Question

Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level?

Answer 1

Let the mass of 0.10 kg be at a distance x from the 2 kg mass and at a distance of (2 − x) from the 4 kg mass.
Force between 0.1 kg mass and 2 kg mass = force between 0.1 kg mass and 4 kg mass

\[\therefore \frac{2 \times 0 . 1}{x^2} = - \frac{4 \times 0 . 1}{\left( 2 - x \right)^2}\]

\[ \Rightarrow \frac{0 . 2}{x^2} = \frac{0 . 4}{\left( 2 - x \right)^2}\]

\[ \Rightarrow \frac{1}{x^2} = \frac{2}{\left( 2 - x \right)^2}\]

\[ \Rightarrow \left( 2 - x \right)^2 = 2 x^2 \]

\[ \Rightarrow 2 - x = \sqrt{2}x\]

\[ \Rightarrow x\left( \sqrt{2} + 1 \right) = 2\]

\[ \Rightarrow x = \frac{2}{2 . 414}\]

\[ = 0 . 83 \ m \text { from the 2 kg mass }\]

Now, gravitation potential energy of the system is given by

\[U = - G\left[ \frac{0 . 1 \times 2}{0 . 83} + \frac{0 . 1 \times 4}{1 . 17} + \frac{2 \times 4}{2} \right]\]

\[ \Rightarrow U = - 6 . 67 \times {10}^{11} \left[ \frac{0 . 1 \times 2}{0 . 83} + \frac{0 . 1 \times 4}{1 . 17} + \frac{2 \times 4}{2} \right]\]

\[ \Rightarrow U = - 3 . 06 \times {10}^{- 10} J\]