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

Hint- The Newton's law of gravity is the gravitational formula. It states or defines the magnitude of force between two objects.

Formula used: To solve this type of questions we have the following formula.
*$V = - \dfrac{{Gm}}{r}$ ; The formula mentioned here is the formula of the gravitational potential. Where, G is universal gravitational constant, m is mass of spheres and r is the distance where gravitational potential is to be calculated.
*$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$ ; this is the formula for gravitational force.

Complete step by step answer:
Due to symmetry in this problem, gravitational force at the midpoint of two equal spheres will be equal and opposite. Therefore, resultant gravitational force will be zero.
Net gravitational force=0
Now we have the following information.
$m = 100kg,r = \dfrac{{10m}}{2} = 0.5m,G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
Now, we can write the gravitational potential at the midpoint using formula$V = - \dfrac{{Gm}}{r}$
${V_{mid - po\operatorname{int} }} = 2 \times \dfrac{{ - Gm}}{r}$
Now let us substitute the value in the above equation.
${V_{mid - po\operatorname{int} }} = 2 \times \dfrac{{ - \left( {6.67 \times {{10}^{ - 11}}} \right)\left( {100} \right)}}{{0.5}}$
Let us further simplify it.
${V_{mid - po\operatorname{int} }} = - \dfrac{{2 \times 667 \times {{10}^{ - 11}}}}{{0.5}}$
On further solving we get the following.
${V_{mid - po\operatorname{int} }} = - 2668 \times {10^{ - 11}}Jk{g^{ - 1}} = - 2.668 \times {10^{ - 8}}Jk{g^{ - 1}}$
Hence, required gravitational potential at the mid-point is $ - 2.668 \times {10^{ - 8}}Jk{g^{ - 1}}$.
At the midpoint net gravitational force is zero therefore if we place an object here it will be in equilibrium. As there is no restoring force, if we displace the object placed at the mid-point will not come back to its position again, hence, it is an unstable equilibrium point.

Note:
*Every conservative force has a potential associated with it, so gravitational potential is the potential associated with gravitational force.
*Gravitational force is the attractive force acting on all matters.
*On the earth we feel that acceleration due to gravity g=9.8 $m{s^{ - 2}}$ .
*The gravitational force is the weakest of all forces of the earth.
*The elementary particles of the gravity force are gravitons.