Figure (1): Arrangement of Bodies

**Where, ****m1 = m2 = 3.0 \ kg,**** m3 = 4.0 \ kg**

The aim of this question is to grasp the concept of **Newton’s law of gravitation**.

According to **Newton’s law of gravitation**, if two masses (say m1 and m2) are placed at some distance (say d) from each other **attract each other** with an **equal and opposite force** given by the following formula:

\[ F = G \dfrac{ m_1 \ m_2 }{ d^2 } \]

where, $ G = 6.67 \times 10^{-11} $ is a universal constant called **gravitational constant**.

## Expert Answer

**Distance $ d_1 $ between $ m_1, \ m_2 $ and origin is given by:**

\[ d_1 = 0.6 \ m \]

**Distance $ d_2 $ between $ m_3 $ and origin is given by:**

\[ d_3 = \sqrt{ (0.6)^2 + (0.6)^2 } \ m \ = \ 0.85 \ m\]

**Force $ F_1 $ acting on 0.055 kg mass (say $ m $) due to mass $ m_1 $ is given by:**

\[ F_1 = G \dfrac{ m \ m_1 }{ d_1^2 } = 6.673 \times 10^{ -11 } \dfrac{ ( 0.055 )( 3 ) }{ (0.6)^2 } = 3 \times 10^{ -11 } \]

In vector form:

\[ F_1 = 3 \times 10^{ -11 } \hat{ j }\]

**Force $ F_2 $ acting on 0.055 kg mass (say $ m $) due to mass $ m_2 $ is given by:**

\[ F_2 = G \dfrac{ m \ m_2 }{ d_1^2 } = 6.673 \times 10^{ -11 } \dfrac{ ( 0.055 )( 3 ) }{ (0.6)^2 } = 3 \times 10^{ -11 } \]

In vector form:

\[ F_2 = 3 \times 10^{ -11 } \hat{ i }\]

**Force $ F_2 $ acting on 0.055 kg mass (say $ m $) due to mass $ m_3 $ is given by:**

\[ F_3 = G \dfrac{ m \ m_3 }{ d_2^2 } = 6.673 \times 10^{ -11 } \dfrac{ ( 0.055 )( 4 ) }{ (0.85)^2 } = 2.04 \times 10^{ -11 } \]

In vector form:

\[ F_3 = 3 \times 10^{ -11 } cos( 45^{ \circ} ) \hat{ i } + 3 \times 10^{ -11 } sin( 45^{ \circ} ) \hat { j }\]

\[ F_3 = 3 \times 10^{ -11 } ( 0.707 ) \hat{ i } + 3 \times 10^{ -11 } ( 0.707 ) \hat { j }\]

\[ F_3 = 2.12 \times 10^{ -11 } \hat{ i } + 2.12 \times 10^{ -11 } \hat { j }\]

**Total force $ F $ acting on 0.055 kg mass (say $ m $) is given by:**

\[ F = F_1 + F_2 + F_3 \]

\[ F = 3 \times 10^{ -11 } \hat{ j } + 3 \times 10^{ -11 } \hat{ i } + 2.12 \times 10^{ -11 } \hat{ i } + 2.12 \times 10^{ -11 } \hat { j } \]

\[ F = 5.12 \times 10^{ -11 } \hat{ i } + 5.12 \times 10^{ -11 } \hat{ j } \]

**Magnitude of $ F $ is given by:**

\[ |F| = \sqrt{ (5.12 \times 10^{ -11 })^2 + (5.12 \times 10^{ -11 })^2 } \]

\[ |F| = 7.24 \times 10^{ -11 } N\]

**Direction of $ F $ is given by:**

\[ F_{\theta} = tan^{-1}( \frac{ 5.12 }{ 5.12 } ) \]

\[ F_{\theta} = tan^{-1}( 1 ) \]

\[ F_{\theta} = 45^{\circ} \]

## Numerical Result

\[ |F| = 7.24 \times 10^{ -11 } N\]

\[ F_{\theta} = 45^{\circ} \]

## Example

**Find the magnitude of the force of gravity acting between 0.055 kg and 1.0 kg masses placed at a distance of 1 m.**

\[ F = G \dfrac{ m_1 \ m_2 }{ d^2 } = 6.673 \times 10^{ -11 } \dfrac{ ( 0.055 )( 1 ) }{ (1)^2 } = 0.37 \times 10^{-11} \ N \]

*All the vector diagrams are constructed by using GeoGebra.*