Question

Two heavy spheres each of mass 100 kg and radius 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 centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable? Question 8.21 Gravitation

Answer 1

Mass of each sphere, M = 100 kg
Separation between the spheres, r= I m

X is the mid point between the spheres. Gravitational force at point X will be zero.
This is because gravitational force exerted by each sphere will act in opposite directions.

Gravitational potential at point x -

-GM/(r/2) - GM/(r/2) = -4GM/r
plug in the values we get :-

4x 6.67 x 10^-11 x100/1

=> -2.67x10^-8 J/kg

Now !!
Any object placed at point X will be in equilibrium state, but the equilibrium is unstable.
This is because any change in the position of the object will change the effective force in
that direction.

Answer 2

here, 
mass of each sphere = 100kg
 radius of each sphere = 0.1m
distance b/w them(r) = 1m
   see the image
let P is the mid point of AB where m' mass appear.
 force act by m1 at point p(Fa) = Gm1m'/r²

force act by m2 at point P(Fb) = Gm2m'/r²

a/c to question,
      m1 = m2 = 100 kg 
hence, 
     Fa = Fb in magnitute but directions are opposite 
hence,  Fnet = 0
gravitational force at midpoint = 0



now gravitational potential at point P 
__________________________________________
potentail at point P due to sphere m1(V1) = -Gm1/(r/2)  [r = 1m
potential at point P due to sphere m2(V2) = -Gm2/(r/2)

resultant gravitational potential at P = V1 + V2 
                                                              = -2Gm1/r -2Gm2/r
                                                             after putting values of m1 , m2 , r and G we got 
   resultant potential = -2.7×10⁻ᵃ j/kg