Conservation of energy and angular momentum in gravitational field: Exercises

Definition The three symbols we reuse everywhere

Before any exercise, pin down what each letter means physically — never use a symbol you can't picture.

$U (r)$ = gravitational potential energy, the "stored" energy of being at distance $r$ from the central body. Derived in the parent as $U (r) = - \frac{GM m}{r}$ (zero far away, more negative as you fall in — you sit in a well).
$θ$ = the polar angle: imagine standing at the central body and pointing an arm at the orbiting object. As the object loops around, that arm sweeps and the angle $θ$ it makes with a fixed reference line grows. Its rate of change $\dot{θ}$ (read "theta-dot") is how fast the arm sweeps — the angular speed. The sideways (tangential) speed is $r \dot{θ}$ .
$L = m r^{2} \dot{θ}$ = angular momentum: mass $\times$ radius $\times$ (radius $\times$ sweep-rate). It stays constant, so when $r$ shrinks $\dot{θ}$ must spike.

Before we start, two pictures we will lean on again and again.

Figure — Conservation of energy and angular momentum in gravitational field

What to look at in figure s01: the black dot is the focus where the Sun/Earth sits — notice it is off-centre, not at the middle of the ellipse. Follow the short orange bar: that is $r_{p}$ , the shortest focus-to-orbit distance (perihelion). The long green bar is $r_{a}$ (aphelion). The gray double-arrow underneath shows the full long axis $2 a$ made of $r_{p} + r_{a}$ end to end. This is one ellipse, and

$r_{p} + r_{a} = 2 a, r_{p} = a (1 - e), r_{a} = a (1 + e),$

where $e$ is the eccentricity (0 = circle, closer to 1 = more stretched). At those two turning points the radial speed $\overset{r}{˙} = 0$ , so the entire velocity is sideways (tangential). That is why the turning points are the friendliest place to use $L$ : there $v = L / (m r)$ exactly.

What to look at in figure s02: it plots speed $v$ against radius $r$ for a single orbit. Trace the blue curve from left to right — it starts high at small $r$ (the orange perihelion dot, fastest) and sinks low at large $r$ (the green aphelion dot, slowest). That falling curve is the vis-viva equation made visible, which we derived in the parent by combining $E = \frac{1}{2} m v^{2} - \frac{GM m}{r}$ with $E = - \frac{GM m}{2 a}$ :

$v^{2} = GM (\frac{2}{r} - \frac{1}{a})$

Read it as a machine: feed in where you are ( $r$ ) and how big the orbit is ( $a$ ), out comes how fast you are moving ( $v$ ). We will use it constantly.

Level 1 — Recognition

Recall Solution

Gravity is central ( $F ∥ r$ ) and conservative. Central $\Rightarrow$ torque zero $\Rightarrow$ $L$ constant. Conservative $\Rightarrow$ $E = K E + U$ constant. (Recall $U = U (r) = - GM m / r$ , the potential energy defined above.)

(a) speed — changes (fast near Sun, slow far away).
(b) kinetic energy — changes (rises as it falls in).
(c) total energy $E$ — constant ✔
(d) angular momentum $L$ — constant ✔
(e) potential energy $U = - GM m / r$ — changes (more negative when close). Answer: only (c) and (d) are constant.

Recall Solution

Use the sign chart $E = - \frac{GM m}{2 a}$ :

$E < 0 \Rightarrow a > 0$ finite $\Rightarrow$ bound ellipse. So $E_{1}$ : ellipse.
$E = 0 \Rightarrow a \to \infty \Rightarrow$ parabola (just barely escapes, arrives at infinity with zero speed). So $E_{2}$ : parabola.
$E > 0 \Rightarrow a < 0 \Rightarrow$ hyperbola (escapes with speed left over). So $E_{3}$ : hyperbola.

Level 2 — Application

Recall Solution

For a circle, the orbit size is the radius: $a = r$ . Plug into vis-viva: $v^{2} = GM (\frac{2}{r} - \frac{1}{a}) = GM (\frac{2}{r} - \frac{1}{r}) = \frac{GM}{r} .$ The two terms partially cancel, leaving the clean circular result. $v = \frac{3.99 \times 1 0 ^{14}}{7.00 \times 1 0 ^{6}} = 5.70 \times 1 0^{7} \approx 7.55 \times 1 0^{3} m/s = 7.55 km/s .$ That's the familiar low-Earth-orbit speed. ✔

Recall Solution

"Just barely escape" means the parabolic case $E = 0$ , i.e. $a \to \infty$ so the $1/ a$ term in vis-viva vanishes. Set total energy to zero at radius $r$ : $0 = \frac{1}{2} m v_{esc}^{2} - \frac{GM m}{r} \Rightarrow v_{esc} = \frac{2 GM}{r} .$ $v_{esc} = \frac{2 ( 3.99 \times 1 0 ^{14} )}{1.00 \times 1 0 ^{7}} = 7.98 \times 1 0^{7} \approx 8.93 \times 1 0^{3} m/s = 8.93 km/s .$

Recall Solution

Compare with the escape speed just found, $v_{esc} = 8.93 km/s$ . Since $9.00 > 8.93$ , we have $v > v_{esc}$ , so $E > 0$ . It escapes on a hyperbolic path. Quick check via energy per kilogram: $ε = \frac{1}{2} v^{2} - \frac{GM}{r} = \frac{1}{2} (9000)^{2} - \frac{3.99 \times 1 0 ^{14}}{1 0 ^{7}} = 4.05 \times 1 0^{7} - 3.99 \times 1 0^{7} = + 6.0 \times 1 0^{5} J/kg > 0.$ Positive $\Rightarrow$ hyperbola. ✔

Level 3 — Analysis

Recall Solution

(a) The long axis is $2 a = r_{p} + r_{a}$ (see figure s01, the gray double-arrow): $a = \frac{r _{p} + r _{a}}{2} = \frac{6.60 \times 1 0 ^{6} + 1.32 \times 1 0 ^{7}}{2} = 9.90 \times 1 0^{6} m .$ (b) From $r_{p} = a (1 - e)$ : $e = 1 - \frac{r _{p}}{a} = 1 - \frac{6.60 \times 1 0 ^{6}}{9.90 \times 1 0 ^{6}} = 0.3333.$ (c) At perihelion, use vis-viva with $r = r_{p}$ : $v_{p}^{2} = GM (\frac{2}{r _{p}} - \frac{1}{a}) = 3.99 \times 1 0^{14} (\frac{2}{6.60 \times 1 0 ^{6}} - \frac{1}{9.90 \times 1 0 ^{6}}) .$ Inside: $\frac{2}{6.60 \times 1 0 ^{6}} = 3.030 \times 1 0^{- 7}$ , $\frac{1}{9.90 \times 1 0 ^{6}} = 1.010 \times 1 0^{- 7}$ , difference $= 2.020 \times 1 0^{- 7}$ . $v_{p}^{2} = 3.99 \times 1 0^{14} \times 2.020 \times 1 0^{- 7} = 8.06 \times 1 0^{7} \Rightarrow v_{p} = 8.98 \times 1 0^{3} m/s .$ (d) At aphelion, $r = r_{a}$ : $v_{a}^{2} = 3.99 \times 1 0^{14} (\frac{2}{1.32 \times 1 0 ^{7}} - \frac{1}{9.90 \times 1 0 ^{6}}) .$ Inside: $\frac{2}{1.32 \times 1 0 ^{7}} = 1.515 \times 1 0^{- 7}$ , minus $1.010 \times 1 0^{- 7}$ $= 5.05 \times 1 0^{- 8}$ . $v_{a}^{2} = 3.99 \times 1 0^{14} \times 5.05 \times 1 0^{- 8} = 2.015 \times 1 0^{7} \Rightarrow v_{a} = 4.49 \times 1 0^{3} m/s .$ (e) $r_{p} v_{p} = 6.60 \times 1 0^{6} \times 8.98 \times 1 0^{3} = 5.93 \times 1 0^{10}$ ; $r_{a} v_{a} = 1.32 \times 1 0^{7} \times 4.49 \times 1 0^{3} = 5.93 \times 1 0^{10}$ . Equal ✔ — that's $L / m$ conserved, since at the turning points $L = m r v$ .

Recall Solution

A minimum is where the slope is zero. Differentiate and set to zero — this is why we use the derivative: it hunts the flat spot of the tug-of-war curve. $\frac{d U _{eff}}{d r} = - \frac{L ^{2}}{m r ^{3}} + \frac{GM m}{r ^{2}} = 0 \Rightarrow \frac{GM m}{r ^{2}} = \frac{L ^{2}}{m r ^{3}} .$ Multiply both sides by $r^{3}$ : $GM m r = \frac{L ^{2}}{m}$ , so $r_{0} = \frac{L ^{2}}{GM m ^{2}} .$ At this radius the object has no radial push either way, so $\overset{r}{˙}$ stays zero forever: it is the circular orbit. Sitting at the bottom of the well means the orbit is stable — nudge $r$ slightly and it oscillates back and forth about $r_{0}$ . That small radial wobble corresponds to a slightly eccentric ellipse (low $e$ ); it is not the same as the full stretched ellipse of a large disturbance — only a tiny nudge stays near the bottom of the well. See figure s03.

What to look at in figure s03: three curves. The dashed red curve is pure gravity $- GM m / r$ (pulls you down and in). The dashed orange curve is the centrifugal barrier $L^{2} /2 m r^{2}$ (pushes you up at small $r$ ). Their sum is the solid blue $U_{eff}$ , which dips to a bowl. The green dot marks the bottom $r_{0}$ — the stable circular orbit. Read the picture as a tug-of-war: gravity wins at large $r$ , the barrier wins at small $r$ , and they tie at $r_{0}$ .

Level 4 — Synthesis

Recall Solution

(a) Energy fixes size: $E = - \frac{GM m}{2 a}$ , so $a = - \frac{GM m}{2 E} = - \frac{( 3.99 \times 1 0 ^{14} ) ( 1000 )}{2 ( - 2.00 \times 1 0 ^{9} )} = \frac{3.99 \times 1 0 ^{17}}{4.00 \times 1 0 ^{9}} = 9.98 \times 1 0^{7} m .$ (b) At a turning point $\overset{r}{˙} = 0$ , so all speed is tangential: $v = \frac{L}{m r}$ . Put this into the energy equation: $E = \frac{1}{2} m (\frac{L}{m r})^{2} - \frac{GM m}{r} = \frac{L ^{2}}{2 m r ^{2}} - \frac{GM m}{r} .$ Multiply through by $r^{2}$ and rearrange into a quadratic in $r$ : $E r^{2} + GM m r - \frac{L ^{2}}{2 m} = 0.$ Numbers: $E = - 2.00 \times 1 0^{9}$ ; $GM m = 3.99 \times 1 0^{17}$ ; $\frac{L ^{2}}{2 m} = \frac{( 4.00 \times 1 0 ^{13} ) ^{2}}{2000} = \frac{1.60 \times 1 0 ^{27}}{2000} = 8.00 \times 1 0^{23}$ . $- 2.00 \times 1 0^{9} r^{2} + 3.99 \times 1 0^{17} r - 8.00 \times 1 0^{23} = 0.$ Divide by $- 2.00 \times 1 0^{9}$ : $r^{2} - 1.995 \times 1 0^{8} r + 4.00 \times 1 0^{14} = 0.$ $r = \frac{1.995 \times 1 0 ^{8} \pm ( 1.995 \times 1 0 ^{8} ) ^{2} - 4 ( 4.00 \times 1 0 ^{14} )}{2} .$ Discriminant: $(1.995 \times 1 0^{8})^{2} = 3.980 \times 1 0^{16}$ ; minus $1.600 \times 1 0^{15}$ gives $3.820 \times 1 0^{16}$ ; square root $= 1.955 \times 1 0^{8}$ . $r = \frac{1.995 \times 1 0 ^{8} \pm 1.955 \times 1 0 ^{8}}{2} .$ The two roots are the two turning points: $r_{a} = \frac{3.950 \times 1 0 ^{8}}{2} = 1.975 \times 1 0^{8} m$ (aphelion) and $r_{p} = \frac{4.00 \times 1 0 ^{6}}{2} = 2.00 \times 1 0^{6} m$ (perihelion). The smaller root is perihelion: $r_{p} \approx 2.00 \times 1 0^{6} m$ . (Sanity check: $r_{p} + r_{a} = 2.00 \times 1 0^{6} + 1.975 \times 1 0^{8} = 1.995 \times 1 0^{8} = 2 a$ ✔.) (c) Perihelion speed: $v_{p} = \frac{L}{m r _{p}} = \frac{4.00 \times 1 0 ^{13}}{1000 \times 2.00 \times 1 0 ^{6}} = 2.00 \times 1 0^{4} m/s = 20.0 km/s .$

Recall Solution

Step 1 — circular speed at $r_{1}$ (initial): $a = r_{1}$ , $v_{circ} = \frac{GM}{r _{1}} = \frac{3.99 \times 1 0 ^{14}}{7.00 \times 1 0 ^{6}} = 7.55 \times 1 0^{3} m/s .$ Step 2 — transfer-orbit size: the transfer ellipse has $a_{t} = \frac{r _{1} + r _{2}}{2} = \frac{7.00 \times 1 0 ^{6} + 4.22 \times 1 0 ^{7}}{2} = 2.46 \times 1 0^{7} m .$ Step 3 — speed needed at perihelion $r_{1}$ of the transfer (vis-viva with $a = a_{t}$ ): $v_{trans}^{2} = GM (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = 3.99 \times 1 0^{14} (\frac{2}{7.00 \times 1 0 ^{6}} - \frac{1}{2.46 \times 1 0 ^{7}}) .$ Inside: $2.857 \times 1 0^{- 7} - 4.065 \times 1 0^{- 8} = 2.451 \times 1 0^{- 7}$ . So $v_{trans}^{2} = 9.78 \times 1 0^{7}$ , $v_{trans} = 9.89 \times 1 0^{3} m/s .$ Step 4 — the boost: both velocities are tangential at $r_{1}$ , so $Δ v = v_{trans} - v_{circ} = 9.89 \times 1 0^{3} - 7.55 \times 1 0^{3} = 2.34 \times 1 0^{3} m/s \approx 2.34 km/s .$ This is exactly the first burn of a Hohmann transfer.

Level 5 — Mastery

Recall Solution

Two expressions describe the same constant $E$ — so set them equal, which lets us eliminate $E$ entirely and keep only geometry: $\frac{1}{2} m v^{2} - \frac{GM m}{r} = - \frac{GM m}{2 a} .$ Step 1 — add $\frac{GM m}{r}$ to both sides to isolate the kinetic term on the left: $\frac{1}{2} m v^{2} = \frac{GM m}{r} - \frac{GM m}{2 a} .$ Step 2 — every term contains the factor $m$ ; divide it out (the trajectory doesn't care about the orbiting mass): $\frac{1}{2} v^{2} = \frac{GM}{r} - \frac{GM}{2 a} = GM (\frac{1}{r} - \frac{1}{2 a}) .$ Step 3 — multiply both sides by $2$ to free $v^{2}$ : $v^{2} = 2 GM (\frac{1}{r} - \frac{1}{2 a}) = GM (\frac{2}{r} - \frac{1}{a}) . ■$ The logic in one breath: energy conservation gives the first equation at any point; the only- $a$ energy law gives the second; equating them trades the invisible constant $E$ for the visible $a$ .

Recall Solution

At perihelion and aphelion $\overset{r}{˙} = 0$ , so all velocity is tangential and $v = \frac{L}{m r}$ . Energy at each turning point: $E = \frac{L ^{2}}{2 m r ^{2}} - \frac{GM m}{r} .$ This must give the same $E$ at $r_{p}$ and $r_{a}$ (energy is conserved), so they are the two roots of $E r^{2} + GM m r - \frac{L ^{2}}{2 m} = 0 (multiply the line above by r^{2}) .$ For a quadratic $E r^{2} + GM m r - \frac{L ^{2}}{2 m} = 0$ , the sum of roots is $r_{p} + r_{a} = - \frac{GM m}{E}$ (from $- b / a$ of $a r^{2} + b r + c$ ). But geometrically $r_{p} + r_{a} = 2 a$ . Equate: $2 a = - \frac{GM m}{E} \Rightarrow E = - \frac{GM m}{2 a} .$ Notice: $L$ appeared only in the product of roots, and $e$ never entered the sum at all. Both cancelled — $E$ depends on $a$ and nothing else. Two orbits with the same $a$ but wildly different $e$ share the same energy.

Recall Solution

(a) Circular speed: $v_{circ} = GM / r = 7.55 \times 1 0^{3} m/s$ (from L2·Q1). Escape speed at the same radius: $v_{esc} = 2 GM / r$ . Compute: $2 GM / r = 2 \cdot 3.99 \times 1 0^{14} /7.00 \times 1 0^{6} = 1.140 \times 1 0^{8}$ , so $v_{esc} = 1.140 \times 1 0^{8} = 1.068 \times 1 0^{4} m/s .$ $Δ v = v_{esc} - v_{circ} = 1.068 \times 1 0^{4} - 7.55 \times 1 0^{3} = 3.13 \times 1 0^{3} m/s \approx 3.13 km/s .$ (b) Algebraically: $v_{esc} = \frac{2 GM}{r} = 2 \cdot \frac{GM}{r} = 2 v_{circ} .$ Therefore $Δ v = v_{esc} - v_{circ} = (2 - 1) v_{circ} \approx 0.414 v_{circ} .$ Check: $0.414 \times 7.55 \times 1 0^{3} = 3.13 \times 1 0^{3} m/s$ ✔. So escaping a circular orbit always costs about 41.4% more speed than you already have — a beautiful mass- and planet-independent constant.

Recall Solution

(a) With $L = 0$ there is no tangential motion at all — the object falls straight down along a line through Earth's centre. This is the degenerate ellipse of eccentricity $e \to 1$ : a needle-thin "orbit" squashed flat onto a radial line. Its turning point (aphelion) is the drop radius $r_{0}$ , and the far focus and the centre coincide with the ends of the line. The semi-major axis is half the fall distance from centre to $r_{0}$ : since the plunge goes from $r_{0}$ down toward $r = 0$ , $a = r_{0} /2 = 1.00 \times 1 0^{7} m$ . (b) Even though $L = 0$ , energy conservation (hence vis-viva) still holds — vis-viva never used $L$ , it used $a$ . Apply it with $a = 1.00 \times 1 0^{7}$ at $r = 7.00 \times 1 0^{6}$ : $v^{2} = GM (\frac{2}{r} - \frac{1}{a}) = 3.99 \times 1 0^{14} (\frac{2}{7.00 \times 1 0 ^{6}} - \frac{1}{1.00 \times 1 0 ^{7}}) .$ Inside: $2.857 \times 1 0^{- 7} - 1.000 \times 1 0^{- 7} = 1.857 \times 1 0^{- 7}$ . So $v^{2} = 3.99 \times 1 0^{14} \times 1.857 \times 1 0^{- 7} = 7.41 \times 1 0^{7}$ , giving $v = 8.61 \times 1 0^{3} m/s \approx 8.61 km/s .$ (Cross-check by raw energy drop: $\frac{1}{2} v^{2} = GM (1/ r - 1/ r_{0}) = 3.99 \times 1 0^{14} (1.429 \times 1 0^{- 7} - 5.00 \times 1 0^{- 8}) = 3.71 \times 1 0^{7}$ , so $v^{2} = 7.41 \times 1 0^{7}$ ✔ — the two methods agree.)

Recall Self-test wrap-up

Without looking, can you: (1) name which two quantities are conserved and why; (2) compute circular and escape speed from vis-viva; (3) turn $(E, L)$ into $(a, r_{p}, v_{p})$ ; (4) derive vis-viva and $E = - GM m /2 a$ ; (5) state the mass-free escape ratio $(2 - 1)$ ; (6) handle the $L = 0$ radial-plunge limit? If any is shaky, revisit that level.