}{)}^2}\right]$
and between $m$ and $m_2$ is $\left|F_2\right|=\left[\frac{Gm{m}_2}{(a/\sqrt{2}{)}^2}\right]$ as shown below
$\because$ Along $BD$ there are two forces of same magnitude $F_2$ acting between $m$ and $m_2$ (on $B$ or $D)$
Net gravitational force along $BD$,
$F_{\text{net }(BD)}=F_2-F_2=0$
Similarly, net gravitational force along $AC$,
$F_{\text{net }(AC)}=F_1-F_1=0$
$\therefore$ Net force on $m$ is $\left|F_{\text{net }}\right|=$ zero