Two heavy spheres each of mass 100 kg andradius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?

Two heavy spheres each of mass 100 kg andradius 0.10 m are placed 1.0

1.	Two heavy spheres each of mass 100 kg andradius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?
Answer» Solution :`implies` Mass of each shpere M = 100 kg O is the midpoind of AB, `:. AO = OB = 1/2 m = 0.5m` Mass of object placed at point O = m `:.`The gravitational FORCE of sphere A and B on a body of mass m. `F_A=(GMm)/((r/2)^2)=(Gxx100 xxm)/((0.5)^2)rarr`in OA direction and `F_B=(GMm)/((r/2)^2)=(Gxx100 xxm)/((0.5)^2)rarr`in OB direction . These both forces are same in value and opposite in direction hence, effective force on the body of mass placed at midpoint is zero. So, the body is in equilibrium. Now, if the body A or B is DISPLACED a little towards either mass body from its equilibrium position it will not return back to its initial position of equilibrium. Hence, body is in unstable equilibrium. Now, gravitational potential `V=-(GM)/2` `V_(A) =-(GM)/(0.5) and V_(B) = - (GM)/(0.5)` `:.` Gravitational potential at point O , V `=V_A +V_B` `:. V_A=-(GM)/(0.5)-(GM)/(0.2)=-(2GM)/(0.5)` `=-(2xx6.67xx10^(-11)xx100)/(0.5) = - 2.7 xx10^(-8) J kg^(-1)` Negative sign INDICATES that equilibrium is unstable.