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Motion In Straight Line

Question
CBSEENPH11017703

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?

Solution
We have,
Mass of the sphere, M1=M2 = 100kg
Radius of the sphere, r = 0.10 m
Distance between the spheres, AB = R = 1m
           
Gravitational field due to mass M1, at P 
stack straight g subscript 1 with rightwards arrow on top space equals space fraction numerator GM subscript 1 over denominator left parenthesis PA right parenthesis squared end fraction space 
    equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 10 squared over denominator left parenthesis 0.5 right parenthesis squared end fraction space along space PA with rightwards arrow on top
    equals 2.67 cross times 10 to the power of negative 8 end exponent straight N divided by kg space along space PA with rightwards arrow on top 
Similarly, gravitational field due to mass M2 at P is, 
stack straight g subscript 2 with rightwards arrow on top equals fraction numerator GM subscript 2 over denominator left parenthesis PB right parenthesis squared end fraction space 
   equals 2.67 cross times 10 to the power of negative 8 end exponent space straight N divided by kg space comma space along space PB with rightwards arrow on top

The gravitational field due to two masses are equal and opposite, therefore net gravitational field at P is zero.
The gravitational potential at P is,
straight V equals negative fraction numerator G M subscript 1 over denominator P A end fraction minus fraction numerator G M subscript 2 over denominator P B end fraction equals negative straight G open square brackets fraction numerator straight M subscript 1 over denominator P A end fraction plus fraction numerator straight M subscript 2 over denominator P B end fraction close square brackets 
  equals negative 6.67 cross times 10 to the power of negative 11 end exponent open square brackets fraction numerator 100 over denominator 0.5 end fraction plus fraction numerator 100 over denominator 0.5 end fraction close square brackets
  equals negative 2.67 cross times 10 to the power of negative 8 end exponent straight J divided by kg 
The net gravitational field at P is zero, therefore the object placed at P will be in equilibrium.
The equilibrium is an unstable equilibrium.