Escape velocity — derivation: Worked examples — Physics deep dive

Everything here rests on the parent: Escape velocity — derivation, and on Gravitational potential energy, Newton's law of universal gravitation and Conservation of mechanical energy.

The scenario matrix

Before solving, let us list every distinct kind of escape-velocity problem. Each later example is tagged with the cell it fills.

Cell	What makes it different	Example
A. Given $g, R$	You know surface gravity + radius, no $G, M$	Ex 1
B. Given $G, M, R$	You know the mass directly	Ex 2
C. Scaling / ratio	"double the mass", "half the radius" — no numbers plugged	Ex 3
D. Launch above the surface	Start at $r = R + h$ , not at $R$	Ex 4
E. Leftover speed (not minimum)	Launched faster than $v_{e}$ : speed at infinity?	Ex 5
F. Real-world word problem	Gas molecule vs. planet — does an atmosphere survive?	Ex 6
G. Limiting / extreme value	Push $R \to$ small until $v_{e} \to c$ (black hole edge)	Ex 7
H. Degenerate input	What breaks when $R \to 0$ or $M \to 0$ ?	Ex 8
I. Exam twist	Same $ρ$ (density), different size — how does $v_{e}$ scale?	Ex 9

Recall Two forms of the formula — which to reach for

If the problem gives you == $g$ and $R$ ==, use $v_{e} = 2 g R$ . If it gives you == $G$ , $M$ , $R$ == (or density), use $v_{e} = 2 GM / R$ . They are the same equation because $g = GM / R^{2}$ , i.e. $GM = g R^{2}$ .

Constants used throughout: $G = 6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}, c = 3.0 \times 1 0^{8} m/s$

The figure below is the map of the whole page: it shows how $v_{e}$ responds to the two levers, mass $M$ and radius $R$ , that every cell of the matrix pulls on in some combination.

Read it now: each curve is a fixed mass; sliding right (bigger $R$ ) lowers $v_{e}$ (Cells D, I), and raising the curve (bigger $M$ ) lifts it (Cells C, G). The steep left edge previews the degenerate $R \to 0$ blow-up of Cell H.

Cell A — given surface gravity and radius

Cell B — given the mass directly

Cell C — scaling with no numbers plugged

Worked example Ex 3 — "Twice the mass, half the radius" (Cell C)

A planet has $2 \times$ Earth's mass and $\frac{1}{2} \times$ Earth's radius. By what factor does $v_{e}$ change relative to Earth's $11.2$ km/s?

Forecast: More mass and smaller radius both make gravity stronger. Do you expect a factor near 1.4, near 2, or near 3?

Step 1. Write $v_{e} \propto M / R$ (the constants $2, G$ cancel in a ratio). Why this step? In a "by what factor" question, only the ratio of the new to old matters, so keep just the variable pieces.

Step 2. Substitute the factors $M \to 2 M$ , $R \to \frac{1}{2} R$ : $\frac{v _{e}^{'}}{v _{e}} = \frac{2 M}{\frac{1}{2} R} \cdot \frac{R}{M} = \frac{2}{1/2} = 4 = 2.$ Why this step? Dividing the new $M / R$ by the old one leaves only the change factors.

Step 3. So $v_{e}^{'} = 2 \times 11.2 = 22.4 km/s$ .

Verify: Sanity — doubling mass alone gives $2 \approx 1.41$ ; halving radius alone gives $2 \approx 1.41$ ; multiplied: $1.41 \times 1.41 = 2$ . ✓ Both effects push the same way, as expected.

Cell D — launching from above the surface

Worked example Ex 4 — Escape from a height

$h$ (Cell D) From what speed can you escape Earth if you start at altitude $h = R$ (i.e. at $r = 2 R$ , twice Earth's radius from the centre)? Take $v_{e, surf} = 11.2$ km/s.

Forecast: Farther out means shallower well. Bigger or smaller than 11.2 km/s?

Step 1. Write the energy balance using the same $U (r) = - GM m / r$ from the definition box, but start at $r = 2 R$ : $\frac{1}{2} m v^{2} - \frac{GM m}{2 R} = 0.$ Why this step? The derivation is identical to the parent; only the launch radius changed, so we just replace $R$ with $2 R$ in the PE term.

Step 2. Solve: $v = \frac{2 GM}{2 R} = \frac{GM}{R}$ . Why this step? The factor of 2 in "twice the radius" cancels the 2 in " $2 GM$ ".

Step 3. Relate to the surface value on this page (no outside fact needed). The surface formula is $v_{e, surf} = 2 GM / R$ , so squaring gives $\frac{GM}{R} = \frac{1}{2} v_{e, surf}^{2}$ . Therefore $v = \frac{GM}{R} = \frac{1}{2} v_{e, surf}^{2} = \frac{v _{e, surf}}{2} = \frac{11.2}{1.414} \approx 7.92 km/s .$

Verify: $7.92 \times 2 = 11.2$ ✓. And it is smaller than the surface value — correct, because from higher up you have already climbed part of the well. (Note: $GM / R$ is exactly the surface orbital speed — a nice cross-check.)

Cell E — launched faster than escape: leftover speed

Worked example Ex 5 — Speed left over at infinity (Cell E)

Fire a probe from Earth's surface at $v_{0} = 15 km/s$ (faster than $v_{e} = 11.2$ ). What speed $v_{\infty}$ does it still have infinitely far away? Ignore air and other bodies.

Forecast: It beats escape by a bit. Will $v_{\infty}$ be close to 15, or much smaller?

Step 1. Energy conservation, now with nonzero final KE (again using $U (r) = - GM m / r$ , which vanishes at infinity): $\frac{1}{2} m v_{0}^{2} - \frac{GM m}{R} = \frac{1}{2} m v_{\infty}^{2} - 0.$ Why this step? At infinity $U = 0$ , but the object still moves, so $K_{\infty} \neq = 0$ . We keep it.

Step 2. Cancel $m$ , and recognise $\frac{2 GM}{R} = v_{e}^{2}$ : $v_{\infty}^{2} = v_{0}^{2} - \frac{2 GM}{R} = v_{0}^{2} - v_{e}^{2} .$ Why this step? The tidy relation $v_{\infty}^{2} = v_{0}^{2} - v_{e}^{2}$ says: "leftover energy = launch energy minus the toll gravity charges."

Step 3. Plug numbers (in km/s, squared): $v_{\infty} = 1 5^{2} - 11. 2^{2} = 225 - 125.44 = 99.56 \approx 9.98 km/s .$

Verify: $9.9 8^{2} + 11. 2^{2} = 99.6 + 125.4 = 225 = 1 5^{2}$ ✓ (a Pythagorean-looking energy sum). Notice $v_{\infty} = 9.98 < 15$ : gravity ate the difference. If we had launched at exactly $v_{e}$ , the formula gives $v_{\infty} = 0$ — the minimum case, consistent with the parent. ✓

The picture below plots this leftover-speed curve $v_{\infty} = v_{0}^{2} - v_{e}^{2}$ and marks the worked point at $v_{0} = 15$ km/s; below $v_{e}$ the object simply falls back (no infinity to reach).

Cell F — real-world word problem

Worked example Ex 6 — Does the Moon keep an atmosphere? (Cell F)

In kinetic theory, molecules do not all share one speed; the standard single number is the root-mean-square (RMS) speed, $v_{rms} = \overline{v^{2}}$ — the speed whose square equals the average of the squared speeds. For air near $300 K$ , $v_{rms} \approx 0.5 km/s$ . The Moon's escape velocity is $2.4 km/s$ (parent note). A rough rule: a body loses its gas over time if $v_{rms}$ is more than about $\frac{1}{6}$ of $v_{e}$ (because the fast tail of the distribution then routinely exceeds $v_{e}$ ). Will the Moon hold an atmosphere?

Forecast: Guess "yes" or "no" before computing the ratio.

Step 1. Compute the retention ratio $v_{e} / v_{rms} = 2.4/0.5 = 4.8$ . Why this step? Atmospheric escape depends on how many times faster $v_{e}$ is than a typical (RMS) molecule — that ratio controls the slow leak.

Step 2. Compare to the rule. The rule wants $v_{e} / v_{rms} ≳ 6$ to safely retain gas. Here $4.8 < 6$ . Why this step? The threshold is the whole point of the word problem — a bare number means nothing until compared to the criterion.

Step 3. Conclude: $4.8 < 6$ , so the fast tail of molecules exceeds $2.4$ km/s often enough to escape → the Moon loses its atmosphere.

Verify: For Earth, $v_{e} / v_{rms} = 11.2/0.5 = 22.4 ≫ 6$ → Earth keeps air. ✓ This is exactly the Why the Moon has no atmosphere story, now made quantitative.

Cell G — limiting value: the black-hole edge

Worked example Ex 7 — Squeeze the Sun until

$v_{e} = c$ (Cell G) Take the Sun's mass $M = 1.99 \times 1 0^{30} kg$ . To what radius must you compress it so its escape velocity equals the speed of light $c$ ? (This Newtonian estimate happens to match the true Black holes — Schwarzschild radius.)

Forecast: A few metres, a few km, or a few thousand km?

Step 1. Set $v_{e} = c$ in $v_{e} = 2 GM / R$ and solve for $R$ : $c = \frac{2 GM}{R} \Rightarrow R = \frac{2 GM}{c ^{2}} .$ Why this step? "Escape needs the speed of light" is precisely the condition $v_{e} = c$ ; we invert the formula because $R$ is now the unknown.

Step 2. Numerator $2 GM = 2 (6.674 \times 1 0^{- 11}) (1.99 \times 1 0^{30}) = 2.656 \times 1 0^{20}$ . Denominator $c^{2} = 9.0 \times 1 0^{16}$ . Why this step? Splitting top and bottom keeps the exponents honest.

Step 3. Divide: $R = \frac{2.656 \times 1 0 ^{20}}{9.0 \times 1 0 ^{16}} \approx 2.95 \times 1 0^{3} m \approx 2.95 km .$

Verify: The known Schwarzschild radius of the Sun is $\approx 2.95$ km. ✓ The Newtonian escape argument, pushed to its limit, lands exactly on the relativistic answer — a lovely coincidence of form.

Cell H — degenerate inputs

Worked example Ex 8 — What breaks as

$R \to 0$ or $M \to 0$ ? (Cell H) Examine the two edge cases of $v_{e} = 2 GM / R$ : (a) $R \to 0$ at fixed $M$ , and (b) $M \to 0$ at fixed $R$ .

Forecast: In which case does $v_{e}$ blow up, and in which does it vanish?

Step 1 — case (a), $R \to 0$ . $v_{e} = 2 GM / R \to 2 GM / 0^{+} \to \infty$ . Why this step? Dividing by a shrinking positive number sends the ratio to infinity; the root of infinity is infinity. Physically: a point mass has a bottomless well — no finite speed escapes, which is again the black-hole intuition.

Step 2 — case (b), $M \to 0$ . $v_{e} = 2 G \cdot 0/ R = 0 = 0$ . Why this step? No mass means no gravity means no well — you escape "a nothing" at zero speed.

Step 3 — the wrong sign trap. Recall the definition box: $U (r) = - GM m / r$ , with the minus. If someone mistakenly wrote $U = + GM m / r$ , the energy balance $\frac{1}{2} m v_{e}^{2} + \frac{GM m}{R} = 0$ would give $v_{e}^{2} = - 2 GM / R < 0$ , i.e. an imaginary speed — a signal you dropped the minus.

Verify: Both limits behave monotonically: $v_{e}$ increases as $R$ shrinks and increases as $M$ grows, so $\partial v_{e} / \partial R < 0$ and $\partial v_{e} / \partial M > 0$ . Testing $M = 0$ gives exactly $0$ , and the trap gives a negative radicand. ✓ Consistent with the parent's "sign error" warning.

Cell I — exam twist: same density, different size

Worked example Ex 9 — Two planets, same density (Cell I)

Planet B is made of the same stuff as planet A (same density $ρ$ ) but has 3× the radius. How does $v_{e}$ change?

Forecast: Bigger planet, but same material. Factor of $3$ ? Of $3$ ? Of something else?

Step 1. Write mass via density: $M = ρ \cdot \frac{4}{3} π R^{3}$ , so $M \propto R^{3}$ at fixed $ρ$ . Why this step? The twist locks density, not mass — so mass is not free; it grows with volume.

Step 2. Substitute into $v_{e} = 2 GM / R$ : $v_{e} \propto \frac{M}{R} \propto \frac{R ^{3}}{R} = R^{2} = R .$ Why this step? Replacing $M$ by $\propto R^{3}$ collapses the whole dependence to a clean $v_{e} \propto R$ — a much stronger growth than the naive feeling suggests.

Step 3. Therefore $v_{e}$ scales by the same factor as $R$ : $\times 3$ .

Verify: Check with numbers — let A have $R = 1, v_{e} = 1$ (arbitrary units). B has $R = 3$ , so $M \propto 27$ , and $27/3 = 9 = 3$ . ✓ The escape speed grows linearly with size at fixed density — the exam-favourite surprise.

Active recall

Escape from

r = 2 R

vs surface

v = v_{e, surf} / 2 \approx 7.92

km/s.

Leftover speed at infinity

v_{\infty} = v_{0}^{2} - v_{e}^{2}

; here

1 5^{2} - 11. 2^{2} \approx 9.98

km/s.

Compress Sun to

v_{e} = c

R = 2 GM / c^{2} \approx 2.95

km (Schwarzschild radius).

Same density,

3 \times

radius

v_{e} \propto R

, so

v_{e}

triples.

Degenerate

M \to 0

v_{e} \to 0

(no gravity, no well).

Degenerate

R \to 0

v_{e} \to \infty

(bottomless well).

Gravitational PE used all through

U (r) = - GM m / r

, negative because the mass is bound.

Connections

Escape velocity — derivation (Hinglish)
Gravitational potential energy
Newton's law of universal gravitation
Conservation of mechanical energy
Orbital velocity & circular motion
Black holes — Schwarzschild radius
Why the Moon has no atmosphere