Classical-Mechanics · Unit 23 · Video 3 · Interactive Practice

What Changes Angular Momentum: Torque, Conservation, and a Meteor Grazing Earth

IKey Formulas

Formula	Name	What it needs
$\vec{\boldsymbol{\tau}}_S = \vec{\mathbf{r}}_S \times \vec{\mathbf{F}} = \dfrac{d\vec{\mathbf{L}}_S}{dt}$	Torque law (one particle)	Differentiate $\vec{\mathbf{L}}_S = \vec{\mathbf{r}}_S \times \vec{\mathbf{p}}$ about a fixed point $S$ : the term $\vec{\mathbf{v}} \times m\vec{\mathbf{v}} = \vec{\mathbf{0}}$ drops out and $\frac{d\vec{\mathbf{p}}}{dt} = \vec{\mathbf{F}}$ , the net force
$\vec{\boldsymbol{\tau}}_S = \vec{\mathbf{0}} \;\Rightarrow\; \vec{\mathbf{L}}_{S,f} = \vec{\mathbf{L}}_{S,i}$	Conservation of angular momentum	A fixed point $S$ about which the total torque vanishes; about another point the torque generally does not
$\vec{\mathbf{J}}_S = \displaystyle\int_{t_i}^{t_f} \vec{\boldsymbol{\tau}}_S\,dt = \Delta\vec{\mathbf{L}}_S$	Angular impulse	The torque about $S$ over the interval; the rotational twin of $\int \vec{\mathbf{F}}\,dt = \Delta\vec{\mathbf{p}}$
$m v_i h = m v_p R_E, \quad h = \sqrt{R_E^2 + \dfrac{2GM_E R_E}{v_i^2}}$	Meteor grazing Earth (Example 19.4)	Angular momentum about Earth's centre plus mechanical energy $\tfrac{1}{2}mv_i^2 = \tfrac{1}{2}mv_p^2 - \tfrac{GmM_E}{R_E}$ ; $h$ is the impact parameter

Key Insight: Gravity on the meteor always points at Earth's centre, so its moment arm about that one point is zero. The meteor's speed and direction both change, yet $\vec{\mathbf{L}}_S$ about Earth's centre does not — and that single fixed number, $-h m v_i\,\hat{\mathbf{k}}$ , ties the far-away approach to the moment of closest approach.

IIVisualization 1 — Choosing the Point

The same gravitational force exerts zero torque about Earth's centre and a nonzero torque about almost any other point.

time 0.42 reference point

IIIVisualization 2 — How Far to the Side Can the Line Be?

For a given speed far away, which impact parameter $h$ makes the meteor just graze the surface?

v_i (km/s) 8.0 h (R_E) 1.30

IVVisualization 3 — Angular Impulse Is the Area Under the Torque

Each second of torque adds $\tau_z\,\Delta t$ to $L_z$ , so the net change is the signed area under $\tau_z(t)$ .

torque profile

VQuiz Questions

Problem 1 · Torque from Angular Momentum

Given: about a fixed point $S$ , a particle's angular momentum is $\vec{\mathbf{L}}_S(t) = (3t^2 + 2)\,\hat{\mathbf{k}}\ \text{kg}\cdot\text{m}^2/\text{s}$ with $t$ in seconds — find the $z$ -component of the total torque about $S$ at $t = 2\ \text{s}$ .

✅ Correct! The torque is the rate of change:

\tau_{S,z} = \frac{dL_{S,z}}{dt} = 6t = 12\ \text{N}\cdot\text{m}

t = 2\ \text{s}

❌ That is $L_{S,z}(2)$ itself.

3(2)^2 + 2 = 14

is the angular momentum at that instant; the torque law asks for its time derivative.

❌ Evaluate the derivative at $t = 2\ \text{s}$ .

\frac{d}{dt}(3t^2 + 2) = 6t

, not the constant

6

❌ Check the power rule.

\frac{d}{dt}(3t^2) = 6t

, which is

12

t = 2

;

24

6t^2

❌ Not quite. Use

\vec{\boldsymbol{\tau}}_S = d\vec{\mathbf{L}}_S/dt

and differentiate

3t^2 + 2

Show solution

The torque law about a fixed point $S$ is $\vec{\boldsymbol{\tau}}_S = \dfrac{d\vec{\mathbf{L}}_S}{dt}$ , so componentwise

\tau_{S,z} = \frac{d}{dt}\left(3t^2 + 2\right) = 6t.

At $t = 2\ \text{s}$ : $\tau_{S,z} = 6(2) = 12\ \text{N}\cdot\text{m}$ . The constant $2$ is the angular momentum at $t = 0$ and contributes nothing to the rate of change.

Problem 2 · Which Quantity Stays Fixed?

Given: a satellite moves on an elliptical orbit around Earth; Earth is fixed, sits at one focus of the ellipse, and exerts the only force — which quantity is constant along the orbit?

✅ Correct! Gravity points along

-\hat{\mathbf{r}}

from Earth's centre, so

\vec{\mathbf{r}}_S \times \vec{\mathbf{F}} = \vec{\mathbf{0}}

about that point and

\vec{\mathbf{L}}_S

is constant — exactly the meteor's argument.

❌ The speed changes. Constant

L_S = r\,m v_\perp

means the satellite moves faster where

r

is small; energy conservation says the same thing, since

U = -GmM_E/r

varies along the orbit.

❌ Wrong point. The force's line of action passes through the focus where Earth sits, not through the centre of the ellipse, so everywhere except at the two ends of the major axis gravity has a nonzero moment arm about the centre and exerts a torque there.

❌ There is a net force.

\vec{\mathbf{F}} = d\vec{\mathbf{p}}/dt \neq \vec{\mathbf{0}}

, so

\vec{\mathbf{p}}

keeps changing direction (and size).

❌ Not quite. Look for the fixed point about which the torque is zero.

Show solution

Take $S$ at Earth's centre. The only force is $\vec{\mathbf{F}} = -\dfrac{GmM_E}{r^2}\,\hat{\mathbf{r}}$ and the position vector is $\vec{\mathbf{r}}_S = r\,\hat{\mathbf{r}}$ , so they are anti-parallel:

\vec{\boldsymbol{\tau}}_S = r\,\hat{\mathbf{r}} \times \left(-\frac{GmM_E}{r^2}\,\hat{\mathbf{r}}\right) = \vec{\mathbf{0}} \;\Rightarrow\; \vec{\mathbf{L}}_S = \text{constant}.

About the centre of the ellipse the line of action generally misses the point, so the torque there is not zero. Linear momentum changes because the net force is not zero, and the speed changes because the potential energy does.

Problem 3 · The Grazing Meteor in Numbers

Given: a meteor approaches from very far away with $v_i = 8.0\ \text{km/s}$ ; $R_E = 6.37 \times 10^{6}\ \text{m}$ and $GM_E = 3.99 \times 10^{14}\ \text{m}^3/\text{s}^2$ — find the impact parameter $h$ for which it just grazes Earth, and its speed $v_p$ at the grazing point.

What is h?

What is v_p?

✅ Correct!

h \approx 1.72\,R_E

: a line aimed well outside Earth's disc still hits, and the meteor arrives at

13.8\ \text{km/s}

with the same angular momentum it started with.

❌ Missing the factor 2. Multiplying the energy equation by 2 gives

v_i^2 = v_p^2 - \frac{2GM_E}{R_E}

, so the extra term is

\frac{2GM_E R_E}{v_i^2}

, not

\frac{GM_E R_E}{v_i^2}

❌ That ignores gravity.

h = R_E

is the answer for a straight line; gravity bends the path inward, so the grazing line lies farther out.

❌ The terms add under the square root.

h^2 = R_E^2 + \frac{2GM_E R_E}{v_i^2}

; you added

R_E

\frac{2GM_E}{v_i^2}

instead.

❌ Check h. Solve

h^2 = R_E^2 + 2GM_ER_E/v_i^2

with

v_i

in m/s.

❌ Ratio upside down.

m v_i h = m v_p R_E

gives

v_p = v_i h / R_E

, and

h > R_E

makes

v_p

larger than

v_i

❌ Constant angular momentum is not constant speed. The moment arm shrinks from

h

R_E

, so

v_p = v_i h/R_E

must grow.

❌ That is the escape speed $\sqrt{2GM_E/R_E}$ . Energy gives

v_p^2 = v_i^2 + \frac{2GM_E}{R_E}

— the meteor also keeps its initial

v_i^2

❌ Check v_p. Use

v_p = v_i h / R_E

from angular momentum.

Show solution

Angular momentum about Earth's centre (zero torque there): $m v_i h = m v_p R_E$ , so $v_p = v_i h/R_E$ .

Energy (potential energy $\to 0$ far away): $\tfrac{1}{2}m v_i^2 = \tfrac{1}{2}m v_p^2 - \dfrac{GmM_E}{R_E}$ . Substituting $v_p$ , cancelling $m$ and multiplying by 2:

h^2 = R_E^2 + \frac{2GM_E R_E}{v_i^2} = (6.37\times10^{6})^2 + \frac{2(3.99\times10^{14})(6.37\times10^{6})}{(8.0\times10^{3})^2}

h^2 \approx 4.06\times10^{13} + 7.94\times10^{13} = 1.20\times10^{14}\ \text{m}^2 \;\Rightarrow\; h \approx 1.10\times10^{7}\ \text{m} \approx 1.72\,R_E.

Then $v_p = v_i\,\dfrac{h}{R_E} \approx (8.0\ \text{km/s})(1.72) \approx 13.8\ \text{km/s}$ .

Check with energy alone: $v_p^2 = v_i^2 + \dfrac{2GM_E}{R_E} \approx 64.0 + 125.3 = 189.3\ \text{km}^2/\text{s}^2$ , so $v_p \approx 13.8\ \text{km/s}$ ✓

Problem 4 · Angular Impulse from a Changing Torque

Given: a particle has $L_{S,z} = 5\ \text{kg}\cdot\text{m}^2/\text{s}$ about a fixed point $S$ at $t = 0$ ; from $t = 0$ to $t = 3\ \text{s}$ the total torque about $S$ is $\tau_{S,z}(t) = (4 - 2t)\ \text{N}\cdot\text{m}$ — find $L_{S,z}$ at $t = 3\ \text{s}$ .

✅ Correct!

J_{S,z} = \int_0^3 (4 - 2t)\,dt = 3\ \text{N}\cdot\text{m}\cdot\text{s}

, so

L_{S,z}

rises from

5

8

❌ The torque is not constant.

\tau_{S,z}(3)\cdot 3 = -6

uses the final torque for the whole interval; integrate instead.

❌ The torque is not constant.

\tau_{S,z}(0)\cdot 3 = 12

uses the initial torque for the whole interval; the torque falls to zero at

t = 2\ \text{s}

and then turns negative.

❌ That is $\Delta L_{S,z}$ , not $L_{S,z}$ . The angular impulse is the change; add it to the initial

5\ \text{kg}\cdot\text{m}^2/\text{s}

❌ Not quite. Integrate the torque over the interval and add the result to

L_{S,z}(0)

Show solution

The angular impulse about $S$ equals the change in angular momentum about $S$ :

J_{S,z} = \int_0^{3} (4 - 2t)\,dt = \Big[4t - t^2\Big]_0^{3} = 12 - 9 = 3\ \text{N}\cdot\text{m}\cdot\text{s}.

L_{S,z}(3) = L_{S,z}(0) + J_{S,z} = 5 + 3 = 8\ \text{kg}\cdot\text{m}^2/\text{s}.

Geometrically: the positive area from $0$ to $2\ \text{s}$ is $4$ , the negative area from $2$ to $3\ \text{s}$ is $-1$ , net $+3$ .

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