Gravitational Potential Energy

Why GPE is always negative for bound objects

Drop a ball from rest 1 m above the ground: it falls, gaining kinetic energy as it loses gravitational potential energy. Raise a satellite from 400 km to 35,900 km orbit: it gains gravitational PE (becomes less negative) and loses kinetic energy (moves more slowly). Both motions obey the same conservation of energy principle — but the formula for GPE is not $mgh$ once you leave Earth's surface. Voyager 1, launched in 1977, used Jupiter's gravity to convert gravitational PE directly into kinetic energy, gaining 35,700 km/h in a single flyby.

Gravitational PE $U = -GMm/r$ — the energy well becomes deeper (more negative) as distance $r$ decreases.

GPE at different orbital radii — bound objects have $U < 0$; escape requires total energy $\geq 0$.

The negative sign is essential. Gravity is attractive, so work must be done against gravity to increase $r$ (pull masses apart). This means:

• As $r$ decreases (object falls), $U$ becomes more negative — the system is more tightly bound
• As $r$ increases (object rises), $U$ becomes less negative — closer to zero
• At $r = \infty$, $U = 0$ by convention — the reference point for zero GPE

Why Zero at Infinity?

Setting $U = 0$ at $r = \infty$ is a physically meaningful convention. At infinite separation there is no gravitational interaction. This choice means:

• Moving an object from finite $r$ to infinity requires positive work (energy must be supplied)
• An object at finite $r$ has less energy than at infinity, hence $U < 0$
• The magnitude $|U|$ tells us the binding energy — the minimum energy needed to free the object

Gravitational PE: $U = -GMm/r$ (J), always negative for finite $r$; zero at infinity. Negative sign = gravity is attractive; the deeper the well (smaller $r$), the more negative $U$. Binding energy = $|U|$. Always use $r = R_\text{planet} + h$ (centre-to-centre distance).

Pause — copy the highlighted GPE definition and the binding energy rule into your book before moving on.

Calculating energy changes when moving between orbital radii

We just saw that gravitational PE is $U = -GMm/r$, always negative. That raises a question: how do we calculate the actual energy change when a satellite moves between two orbital radii? This card answers it → use $\Delta U = GMm(1/r_1 - 1/r_2)$, with sign indicating whether work is done by or against gravity.

When an object moves from one point to another in a gravitational field, the work done by the gravitational force equals the change in gravitational potential energy. The key formula connects $\Delta U$ to the two radii directly.

Physical Interpretation

• Moving outward ($r_2 > r_1$): $(1/r_1 - 1/r_2) > 0$, so $W > 0$. Work is done against gravity (by an external agent). Gravity does negative work.
• Moving inward ($r_2 < r_1$): $(1/r_1 - 1/r_2) < 0$, so $W < 0$. Gravity does positive work. The object speeds up as GPE is converted to kinetic energy.

Work done in a gravitational field: $W = \Delta U = GMm(1/r_1 - 1/r_2)$. Outward ($r_2 > r_1$): $\Delta U > 0$ (work done against gravity). Inward ($r_2 < r_1$): $\Delta U < 0$ (gravity does positive work; object speeds up). Always convert altitude to $r = R + h$.

Add the highlighted work formula and sign rules to your notes before the check below.

The near-surface approximation derived from the exact formula

We just saw that $\Delta U = GMm(1/r_1 - 1/r_2)$ is the exact formula for work in a gravitational field. That raises a question: how does the familiar $mgh$ formula relate to this, and when does it fail? This card answers it → $mgh$ is derived by assuming $h \ll R$, making $R + h \approx R$; it breaks down at satellite altitudes.

For small heights above Earth's surface, the exact gravitational PE formula reduces to the familiar $mgh$. Understanding how this approximation is derived — and when it breaks down — is essential for the HSC.

The $mgh$ formula is an approximation valid only when the height $h$ is much smaller than Earth's radius $R \approx 6.37 \times 10^6 \text{ m}$. For large altitudes (satellites, spacecraft), the exact formula must be used.

Near-surface approximation: $\Delta U \approx mgh$ (valid only when $h \ll R$). Derivation: $\Delta U = GMm \cdot h/(R(R+h)) \approx GMmh/R^2 = mgh$ when $R+h \approx R$. For satellite altitudes $h \sim R$, error exceeds 100% — use the exact formula $\Delta U = GMm(1/r_1 - 1/r_2)$.

Pause — write the highlighted approximation condition and derivation into your book before moving on.

Misconceptions — final check

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