Three identical point masses, each of mass 1kg lie in the x-y plane at points (0, 0), (0, 0.2m) and (0.2m, 0). The net gravitational force on the mass at the origin is

To find the net gravitational force on the mass at the origin, we can use Newton's law of universal gravitation and the principle of superposition.

Let the three identical point masses be:
- $m_1=1\mathrm{kg}$ at the origin $A_1(0,0)$
- $m_2=1\mathrm{kg}$ at point $A_2(0,0.2m)$ (along the y-axis)
- $m_3=1\mathrm{kg}$ at point $A_3(0.2m,0)$ (along the x-axis)

The gravitational force is attractive. Thus, the force exerted on the mass at the origin $m_1$ by the mass $m_2$ (which lies along the positive y-axis) will be directed towards $m_2$, i.e., in the positive y-direction ($\hat{j}$).

The magnitude of this force $F_{12}$ is given by:
$F_{12}=\frac{G{m}_1{m}_2}{r^2}$

Substituting the given values, where $G\approx 6.67\times {10}^{-11}{\mathrm{N\; m}}^2{\mathrm{kg}}^{-2}$ and $r=0.2m$:
$F_{12}=\frac{6.67\times {10}^{-11}\times 1\times 1}{{0.2}^2}=\frac{6.67\times {10}^{-11}}{0.04}=1.67\times {10}^{-9}N$

Expressing this as a vector:
${\overrightarrow{F}}_{12}=1.67\times {10}^{-9}\hat{j}N$

Similarly, the force exerted on the mass at the origin $m_1$ by the mass $m_3$ (which lies along the positive x-axis) will be directed towards $m_3$, i.e., in the positive x-direction ($\hat{i}$).

Since the masses and the distance ($r=0.2m$) are the same, the magnitude of this force $F_{13}$ is:
$F_{13}=1.67\times {10}^{-9}N$

Expressing this as a vector:
${\overrightarrow{F}}_{13}=1.67\times {10}^{-9}\hat{i}N$

Using the principle of superposition, the net gravitational force ${\overrightarrow{F}}_{\text{net}}$ on the mass at the origin is the vector sum of these two forces:
${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{12}$

${\overrightarrow{F}}_{\text{net}}=1.67\times {10}^{-9}\hat{i}+1.67\times {10}^{-9}\hat{j}$

Factoring out the common terms:
${\overrightarrow{F}}_{\text{net}}=1.67\times {10}^{-9}(\hat{i}+\hat{j})N$

Therefore, the net gravitational force on the mass at the origin is 1.67 x 10⁻⁹( î + ĵ )N.