Gravitational potential energy — U = −GMm - r (not mgh): Question bank

Before we start, look at the actual shape every trap below refers to — the gravitational potential well:

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

Recall The single rule behind all these traps

$U = - \frac{GM m}{r}$ is negative, and getting larger means moving toward zero (rightward and upward on the well curve). So "more energy" = "less negative" = "bigger $r$ ". Which way is higher energy on the well? ::: Rightward and upward — toward the $U = 0$ ceiling. Moving away from the planet climbs out of the well, raising energy.

True or false — justify

Each item is a statement. Decide true/false, then give the reason — a bare T/F earns nothing.

U = - \frac{GM m}{r}

can never be positive for real masses at finite

r

True.

G, M, m, r

are all positive, so

- GM m / r < 0

always; it only reaches

0

in the limit

r \to \infty

, never crosses above.

If satellite A is at

r

and satellite B is at

2 r

, then B has more (less negative) gravitational PE.

True.

U_{B} = - GM m /2 r

is only half the magnitude of

U_{A} = - GM m / r

, and half of a negative number is less negative, so

U_{B} > U_{A}

. Higher orbit = higher potential energy. See the two-orbit figure below.

Doubling the height

h

above the surface always doubles the PE gained.

False. Only in the

m g h

(constant-

g

) approximation. The exact

Δ U = GM m (\frac{1}{R} - \frac{1}{R + h})

is nonlinear in

h

; the increase per metre shrinks as you rise.

The zero of gravitational PE is a law of nature fixed at infinity.

False. The zero is a choice. We pick

U (\infty) = 0

because it's the only reference that works for all

r

at once for an inverse-square force, but any constant could be added — only

Δ U

is physical.

For a bound circular orbit, the total mechanical energy is negative.

True.

E = - GM m /2 r < 0

. Being negative (below the

r = \infty

zero) is exactly what "bound" means — you'd need to add energy to reach freedom.

Escape speed depends on the mass of the escaping object.

False.

m

cancels in

\frac{1}{2} m v^{2} = GM m / R

, giving

v_{esc} = 2 GM / R

. A pebble and a probe need the same launch speed.

m g h

and

- GM m / r

are two rival, competing formulas.

False.

m g h

- GM m / r

expanded for

h ≪ R

with

g = GM / R^{2}

. They're the same physics at two zoom levels, not rivals.

Since

U < 0

near a planet, a nearby object has less energy than one far away.

True (for PE). Closer means more negative means less PE. But total energy also includes kinetic energy, so "less PE" ≠ "less total energy" without checking speed.

The force is strongest where

U

is most negative.

True.

U

is most negative at small

r

, and

F = GM m / r^{2}

is also largest at small

r

. Deep in the well, both the slope of

U

and the force magnitude are biggest.

Spot the error

Each line contains a flawed statement or step. Name what's wrong and correct it.

"Going up increases

r

, and

- GM m / r

gets more negative, so energy decreases going up."

Error: bigger

r

makes

- GM m / r

go toward zero, i.e. less negative. Energy increases going up. The mistake is confusing magnitude with signed value.

g = 6.67 \times 1 0^{- 11}

Error: that's

G

, the universal constant. The local field is

g = GM / R^{2} \approx 9.8 m/s^{2}

G

is universal;

g

is derived and planet-specific.

"At infinity the potential energy is at its minimum since the object is farthest away."

Error: at infinity

U = 0

, which is the maximum value

U

ever takes. The minimum is at the smallest

r

, deepest in the well.

"To find PE of a satellite we use

m g h

with

h

its orbital altitude."

Error:

m g h

needs constant

g

, valid only for

h ≪ R

. At orbital altitudes

g

has dropped a lot; you must use

- GM m / r

U = - GM m / r

, so if I drop

r

to zero the energy is zero."

Error: as

r \to 0

U \to - \infty

, not zero. The formula blows up (idealised point mass);

U = 0

happens at

r \to \infty

, the opposite end.

"Escape means reaching infinity with some leftover speed, so set

E > 0

Error: minimum escape means arriving at infinity with zero speed, so

E = 0

(

\frac{1}{2} m v^{2} + U = 0

E > 0

gives a faster-than-minimum escape, not the threshold.

"Work done by gravity as a mass falls inward is negative, since gravity is a negative force."

Error: falling inward, force and displacement point the same way (both inward), so gravity does positive work. The minus in

F = - \frac{GM m}{r ^{2}} \overset{r}{^}

only encodes direction relative to

\overset{r}{^}

, not the sign of the work. The vector figure below shows the two arrows lined up.

Why questions

Answer the reasoning, not just the result.

Why is the reference for gravitational PE chosen at infinity, not at the surface?

Infinity is the only distance where the inverse-square force vanishes for every planet at once, giving one universal, natural zero. A surface reference would be arbitrary and different for each body.

Why must the potential-energy formula have a minus sign out front?

Because

Δ U = - \int F \cdot d r

: PE change is minus the work gravity does. Gravity does positive work pulling things in, so PE drops going in — the minus builds that in.

Why does total orbital energy being negative guarantee the orbit is bound?

Negative means below the

r = \infty

zero. To reach freedom (where

E \geq 0

) you'd have to add energy, so the object can't escape on its own — it's trapped.

Why can we even define a single-number

U (r)

for gravity at all?

Because gravity is a conservative force: work depends only on start and end points, not the path. That path-independence is what lets one number encode position.

Why does

m g h

give a positive number while

- GM m / r

is always negative?

m g h

measures a change from a local zero you place at

h = 0

, so it can be positive.

- GM m / r

is an absolute value against the infinity zero, so it stays negative. Different references, not different physics.

Why is escape speed independent of launch angle (ignoring the planet's surface/air)?

Energy conservation only cares about speed (through

\frac{1}{2} m v^{2}

) and

r

, not direction. Any direction that eventually reaches infinity needs the same

\frac{1}{2} m v^{2} = GM m / R

Edge cases

The scenarios that break naive intuition. Cover every limit.

What is

U

exactly at the surface,

r = R

U (R) = - GM m / R

, a finite negative number — the deepest PE an object outside the planet can have. It's the "bottom of the well" for surface launches, the starting point for escape-speed calculations.

What happens to

U

r \to \infty

U \to 0

from below. This is the chosen zero and the ceiling of the well; an object here is exactly at the free/bound boundary.

What happens to

U

r \to 0

for an idealised point mass?

U \to - \infty

. The formula diverges because a true point mass concentrates infinite depth; real bodies have finite radius, so the

- GM m / r

branch only applies for

r \geq R

For a real planet of radius

R

, what does

U (r)

do inside the body (

r < R

It does not go to

- \infty

. Inside a uniform sphere the enclosed mass shrinks as

r^{3}

, giving

U (r) = - \frac{GM m}{2 R ^{3}} (3 R^{2} - r^{2})

— a smooth parabola that bottoms out at a finite value

- \frac{3 GM m}{2 R}

at the centre. The point-mass formula is only the outside branch; see the finite-bottom well figure.

If total energy

E

is exactly

0

, what is the trajectory?

The threshold case: the object just barely escapes, arriving at infinity with zero speed.

E < 0

is bound (falls back),

E > 0

is unbound with leftover speed — see Escape Velocity.

Two identical objects: one at rest at

r = R

, one orbiting at

r = R

. Which has more total energy?

The orbiting one — it has the same

U

but extra kinetic energy

\frac{1}{2} m v^{2} > 0

. Same PE, more KE, so higher total energy.

Can gravitational PE be exactly zero at a finite distance for two point masses?

No.

- GM m / r = 0

requires

r \to \infty

. At any finite separation with real positive masses,

U

is strictly negative.

m g h

ever exactly equal to

Δ U

, or only approximately?

Only approximately, and only in the limit

h \to 0

. The exact

Δ U = GM m h / [R (R + h)]

equals

m g h

precisely when

R + h = R

, which never truly holds for

h > 0

— it's a leading-order match.

Connections

Parent topic — the derivations these traps test.
Conservative Forces and Potential Energy — why a single-number $U$ exists.
Work-Energy Theorem — the sign-of-work traps.
Escape Velocity — the $E = 0$ threshold edge cases.
Kepler's Laws & Orbital Energy — the "negative = bound" reasoning.
Acceleration due to gravity g — the $g$ vs $G$ error.
Newton's Law of Universal Gravitation — the force behind the minus sign.