Gravitational potential energy — U = −GMm - r (not mgh): Worked examples

Two quantities we will use in every single problem — define them once, out loud:

$U (r) = - \frac{GM m}{r}$ — the stored energy of the two-mass system when they are a distance $r$ apart. Always $\leq 0$ . Bottom of the pit is very negative; far away ( $r \to \infty$ ) it climbs to $0$ .
$K = \frac{1}{2} m v^{2}$ — kinetic energy, the energy of motion. Always $\geq 0$ .
$E = K + U$ — the total mechanical energy. Because gravity is a conservative force, $E$ does not change as the object coasts (no engines, no air). That single fact — " $E$ before = $E$ after" — solves almost everything below.

See the parent: (parent topic).

The scenario matrix

Every gravity-energy question is one of these cells. Columns tell you what is fixed and what you solve for.

#	Cell class	What's special	Solved in
A	Sign of $E$ negative — bound object	$E < 0$ : object cannot escape	Ex 1
B	Sign of $E$ zero — exact escape	$E = 0$ : just barely reaches $\infty$	Ex 2
C	Sign of $E$ positive — unbound / excess speed	$E > 0$ : arrives at $\infty$ still moving	Ex 3
D	Small- $h$ limit — degenerate to $m g h$	check $- GM m / r \to m g h$ , $v \to 2 g h$	Ex 4
E	Large- $r$ / limiting value	$r \to \infty$ , what happens to $U, g$ ?	Ex 5
F	Circular orbit energy split	$K, U, E$ relationship, virial	Ex 6
G	Changing orbit (radius $r_{1} \to r_{2}$ )	energy difference, sign of work	Ex 7
H	Real-world word problem	messy numbers, satellite launch	Ex 8
I	Exam twist — two bodies / non-radial	conservation with a hidden trick	Ex 9
J	Degenerate input $r \to 0$ / $M = 0$	what the formula says at edges	Ex 10

Constants used throughout (memorise these three): $G = 6.67 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}, M_{\oplus} = 5.97 \times 1 0^{24} kg, R_{\oplus} = 6.37 \times 1 0^{6} m .$

Ex 1 — Cell A: a bound object (negative total energy)

Worked example

A $1 kg$ probe sits on Earth's surface and is thrown straight up at $v = 5000 m/s$ . Is it bound (will fall back) or free? Find its total energy $E$ .

Forecast: $5 km/s$ is less than half of escape speed. Guess the sign of $E$ before reading on.

Step 1 — Write both energies at the surface. Why this step? $E = K + U$ ; we only need the state at one moment because $E$ is conserved. $K = \frac{1}{2} (1) (5000)^{2} = 1.25 \times 1 0^{7} J .$ $U = - \frac{G M _{\oplus} m}{R _{\oplus}} = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1 )}{6.37 \times 1 0 ^{6}} = - 6.25 \times 1 0^{7} J .$

Step 2 — Add them. Why this step? The sign of $E$ is the entire answer: $E < 0$ means "in the pit, will return." $E = 1.25 \times 1 0^{7} - 6.25 \times 1 0^{7} = - 5.00 \times 1 0^{7} J .$

Step 3 — Interpret. $E < 0 \Rightarrow$ bound. The probe rises, stops, falls back. This is Cell A.

Verify: $∣ U ∣$ (the "depth of the pit," $6.25 \times 1 0^{7}$ ) exceeds $K$ ( $1.25 \times 1 0^{7}$ ), so the probe cannot climb out — consistent with $E < 0$ . Units: all joules. ✔

Ex 2 — Cell B: exact escape (total energy zero)

Ex 3 — Cell C: unbound object (positive total energy, leftover speed)

Worked example

A meteoroid is fired from Earth's surface at $v_{0} = 15 km/s$ (faster than escape). How fast is it moving very far away ( $r \to \infty$ )?

Forecast: Will $v_{\infty}$ be more or less than $15 km/s$ ? By roughly how much?

Step 1 — Conserve energy, surface to infinity. Why this step? At $\infty$ , $U \to 0$ , so all remaining energy is kinetic — that leftover is the "excess" over escape. $\frac{1}{2} m v_{0}^{2} - \frac{G M _{\oplus} m}{R _{\oplus}} = \frac{1}{2} m v_{\infty}^{2} + 0.$

Step 2 — Cancel $m$ , isolate $v_{\infty}$ . $v_{\infty}^{2} = v_{0}^{2} - \frac{2 G M _{\oplus}}{R _{\oplus}} = v_{0}^{2} - v_{esc}^{2} .$ Why this step? This is the neat identity $v_{\infty}^{2} = v_{0}^{2} - v_{esc}^{2}$ : escape speed is exactly the amount "eaten" by the gravity pit.

Step 3 — Plug in. $v_{0} = 1.5 \times 1 0^{4}$ , $v_{esc} = 1.118 \times 1 0^{4}$ : $v_{\infty} = (1.5 \times 1 0^{4})^{2} - (1.118 \times 1 0^{4})^{2} = 2.25 \times 1 0^{8} - 1.250 \times 1 0^{8} \approx 10.0 \times 1 0^{3} m/s .$

Verify: $v_{\infty} \approx 10.0 km/s < 15 km/s$ — it slowed down climbing out, as forecast. And $E = \frac{1}{2} m v_{\infty}^{2} > 0$ confirms Cell C (unbound). ✔

Ex 4 — Cell D: the degenerate small-height limit ( $m g h$ )

Worked example

A stone is dropped from a $45 m$ cliff. Find its landing speed two ways — exact $- GM m / r$ and the approximation $2 g h$ — and show they agree.

Notation first: let $R = R_{\oplus} = 6.37 \times 1 0^{6} m$ be Earth's radius (the distance from Earth's centre to the ground), and $h = 45 m$ the drop height. So the stone falls from $r = R + h$ down to $r = R$ .

Forecast: Will the exact answer be slightly bigger or smaller than $2 g h$ ? (Think: is $g$ slightly stronger lower down?)

Step 1 — Exact energy conservation. Why this step? We test whether the full formula collapses to the schoolbook one when $h ≪ R$ . $v = 2 G M_{\oplus} (\frac{1}{R} - \frac{1}{R + h}) .$ With $R = 6.37 \times 1 0^{6} m$ and $h = 45 m$ : $v = 2 (3.98 \times 1 0^{14}) (\frac{1}{6.37 \times 1 0 ^{6}} - \frac{1}{6.37 \times 1 0 ^{6} + 45}) \approx 29.7 m/s .$

Step 2 — Approximate with $g = 9.8$ . Why this step? $m g h$ comes from $R + h \approx R$ ; here $h / R \sim 7 \times 1 0^{- 6}$ is tiny. $v \approx 2 (9.8) (45) = 882 \approx 29.7 m/s .$

Step 3 — Compare. They match to the digits shown; the difference is one part in $1 0^{5}$ .

Verify: The exact value is microscopically larger (gravity is very slightly stronger at the ground than at $45 m$ ), matching the forecast. For everyday drops, use $g$. ✔

Ex 5 — Cell E: limiting values as $r \to \infty$ and as $r$ doubles

Worked example

(a) What are $U$ and the field $g (r) = GM / r^{2}$ as $r \to \infty$ ? (b) Compare $U$ at $r = 2 R_{\oplus}$ vs $r = R_{\oplus}$ for a $100 kg$ satellite.

Forecast: As $r \to \infty$ does $U$ approach $0$ from above or below?

Step 1 — Limits. Why this step? Checking end-behaviour is how we know the reference is sensible. $lim_{r \to \infty} (- \frac{GM m}{r}) = 0^{-}, lim_{r \to \infty} \frac{GM}{r ^{2}} = 0.$ Both vanish — infinitely far, no stored energy and no pull. $U$ approaches $0$ from below (always negative).

Step 2 — $U$ at $R_{\oplus}$ . Why this step? We need a concrete reference value at the surface so the "doubling" comparison in Step 3 has something to halve. $U (R) = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 100 )}{6.37 \times 1 0 ^{6}} = - 6.25 \times 1 0^{9} J .$

Step 3 — $U$ at $2 R_{\oplus}$ . Why this step? Since $U \propto 1/ r$ , doubling $r$ must halve the magnitude — this step makes the abstract " $1/ r$ " scaling numerically visible. $U (2 R) = - \frac{G M _{\oplus} m}{2 R _{\oplus}} = - 3.13 \times 1 0^{9} J .$

Verify: $U (2 R) = \frac{1}{2} U (R)$ exactly (formula is $\propto 1/ r$ ). Higher up ⇒ less negative ⇒ larger $U$ , matching "going up raises energy." ✔

Ex 6 — Cell F: energy split of a circular orbit

Worked example

A $500 kg$ satellite circles Earth at radius $r = 7.0 \times 1 0^{6} m$ . Find $K$ , $U$ , and $E$ , and confirm the virial relations.

Forecast: Guess the sign of $E$ and whether $∣ K ∣$ or $∣ U ∣$ is bigger.

Step 1 — Use "gravity = centripetal." Why this step? For a circular orbit the inward pull supplies exactly the centripetal force, which fixes $v$ (and hence $K$ ) purely from $r$ . $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow K = \frac{1}{2} m v^{2} = \frac{GM m}{2 r} .$ $K = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 500 )}{2 ( 7.0 \times 1 0 ^{6} )} = 1.42 \times 1 0^{10} J .$

Step 2 — Compute $U$ . Why this step? We need $U$ separately to add it to $K$ and see the total's sign. $U = - \frac{GM m}{r} = - 2.84 \times 1 0^{10} J .$

Step 3 — Add for $E$ . Why this step? The total's sign classifies the orbit as bound, and its value exposes the virial pattern. $E = K + U = 1.42 \times 1 0^{10} - 2.84 \times 1 0^{10} = - 1.42 \times 1 0^{10} J = - \frac{GM m}{2 r} .$

Verify: $E = - 1.42 \times 1 0^{10}$ equals $\frac{1}{2} U$ and equals $- K$ — the virial relations $E = \frac{1}{2} U = - K$ . $E < 0$ (bound). See Kepler's Laws & Orbital Energy. ✔

The figure below plots this split as curves in $r$ : the cyan potential $U$ sits below zero, the amber kinetic $K$ above it, and the dashed white total $E = - GM m /2 r$ threads exactly halfway down. Read off the orbit radius (marked): at that vertical line the dashed $E$ curve is at half the cyan $U$ value and the negative of the amber $K$ value — the picture of $E = \frac{1}{2} U = - K$ .

Figure — Gravitational potential energy — U = −GMm - r (not mgh)

Ex 7 — Cell G: raising an orbit (energy difference and its sign)

Worked example

Move the $500 kg$ satellite of Ex 6 from $r_{1} = 7.0 \times 1 0^{6} m$ up to $r_{2} = 1.4 \times 1 0^{7} m$ . How much energy must the thrusters add?

Forecast: Since higher orbits are higher-energy, will $Δ E$ be positive? And is it more or less than the full $∣ E_{1} ∣$ ?

Step 1 — Use $E = - \frac{GM m}{2 r}$ at each radius. Why this step? Only the difference of total energies matters; the work you must supply is $W = E_{2} - E_{1}$ . $E_{1} = - \frac{GM m}{2 r _{1}} = - 1.42 \times 1 0^{10} J, E_{2} = - \frac{GM m}{2 r _{2}} = - 7.11 \times 1 0^{9} J .$

Step 2 — Subtract. Why this step? The sign of $E_{2} - E_{1}$ tells us whether the engine adds or removes energy. $W = E_{2} - E_{1} = (- 7.11 \times 1 0^{9}) - (- 1.42 \times 1 0^{10}) = + 7.11 \times 1 0^{9} J .$

Step 3 — Interpret. Positive ⇒ you must add energy (fire the engine), consistent with the Work-Energy Theorem.

Verify: $W > 0$ as forecast; and since $r_{2} = 2 r_{1}$ , $E_{2} = \frac{1}{2} E_{1}$ , so $W = - \frac{1}{2} E_{1} = 7.11 \times 1 0^{9} J$ . Curiously the satellite's speed drops even though energy rose — the pit got shallower faster than $K$ fell. ✔

Ex 8 — Cell H: real-world word problem (satellite launch budget)

Worked example

A telecom company wants a $1200 kg$ satellite in circular orbit at altitude $800 km$ ( $r = R_{\oplus} + 8.0 \times 1 0^{5} = 7.17 \times 1 0^{6} m$ ). Starting from rest on Earth's surface, what is the minimum energy the launch must deliver? (Ignore Earth's rotation and air.)

Forecast: Will most of the energy go into lifting (raising $U$ ) or into speeding up (giving orbital $K$ )?

Step 1 — Energy on the launch pad. Why this step? Start state: at rest ( $K_{i} = 0$ ) on the surface. $E_{i} = 0 - \frac{G M _{\oplus} m}{R _{\oplus}} = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{6.37 \times 1 0 ^{6}} = - 7.50 \times 1 0^{10} J .$

Step 2 — Energy in the final orbit. Why this step? Circular orbit total energy is $- GM m /2 r$ . $E_{f} = - \frac{G M _{\oplus} m}{2 r} = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 7.17 \times 1 0 ^{6} )} = - 3.33 \times 1 0^{10} J .$

Step 3 — Required energy = difference. $W = E_{f} - E_{i} = (- 3.33 \times 1 0^{10}) - (- 7.50 \times 1 0^{10}) = + 4.17 \times 1 0^{10} J .$

Verify: Positive (you must supply it). Split: lifting cost $Δ U = U_{f} - U_{i} \approx + 8.3 \times 1 0^{9} J$ ; orbital $K_{f} \approx 3.33 \times 1 0^{10} J$ . So most energy becomes orbital speed, not height — matching the forecast for most physics students' surprise. Units: joules throughout. ✔

Ex 9 — Cell I: exam twist (two comparable masses, non-radial launch)

Worked example

Two identical asteroids, each mass $m = 2.0 \times 1 0^{12} kg$ , are released from rest a distance $d = 1.0 \times 1 0^{5} m$ apart in deep space. How fast is each moving when they are half that distance apart, $d /2$ ?

Forecast: The trap: which $M$ goes in $U = - GM m / r$ when both masses move? Guess the speed — will both share the kinetic energy equally?

Step 1 — Write the shared potential energy. Why this step? The pair has ONE potential energy $U = - G m_{1} m_{2} / r$ with $r =$ their separation — no "central body." $U_{i} = - \frac{G m ^{2}}{d}, U_{f} = - \frac{G m ^{2}}{d /2} = - \frac{2 G m ^{2}}{d} .$

Step 2 — Conserve energy (start at rest). Why this step? $E$ conserved: initial $K = 0$ , so released potential becomes kinetic of both bodies. $0 + U_{i} = both asteroids 2 \cdot \frac{1}{2} m v^{2} + U_{f} \Rightarrow m v^{2} = U_{i} - U_{f} = \frac{G m ^{2}}{d} .$ By symmetry each has the same speed $v$ ; that's why $K_{total} = 2 \cdot \frac{1}{2} m v^{2} = m v^{2}$ .

Step 3 — Isolate $v$ and compute. Why this step? Divide the last equation by $m$ to get $v^{2} = G m / d$ , then take the square root — this is the final numerical answer. $v = \frac{G m}{d} = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 2.0 \times 1 0 ^{12} )}{1.0 \times 1 0 ^{5}} = 1.334 \times 1 0^{- 3} \approx 3.65 \times 1 0^{- 2} m/s .$

Verify: Momentum is zero throughout (started at rest, symmetric), so the two velocities are equal and opposite — each $3.65 \times 1 0^{- 2} m/s$ . If you'd wrongly assumed one asteroid fixed, you'd have doubled the KE and got the wrong $v$ . See Newton's Law of Universal Gravitation. ✔

Ex 10 — Cell J: degenerate inputs ( $r \to 0$ , $M \to 0$ )

Worked example

What does $U = - GM m / r$ predict as (a) $r \to 0$ and (b) $M \to 0$ ? Are these physical?

Forecast: Guess whether $U \to 0$ or $U \to - \infty$ as the two masses touch.

Step 1 — Let $r \to 0$ . Why this step? Edge cases test where a formula stops being trusted — pushing $r$ to zero probes the "masses touch" limit. $lim_{r \to 0^{+}} (- \frac{GM m}{r}) = - \infty.$ The pit becomes infinitely deep. Interpretation: real masses have finite size $R$ , so $r$ never actually reaches $0$ ; the formula is for point masses / distances $\geq$ the surface. Below the surface a different expression applies, so the $- \infty$ is a modelling artefact, not a real infinite energy.

Step 2 — Let $M \to 0$ . Why this step? Sending the source mass to zero probes the "no gravity" limit and checks the formula degrades gracefully. $lim_{M \to 0} (- \frac{GM m}{r}) = 0.$ Interpretation: no central mass ⇒ no gravity ⇒ no stored energy. Perfectly sensible — the pit flattens to nothing.

Step 3 — Final interpretation. Cell J is the "sanity of the edges" check: the $M \to 0$ limit is physically clean ( $U \to 0$ ), while the $r \to 0$ limit ( $U \to - \infty$ ) is a warning flag that the point-mass model has left its valid range. Trust $U = - GM m / r$ only for $r \geq$ the body's radius and $M > 0$ .

Verify: At $M = 0$ the answer is exactly $0$ (checkable); at $r \to 0$ the magnitude grows without bound. Both are consistent with $U \propto M$ and $U \propto 1/ r$ . ✔

Recall Quick self-test across the matrix

Which cell is each question? "Satellite drops from 400 km to 300 km — energy change?" ::: Cell G (changing orbit). "Rock thrown at 20 km/s — speed at infinity?" ::: Cell C (unbound, $E > 0$ ). "Ball off a 10 m roof — landing speed?" ::: Cell D (small- $h$ , use $2 g h$ ).

Connections

Conservative Forces and Potential Energy — why one number $E$ stays fixed.
Work-Energy Theorem — thruster work $= Δ E$ in Ex 7–8.
Escape Velocity — Cells B and C.
Kepler's Laws & Orbital Energy — Cell F virial split.
Newton's Law of Universal Gravitation — the force behind $U$ , Ex 9.
Acceleration due to gravity g — the local limit $g = GM / R^{2}$ , Cell D.