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

Explanation:

• It is given that two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m.  
• We know that the force between 0.1kg mass and 2kg mass is equal to the force between 0.1 kg mass and 4 kg mass.  

• Therefore,$\frac{2+0.1}{x^{2}}=-\frac{4+0.1}{(2-x)^{2}}$. Thus, on solving we get, $x(\sqrt{2}+1)=2$.  

• This implies that $x=\frac{2}{2.414}=0.83 m$ from the 2kg mass. The gravitational potential energy of the system of three particles with usual reference level, can be written as,  $U=-G\left[\left(0.1 * \frac{2}{0.83}\right)+\left(0.1 * \frac{4}{1.17}\right)+\left(2 * \frac{4}{2}\right)\right]$                            

• Thus, on solving, we get, $U=-3.06^{*} 10^{-10} \mathrm{J}$.