An object of mass $m$ is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2R$ from the centre of the earth. What is the gain in its potential energy?

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that

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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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Gravitational potential energy depends on distance from Earth's center, not height above the surface. Moving from $R$ to $2R$ increases the potential energy by $\frac{1}{2}mgR$ .

When we lift an object near Earth's surface, the familiar $mgh$ formula works beautifully because gravity is nearly constant. But when the height becomes comparable to Earth's radius, gravity weakens noticeably as we climb. The potential energy is no longer linear in height; we must use the universal expression for gravitational potential energy.

The gravitational potential energy of a mass $m$ at distance $r$ from Earth's center is

U(r) = -\frac{GMm}{r}

where $G$ is the gravitational constant and $M$ is Earth's mass. The negative sign reflects that we define zero potential energy at infinite separation; any finite distance corresponds to a bound state with negative energy.

U(r) = -\frac{GMm}{r}

The gain in potential energy is simply the difference between final and initial states:

Initial potential energy at the surface (distance $r = R$ ):

U_i = -\frac{GMm}{R}

Final potential energy at height $R$ above the surface (distance $r = 2R$ ):

U_f = -\frac{GMm}{2R}

Calculate the gain:

\Delta U = U_f - U_i = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right)

$\Delta U = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{R}\left(1 - \frac{1}{2}\right) = \frac{GMm}{2R}$ …

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Q.An object of mass mmm is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance RRR to 2R2R2R from the centre of the earth. What is the gain in its potential energy?

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Q.An object of mass $m$ is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2R$ from the centre of the earth. What is the gain in its potential energy?

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