Physics I: Mechanics › Oscillations & gravitation › Gravitational potential energy

U = −GMm/rSay it: “the gravitational potential energy equals minus G M m over r”

Gravitational potential energy

The energy ledger of gravity — negative because bound things have less than nothing, with zero parked at infinity.

Notation on this page: U is the potential energy (joules), r the centre-to-centre distance, with U = 0 at r → ∞ by convention.

Before this lesson: Newton's gravitation

Where it comes from

Potential energy is stored work: U(r) = −(work gravity does bringing the mass in from infinity). Gravity pulls inward, so as a mass falls from far away toward a planet, gravity does positive work on it — and the potential, defined as minus that work, goes negative:

U(r) = −GMm/rzero at infinity, increasingly negative as masses come together
Before reading on: two asteroids drift toward each other from far apart, speeding up as they fall together. Is their potential energy increasing or decreasing — and what does the sign of U tell you about whether they're bound?

Decreasing — more negative. Falling together converts potential into kinetic (they speed up), and the negative U marks them as bound: separated to infinity, they'd have zero total energy, so negative total energy means they can't escape without added energy.

Derivation

We build U by integrating the gravitational force from infinity inward. The integral of 1/r² is −1/r — that single antiderivative is the whole derivation.

U(r)
=
−∫∞r (GMm/r′²) dr′
Step 1 — definition. U = −(work by gravity from ∞ to r). The force GMm/r′² points inward, along the displacement — positive work, hence the overall minus.
=
−GMm · [−1/r′]∞r
Step 2 — integrate. ∫ r′−2 dr′ = −1/r′. GMm is constant — pull it out.
=
−GMm · (−1/r − 0)
Step 3 — evaluate. At r′ = r: −1/r. At r′ → ∞: −1/∞ = 0. Infinity contributes nothing — that is why zero lives there.
U
=
−GMm/r
Step 4 — simplify. −GMm × (−1/r) = −GMm/r. Negative, as promised: bound systems sit below zero. ∎

Check against mgh: near Earth's surface, ΔU = GMm(1/R − 1/(R+h)) ≈ GMmh/R² = mgh for h ≪ R. The familiar mgh is just this formula zoomed in — Example 3 proves it numerically.

How to use it

The procedure, every time:

  1. r from the centre. U = −GMm/r with centre-to-centre r — altitude h means r = R + h.
  2. Work with differences. ΔU = U2 − U1 = GMm(1/r1 − 1/r2). The minus signs sort themselves out.
  3. Near Earth, use mgh. For h ≪ R (metres, kilometres), ΔU ≈ mgh — the full formula is overkill.
  4. Read the sign of E = K + U. E < 0: bound (orbits, resting on the surface). E = 0: just barely escapes. E > 0: unbound, leaves with speed to spare.
  5. Conservation does the work. K1 + U1 = K2 + U2 — speeds from energy, no trajectories needed.

The difference form

ΔU = GMm (1/r1 − 1/r2)energy cost of moving from r1 to r2 — positive when climbing outSay it: “delta U equals G M m times one over r-one minus one over r-two”
Common mistake: dropping the minus and writing U = +GMm/r, then concluding satellites have positive energy. The minus is the physics: bound means below zero. Keep it.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: 1000 kg satellite at 400 km altitude

  1. r from the centre. r = 6.371×106 + 0.400×106 = 6.771×106 m.
  2. Substitute. U = −GMm/r = −6.674×10−11 × 5.972×1024 × 1000 / 6.771×106.
  3. Evaluate. = −3.986×1017 / 6.771×106 ≈ −5.89×1010 J.
  4. Read the sign. Negative: the satellite is bound — it would take +5.89×1010 J to lift it to infinity at rest ✓
Common mistake: r = 4.0×105 m (the altitude) → U ≈ −1012 J, off by 17×. r is measured from Earth's centre — always add R.
Your turn — 500 kg satellite at 800 km altitude. U?

Answer: ≈ −2.78×1010 J. r = 7.171×106 m; U = −6.674×10−11 × 5.972×1024 × 500/7.171×106 = −1.993×1017/7.171×106 ≈ −2.78×1010 J. Higher orbit → less negative → less bound.

Example 2 — the climb: energy to lift that 1000 kg satellite from the surface to 400 km

  1. Use differences. ΔU = GMm(1/R − 1/r) = 3.986×1014 × 1000 × (1/6.371×106 − 1/6.771×106).
  2. Evaluate. (1.5696 − 1.4769)×10−7 = 9.273×10−9; ΔU = 3.986×1017 × 9.273×10−9 ≈ 3.70×109 J.
  3. Sanity check. Positive (climbing costs energy) and far less than |U| ≈ 5.89×1010 J — most of the satellite's binding energy is the trip from infinity to the surface, not the last 400 km ✓
Common mistake: ΔU = mgh = 1000 × 9.8 × 400,000 = 3.92×109 J — close here (6% high) but wrong in principle at 400 km: g has already fallen to 8.69 m/s². mgh assumes constant g.
Your turn — lift the 500 kg satellite from the surface to 800 km. ΔU?

Answer: ≈ 3.49×109 J. ΔU = 6.674×10−11 × 5.972×1024 × 500 × (1/6.371×106 − 1/7.171×106) = 1.993×1017 × 1.751×10−8 ≈ 3.49×109 J.

Example 3 — the mgh audit: 70 kg person climbs 10 m

  1. Exact formula. ΔU = GMm(1/R − 1/(R+10)) with GMm = 3.986×1014 × 70 = 2.790×1016.
  2. Evaluate. 1/6.371×106 − 1/6.37101×106 ≈ 2.464×10−13; ΔU ≈ 2.790×1016 × 2.464×10−13 ≈ 6874 J.
  3. Compare mgh. 70 × 9.8 × 10 = 6860 J — the two agree to 0.2% ✓. mgh is −GMm/r, zoomed in.
Common mistake: concluding mgh and −GMm/r are rival formulas. They're the same law at different zoom: mgh is the tangent-line approximation, valid while h ≪ R.
Your turn — 500 kg climbs 100 m: exact-style estimate via mgh, and is it trustworthy?

Answer: ≈ 4.91×105 J, trustworthy. mgh = 500 × 9.8 × 100 = 4.90×105 J (using g = 9.82: 4.91×105 J). h/R ≈ 1.6×10−5 — utterly negligible, so mgh is exact for all practical purposes.

Before reading on: the 1000 kg satellite circles at 400 km with U ≈ −5.89×1010 J. Its kinetic energy is K = GMm/(2r) ≈ 2.94×1010 J. Is the total E positive or negative — and what does that say about the orbit?

Example 4 — judgment call: bound or free?

  1. Total it. E = K + U = 2.94×1010 − 5.89×1010 = −2.94×1010 J.
  2. Read the sign. E < 0 → bound. The satellite can't reach infinity: it lacks 2.94×1010 J.
  3. The pattern. For a circular orbit K = −U/2 always (virial theorem), so E = U/2 < 0 — every circular orbit is bound, with total energy exactly half its (negative) potential.
Common mistake: “it's moving fast, so E > 0 and it will fly away.” Speed alone says nothing — only K + U decides. Fast + deeply negative U is still bound (that's what an orbit is).
Your turn — a comet has K = 6.0×1010 J and U = −5.0×1010 J far from the Sun. Bound or free?

Answer: free (unbound). E = 6.0×1010 − 5.0×1010 = +1.0×1010 J > 0 — it will leave the solar system with energy to spare. Positive total energy can never be trapped.

Memorization tips

  • Chant it: “U equals minus G M m over r.” Say the minus — it's the physics.
  • Zero at infinity, negative inside. The sign convention in one line: far = 0, together = below zero.
  • E < 0 means bound. Negative total energy can't reach infinity. This one fact unlocks orbits, escape speed, and the virial theorem.
  • Differences, not absolutes: ΔU = GMm(1/r1 − 1/r2). Climbing out (r2 > r1) costs positive energy — check the sign every time.
  • mgh is the zoom-in. If h ≪ R, skip the big formula: ΔU ≈ mgh. Same law, less arithmetic.
  • r from the centre. The eternal rule of gravitation pages: altitude is not r. Add R first.

Final challenge

Five mixed questions — signs, differences, and the traps, all in one. Score 5/5 and gravitational potential energy is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Why is gravitational potential energy negative?

Because zero is defined at infinite separation, and gravity does positive work as masses fall together — so bound systems have less energy than the zero reference. Negative U means bound; zero or positive means free.

Why is zero potential at infinity?

It's a convention: at infinite separation gravity's influence vanishes, so that's the natural ‘no interaction’ reference. Any finite separation then has negative U, measuring how deeply bound the system is.

When can I use mgh instead of −GMm/r?

When h is tiny compared to Earth's radius (metres to kilometres): then ΔU = GMm(1/R − 1/(R+h)) is approximately mgh. For satellites and interplanetary work, use the full −GMm/r.

Is potential energy real or just bookkeeping?

The differences are real: ΔU converts exactly into kinetic energy and work. The absolute value depends on where you put zero, but every measurable prediction uses differences, which are convention-free.

Does U depend on the path taken?

No — gravity is conservative, so U depends only on the endpoints r1 and r2, never the path. That's what makes potential energy a useful concept at all.

More from the codex

Live

Physics I formula sheet

All the mechanics formulas — kinematics, forces, energy — printable and quiz-ready.

Open sheet → Live

Formula Sheet Builder

Mix and match any sections into your own printable sheet.

Open tool → Live

Prompt Simulator

Practice prompt engineering with deterministic scoring.

Open tool →