Exercises — Specific angular momentum h = √(GMp)

WHAT: circle means $e=0$, so $p=a=r$. WHY: the semi-latus rectum collapses to the radius when the orbit is a perfect circle. $h=\sqrt{GMp}=\sqrt{GMr}=\sqrt{3.986\times 10^{14}\times 6.6\times 10^6}$ $=\sqrt{2.631\times 10^{21}}\approx 5.13\times 10^{10}{\mathrm{m}}^2/s.$

Units are ${\mathrm{m}}^2/s$. Because $h$ is angular momentum per unit mass ($h=L/m$); we deliberately divided the orbiting body's mass out, so the $\mathrm{kg}$ cancels.

Direct plug-in, no shape info needed beyond $p$ — that is the whole power of the formula. $h=\sqrt{3.986\times 10^{14}\times 1.30\times 10^7}=\sqrt{5.182\times 10^{21}}\approx 7.20\times 10^{10}{\mathrm{m}}^2/s.$

WHAT: at periapsis the velocity is purely tangential, so $\vec{v}\perp \vec{r}$ and $\varphi =90^{\circ }$. WHY: then $\sin \varphi =1$, so $h=r_p{v}_p$ with no angle loss. $h=r_p{v}_p=8.0\times 10^6\times 9.0\times 10^3=7.20\times 10^{10}{\mathrm{m}}^2/s.$ Invert $h=\sqrt{GMp}$: $p=\frac{h^2}{GM}=\frac{(7.20\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{5.184\times 10^{21}}{3.986\times 10^{14}}\approx 1.30\times 10^7\mathrm{m}.$

Use $p=a(1-e^2)$ first — the ellipse's half-width, not $a$ itself. $p=10^7\times (1-0.09)=9.10\times 10^6\mathrm{m}.$ $h=\sqrt{GMp}=\sqrt{3.986\times 10^{14}\times 9.10\times 10^6}=\sqrt{3.627\times 10^{21}}\approx 6.02\times 10^{10}{\mathrm{m}}^2/s.$

Rearrange the orbit equation $r=\frac{p}{1+e\cos \theta }$ for $p$: $p=r(1+e\cos \theta )=9.0\times 10^6(1+0.20\times \cos 60^{\circ }).$ Since $\cos 60^{\circ }=0.5$: $p=9.0\times 10^6\times (1+0.10)=9.90\times 10^6\mathrm{m}.$ $h=\sqrt{3.986\times 10^{14}\times 9.90\times 10^6}=\sqrt{3.946\times 10^{21}}\approx 6.28\times 10^{10}{\mathrm{m}}^2/s.$

Step 1 (WHAT): use the harmonic-mean relation $\frac{2}{p}=\frac{1}{r_p}+\frac{1}{r_a}$. WHY: it lets us get $p$ straight from the two apsis radii without first finding $a$ or $e$. $\frac{2}{p}=\frac{1}{7.0\times 10^6}+\frac{1}{2.1\times 10^7}=1.4286\times 10^{-7}+4.762\times 10^{-8}=1.9048\times 10^{-7}.$ $p=\frac{2}{1.9048\times 10^{-7}}=1.050\times 10^7\mathrm{m}.$ Step 2: eccentricity from $r_p=\frac{p}{1+e}\Rightarrow e=\frac{p}{r_p}-1$: $e=\frac{1.050\times 10^7}{7.0\times 10^6}-1=1.500-1=\mathrm{0.500.}$ Step 3: $h=\sqrt{GMp}=\sqrt{3.986\times 10^{14}\times 1.050\times 10^7}=\sqrt{4.185\times 10^{21}}\approx 6.47\times 10^{10}{\mathrm{m}}^2/s.$

At both apsides $\vec{v}\perp \vec{r}$, so $h=rv$ and $v=h/r$. $v_p=\frac{h}{r_p}=\frac{6.47\times 10^{10}}{7.0\times 10^6}\approx 9.24\times 10^{3}m/s.$ $v_a=\frac{h}{r_a}=\frac{6.47\times 10^{10}}{2.1\times 10^7}\approx 3.08\times 10^{3}m/s.$ Sanity: $v_p/v_a=r_a/r_p=3.0$ — the body moves faster where the orbit is tighter, exactly as constant $h$ demands.

WHAT: here $\vec{v}$ is not perpendicular to $\vec{r}$, so we must use the full $h=rv\sin \varphi$. WHY: the cross product $|\vec{r}\times \vec{v}|=rv\sin \varphi$ only picks out the component of $\vec{v}$ perpendicular to $\vec{r}$. See Flight-path angle. $h=rv\sin \varphi =10^7\times 6000\times \sin 30^{\circ }=10^7\times 6000\times 0.5=3.0\times 10^{10}{\mathrm{m}}^2/s.$ $p=\frac{h^2}{GM}=\frac{(3.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{9.0\times 10^{20}}{3.986\times 10^{14}}\approx 2.258\times 10^6\mathrm{m}.$

Step 1: get $p$ from $h$: $p=\frac{h^2}{GM}=\frac{(6.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{3.6\times 10^{21}}{3.986\times 10^{14}}=9.031\times 10^6\mathrm{m}.$ Step 2: invert $p=a(1-e^2)$ for $e$: $1-e^2=\frac{p}{a}=\frac{9.031\times 10^6}{1.2\times 10^7}=0.7526\Rightarrow e^2=0.2474\Rightarrow e\approx \mathrm{0.497.}$ Step 3: $r_p=a(1-e)=1.2\times 10^7\times (1-0.497)=6.03\times 10^6\mathrm{m};r_a=a(1+e)=1.2\times 10^7\times 1.497=1.797\times 10^7\mathrm{m}.$

WHAT: Kepler's second law says the radius vector sweeps equal areas in equal times, at rate $\frac{dA}{dt}=\frac{h}{2}$. See Kepler's Second Law. $\frac{dA}{dt}=\frac{h}{2}=\frac{6.0\times 10^{10}}{2}=3.0\times 10^{10}{\mathrm{m}}^2/s.$ Total area: $b=a\sqrt{1-e^2}=1.2\times 10^7\times \sqrt{0.7526}=1.2\times 10^7\times 0.8675=1.041\times 10^7\mathrm{m}.$ $A=\pi ab=\pi \times 1.2\times 10^7\times 1.041\times 10^7=3.925\times 10^{14}{\mathrm{m}}^2.$ Quarter area: $A/4=9.812\times 10^{13}{\mathrm{m}}^2.$ $t=\frac{A/4}{dA/dt}=\frac{9.812\times 10^{13}}{3.0\times 10^{10}}\approx 3.27\times 10^3\mathrm{s}(\approx 54.5\text{min}).$

Circular speed comes from balancing gravity and the centripetal requirement: $\frac{GM}{r^2}=\frac{v^2}{r}\Rightarrow v_{\text{circ}}=\sqrt{GM/r}$. Then $v_{\text{circ}}r=\sqrt{GM/r}\cdot r=\sqrt{GMr}=h$. Same thing. ✓ Numbers: $v_{\text{circ}}=\sqrt{3.986\times 10^{14}/7.0\times 10^6}=\sqrt{5.694\times 10^7}=7546m/s.$ $h=v_{\text{circ}}r=7546\times 7.0\times 10^6=5.28\times 10^{10}{\mathrm{m}}^2/s$, matching $\sqrt{GMr}$.

Step 1: at periapsis $\vec{v}\perp \vec{r}$, so $h=r_p{v}_p=1.5\times 10^{11}\times 5.0\times 10^4=7.5\times 10^{15}{\mathrm{m}}^2/s.$ Step 2: $p=\frac{h^2}{G{M}_{\odot }}=\frac{(7.5\times 10^{15}{)}^2}{1.327\times 10^{20}}=\frac{5.625\times 10^{31}}{1.327\times 10^{20}}\approx 4.239\times 10^{11}\mathrm{m}.$ Step 3: from $r_p=\frac{p}{1+e}$, $e=\frac{p}{r_p}-1=\frac{4.239\times 10^{11}}{1.5\times 10^{11}}-1=2.826-1=\mathrm{1.826.}$ Since $e>1$, the orbit is indeed hyperbolic — the comet is not bound and will escape.

Same $h$ ⟹ same $p=\frac{h^2}{GM}=\frac{(5.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{2.5\times 10^{21}}{3.986\times 10^{14}}=6.272\times 10^6\mathrm{m}.$ Circle ($e=0$): $a_1=p=6.272\times 10^6\mathrm{m}.$ Ellipse ($e=0.6$): $a_2=\frac{p}{1-e^2}=\frac{6.272\times 10^6}{1-0.36}=\frac{6.272\times 10^6}{0.64}=9.800\times 10^6\mathrm{m}.$ The ellipse has the larger $a$. Key insight: equal $h$ fixes the half-width $p$, not the length $a$. Squeezing eccentricity in (fixed $p$) stretches the orbit outward along the major axis, so $a$ grows as $1/(1-e^2)$.

At periapsis $h=r_{p}v$ (since $\vec{v}\perp \vec{r}$). With $r_p$ fixed, $h$ is directly proportional to $v$, so $v_2>v_1\Rightarrow h_2>h_1$. Then $p=\frac{h^2}{GM}$ increases with $h$, so $p_2>p_1$. Because $h$ and $p$ rise together (that is exactly $h=\sqrt{GMp}$), a faster shove at the same closest point produces a wider, higher-energy orbit. This is the essence of why a prograde burn at periapsis raises the far side of the orbit — it pumps up $h$ and hence $p$. Links to Vis-viva equation and Specific orbital energy.