Classical-Mechanics · Unit 18 · Video 1 · Interactive Practice

Escape Velocity: Zero Total Energy on Asteroid Toro

IKey Formulas

Formula	Name	What it says
$U^G(r) = -\dfrac{G m_t m}{r}, \qquad U^G(\infty) \equiv 0$	Gravitational potential energy	Zero at infinite separation, negative at every finite $r$
$K_f = 0 \text{ at } r \to \infty \;\Longrightarrow\; E_f = 0$	The escape condition	Barely makes it: nothing left over at infinity
$\tfrac{1}{2} m v_{\text{esc}}^2 - \dfrac{G m_t m}{R} = 0$	Zero mechanical energy at launch	$W_{\text{nc}} = 0$ , so $E_i = E_f = 0$
$v_{\text{esc}} = \sqrt{\dfrac{2 G m_t}{R}}$	Escape velocity	Set by the asteroid's $m_t$ and $R$ alone

Key Insight: The escaping object's own mass cancels, so a pebble and a person both leave Toro at $7.3\ \text{m}\cdot\text{s}^{-1}$ . What differs is the kinetic energy each must be given: $\tfrac{1}{2}mv_{\text{esc}}^2$ still carries the $m$ .

IIVisualization 1 — Three Fates, One Sign

Toro's grip is decided by a sign: whether $E = K + U$ at launch is negative, zero, or positive.

Launch speed v₀ (m/s) 7.30 Distance out (r/R) 1.00

IIIVisualization 2 — From $E_f = 0$ to $7.3\ \text{m/s}$

One equation does all the work: the launch kinetic energy exactly fills the well it has to climb out of.

Step 1

Step 1 — The final state fixes the total

K_f = 0

at infinite separation is what "escape" means, and

U_f = 0

is where we put the zero of potential energy.

E_f = K_f + U_f = 0

Step 2 — Conservation carries that zero back to the surface

Gravity is conservative and nothing else does work, so

W_{\text{nc}} = 0

and

\Delta E_m = 0

E_i = E_f = 0 \quad\Longrightarrow\quad \tfrac{1}{2} m v_{\text{esc}}^2 - \frac{G m_t m}{R} = 0

Step 3 — The object's mass cancels

\tfrac{1}{2} \cancel{m} v_{\text{esc}}^2 = \frac{G m_t \cancel{m}}{R} \quad\Longrightarrow\quad v_{\text{esc}} = \sqrt{\frac{2 G m_t}{R}}

Per kilogram both terms are

G m_t / R = 26.68\ \text{J}\cdot\text{kg}^{-1}

Step 4 — Toro's numbers

2 G m_t = 2(6.67 \times 10^{-11})(2.0 \times 10^{15}) = 2.67 \times 10^{5}\ \text{m}^3\cdot\text{s}^{-2}

\frac{2 G m_t}{R} = \frac{2.67 \times 10^{5}}{5.0 \times 10^{3}} = 53.4\ \text{m}^2\cdot\text{s}^{-2}

v_{\text{esc}} = \sqrt{53.4} = 7.3\ \text{m}\cdot\text{s}^{-1}

IVVisualization 3 — How Big Before You Cannot Outrun It?

At Toro's density escape speed grows in proportion to radius, so a bigger rock eventually beats a sprint.

Your guess: radius (km) 6.0

💡 Toro's surface gravity is only $5.3\ \text{mm}\cdot\text{s}^{-2}$ , so a stride that lifts you at $0.15\ \text{m}\cdot\text{s}^{-1}$ leaves you floating for nearly a minute before the next push-off. Reaching $7.3\ \text{m}\cdot\text{s}^{-1}$ is easy on earth; generating it on Toro is the hard part.

VQuiz Questions

Problem 1 · Escape Velocity from a Given Body

Given: a uniform spherical asteroid of mass $m_a = 3.0 \times 10^{15}\ \text{kg}$ and radius $R = 4.0 \times 10^{3}\ \text{m}$ , with $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2\cdot\text{kg}^{-2}$ — find the escape velocity from its surface.

✅ Correct!

2Gm_a = 4.00 \times 10^{5}\ \text{m}^3\cdot\text{s}^{-2}

, and dividing by

R

gives

100.1\ \text{m}^2\cdot\text{s}^{-2}

, whose square root is

10.0\ \text{m}\cdot\text{s}^{-1}

❌ The factor of 2 is missing.

\sqrt{Gm_a/R} = 7.1

comes from

\tfrac{1}{2}mv^2 = Gm_am/R

without multiplying both sides by

2

before taking the root.

❌ That is $v_{\text{esc}}^2$ , not $v_{\text{esc}}$ .

2Gm_a/R = 100.1\ \text{m}^2\cdot\text{s}^{-2}

has units of speed squared — take the square root.

❌ Check the units of $R$ .

316

comes from

R = 4.0

instead of

R = 4.0 \times 10^{3}\ \text{m}

;

G

is in SI units, so the radius must be in metres.

Show solution

Set the total mechanical energy at launch to zero, since escape means arriving at infinity with $K_f = 0$ and $U_f = 0$ :

\tfrac{1}{2} m v_{\text{esc}}^2 - \frac{G m_a m}{R} = 0 \quad\Longrightarrow\quad v_{\text{esc}} = \sqrt{\frac{2 G m_a}{R}}

Substituting, with $R$ in metres:

2 G m_a = 2(6.67 \times 10^{-11})(3.0 \times 10^{15}) = 4.00 \times 10^{5}\ \text{m}^3\cdot\text{s}^{-2}

\frac{2 G m_a}{R} = \frac{4.00 \times 10^{5}}{4.0 \times 10^{3}} = 100.1\ \text{m}^2\cdot\text{s}^{-2}

v_{\text{esc}} = \sqrt{100.1} = 10.0\ \text{m}\cdot\text{s}^{-1}

Note that the escaping object's mass never entered — it cancelled in the first line.

Problem 2 · Pebble and Astronaut

Given: a $0.20\ \text{kg}$ pebble and a $60\ \text{kg}$ astronaut, both on Toro's surface, where $v_{\text{esc}} = 7.3\ \text{m}\cdot\text{s}^{-1}$ — decide what each one needs in order to escape.

✅ Correct! The

m

\tfrac{1}{2}mv^2 = Gm_tm/R

cancels, so the speed is common; but

K = \tfrac{1}{2}mv_{\text{esc}}^2

still carries it:

5.3\ \text{J}

for the pebble against

1.6 \times 10^{3}\ \text{J}

for the astronaut.

❌ The speed is right, the energy is not.

K = \tfrac{1}{2}mv_{\text{esc}}^2

is proportional to

m

\tfrac{1}{2}(0.20)(53.4) = 5.3\ \text{J}

versus

\tfrac{1}{2}(60)(53.4) = 1.6 \times 10^{3}\ \text{J}

❌ The well really is deeper — but so is the tank.

U

and

K

both scale with

m

, so

m

divides out of

\tfrac{1}{2}mv^2 = Gm_tm/R

and the required speed is unchanged.

❌ Inertia does not enter.

v_{\text{esc}} = \sqrt{2Gm_t/R}

contains only the asteroid's mass and radius; the escaping object's mass has already cancelled.

Show solution

Zero total mechanical energy at launch gives, for an object of any mass $m$ :

\tfrac{1}{2} m v_{\text{esc}}^2 = \frac{G m_t m}{R} \quad\Longrightarrow\quad v_{\text{esc}} = \sqrt{\frac{2 G m_t}{R}} = 7.3\ \text{m}\cdot\text{s}^{-1}

Both terms are proportional to $m$ , so $m$ cancels and the speed is identical for both objects.

The energy needed is not. With $v_{\text{esc}}^2 = 53.4\ \text{m}^2\cdot\text{s}^{-2}$ :

K_{\text{pebble}} = \tfrac{1}{2}(0.20)(53.4) = 5.3\ \text{J}, \qquad K_{\text{astronaut}} = \tfrac{1}{2}(60)(53.4) = 1.6 \times 10^{3}\ \text{J}

The ratio is exactly the mass ratio, $60/0.20 = 300$ .

Problem 3 · Launched Too Slowly

Given: an object leaves Toro's surface radially at $v_0 = 5.0\ \text{m}\cdot\text{s}^{-1}$ , with $G m_t = 1.334 \times 10^{5}\ \text{m}^3\cdot\text{s}^{-2}$ and $R = 5.0 \times 10^{3}\ \text{m}$ — find its total mechanical energy per kilogram and the greatest distance from Toro's centre that it reaches.

Total mechanical energy per kilogram at launch

Greatest distance from Toro's centre

✅ Correct!

E/m = 12.5 - 26.68 = -14.18\ \text{J}\cdot\text{kg}^{-1}

is negative, so the object is bound; at the turning point

K = 0

and

-Gm_t/r = -14.18

gives

r = 9.41 \times 10^{3}\ \text{m}

❌ That is $K/m$ alone. The gravitational potential energy at the surface,

-Gm_t/R = -26.68\ \text{J}\cdot\text{kg}^{-1}

, still has to be added in.

❌ That is $U/m$ alone. Add the launch kinetic energy

\tfrac{1}{2}v_0^2 = 12.5\ \text{J}\cdot\text{kg}^{-1}

to the well depth.

❌ Check the sign of $K$ . Kinetic energy is positive:

E/m = +12.5 - 26.68

, not

-12.5 - 26.68

❌ Not quite.

E/m = \tfrac{1}{2}v_0^2 - Gm_t/R

, with

\tfrac{1}{2}(5.0)^2 = 12.5

and

Gm_t/R = 26.68\ \text{J}\cdot\text{kg}^{-1}

❌ That is the altitude, not the distance from the centre.

4.4\ \text{km}

is the height above the surface; add Toro's radius

R = 5.0\ \text{km}

❌ That assumes constant gravity. Using

h = v_0^2/2g

with

g = 5.34 \times 10^{-3}\ \text{m}\cdot\text{s}^{-2}

gives

2.3\ \text{km}

, but

g

falls off badly over that distance — use

U = -Gm_t m/r

instead.

❌ The initial potential energy was dropped.

r = 2Gm_t/v_0^2 = 10.7\ \text{km}

comes from setting

\tfrac{1}{2}v_0^2 = Gm_t/r

, which forgets that the object started already deep in the well.

❌ Not quite. At the highest point

K = 0

, so the total energy is entirely potential:

-Gm_t/r_{\max} = E/m

Show solution

Step 1 — Total mechanical energy per kilogram at launch.

\frac{E}{m} = \tfrac{1}{2}v_0^2 - \frac{G m_t}{R} = \tfrac{1}{2}(5.0)^2 - \frac{1.334 \times 10^{5}}{5.0 \times 10^{3}} = 12.5 - 26.68 = -14.18\ \text{J}\cdot\text{kg}^{-1}

It is negative, so $5.0\ \text{m}\cdot\text{s}^{-1}$ is below the $7.3\ \text{m}\cdot\text{s}^{-1}$ escape velocity and the object is bound.

Step 2 — The turning point. At the greatest distance the object is momentarily at rest, so $K = 0$ and all of $E$ is potential. Energy is conserved, so that value is still $-14.18\ \text{J}\cdot\text{kg}^{-1}$ :

-\frac{G m_t}{r_{\max}} = -14.18 \quad\Longrightarrow\quad r_{\max} = \frac{1.334 \times 10^{5}}{14.18} = 9.41 \times 10^{3}\ \text{m}

So $r_{\max} = 9.4\ \text{km}$ from the centre — about $1.9R$ , or $4.4\ \text{km}$ above the surface — and then it falls back.

Problem 4 · The Same Rock, Twice the Size

Given: a second asteroid made of the same material as Toro (same uniform density) but with twice the radius, $R' = 10.0\ \text{km}$ , while Toro has $m_t = 2.0 \times 10^{15}\ \text{kg}$ , $R = 5.0\ \text{km}$ and $v_{\text{esc}} = 7.3\ \text{m}\cdot\text{s}^{-1}$ — find the escape velocity from the larger body.

✅ Correct! Equal density means

m \propto R^3

, so

m' = 8m_t = 1.6 \times 10^{16}\ \text{kg}

and

v'_{\text{esc}} = \sqrt{2Gm'/R'} = \sqrt{213.4} = 14.6\ \text{m}\cdot\text{s}^{-1}

— twice Toro's, and now beyond a

12\ \text{m}\cdot\text{s}^{-1}

sprint.

❌ Escape velocity does not fall with size here.

v_{\text{esc}} = \sqrt{2Gm/R}

drops with

R

only at fixed mass; at fixed density the mass grows as

R^3

, which wins.

❌ The mass was held fixed.

\sqrt{2Gm_t/R'} = 5.2

doubles the radius but keeps Toro's mass; the same rock at twice the radius is

8

times as massive.

❌ Escape velocity is not a property of the material alone. With

m \propto R^3

v_{\text{esc}} = \sqrt{2Gm/R} \propto R

— it doubles when the radius doubles.

Show solution

Step 1 — Same density, so the mass scales as the volume.

m' = m_t \left(\frac{R'}{R}\right)^3 = (2.0 \times 10^{15})(2)^3 = 1.6 \times 10^{16}\ \text{kg}

Step 2 — Escape velocity of the bigger body.

v'_{\text{esc}} = \sqrt{\frac{2 G m'}{R'}} = \sqrt{\frac{2(6.67 \times 10^{-11})(1.6 \times 10^{16})}{1.0 \times 10^{4}}} = \sqrt{213.4} = 14.6\ \text{m}\cdot\text{s}^{-1}

The scaling directly. With $\rho$ fixed, $m = \tfrac{4}{3}\pi\rho R^3$ , so

v_{\text{esc}} = \sqrt{\frac{2G}{R} \cdot \tfrac{4}{3}\pi\rho R^3} = R\sqrt{\tfrac{8}{3}\pi G \rho} \;\propto\; R

Doubling the radius doubles the escape velocity: $2 \times 7.3 = 14.6\ \text{m}\cdot\text{s}^{-1}$ . Since $14.6 > 12$ , an Olympic sprinter could no longer outrun this asteroid's gravity.

Solved: 0 / 4

Escape Velocity: Zero Total Energy on Asteroid Toro

IKey Formulas

IIVisualization 1 — Three Fates, One Sign

IIIVisualization 2 — From Ef=0E_f = 0Ef​=0 to 7.3 m/s7.3\ \text{m/s}7.3 m/s

IVVisualization 3 — How Big Before You Cannot Outrun It?

VQuiz Questions

IIIVisualization 2 — From $E_f = 0$ to $7.3\ \text{m/s}$