Escape velocity — derivation: Foundations — Physics deep dive

This is the ground-floor page for Escape velocity — derivation. If any letter in that note made you pause, it is defined here from nothing. Read top to bottom: each symbol is earned before the next uses it.

The cast of characters

Before formulas, meet every symbol as a plain-language character with a picture.

Figure 1 — The five symbols in one scene: the big mass $M$ (lavender ball) at the centre; the escaping object $m$ (coral dot); the distance $r$ (grey arrow from centre to $m$ ); the radius $R$ (mint arrow from centre to surface); and the speed $v$ (yellow arrow showing motion). Notice $r$ and $R$ both start at the centre, not the surface.

Building the two big ideas

Now the characters combine into the two concepts the derivation actually runs on.

Idea 1 — Gravity is a force that pulls, and it weakens with distance

Let us earn every piece:

$M$ and $m$ on top: the more mass on either side, the stronger the pull — makes sense, more stuff, more grip.
$r^{2}$ on the bottom: pull gets weaker as you move away, and specifically as the square of distance. Double the distance → one-quarter the pull.
$G$ : a fixed number of nature, the gravitational constant, $G \approx 6.67 \times 1 0^{- 11}$ . It is just the "conversion rate" that turns kilograms and metres into newtons. You never change it; it is the same everywhere in the universe.

Figure 2 — Gravity's strength $F$ (vertical axis) plotted against distance $r$ (horizontal axis). At distance $r$ the pull is full (coral dot); at $2 r$ it has dropped to one quarter (mint dot) — the signature of a $1/ r^{2}$ law. The curve flattens onto zero far to the right: at infinity, gravity vanishes.

Idea 2 — Energy: kinetic and potential

Energy is "the ability to make things happen," measured in joules. Two flavours matter here. We define both pieces first, then combine them into the single total $E$ .

Why $v^{2}$ and not just $v$ ? Because stopping a thing going twice as fast takes four times the work — this is a measured fact of nature, and the square captures it. We need $K$ because escape is fundamentally a race: does your motion-energy last all the way to infinity?

Intuition The valley picture — why

$U$ is negative Imagine a smooth bowl whose walls we only draw from the surface outward ( $r \geq R$ ). The lowest point we allow is the planet's surface $r = R$ ; the flat rim infinitely far away is "space." A marble at the surface is trapped — it has less energy than a marble on the rim. We choose the rim to be the zero of energy ( $U = 0$ at $r \to \infty$ ). Everything below the rim is therefore negative. That is why you will always see $U (r) = - \frac{GM m}{r}$ with a minus sign: being bound = being below zero.

A subtlety we deliberately fence off: the plain formula $U = - GM m / r$ would keep dropping for imaginary points inside the planet ( $r < R$ ), because the mass law changes there. We never launch from inside a planet, so we simply restrict everything to $r \geq R$ — the surface is the floor of our story.

Figure 3 — The gravity valley: potential energy $U (r)$ (vertical axis) against distance $r$ (horizontal axis), drawn only for $r \geq R$ . The dashed line at the top is the rim $U = 0$ at infinity. The lavender floor at $r = R$ (surface) is the deepest, most negative point we allow; far to the right $U$ climbs back toward zero — the object is nearly free.

The two tools that turn ideas into a formula

The parent note uses two mathematical moves. Here is what each one is and why it is the right tool.

Formula Carrying out the sum — the antiderivative step, shown

We must sum $\frac{GM m}{r ^{'2}} = GM m r^{' - 2}$ over the slivers. Summing a power $r^{' n}$ (for $n \neq = - 1$ ) is undone by the rule $\int r^{' n} d r^{'} = \frac{r ^{' n + 1}}{n + 1} .$ Why this rule? It is the reverse of the pattern "the slope of $r^{' k}$ is $k r^{' k - 1}$ ." Here $n = - 2$ , so $n + 1 = - 1$ , giving $\int r^{' - 2} d r^{'} = \frac{r ^{' - 1}}{- 1} = - \frac{1}{r ^{'}} .$ Check it backwards: the slope of $- \frac{1}{r ^{'}} = - r^{' - 1}$ is $(- 1) (- 1) r^{' - 2} = r^{' - 2}$ — exactly what we summed. Good. Now evaluate between the limits: $U (r) = - \int_{\infty}^{r} \frac{GM m}{r ^{'2}} d r^{'} = - GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = - GM m (- \frac{1}{r} + \frac{1}{\infty}) = - \frac{GM m}{r} .$ The term $1/\infty = 0$ is exactly our choice $U (\infty) = 0$ . So the minus sign records that we measure energy relative to the rim at infinity — this is the $U (r) = - GM m / r$ we drew as the valley.

How the foundations feed the topic

The map below reads top-down: the raw symbols (top row) build Newton's force law; summing that force with the integral produces the potential energy $U$ ; kinetic energy $K$ and potential energy $U$ pour into one total $E$ that conservation keeps constant; evaluating that total at the surface gives $E (R)$ , and equating it to the value at infinity finally yields escape velocity $v_{e}$ . In words: symbols → force → energy → conservation → answer.

Read it top-down: the raw symbols build the force law, the force law is summed into potential energy, kinetic and potential energy combine into one total energy, conservation freezes that total, evaluating it at the surface gives $E (R)$ , and equating that to the value at infinity gives escape velocity.

A first sanity check with numbers

Worked example Do the units even make sense? (dimensional analysis first)

Before plugging numbers, check that $2 GM / R$ actually comes out in metres per second. Write each symbol's units in the square bracket $[]$ :

$G$ has units $[m^{3} kg^{- 1} s^{- 2}]$ (this is fixed by Newton's law: force $kg m s^{- 2}$ equals $G kg^{2} / m^{2}$ , solve for $G$ ).
$M$ has units $[kg]$ .
$R$ has units $[m]$ .

Combine inside the root (the pure number $2$ has no units): $[\frac{GM}{R}] = \frac{( m ^{3} kg ^{- 1} s ^{- 2} ) ( kg )}{m} = \frac{m ^{3} s ^{- 2}}{m} = m^{2} s^{- 2} .$ Take the square root: $m^{2} s^{- 2} = m s^{- 1} = m/s$ . A speed — exactly what we wanted. The formula is unit-consistent, so now we may trust the numbers.

Now the numbers (Earth: $G = 6.67 \times 1 0^{- 11}$ , $M = 5.97 \times 1 0^{24} kg$ , $R = 6.37 \times 1 0^{6} m$ ): $v_{e} = \frac{2 ( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.37 \times 1 0 ^{6}} \approx 1.12 \times 1 0^{4} m/s \approx 11.2 km/s,$ the famous "11 km/s." Units check and value check — the foundations are wired up correctly. The derivation of this formula lives in the parent note; here we only confirm the tools produce the right ballpark.

Active recall

Centre or surface for

r

The centre of the big mass

M

Why is

U (r)

negative?

We set

U = 0

at infinity (the rim); anything trapped below the rim has less energy, so it is negative.

What does the prime on

r^{'}

inside the integral mean?

r^{'}

is a dummy "walking-foot" variable marching through every distance being summed; plain

r

is the fixed endpoint where the walk stops.

Which antiderivative rule gives

- 1/ r

\int r^{' - 2} d r^{'} = \frac{r ^{' - 1}}{- 1} = - \frac{1}{r ^{'}}

, checked because the slope of

- 1/ r^{'}

r^{' - 2}

Why is total energy

E

conserved during escape?

Gravity is conservative — its work depends only on start and end, not the path — so

K

and

U

trade perfectly with no leak (given no drag/engine).

Why does gravity reach exactly zero at infinity?

Because

F \propto 1/ r^{2}

, and

1/ r^{2} \to 0

r \to \infty

Equipment checklist

Each line tests a piece of understanding you could not have stated before reading this page — not a bare restatement.

Given

G, M, R

, argue in one line why

2 GM / R

must be a speed

Its units reduce to

m^{2} s^{- 2}

inside the root, whose square root is

m/s

Explain why

m

can be expected to vanish from the escape speed

Both

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

and

U = - GM m / r

carry one factor of

m

, so it divides out of any energy balance.

State the antiderivative rule you must invoke and check it

\int r^{' n} d r^{'} = \frac{r ^{' n + 1}}{n + 1}

; with

n = - 2

it gives

- 1/ r^{'}

, verified since the slope of

- 1/ r^{'}

r^{' - 2}

Say what makes the energy method valid at all

Gravity is conservative (work path-independent), so

E = K + U

is constant with no air drag or engine.

Name what is deferred to the parent note versus done here

Here: defining every symbol,

F

K

U

, the integral, conservation. There: writing

E_{i}

, setting

E_{f} = 0

, equating, solving for

v_{e}

Foundations — Escape velocity — derivation