Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles are their mutual gravitation find the initial acceleration of the two particles.

Explanation

The force of gravitastion exerted by one particle on another is
`F_1=(Gm_1m_2)/r^2`
`=96.67xx10^-11(N-m^2)/(kg^2)(1.0kg)xx(2.0kg))/((0.5m)^2)`
`=5.3xx10^-10N`
The acceleration of 1.0 kg particle is
`a_`=F/m_1=(5.3xx10^-10N)/(1.0kg)`
`=5.3xx10^-10N ms^2-`
This acceleration is towards the 2.0 kg particle. The acceleration of the 2.0 kg particle is
`a_2=F/m_2=(5.3xx10^-10N)/(2.0kg)`
`=2.65xx10^-10ms^-2`
This acceleration is towards the 1.0 kg particle.