Consider two point masses `m_(1) and m_(2)` which are separated by a distance of 10 metre as shown in the following figure. Calculate the force of attraction between them and draw the directions of forces on each of them. Take `m_(1) = 1 "kg " and m_(2) = 2 ` kg

Explanation

The force of attraction is `F = (Gm_(1)m_(2))/(r^(2))`
`= (6.67 xx10^(11)xx1xx2)/((10)^(2))`
`=(13.34xx10^-11)/(100)`
`=13.34 xx10^(-11)xx10^(-2)`
`F = 13.34 xx10^(-13) N rArr F = 1.334 xx10^(-12)N`
Direction ,
The force attraction `vec(F)_(12)` experienced by the mass `m_(2)` due to `m_(1)` is negative y - direction
`vec(F)_(12) = - 13.34 xx10^(-13) N = -1.334 xx10^(-12)N`
The force of attraction `vec(F)_(21)` experienced by the mass `m_(1)` due to `m_(2)` is positive y - direction
`vec(F)_(21) = 13.34 xx10^(-13) N = 1..334 xx10^(-12)N`
` vec(F)_(12) = - vec(F)_(21) `which confirms Newton.s third law .