Q.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 centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

The key idea is that gravitational force is a vector sum, while gravitational potential is a scalar sum. Equilibrium depends on the net force, and its stability depends on the nature of the potential energy extremum.

1. Gravitational Force: Let the midpoint be $P$. The gravitational force exerted by each sphere on a test mass at $P$ has the same magnitude $F = \frac{GM}{(d/2)^2}$ but acts in opposite directions along the line joining the centers. Due to symmetry, these forces cancel out.

2. Gravitational Potential: The gravitational potential at $P$ is the scalar sum of potentials due to each sphere.

Substituting $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$, $M = 100 \text{ kg}$, and $d = 1.0 \text{ m}$:

3. Equilibrium: Since the net gravitational force at the midpoint is zero, an object placed there will be in equilibrium. …

When dealing with gravitational interactions, it is crucial to distinguish between gravitational force and gravitational potential. Gravitational force is a vector quantity, meaning it has both magnitude and direction. When multiple masses exert forces, these forces must be added vectorially. Gravitational potential, on the other hand, is a scalar quantity, meaning it only has magnitude. When multiple masses contribute to the potential at a point, their individual potentials are added algebraically. This distinction is fundamental to solving problems like this.

The concept of equilibrium relates directly to the net force. If the net force on an object is zero, it is in equilibrium. The stability of this equilibrium depends on how the potential energy (or gravitational potential, for a unit mass) changes if the object is slightly displaced. A local minimum in potential energy corresponds to stable equilibrium, while a local maximum corresponds to unstable equilibrium.

Let's break down the problem step by step.

Given:

• Mass of each sphere, $M = 100 \text{ kg}$
• Radius of each sphere, $R = 0.10 \text{ m}$
• Distance between centres of spheres, $D = 1.0 \text{ m}$
• Gravitational constant, $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$

The midpoint of the line joining the centres is at a distance $r = D/2 = 1.0 \text{ m} / 2 = 0.5 \text{ m}$ from the centre of each sphere. Since $r = 0.5 \text{ m}$ is greater than the radius $R = 0.10 \text{ m}$, we can treat the spheres as point masses located at their centres for calculations outside their volume.

1. Gravitational Force at the Midpoint

Gravitational force is a vector. Consider a small test mass $m$ placed at the midpoint.

• The first sphere exerts a gravitational force $F_1$ on the test mass, directed towards the centre of the first sphere.
• The second sphere exerts a gravitational force $F_2$ on the test mass, directed towards the centre of the second sphere.

The magnitude of the force exerted by each sphere is given by Newton's Law of Universal Gravitation:

$F = \frac{GMm}{r^2}$

For sphere 1: $F_1 = \frac{GMm}{(D/2)^2}$

For sphere 2: $F_2 = \frac{GMm}{(D/2)^2}$

Since both spheres have the same mass $M$ and the midpoint is equidistant from their centres ($D/2$), the magnitudes of the forces $F_1$ and $F_2$ are equal.

$F_1 = F_2 = \frac{(6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (100 \text{ kg}) \times m}{(0.5 \text{ m})^2}$

$F_1 = F_2 = \frac{6.67 \times 10^{-9} \times m}{0.25} = 2.668 \times 10^{-8} \times m \text{ N}$

At the midpoint, these two forces act in opposite directions along the line joining the centres. Therefore, the net gravitational force $F_{net}$ on the test mass $m$ is:

$F_{net} = F_1 - F_2 = 0$

This means the gravitational field strength (force per unit mass) at the midpoint is also zero.

2. Gravitational Potential at the Midpoint

Gravitational potential is a scalar. The total potential at a point due to multiple masses is the algebraic sum of the potentials due to each individual mass.

The gravitational potential $V$ due to a point mass $M$ at a distance $r$ is given by:

For sphere 1, the potential at the midpoint is $V_1 = -\frac{GM}{D/2}$.

For sphere 2, the potential at the midpoint is $V_2 = -\frac{GM}{D/2}$.

The total gravitational potential $V_{total}$ at the midpoint is the sum of these individual potentials:

$V_{total} = V_1 + V_2 = -\frac{GM}{D/2} - \frac{GM}{D/2} = -\frac{2GM}{D/2} = -\frac{4GM}{D}$

Now, substitute the given values:

$V_{total} = -\frac{4 \times (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (100 \text{ kg})}{1.0 \text{ m}}$

$V_{total} = -4 \times 6.67 \times 10^{-9} \text{ J/kg}$

$V_{total} = -26.68 \times 10^{-9} \text{ J/kg}$

$V_{total} = -2.668 \times 10^{-8} \text{ J/kg}$

3. Is an Object Placed at That Point in Equilibrium?

An object is in equilibrium if the net force acting on it is zero. From our calculation in Step 1, the net gravitational force at the midpoint is zero.

Therefore, an object placed at the midpoint of the line joining the centres of the spheres is in equilibrium.

4. Is the Equilibrium Stable or Unstable?

To determine the stability of equilibrium, we consider what happens if the object is slightly displaced from the equilibrium position.

• Stable equilibrium: If, upon slight displacement, the net force tends to restore the object to its original position. This corresponds to a local minimum in potential energy. …

Step 1: Force is a vector; at the midpoint both spheres are equidistant (each at $d/2$) and pull in opposite directions along the line joining the centres, so the vector sum is exactly zero: $F_{net}=0$.

Step 2: Potential is a scalar; both spheres contribute negative potentials that ADD rather than cancel: $V = -\dfrac{GM}{d/2} - \dfrac{GM}{d/2} = -\dfrac{4GM}{d}$.

Step 3: Substitute $G=6.67\times10^{-11}$, $M=100$ kg, $d=1.0$ m: $V = -\dfrac{4(6.67\times10^{-11})(100)}{1.0} \approx -2.67\times10^{-8}$ J/kg. …