Escape velocity — derivation: Question bank — Physics deep dive

Reminders you may lean on while answering:

Escape speed: $v_{e} = \frac{2 GM}{R} = 2 g R$ .
Gravitational PE with the $U (\infty) = 0$ convention: $U (r) = - \frac{GM m}{r}$ (negative ⇒ bound).
Energy bookkeeping: $E = \frac{1}{2} m v^{2} - \frac{GM m}{r}$ , conserved once only gravity acts.

True or false — justify

A tennis ball and a battleship, launched from the same planet, need the same escape speed.

True. Both the kinetic energy supplied and the gravitational energy owed scale with the object's mass

m

, so

m

cancels in

\frac{1}{2} m v_{e}^{2} = \frac{GM m}{R}

— the threshold is a speed, not an energy.

Escape velocity is a vector, so pointing the launch straight up matters.

False. It is really an energy condition, hence a speed. Ignoring air and rotation, only the magnitude

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

enters the energy equation, so any launch direction (except into the ground) needs the same speed.

An object launched exactly at escape speed comes to rest at infinity and then gets pulled back.

False. Gravity's force

\propto 1/ r^{2} \to 0

r \to \infty

, so at infinity there is nothing left to pull it back; reaching infinity with

v = 0

is the just-barely-forever case.

Total mechanical energy of a just-escaping object equals exactly zero.

True. At infinity

K = 0

and

U (\infty) = 0

, so

E_{f} = 0

; since energy is conserved,

E

is zero everywhere along the path, including at launch.

A bound object (never escapes) has negative total mechanical energy.

True. If

E < 0

, then

\frac{1}{2} m v^{2} = E + \frac{GM m}{r}

hits zero at a finite

r

(the turning point), so the object stops and returns — that is what "bound" means.

Escape velocity is faster than orbital velocity by a fixed ratio, whatever the planet.

True.

v_{e} = 2 GM / R

and orbital

v_{o} = GM / R

, so

v_{e} = 2 v_{o}

always; the

2

is geometry-free and planet-independent (see Orbital velocity & circular motion).

Doubling a planet's mass at fixed radius doubles its escape velocity.

False.

v_{e} \propto M

, so doubling

M

multiplies

v_{e}

2 \approx 1.41

, not

2

Two planets with the same density but different sizes have the same escape velocity.

False. With fixed density,

M \propto R^{3}

, so

v_{e} = 2 GM / R \propto R^{3} / R = R

— escape speed grows in direct proportion to radius, so the bigger planet is harder to escape.

At the surface, a body already moving at exactly

v_{e}

needs no further push to leave forever.

True. Once it possesses speed

v_{e}

with no drag, energy conservation guarantees arrival at infinity with

v \to 0

; no engine is needed after that instant.

Spot the error

"Since

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

, setting

\frac{1}{2} m v_{e}^{2} = U

gives the escape speed."

The sign is wrong: with

U (\infty) = 0

the bound PE is

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

. A positive

U

would force

\frac{1}{2} v_{e}^{2} = - GM / R

, an imaginary speed — a red flag that the minus was dropped.

"Escape needs enough kinetic energy to cancel gravity, so

\frac{1}{2} m v_{e}^{2} = F R = \frac{GM m}{R ^{2}} \cdot R

Confuses force with energy. You must pay the work integral

\int_{R}^{\infty} F d r = \frac{GM m}{R}

, not force

\times

one radius; the correct condition is

\frac{1}{2} m v_{e}^{2} = \frac{GM m}{R}

, giving the extra factor of

2

under the root.

"Air resistance doesn't matter, so the real rocket also just needs

11.2

km/s at the ground."

In vacuum the energy threshold is

11.2

km/s, but real rockets fight drag and gravity while climbing under continuous thrust, so they never need that single ground speed at once — they trade the problem for sustained acceleration.

"The Moon has no atmosphere because it has no gravity."

The Moon has gravity (

g \approx 1.62

m/s²); the real reason is that its escape speed (

\approx 2.4

km/s) is low enough that gas molecules routinely exceed it and leak away (see Why the Moon has no atmosphere).

"At

v_{e}

the object is in orbit, just a very high one."

An object at escape speed follows an unbound parabolic path and never returns; an orbit is a bound closed ellipse with

E < 0

. The line

E = 0

separates the two — orbit is below it, escape is on it.

"Because

v_{e} = 2 g R

, and

g

is the same everywhere on Earth's surface, escape speed is unaffected by launch altitude."

Wrong on two counts: launching from height

h

replaces

R

with

R + h

(and

g

drops), so escape speed decreases with altitude. The mistake treats

R

as fixed regardless of where you actually start.

"Setting

v_{e} = c

for a black hole is nonsense because nothing about light entered the derivation."

The Newtonian estimate

v_{e} = c

does give the correct Schwarzschild radius

R_{s} = 2 GM / c^{2}

numerically (a lucky coincidence), which is why it's a standard heuristic for Black holes — Schwarzschild radius — though the real justification is relativistic.

Edge cases

What is the escape velocity from a point infinitely far from all mass?

Effectively zero: as

R \to \infty

v_{e} = 2 GM / R \to 0

. Infinitely far out you are already at the top of the well, so no speed is needed to stay free.

If launched at exactly

v_{e}

, how long does the escape take?

Infinitely long. Reaching

r = \infty

requires unbounded time even though total energy is finite, because the object keeps slowing toward

v = 0

as it climbs.

What happens for a launch speed just

1 m/s

below

v_{e}

The object is bound (

E < 0

): it climbs to a huge but finite maximum radius, stops, and falls back. There is no gradual failure — crossing

E = 0

is a sharp threshold.

Does escape velocity make sense for an object below the surface (inside the planet)?

The surface formula no longer applies; inside, only the mass interior to your radius pulls you (shell theorem), so the enclosed

M

and the potential change, altering the energy needed to leave.

Does the planet's own rotation change the required launch speed?

Yes, slightly, in the idealized problem we usually ignore it: launching eastward at the equator you already carry the surface's spin speed, so the rocket must supply less additional speed to reach

v_{e}

— but the energy threshold to infinity itself is unchanged.

What if two bodies (like Earth and Moon) both pull on the object at once?

The single-body formula breaks down; you must sum both potentials

U = - \frac{G M _{1} m}{r _{1}} - \frac{G M _{2} m}{r _{2}}

and "escape" means clearing the combined well, which has saddle points, not one clean

v_{e}

For a black hole, what does setting

v_{e} = c

physically claim?

That at the Schwarzschild radius

R_{s} = 2 GM / c^{2}

even light-speed is insufficient to escape — nothing gets out — which is exactly the defining property of an event horizon in Black holes — Schwarzschild radius.

Recall One-line summary of the traps

Escape velocity is a speed (not a vector, not an energy), set by the condition $E = 0$ , paid via a work integral with negative bound PE, independent of $m$ , scaling as $M$ and $R \cdot g$ — and it is a sharp threshold, not a gradual one.

Question bank — Escape velocity — derivation

True or false — justify

Spot the error

Why questions

Edge cases

Connections