Exercises — Escape velocity — derivation

$v_e=\sqrt{\frac{2GM}{R}}\text{and}{v}_e=\sqrt{2gR}.$ The escaping object's mass $m$ cancels because both the kinetic energy you must supply ($\frac{1}{2}m{v}_e^2$) and the potential energy you must pay ($GMm/R$) are each proportional to $m$. Divide the energy equation $\frac{1}{2}m{v}_e^2=GMm/R$ by $m$ and $m$ is gone. (See Conservation of mechanical energy.)

Use $v_e=\sqrt{2gR}$ — chosen because we were handed $g$ and $R$ directly, so no need for $G$ or $M$. $v_e=\sqrt{2(4.0)(3.0\times 10^6)}=\sqrt{2.4\times 10^7}\approx 4.9\times 10^3\text{m/s}\approx 4.9\text{km/s}.$

From the parent note, $v_e=\sqrt{2}{v}_o$, so $v_o=\frac{v_e}{\sqrt{2}}=\frac{11.2}{1.41}\approx 7.9\text{km/s}.$ This is the low-Earth-orbit speed (see Orbital velocity & circular motion).

Start from $v_e=\sqrt{2gR}$ and solve for $g$ (this is the unknown, so we isolate it): $v_e^2=2gR\Rightarrow g=\frac{v_e^2}{2R}.$ $g=\frac{(2.4\times 10^3{)}^2}{2(1.74\times 10^6)}=\frac{5.76\times 10^6}{3.48\times 10^6}\approx 1.66{\text{m/s}}^2.$ This matches the Moon's $g\approx 1.62{\text{m/s}}^2$ — the small rounding is from using $2.4$ instead of $2.37$ km/s.

Here we know $v_e$ and $R$ but not $g$, so we use the $G,M$ form and solve for $M$: $v_e=\sqrt{\frac{2GM}{R}}\Rightarrow v_e^2=\frac{2GM}{R}\Rightarrow M=\frac{v_e^{2}R}{2G}.$ $M=\frac{(5.0\times 10^3{)}^2(2.0\times 10^6)}{2(6.674\times 10^{-11})}=\frac{(2.5\times 10^7)(2.0\times 10^6)}{1.3348\times 10^{-10}}.$ $M=\frac{5.0\times 10^{13}}{1.3348\times 10^{-10}}\approx 3.75\times 10^{23}\text{kg}.$

$v_e=\sqrt{2(9.8)(6.4\times 10^6)}=\sqrt{1.2544\times 10^8}\approx 1.12\times 10^4\text{m/s}.$ Energy per kilogram (set $m=1$ in $\frac{1}{2}m{v}_e^2$): $\frac{K}{m}=\frac{1}{2}{v}_e^2=\frac{1}{2}(1.12\times 10^4{)}^2\approx 6.27\times 10^7\text{J/kg}.$ That's $\approx 62.7\text{MJ}$ to throw one kilogram off the planet — a useful feel for why rockets are enormous.

Read the figure alongside the algebra. In the figure, Planet A (blue) has radius $R$ with a short yellow escape arrow of length $v_e$; Planet B (pink) has radius $2R$ with a yellow escape arrow drawn twice as long, length $2v_e$. That arrow-length ratio is exactly what the algebra below produces — watch how each factor of $R$ in the picture turns into a factor in the formula.

Density link: mass $M=\rho \cdot \frac{4}{3}\pi R^3\propto R^3$. Substitute into $v_e$: $v_e=\sqrt{\frac{2GM}{R}}\propto \sqrt{\frac{R^3}{R}}=\sqrt{R^2}=R.$ So $v_e\propto R$: the escape arrow's length is proportional to the planet's radius. In the figure B's radius is drawn $2\times$ A's radius, so its escape arrow is drawn $2\times$ as long — the diagram is a faithful ruler for this result. Doubling the radius doubles $v_e$: $\frac{v_{e,B}}{v_{e,A}}=2.$ Why not $\sqrt{2}$? Because here $M$ is not fixed — a bigger same-density planet is much heavier ($M\propto R^3$), and that extra mass wins over the larger radius. (If we had instead scaled the radius of the circle without lengthening the arrow proportionally, the figure would be lying about the physics.)

With $R$ fixed, $v_e=\sqrt{2GM/R}\propto \sqrt{M}$. Tripling $M$ gives $\frac{v_e^{\prime }}{v_e}=\sqrt{3}\approx \mathrm{1.73.}$ So $v_e^{\prime }\approx 11.2\times 1.73\approx 19.4\text{km/s}$. (Contrast with 3.1 where density was held fixed instead of radius.)

It depends on what is held constant — this is the whole point.

• X is right when density is fixed (Problem 3.1): $M\propto R^3$, so $v_e\propto R$ ⇒ doubling $R$ doubles $v_e$ (factor $2$).
• Y is right when mass $M$ is fixed: then $v_e=\sqrt{2GM/R}\propto 1/\sqrt{R}$. Doubling $R$ multiplies $v_e$ by $1/\sqrt{2}\approx 0.71$ — it decreases, but the exact factor is $1/\sqrt{2}\approx 0.71$, not one-half. Any student who says "$v_e$ halves" has the direction right but the number wrong.

Moral: you cannot say how $v_e$ scales with $R$ until you state what else stays fixed — and even then, "smaller" is $1/\sqrt{2}$, not $1/2$.

Use energy conservation between launch ($r=R$, speed $\frac{1}{2}{v}_e$) and top ($r=R+h$, speed $0$), with the sign convention $U(r)=-GMm/r$ stated at the top of this sheet: $\frac{1}{2}m{\left(\frac{1}{2}{v}_e\right)}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}.$ Cancel $m$; use $v_e^2=2GM/R$ so ${\left(\frac{1}{2}{v}_e\right)}^2=\frac{1}{4}\cdot \frac{2GM}{R}=\frac{GM}{2R}$: $\frac{GM}{4R}-\frac{GM}{R}=-\frac{GM}{R+h}.$ (The left kinetic term is $\frac{1}{2}\cdot \frac{GM}{2R}=\frac{GM}{4R}$.) Divide through by $GM$: $\frac{1}{4R}-\frac{1}{R}=-\frac{1}{R+h}\Rightarrow -\frac{3}{4R}=-\frac{1}{R+h}\Rightarrow R+h=\frac{4R}{3}.$ $\boxed{h=\frac{R}{3}.}$ Half the escape speed carries you only one-third of an Earth-radius high — far short of "halfway to infinity", proving that speed does not map linearly to distance in a $1/r$ potential.

Orbital condition (gravity provides centripetal force, from Orbital velocity & circular motion): $\frac{GMm}{R^2}=\frac{m{v}_o^2}{R}\Rightarrow v_o^2=\frac{GM}{R}.$ Escape result: $v_e^2=\frac{2GM}{R}=2v_o^2$. Taking square roots: $v_e=\sqrt{2}{v}_o.■$ This is independent of $M$ and $R$ — it holds for every body, which is why it's worth memorising.

Set $v_e=c$ in $v_e=\sqrt{2GM/R}$ and solve for the radius, now called $R_s$: $c=\sqrt{\frac{2GM}{R_s}}\Rightarrow c^2=\frac{2GM}{R_s}\Rightarrow R_s=\frac{2GM}{c^2}.$ For the Sun: $R_s=\frac{2(6.674\times 10^{-11})(1.99\times 10^{30})}{(3.0\times 10^8{)}^2}=\frac{2.656\times 10^{20}}{9.0\times 10^{16}}\approx 2.95\times 10^3\text{m}.$ About 3 km. (This Newtonian shortcut lands on the correct General-Relativity answer — a happy coincidence. See Black holes — Schwarzschild radius.)

The derivation logic is unchanged — only the starting radius differs. Using the sign convention $U(r)=-GMm/r$ (stated at the top of this sheet), apply energy conservation from $r_0$ (speed $v_e^{\prime }$) to infinity (speed $0$, where $U=0$): $\frac{1}{2}m{v}_e^{\prime 2}-\frac{GMm}{r_0}=0\Rightarrow v_e^{\prime }(r_0)=\sqrt{\frac{2GM}{r_0}}=\sqrt{\frac{2GM}{R+H}}.$ Degenerate check ($H=0\Rightarrow r_0=R$): $v_e^{\prime }(R)=\sqrt{\frac{2GM}{R}}=v_e,$ the surface value — as it must. Reasoning: higher up, you have already climbed part-way out of the well, so less speed is needed — $v_e^{\prime }$ shrinks with increasing $H$, and $\to 0$ as $H\to \infty$.

Using the altitude result, we want $v_e^{\prime }(r_0)=\frac{1}{2}{v}_e$: $\sqrt{\frac{2GM}{r_0}}=\frac{1}{2}\sqrt{\frac{2GM}{R}}.$ Square both sides (undo the roots together): $\frac{2GM}{r_0}=\frac{1}{4}\cdot \frac{2GM}{R}\Rightarrow \frac{1}{r_0}=\frac{1}{4R}\Rightarrow r_0=4R.$ Since $r_0=R+H$: $H=4R-R=3R.$ You must be three Earth-radii above the surface for escape speed to fall to half. Because $v_e\propto 1/\sqrt{r}$, halving $v_e$ needs quadrupling $r$ — a nice consistency check on the $1/\sqrt{r}$ scaling.

Step 1 — compute the molecular speed. Plug the given numbers into $v_{rms}=\sqrt{3kT/m_{mol}}$: $v_{rms}=\sqrt{\frac{3(1.38\times 10^{-23})(400)}{4.65\times 10^{-26}}}=\sqrt{\frac{1.656\times 10^{-20}}{4.65\times 10^{-26}}}=\sqrt{3.56\times 10^5}\approx 597\text{m/s}.$ Step 2 — compare to escape speed. Form the ratio $v_{rms}/v_e$: $\frac{v_{rms}}{v_e}=\frac{597}{2370}\approx \mathrm{0.25.}$ So a typical nitrogen molecule already moves at a quarter of the Moon's escape speed.

Step 3 — the atmospheric-loss argument. A temperature does not give a single speed; it gives a whole spread of speeds, with a long high-speed tail where many molecules move much faster than the average $v_{rms}$. When $v_{rms}$ is as large as $\frac{1}{4}$ of $v_e$, that fast tail routinely pokes above the escape threshold $2.37\text{km/s}$, so molecules leak away continuously. Over billions of years the atmosphere bleeds off entirely — this is why the Moon has essentially no atmosphere. For contrast, Earth's $v_e=11.2\text{km/s}$ is about $19\times$ this molecular speed ($11200/597\approx 18.8$), placing the escape threshold far out in the negligible tail — so Earth keeps its air. (See Why the Moon has no atmosphere.)