Intuition The one-line ideaEvery orbit has a conserved quantity called specific angular momentum h h h per unit mass ). It encodes how "wide" the orbit is. The relation h = G M p h=\sqrt{GMp} h = GM p semi-latus rectum p p p measure the shape (p p p h h h
Definition Specific angular momentumFor a body of mass m m m M M M L ⃗ = r ⃗ × m v ⃗ \vec L = \vec r \times m\vec v L = r × m v m m m h ⃗ ≡ r ⃗ × v ⃗ , h = ∣ h ⃗ ∣ \vec h \equiv \vec r \times \vec v, \qquad h=|\vec h| h ≡ r × v , h = ∣ h ∣ m 2 / s \mathrm{m^2/s} m 2 /s not k g m 2 / s \mathrm{kg\,m^2/s} kg m 2 /s
WHY divide by m m m Because in the two-body problem the small mass cancels from the equation of motion (r ⃗ ¨ = − G M r 2 r ^ \ddot{\vec r} = -\frac{GM}{r^2}\hat r r ¨ = − r 2 GM r ^ r ⃗ , v ⃗ \vec r,\vec v r , v natural conserved quantities are also per-mass.
Intuition No torque ⟹ constanth ⃗ \vec h h r ⃗ \vec r r τ ⃗ = r ⃗ × F ⃗ \vec\tau=\vec r\times\vec F τ = r × F r ⃗ × ( something ∥ r ⃗ ) = 0 \vec r\times(\text{something}\parallel\vec r)=0 r × ( something ∥ r ) = 0
Derivation from scratch.
d h ⃗ d t = d d t ( r ⃗ × v ⃗ ) = v ⃗ × v ⃗ ⏟ = 0 + r ⃗ × a ⃗ \frac{d\vec h}{dt}=\frac{d}{dt}(\vec r\times\vec v)=\underbrace{\vec v\times\vec v}_{=0}+\vec r\times\vec a d t d h = d t d ( r × v ) = = 0 v × v + r × a v ⃗ × v ⃗ = 0 \vec v\times\vec v=0 v × v = 0
The acceleration is purely radial: a ⃗ = − G M r 2 r ^ \vec a=-\frac{GM}{r^2}\hat r a = − r 2 GM r ^ r ⃗ × a ⃗ = − G M r 2 ( r ⃗ × r ^ ) = 0 \vec r\times\vec a = -\frac{GM}{r^2}\,(\vec r\times\hat r)=0 r × a = − r 2 GM ( r × r ^ ) = 0 r ⃗ = r r ^ \vec r = r\hat r r = r r ^ r ^ × r ^ = 0 \hat r\times\hat r=0 r ^ × r ^ = 0 d h ⃗ d t = 0 ⇒ h ⃗ = const. \boxed{\dfrac{d\vec h}{dt}=0}\quad\Rightarrow\quad \vec h=\text{const.} d t d h = 0 ⇒ h = const.
Constant h ⃗ \vec h h fixed plane (motion stays perpendicular to h ⃗ \vec h h Kepler's 2nd law holds (d A d t = h 2 \frac{dA}{dt}=\frac h2 d t d A = 2 h
We need the orbit equation . Start from the conic-section solution of the two-body problem (derived once below).
Use polar coordinates. The radial equation of motion is
r ¨ − r θ ˙ 2 = − G M r 2 \ddot r - r\dot\theta^2 = -\frac{GM}{r^2} r ¨ − r θ ˙ 2 = − r 2 GM
Angular momentum gives h = r 2 θ ˙ h=r^2\dot\theta h = r 2 θ ˙ z z z r ⃗ × v ⃗ \vec r\times\vec v r × v θ ˙ = h / r 2 \dot\theta = h/r^2 θ ˙ = h / r 2
Trick: substitute u = 1 / r u=1/r u = 1/ r θ \theta θ r ˙ = d r d θ θ ˙ = − 1 u 2 d u d θ ⋅ h u 2 = − h d u d θ \dot r = \frac{dr}{d\theta}\dot\theta = -\frac{1}{u^2}\frac{du}{d\theta}\cdot hu^2 = -h\frac{du}{d\theta} r ˙ = d θ d r θ ˙ = − u 2 1 d θ d u ⋅ h u 2 = − h d θ d u r = 1 / u ⇒ d r d θ = − u − 2 d u d θ r=1/u\Rightarrow \frac{dr}{d\theta}=-u^{-2}\frac{du}{d\theta} r = 1/ u ⇒ d θ d r = − u − 2 d θ d u θ ˙ = h u 2 \dot\theta=hu^2 θ ˙ = h u 2
Differentiate again:
r ¨ = − h d 2 u d θ 2 θ ˙ = − h 2 u 2 d 2 u d θ 2 \ddot r = -h\frac{d^2u}{d\theta^2}\dot\theta = -h^2u^2\frac{d^2u}{d\theta^2} r ¨ = − h d θ 2 d 2 u θ ˙ = − h 2 u 2 d θ 2 d 2 u
Substitute into the radial equation (with r θ ˙ 2 = 1 u ( h u 2 ) 2 = h 2 u 3 r\dot\theta^2 = \frac1u(hu^2)^2 = h^2u^3 r θ ˙ 2 = u 1 ( h u 2 ) 2 = h 2 u 3 − h 2 u 2 u ′ ′ − h 2 u 3 = − G M u 2 -h^2u^2 u'' - h^2u^3 = -GM\,u^2 − h 2 u 2 u ′′ − h 2 u 3 = − GM u 2 − h 2 u 2 -h^2u^2 − h 2 u 2 u ′ ′ + u = G M h 2 u''+u = \frac{GM}{h^2} u ′′ + u = h 2 GM
Its solution: particular u p = G M / h 2 u_p=GM/h^2 u p = GM / h 2 A cos ( θ − θ 0 ) A\cos(\theta-\theta_0) A cos ( θ − θ 0 ) u = G M h 2 ( 1 + e cos θ ) ⇒ r = h 2 / G M 1 + e cos θ u=\frac{GM}{h^2}\big(1+e\cos\theta\big)\;\Rightarrow\; r=\frac{h^2/GM}{1+e\cos\theta} u = h 2 GM ( 1 + e cos θ ) ⇒ r = 1 + e c o s θ h 2 / GM
Comparing with r = p 1 + e cos θ r=\dfrac{p}{1+e\cos\theta} r = 1 + e cos θ p p = h 2 G M ⟹ h = G M p p=\frac{h^2}{GM}\quad\Longrightarrow\quad \boxed{h=\sqrt{GM\,p}} p = GM h 2 ⟹ h = GM p
Intuition What IS the semi-latus rectum?Set θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ cos 90 ∘ = 0 \cos90^\circ=0 cos 9 0 ∘ = 0 r = p r=p r = p p p p perpendicular to the major axis from the focus — the "half-width" of the ellipse at the focus. Bigger p p p h h h
Key checkpoints (all from r = p 1 + e cos θ r=\frac{p}{1+e\cos\theta} r = 1 + e c o s θ p
Periapsis (θ = 0 \theta=0 θ = 0 r p = p 1 + e r_p=\dfrac{p}{1+e} r p = 1 + e p
Apoapsis (θ = 180 ∘ \theta=180^\circ θ = 18 0 ∘ r a = p 1 − e r_a=\dfrac{p}{1-e} r a = 1 − e p
Harmonic-mean relation: 2 p = 1 r p + 1 r a \dfrac{2}{p}=\dfrac1{r_p}+\dfrac1{r_a} p 2 = r p 1 + r a 1
Worked example 1 — Circular Low-Earth orbitG M ⊕ = 3.986 × 10 14 m 3 / s 2 GM_\oplus=3.986\times10^{14}\,\mathrm{m^3/s^2} G M ⊕ = 3.986 × 1 0 14 m 3 / s 2 r = 7000 k m = 7 × 10 6 m r=7000\,\mathrm{km}=7\times10^6\,\mathrm m r = 7000 km = 7 × 1 0 6 m p = r p=r p = r h = G M p = 3.986 × 10 14 × 7 × 10 6 = 2.79 × 10 21 ≈ 5.28 × 10 10 m 2 / s h=\sqrt{GM\,p}=\sqrt{3.986\times10^{14}\times7\times10^6}=\sqrt{2.79\times10^{21}}\approx5.28\times10^{10}\,\mathrm{m^2/s} h = GM p = 3.986 × 1 0 14 × 7 × 1 0 6 = 2.79 × 1 0 21 ≈ 5.28 × 1 0 10 m 2 /s Why this step? Circle ⟹ e = 0 e=0 e = 0 p = a = r p=a=r p = a = r v = G M / r = 7547 m / s v=\sqrt{GM/r}=7547\,\mathrm{m/s} v = GM / r = 7547 m/s h = v r = 7547 × 7 × 10 6 = 5.28 × 10 10 h=vr=7547\times7\times10^6=5.28\times10^{10} h = v r = 7547 × 7 × 1 0 6 = 5.28 × 1 0 10
Worked example 2 — From periapsis speed and radiusA probe at periapsis has r p = 8000 k m r_p=8000\,\mathrm{km} r p = 8000 km v p = 9.0 k m / s v_p=9.0\,\mathrm{km/s} v p = 9.0 km/s h = r p v p h=r_p v_p h = r p v p h = 8 × 10 6 × 9000 = 7.2 × 10 10 m 2 / s h=8\times10^6\times9000=7.2\times10^{10}\,\mathrm{m^2/s} h = 8 × 1 0 6 × 9000 = 7.2 × 1 0 10 m 2 /s p = h 2 G M = ( 7.2 × 10 10 ) 2 3.986 × 10 14 = 1.30 × 10 7 m = 13,000 k m p=\dfrac{h^2}{GM}=\dfrac{(7.2\times10^{10})^2}{3.986\times10^{14}}=1.30\times10^7\,\mathrm m=13{,}000\,\mathrm{km} p = GM h 2 = 3.986 × 1 0 14 ( 7.2 × 1 0 10 ) 2 = 1.30 × 1 0 7 m = 13 , 000 km Why this step? v ⃗ ⊥ r ⃗ \vec v\perp\vec r v ⊥ r ∣ r ⃗ × v ⃗ ∣ = r v |\vec r\times\vec v|=rv ∣ r × v ∣ = r v sin θ \sin\theta sin θ
Worked example 3 — Find eccentricitySame probe: from r p = p / ( 1 + e ) r_p=p/(1+e) r p = p / ( 1 + e ) e = p r p − 1 = 13000 8000 − 1 = 0.625 \;e=\frac{p}{r_p}-1=\frac{13000}{8000}-1=0.625 e = r p p − 1 = 8000 13000 − 1 = 0.625 Why this step? Use the periapsis form of the orbit equation, now that we know p p p
h h h Why it feels right: ordinary angular momentum L = m v r L=mvr L = m v r Fix: h h h specific (per unit mass), h = L / m h=L/m h = L / m m 2 / s \mathrm{m^2/s} m 2 /s
h = r v h=rv h = r v Why it feels right: it's true at peri/apoapsis so it looks general. Fix: generally h = r v sin ϕ h=rv\sin\phi h = r v sin ϕ ϕ \phi ϕ r ⃗ \vec r r v ⃗ \vec v v ϕ = 90 ∘ \phi=90^\circ ϕ = 9 0 ∘ sin ϕ = 1 \sin\phi=1 sin ϕ = 1
p = a p=a p = a Why it feels right: true for circles, and a a a Fix: p = a ( 1 − e 2 ) p=a(1-e^2) p = a ( 1 − e 2 ) a a a e = 0 e=0 e = 0
h 2 = G M p h^2=GMp h 2 = GM p h = G M p / 2 h=GMp/2 h = GM p /2 Why it feels right: sloppy algebra. Fix: h = G M p h=\sqrt{GMp} h = GM p
Recall Feynman: explain to a 12-year-oldTie a ball to a string and whirl it. The "spin amount" of the ball stays the same as long as you don't push it sideways — gravity is like a string that only pulls toward the centre, never sideways, so the planet keeps the same spin amount forever. That spin amount is h h h p p p G M GM GM
"He Got My Pull" → h = G M p h=\sqrt{GM\,p} h = GM p He(h) Got(√) My(GM) Pull(p).
And for the geometry: "p is the perpendicular peek" — p p p 90 ∘ 90^\circ 9 0 ∘
Why is h ⃗ \vec h h
What are the units of h h h
At which orbital points does h = r v h=rv h = r v
Derive p = h 2 / G M p=h^2/GM p = h 2 / GM
What is specific angular momentum h ⃗ \vec h h h ⃗ = r ⃗ × v ⃗ \vec h=\vec r\times\vec v h = r × v , angular momentum per unit mass, units
m 2 / s \mathrm{m^2/s} m 2 /s .
Why is h ⃗ \vec h h Gravity is central (parallel to
r ⃗ \vec r r ), so torque
r ⃗ × F ⃗ = 0 \vec r\times\vec F=0 r × F = 0 , hence
d h ⃗ / d t = 0 d\vec h/dt=0 d h / d t = 0 .
State h h h p p p h = G M p h=\sqrt{GM\,p} h = GM p , where
p p p is the semi-latus rectum.
Express p p p a a a e e e p = a ( 1 − e 2 ) p=a(1-e^2) p = a ( 1 − e 2 ) , so
h = G M a ( 1 − e 2 ) h=\sqrt{GM\,a(1-e^2)} h = GM a ( 1 − e 2 ) .
What is the semi-latus rectum geometrically? Orbital radius at true anomaly
90 ∘ 90^\circ 9 0 ∘ :
r = p r=p r = p when
cos θ = 0 \cos\theta=0 cos θ = 0 .
When does h = r v h=rv h = r v At periapsis and apoapsis, where
v ⃗ ⊥ r ⃗ \vec v\perp\vec r v ⊥ r (flight-path angle
= 90 ∘ =90^\circ = 9 0 ∘ ).
General relation between h h h r r r v v v h = r v sin ϕ h=rv\sin\phi h = r v sin ϕ with
ϕ \phi ϕ the angle between
r ⃗ \vec r r and
v ⃗ \vec v v .
h h h r r r h = G M r = v c i r c r h=\sqrt{GMr}=v_{circ}\,r h = GM r = v c i r c r , since
p = a = r p=a=r p = a = r .
What ODE does u = 1 / r u=1/r u = 1/ r Binet's equation
u ′ ′ + u = G M / h 2 u''+u=GM/h^2 u ′′ + u = GM / h 2 .
Kepler's 2nd law via h h h d A / d t = h / 2 = dA/dt=h/2= d A / d t = h /2 = const (equal areas in equal times).
Two-body problem — where the orbit equation r = p / ( 1 + e cos θ ) r=p/(1+e\cos\theta) r = p / ( 1 + e cos θ ) Kepler's Second Law — d A / d t = h / 2 dA/dt=h/2 d A / d t = h /2 h h h Specific orbital energy — pairs with h h h a a a e e e Eccentricity vector — e ⃗ = v ⃗ × h ⃗ G M − r ^ \vec e=\frac{\vec v\times\vec h}{GM}-\hat r e = GM v × h − r ^ h ⃗ \vec h h Flight-path angle — explains the sin ϕ \sin\phi sin ϕ h = r v sin ϕ h=rv\sin\phi h = r v sin ϕ Vis-viva equation — v 2 = G M ( 2 / r − 1 / a ) v^2=GM(2/r-1/a) v 2 = GM ( 2/ r − 1/ a ) h h h
Angular momentum L = r x m v
Specific ang. mom. h = r x v
Kepler 2nd law dA/dt = h/2
Orbit equation r = p / 1+e cos theta
Intuition Hinglish mein samjhoDekho, specific angular momentum h h h h ⃗ = r ⃗ × v ⃗ \vec h=\vec r\times\vec v h = r × v m 2 / s \mathrm{m^2/s} m 2 /s
h h h r ⃗ × F ⃗ = 0 \vec r\times\vec F=0 r × F = 0 d h ⃗ / d t = 0 d\vec h/dt=0 d h / d t = 0 d A / d t = h / 2 dA/dt=h/2 d A / d t = h /2
Ab asli formula: orbit equation r = p / ( 1 + e cos θ ) r=p/(1+e\cos\theta) r = p / ( 1 + e cos θ ) p = h 2 / G M p=h^2/GM p = h 2 / GM p p p θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ h = G M p h=\sqrt{GMp} h = GM p p = r p=r p = r h = G M r = v r h=\sqrt{GMr}=v\,r h = GM r = v r
Practical baat: peri-apsis aur apoapsis par velocity radius ke perpendicular hoti hai, toh wahan h = r v h=rv h = r v h = r v sin ϕ h=rv\sin\phi h = r v sin ϕ p p p a a a p = a ( 1 − e 2 ) p=a(1-e^2) p = a ( 1 − e 2 ) p p p h h h
