Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mi

NCERT Exemplar · Q29

Q. Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.

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Concept understanding — Gravitational Potential Energy

Gravitational Potential Energy

The Intuition: Energy Stored by Height

Imagine holding a heavy book above the floor. Your arm feels tired — that's because you're working against gravity. If you let go, the book falls and gains speed. Where did that motion come from? It came from the position of the book. By lifting it, you stored energy in the Earth–book system. That stored energy is gravitational potential energy.

The higher you lift, the more energy you store. The heavier the object, the more energy you store. This is the core idea: Gravitational potential energy is the energy an object has because of its position in a gravitational field.

The Precise Definition

Gravitational potential energy ( $U$ ) is the work done against gravity to bring an object from a reference point (usually the ground) to its current position.

For objects near the Earth's surface (where gravity is roughly constant), the formula is beautifully simple:

U = mgh

Where:

$U$ = gravitational potential energy (joules, J)
$m$ = mass of the object (kg)
$g$ = acceleration due to gravity (≈ 9.8 m/s² on Earth)
$h$ = height above the reference point (m)

Why "Potential"?

The word "potential" means "stored and ready to be used." The book at height $h$ has the potential to do work — it can smash a table, compress a spring, or generate sound when it hits the ground. That energy was put in when you lifted it.

The Reference Point is Arbitrary

Here's a crucial point: Only changes in gravitational potential energy matter. You can choose any height as $h = 0$ . In most problems, we take the ground as zero, but you could take the floor, the tabletop, or even the ceiling.

If you lift a 2 kg book from the floor ( $h = 0$ ) to a shelf ( $h = 2$ m), the change in potential energy is:

\Delta U = mg\Delta h = 2 \times 9.8 \times 2 = 39.2 \text{ J}

If you instead took the shelf as $h = 0$ , the book on the floor would have negative potential energy ( $-39.2$ J). The difference between the two positions is still 39.2 J — that's what matters.

Watch out

Never say "the object has $mgh$ energy" without specifying the reference level. The value is meaningless without a zero point.

The Bigger Picture: Variable Gravity

The formula $U = mgh$ works only when $g$ is constant — that is, near Earth's surface. For large distances (like a rocket leaving Earth), gravity weakens with distance. The general formula for gravitational potential energy between two masses $M$ and $m$ separated by distance $r$ is:

U = -\frac{GMm}{r}

The negative sign means that potential energy is zero at infinite separation and becomes more negative as objects come closer. This is the true definition, and $U = mgh$ is a special case of it (derived by approximating near the surface).

Key Takeaways for Exams

Gravitational potential energy is always relative — you must state or imply a reference level. …

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The gravitational potential energy at the midpoint is a local maximum along the line joining the centres, so the equilibrium is unstable — a small displacement causes the object to fall toward one sphere.

Why gravitational potential energy decides stability

For any conservative force (like gravity), equilibrium occurs where the net force is zero. But the type of equilibrium — stable or unstable — depends on the shape of the potential energy curve. If the potential energy is at a minimum, a small displacement produces a restoring force (stable). If it is at a maximum, the force pushes the object further away (unstable).

Here, the object is at the midpoint of two identical spheres. By symmetry, the gravitational pulls from the two spheres cancel exactly — so the net force is zero. That is the equilibrium condition. The question is: what happens if you nudge it slightly along the line joining the centres?

Step-by-step reasoning

Set up the geometry.

Let each sphere have mass $M$ and radius $R$ . The distance between their centres is $10R$ . The midpoint is at a distance $5R$ from each centre.
Write the gravitational potential energy of the object.

For a small test mass $m$ placed at a distance $x$ from the left sphere (measured along the line joining centres), the distance to the right sphere is $10R - x$ . The total gravitational potential energy is

U(x) = -\frac{GMm}{x} - \frac{GMm}{10R - x}.

The constant $-\frac{GMm}{10R}$ from the spheres interacting with each other is irrelevant here — we only care about the shape of $U(x)$ for the test mass.

Check equilibrium at the midpoint. At $x = 5R$ , the derivative is

U'(x) = \frac{GMm}{x^2} - \frac{GMm}{(10R - x)^2}.

Substituting $x = 5R$ gives $U'(5R) = 0$ — indeed, net force is zero.

Find the second derivative to test stability.

U''(x) = -\frac{2GMm}{x^3} - \frac{2GMm}{(10R - x)^3}.

At $x = 5R$ ,

U''(5R) = -\frac{2GMm}{(5R)^3} - \frac{2GMm}{(5R)^3} = -\frac{4GMm}{125 R^3}.

This is negative. …

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Q.Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.

Why gravitational potential energy decides stability

Step-by-step reasoning

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