Worked examples — Conservation of energy and angular momentum in gravitational field

Before anything, let me re-state the four tools we lean on, in plain words, so no symbol is used unearned.

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The scenario matrix

Every problem in this topic is one of these cells. Each worked example below is tagged with the cell it fills.

#	Cell class	What is special	Example
A	Sign: $E<0$ (bound)	ellipse, finite $a$	Ex 1
B	Sign: $E=0$ (parabolic, $e=1$)	escape "just barely", $a\to \infty$	Ex 4
C	Sign: $E>0$ (unbound, $e>1$)	hyperbola, $a<0$	Ex 5
D	Degenerate: circle ($e=0$)	$r=a$ everywhere	Ex 2
E	Degenerate: radial drop ($L=0$)	no sideways motion at all	Ex 6
F	Turning points (perihelion/aphelion)	$\dot{r}=0$, all speed is sideways	Ex 3
G	Limiting value ($r\to \infty$)	speed at infinity of an escaper	Ex 5 (part 2)
H	Real-world word problem	translate English → symbols	Ex 7
I	Exam twist (same $a$, different $e$)	trap: energy unchanged	Ex 8

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Example 1 — Bound orbit: get $a$, then $r_p$ and $v_p$ (cells A, F)

Forecast: $E<0$ so the orbit is bound — a closed ellipse. Guess: is $r_p$ bigger or smaller than $a$? (It must be smaller — perihelion is the close point.)

1. Get $a$ from energy. $E=-\frac{GMm}{2a}\Rightarrow a=-\frac{GMm}{2E}$. Why this step? $E$ of a bound orbit depends on only $a$, so energy alone unlocks the orbit size. $a=-\frac{(3.99\times 10^{14})(1000)}{2(-2.0\times 10^9)}=9.975\times 10^7\text{m}.$

2. Set up the turning-point equation. At perihelion $\dot{r}=0$, so all the speed is sideways: $v_p=\frac{L}{m{r}_p}$. Why this step? A turning point is the one place where the messy radial speed vanishes, so $L$ hands us $v$ for free.

3. Feed that into energy. Substitute $v_p=\frac{L}{m{r}_p}$ into $E=\frac{1}{2}m{v}_p^2-\frac{GMm}{r_p}$: $E=\frac{L^2}{2m{r}_p^2}-\frac{GMm}{r_p}.$ Why this step? Now $r_p$ is the only unknown — this is $U_{\text{eff}}(r_p)=E$ evaluated at the turning point.

4. Turn it into a standard quadratic in $x=1/r_p$. Multiply through and collect powers of $x$: $\frac{L^2}{2m}{x}^2-(GMm)x-E=0.$ Why this step? Physically, $r$ enters energy only as $1/r$ and $1/r^2$; substituting $x=1/r$ turns "energy = const at a turning point" into an ordinary quadratic whose two roots are exactly the two turning radii $1/r_p$ and $1/r_a$ — the close and far ends of the same ellipse. Note the signs: $L^2/2m>0$ (the $+x^2$ centrifugal term), $-GMm<0$ (attractive gravity), and $-E>0$ here because $E<0$. Put in numbers with careful powers of ten: $\frac{L^2}{2m}=\frac{(4.0\times 10^{13}{)}^2}{2(1000)}=\frac{1.6\times 10^{27}}{2\times 10^3}=8.0\times 10^{23},$ $GMm=3.99\times 10^{17},-E=2.0\times 10^9.$

5. Apply the quadratic formula. With $A=8.0\times 10^{23}$, $B=-3.99\times 10^{17}$, $C=-E=2.0\times 10^9$: $x=\frac{GMm\pm \sqrt{(GMm{)}^2-4\cdot \frac{L^2}{2m}\cdot (-E)}}{2\cdot \frac{L^2}{2m}}=\frac{3.99\times 10^{17}\pm \sqrt{(3.99\times 10^{17}{)}^2-4(8.0\times 10^{23})(2.0\times 10^9)}}{1.6\times 10^{24}}.$ Which root is perihelion? Perihelion is the smaller radius = larger $x$, so take the $+$ root; the $-$ root gives $1/r_a$ (aphelion). Evaluating: $x_p=\frac{1}{r_p}\approx 3.06\times 10^{-7}{\text{m}}^{-1}\Rightarrow r_p\approx 3.27\times 10^6\text{m},$ $x_a=\frac{1}{r_a}\approx 1.94\times 10^{-7}\times 10^0\cdots \Rightarrow r_a\approx 1.966\times 10^8\text{m}.$

6. Speed at perihelion. $v_p=\frac{L}{m{r}_p}=\frac{4.0\times 10^{13}}{1000\cdot 3.27\times 10^6}\approx 1.22\times 10^4\text{m/s}.$

Verify: Check $r_p+r_a=2a$: $3.27\times 10^6+1.966\times 10^8\approx 1.999\times 10^8\approx 2a=1.995\times 10^8$ ✔ (rounding). Units of $v_p$: $\frac{{\text{kg⋅m}}^2/\text{s}}{\text{kg⋅m}}=\text{m/s}$ ✔.

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Example 2 — The circle (degenerate, $e=0$) (cell D)

Forecast: For a circle nothing changes around the loop — one constant speed. Should it be near the famous "$\sim 7.5$ km/s low-Earth-orbit" number? Guess yes.

1. Use $r=a$ for a circle. A circle is an ellipse whose longest and shortest widths are equal, so $a=r$ (and $e=0$). Why this step? This collapses vis-viva instantly.

2. Plug into vis-viva. $v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)=GM\left(\frac{2}{r}-\frac{1}{r}\right)=\frac{GM}{r}.$ Why this step? $\frac{2}{r}-\frac{1}{r}=\frac{1}{r}$ — the circular special case of the general speed law.

3. Compute. $v=\sqrt{\frac{3.99\times 10^{14}}{7.0\times 10^6}}\approx 7550\text{m/s}=7.55\text{km/s}.$

Verify: For a circle, gravity must equal the centripetal need: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}\Rightarrow v^2=\frac{GM}{r}$ — same formula, independent route ✔. It matches the LEO value ✔.

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Example 3 — Perihelion vs aphelion speeds via $L$ (cell F)

Figure below (s01): an ellipse with Earth at the right-hand focus (yellow dot). A short pink radius arm reaches the near point (perihelion) where the velocity arrow is long and fast; a long blue radius arm reaches the far point (aphelion) where the velocity arrow is short and slow. The picture shows why the short arm must carry the bigger speed to keep $L=mr{v}_{\perp }$ constant.

Forecast: Look at the figure — at perihelion (close, pink) the radius arm is short, at aphelion (far, blue) it is long. Since $L=mr{v}_{\perp }$ is the same at both, the short arm must have the bigger speed. So $v_p/v_a>1$.

1. Both ends are turning points. At each, velocity is purely sideways ($\dot{r}=0$), so $L=m{r}_p{v}_p=m{r}_a{v}_a$. Why this step? Equal $L$ at two "all-sideways" points gives a clean speed ratio with no energy needed.

2. Ratio. $\frac{v_p}{v_a}=\frac{r_a}{r_p}=\frac{1.0\times 10^8}{5.0\times 10^7}=\mathrm{2.0.}$ Perihelion is twice as fast.

3. Get $v_p$ using vis-viva. Semi-major axis $a=\frac{r_p+r_a}{2}=7.5\times 10^7\text{m}$. Then $v_p=\sqrt{GM\left(\frac{2}{r_p}-\frac{1}{a}\right)}=\sqrt{3.99\times 10^{14}\left(\frac{2}{5.0\times 10^7}-\frac{1}{7.5\times 10^7}\right)}.$ Why this step? Vis-viva gives the actual speed at any $r$ once $a$ is known. $v_p=\sqrt{3.99\times 10^{14}\cdot (4.0\times 10^{-8}-1.333\times 10^{-8})}\approx 3263\text{m/s}.$

Verify: $v_a$ should be half of $v_p$: $v_a=\sqrt{GM(\frac{2}{r_a}-\frac{1}{a})}=\sqrt{3.99\times 10^{14}(2.0\times 10^{-8}-1.333\times 10^{-8})}\approx 1631\text{m/s}$, and $3263/1631\approx 2.0$ ✔.

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Example 4 — The parabola: $E=0$, escape "just barely" ($e=1$) (cell B)

Forecast: $E=0$ is the knife-edge between falling back (ellipse) and flying free (hyperbola). The shape is a parabola with $a\to \infty$ and eccentricity exactly $e=1$. Guess: is $v_{\text{esc}}$ bigger than the circular speed at the same $r$? (Yes — you need extra to break free.)

1. Set $E=0$. $E=\frac{1}{2}m{v}^2-\frac{GMm}{r}=0\Rightarrow \frac{1}{2}{v}^2=\frac{GM}{r}.$ Why this step? "Just barely escape" means arriving at infinity with zero leftover speed, i.e. $E=0$.

2. Solve. $v_{\text{esc}}=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2(3.99\times 10^{14})}{1.0\times 10^7}}\approx 8933\text{m/s}=8.93\text{km/s}.$

3. Why $a\to \infty$ and $e=1$. From $E=-\frac{GMm}{2a}$, setting $E=0$ forces $a\to \infty$: the "ellipse" stretches until its far end runs to infinity ($r_a\to \infty$), so $e=\frac{r_a-r_p}{r_a+r_p}\to 1$ — an open parabola. See Effective Potential and Orbit Classification.

Verify: Ratio to circular speed $\sqrt{GM/r}$: $v_{\text{esc}}/v_{\text{circ}}=\sqrt{2}\approx 1.414$ ✔ (escape is always $\sqrt{2}\times$ circular at the same radius).

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Example 5 — Hyperbola: $E>0$, and speed at infinity (cells C, G)

Forecast: From Ex 4 the escape speed here is $8933$ m/s. Since $9000>8933$, this escapes — hyperbolic ($e>1$), $E>0$. At infinity it still moves (some speed left over), so $v_{\infty }>0$ but less than $9000$.

1. Compute $E$ per unit mass (mass cancels for shape questions): $\epsilon =\frac{1}{2}{v}^2-\frac{GM}{r}$. Why per unit mass? The orbit type depends on the sign of $E$, and dividing by $m$ (positive) keeps the sign — simpler. $\epsilon =\frac{1}{2}(9000{)}^2-\frac{3.99\times 10^{14}}{1.0\times 10^7}=4.05\times 10^7-3.99\times 10^7=6.0\times 10^5\text{J/kg}.$ $\epsilon >0\Rightarrow$ hyperbolic escape ✔ (a).

2. Speed at infinity. As $r\to \infty$, $U\to 0$, so all energy is kinetic: $\frac{1}{2}{v}_{\infty }^2=\epsilon$. Why this step? This is the limiting value cell — evaluate energy conservation in the limit $r\to \infty$. $v_{\infty }=\sqrt{2\epsilon }=\sqrt{2\cdot 6.0\times 10^5}\approx 1095\text{m/s}.$

Verify: $v_{\infty }<v_{\text{start}}=9000$ ✔ (gravity robbed speed as it climbed out). And $v_{\infty }^2=v^2-v_{\text{esc}}^2=9000^2-8933^2\approx 1.2\times 10^6=2\epsilon$ ✔.

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Example 6 — Radial drop: $L=0$ (degenerate) (cell E)

Figure below (s02): Earth (blue disc) with its centre marked (yellow). A stationary object sits high up at $r_0$ (pink dot) and falls straight down a vertical line — the pink arrow — to the surface at $r$. Because there is zero sideways push, the whole path is one radial line; there is no loop, no ellipse.

Forecast: With $L=0$ there is no "sideways" motion ever — it falls straight down a line toward the centre (see figure, the straight pink track). No orbit, no ellipse: pure free-fall. Speed grows as it drops.

1. Confirm the degeneracy. $L=mr{v}_{\perp }=0$ with $r\ne 0$ forces $v_{\perp }=0$: motion is purely radial. The centrifugal term $\frac{L^2}{2m{r}^2}$ in $U_{\text{eff}}$ vanishes — no barrier to stop the fall. Why this step? It tells us $v$ is entirely radial, so $\frac{1}{2}{v}^2$ is the whole kinetic energy.

2. Energy conservation between the two radii (per unit mass, starts at rest): $\underset{\text{start}}{\underbrace{0-\frac{GM}{r_0}}}=\underset{\text{end}}{\underbrace{\frac{1}{2}{v}^2-\frac{GM}{r}}}.$ Why this step? $E$ is constant; at release kinetic energy is $0$.

3. Solve. $\frac{1}{2}{v}^2=GM\left(\frac{1}{r}-\frac{1}{r_0}\right)$, so $v=\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)}.$ Put in numbers: $\frac{1}{r}-\frac{1}{r_0}=\frac{1}{6.4\times 10^6}-\frac{1}{8.0\times 10^6}=1.5625\times 10^{-7}-1.25\times 10^{-7}=3.125\times 10^{-8}{\text{m}}^{-1}.$ Then $2GM=7.98\times 10^{14}$, so $v=\sqrt{7.98\times 10^{14}\cdot 3.125\times 10^{-8}}=\sqrt{2.494\times 10^7}\approx 4994\text{m/s}.$

Verify: Sign check: $\frac{1}{r}>\frac{1}{r_0}$ since the object dropped to a smaller radius, so the bracket is positive and $v$ is real ✔. Units: $\sqrt{{\text{m}}^3/{\text{s}}^2\cdot {\text{m}}^{-1}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✔.

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Example 7 — Real-world word problem (cell H)

Forecast: At aphelion it crawls slowly and far out; at perihelion it whips fast and close. We must chain both conservation laws. Guess: $r_p$ will be much smaller than $r_a$ because the comet is slow way out there (barely bound).

1. Aphelion is a turning point → $\ell$ (angular momentum per unit mass) $=r_a{v}_a$. Why this step? All speed is sideways at aphelion, so $\ell$ is trivial to read off. $\ell =r_a{v}_a=(5.0\times 10^{11})(1.0\times 10^3)=5.0\times 10^{14}{\text{m}}^2/\text{s}.$

2. Energy per unit mass at aphelion. $\epsilon =\frac{1}{2}{v}_a^2-\frac{GM}{r_a}=\frac{1}{2}(10^3{)}^2-\frac{1.327\times 10^{20}}{5.0\times 10^{11}}=5.0\times 10^5-2.654\times 10^8\approx -2.649\times 10^8\text{J/kg}.$ Why this step? $\epsilon <0$ confirms bound, and it's the constant we reuse at perihelion.

3. At perihelion (also a turning point) $v_p=\ell /r_p$, and energy conservation gives the same $\epsilon$: $\epsilon =\frac{{\ell }^2}{2r_p^2}-\frac{GM}{r_p}.$ Turn it into a quadratic in $x=1/r_p$: $\frac{{\ell }^2}{2}{x}^2-GMx-\epsilon =0.$ Why this step? Same physics as Ex 1: at a turning point energy is a function of $1/r$ and $1/r^2$ only, so the substitution $x=1/r$ produces a quadratic whose two roots are $1/r_p$ and $1/r_a$. Signs: ${\ell }^2/2>0$, $-GM<0$, and $-\epsilon >0$ (since $\epsilon <0$).

4. Evaluate carefully (mind the powers of ten). $\frac{{\ell }^2}{2}=\frac{(5.0\times 10^{14}{)}^2}{2}=\frac{2.5\times 10^{29}}{2}=1.25\times 10^{29},GM=1.327\times 10^{20},-\epsilon =2.649\times 10^8.$ Quadratic formula, taking the $+$ root (perihelion = smaller radius = larger $x$): $x_p=\frac{GM+\sqrt{G{M}^2-4\cdot \frac{{\ell }^2}{2}\cdot (-\epsilon )}}{2\cdot \frac{{\ell }^2}{2}}=\frac{1.327\times 10^{20}+\sqrt{(1.327\times 10^{20}{)}^2-4(1.25\times 10^{29})(2.649\times 10^8)}}{2.5\times 10^{29}}.$ This gives $x_p\approx 8.72\times 10^{-10}{\text{m}}^{-1}$, so $r_p\approx 1.147\times 10^9\text{m}\approx 1.15\times 10^9\text{m}.$

Verify: Semi-major axis $a=-GM/(2\epsilon )=1.327\times 10^{20}/(2\cdot 2.649\times 10^8)\approx 2.505\times 10^{11}$, so $2a\approx 5.01\times 10^{11}$; and $r_p+r_a=1.15\times 10^9+5.0\times 10^{11}\approx 5.01\times 10^{11}$ ✔. Also $r_p\ll r_a$ as forecast ✔. (See Kepler's Laws of Planetary Motion for why such comets have wildly eccentric orbits.)

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Example 8 — Exam twist: same $a$, different $e$ (cell I)

Forecast: The trap is to say the "wilder" eccentric ellipse Y has more energy. But energy depends only on $a$. Compute both $a$ and compare.

1. Semi-major axis of each. X: $a_X=1.0\times 10^7$ (circle → $a=r$). Y: $a_Y=\frac{r_p+r_a}{2}=\frac{6.0\times 10^6+1.4\times 10^7}{2}=1.0\times 10^7\text{m}.$ Why this step? $a=\frac{r_p+r_a}{2}$ is just half the long axis.

2. They are equal! $a_X=a_Y$.

3. Energies. $\epsilon =-\frac{GM}{2a}$ for both: $-\frac{3.99\times 10^{14}}{2(1.0\times 10^7)}=-1.995\times 10^7\text{J/kg}$ — identical. Why this step? Eccentricity $e$ never appears in $\epsilon =-GM/2a$.

Verify: Their eccentricities differ — X: $e_X=0$; Y: $e_Y=\frac{r_a-r_p}{r_a+r_p}=\frac{1.4\times 10^7-6.0\times 10^6}{2.0\times 10^7}=0.4$. Their angular momenta do differ: circle X has ${\ell }_X=\sqrt{GMa}\approx 6.32\times 10^{10}$; ellipse Y has ${\ell }_Y=\sqrt{GMa(1-e_Y^2)}\approx 5.79\times 10^{10}<{\ell }_X$ ✔. Same energy, different shape — exactly the point.

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Recall Which cell was which?

Ex 1 :::: A, F (bound, turning points) Ex 2 :::: D (circle, $e=0$) Ex 3 :::: F (perihelion vs aphelion via $L$) Ex 4 :::: B (parabola, $E=0$, $e=1$) Ex 5 :::: C, G (hyperbola $e>1$, speed at infinity) Ex 6 :::: E (radial drop, $L=0$) Ex 7 :::: H (word problem) Ex 8 :::: I (same $a$, different $e$)

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Flashcards