Intuition The big pictureAn orbit is a fixed ellipse floating in 3D space , with a satellite moving along it. To pin it down completely you need to answer three questions :
What is the shape & size of the ellipse? → a a a e e e How is the ellipse tilted/oriented in space? → i i i Ω \Omega Ω ω \omega ω Where on the ellipse is the satellite right now? → ν \nu ν
That's exactly 6 numbers , because a 3D position + 3D velocity is also 6 numbers. The Keplerian elements are just a geometrically meaningful repackaging of ( r ⃗ , v ⃗ ) (\vec r,\vec v) ( r , v )
Intuition Degrees of freedomA satellite's full state at one instant is position r ⃗ = ( x , y , z ) \vec r=(x,y,z) r = ( x , y , z ) v ⃗ = ( v x , v y , v z ) \vec v=(v_x,v_y,v_z) v = ( v x , v y , v z ) 6 numbers . Newton's equation r ⃗ ¨ = − μ r ⃗ / r 3 \ddot{\vec r}=-\mu\vec r/r^3 r ¨ = − μ r / r 3 exactly 6 independent quantities . Keplerian elements are the choice where 5 are constant (the orbit doesn't change shape/orientation under ideal two-body gravity) and only 1 (ν \nu ν . That's why they're loved: they separate the unchanging geometry from the moving point .
Definition Semi-major axisa a a half of the longest diameter of the orbital ellipse. It sets the size of the orbit and, through the vis-viva relation, the total energy .
Intuition WHY energy fixesa a a 1 / r 2 1/r^2 1/ r 2 size of the orbit is locked — no matter where the satellite sits.
HOW. Specific orbital energy (energy per unit mass):
ε = v 2 2 − μ r . \varepsilon = \frac{v^2}{2}-\frac{\mu}{r}. ε = 2 v 2 − r μ . h = r v h=rv h = r v v = h / r v=h/r v = h / r ε \varepsilon ε h 2 2 r p 2 − μ r p = h 2 2 r a 2 − μ r a . \frac{h^2}{2r_p^2}-\frac{\mu}{r_p}=\frac{h^2}{2r_a^2}-\frac{\mu}{r_a}. 2 r p 2 h 2 − r p μ = 2 r a 2 h 2 − r a μ . r p = a ( 1 − e ) , r a = a ( 1 + e ) , r p + r a = 2 a r_p=a(1-e),\;r_a=a(1+e),\;r_p+r_a=2a r p = a ( 1 − e ) , r a = a ( 1 + e ) , r p + r a = 2 a ε = − μ 2 a \boxed{\varepsilon=-\frac{\mu}{2a}} ε = − 2 a μ ε = v 2 / 2 − μ / r \varepsilon=v^2/2-\mu/r ε = v 2 /2 − μ / r vis-viva equation :
v 2 = μ ( 2 r − 1 a ) \boxed{v^2=\mu\left(\frac{2}{r}-\frac{1}{a}\right)} v 2 = μ ( r 2 − a 1 )
We work in an Earth-Centered Inertial (ECI) frame: X X X Z Z Z X X X Y Y Y
i i i tilt of the orbital plane relative to the equatorial plane: the angle between the orbit's angular-momentum vector h ⃗ \vec h h Z Z Z i = 0 i=0 i = 0 i = 90 ∘ i=90^\circ i = 9 0 ∘ i > 90 ∘ i>90^\circ i > 9 0 ∘
Ω \Omega Ω X X X ascending node is where the satellite crosses the equator going south→north . RAAN says which way the tilted plane is swung around .
Definition Argument of perigeeω \omega ω angle, measured in the orbital plane, from the ascending node to perigee . It tells you where in the plane the ellipse points (where the closest approach sits).
ν \nu ν angle from perigee to the satellite's current position , measured at the focus (Earth's center), in the direction of motion. This is the only element that changes with time .
Intuition WHY a rotation matrixIn the orbit's own "perifocal" frame (P toward perigee, Q 90 ∘ 90^\circ 9 0 ∘ rotate that frame into ECI using the three angles — that's the geometric meaning of Ω , i , ω \Omega,i,\omega Ω , i , ω
Perifocal position (P–Q–W axes):
r ⃗ P Q W = r ( cos ν sin ν 0 ) , r = a ( 1 − e 2 ) 1 + e cos ν . \vec r_{PQW}=r\begin{pmatrix}\cos\nu\\ \sin\nu\\ 0\end{pmatrix},\qquad r=\frac{a(1-e^2)}{1+e\cos\nu}. r P Q W = r cos ν sin ν 0 , r = 1 + e c o s ν a ( 1 − e 2 ) . Transform to ECI with three elementary rotations:
r ⃗ E C I = R z ( Ω ) R x ( i ) R z ( ω ) r ⃗ P Q W . \vec r_{ECI}=R_z(\Omega)\,R_x(i)\,R_z(\omega)\,\vec r_{PQW}. r E C I = R z ( Ω ) R x ( i ) R z ( ω ) r P Q W . R z ( θ ) R_z(\theta) R z ( θ ) Z Z Z R x ( i ) R_x(i) R x ( i ) is the operational definition of all three orientation angles.
a a a e e e 400 400 400 1600 1600 1600 R E = 6378 R_E=6378 R E = 6378 a a a e e e
Step 1: Convert to radii from Earth's center.
r p = 6378 + 400 = 6778 r_p=6378+400=6778 r p = 6378 + 400 = 6778 r a = 6378 + 1600 = 7978 r_a=6378+1600=7978 r a = 6378 + 1600 = 7978 Why? Orbital elements use distance from the focus (Earth's center), not altitude.
Step 2: a = r p + r a 2 = 6778 + 7978 2 = 7378 a=\dfrac{r_p+r_a}{2}=\dfrac{6778+7978}{2}=7378 a = 2 r p + r a = 2 6778 + 7978 = 7378 Why? The major axis = r p + r a = 2 a =r_p+r_a=2a = r p + r a = 2 a
Step 3: e = r a − r p r a + r p = 1200 14756 = 0.0813 e=\dfrac{r_a-r_p}{r_a+r_p}=\dfrac{1200}{14756}=0.0813 e = r a + r p r a − r p = 14756 1200 = 0.0813 Why? From r a − r p = 2 a e r_a-r_p=2ae r a − r p = 2 a e r a + r p = 2 a r_a+r_p=2a r a + r p = 2 a
Worked example 2 — Speed at perigee (vis-viva)Same orbit, μ E = 3.986 × 10 5 \mu_E=3.986\times10^5 μ E = 3.986 × 1 0 5 v p v_p v p
Step 1: Use vis-viva at r = r p r=r_p r = r p v 2 = μ ( 2 r p − 1 a ) v^2=\mu\!\left(\frac{2}{r_p}-\frac1a\right) v 2 = μ ( r p 2 − a 1 ) Why? vis-viva connects speed to position using only a a a
Step 2: v p 2 = 3.986 × 10 5 ( 2 6778 − 1 7378 ) = 3.986 × 10 5 ( 2.951 × 10 − 4 − 1.355 × 10 − 4 ) v_p^2=3.986\times10^5\!\left(\frac{2}{6778}-\frac{1}{7378}\right)=3.986\times10^5(2.951\times10^{-4}-1.355\times10^{-4}) v p 2 = 3.986 × 1 0 5 ( 6778 2 − 7378 1 ) = 3.986 × 1 0 5 ( 2.951 × 1 0 − 4 − 1.355 × 1 0 − 4 )
Step 3: v p 2 = 3.986 × 10 5 ( 1.596 × 10 − 4 ) = 63.6 v_p^2=3.986\times10^5(1.596\times10^{-4})=63.6 v p 2 = 3.986 × 1 0 5 ( 1.596 × 1 0 − 4 ) = 63.6 v p ≈ 7.98 v_p\approx 7.98 v p ≈ 7.98 Why this is sensible? Slightly above circular LEO speed (~7.6 km/s) because perigee is the fastest point.
Worked example 3 — Current radius fromν \nu ν a = 7378 a=7378 a = 7378 e = 0.0813 e=0.0813 e = 0.0813 r r r ν = 90 ∘ \nu=90^\circ ν = 9 0 ∘
Step 1: p = a ( 1 − e 2 ) = 7378 ( 1 − 0.00661 ) = 7329 p=a(1-e^2)=7378(1-0.00661)=7329 p = a ( 1 − e 2 ) = 7378 ( 1 − 0.00661 ) = 7329 Step 2: r = p 1 + e cos ν = 7329 1 + 0.0813 ⋅ 0 = 7329 r=\dfrac{p}{1+e\cos\nu}=\dfrac{7329}{1+0.0813\cdot 0}=7329 r = 1 + e cos ν p = 1 + 0.0813 ⋅ 0 7329 = 7329 Why? At ν = 90 ∘ \nu=90^\circ ν = 9 0 ∘ cos ν = 0 \cos\nu=0 cos ν = 0 r = p r=p r = p
Common mistake "Inclination tells you which direction the plane is rotated around."Why it feels right: both i i i Ω \Omega Ω Fix: i i i tilt (how far the plane leans from the equator). Ω \Omega Ω swing (which longitude the tilted plane's node sits at). One leans, the other spins around Z Z Z
Common mistake "Semi-major axis = altitude of the orbit."Why it feels right: for a circular LEO a ≈ a\approx a ≈ Fix: a a a distance from Earth's center , and it's the average of apogee and perigee radii, not an altitude. Subtract R E R_E R E
Common mistake "True anomalyν \nu ν Why it feels right: it's the obvious "where am I" angle.
Fix: ν \nu ν fastest at perigee, slowest at apogee (Kepler's 2nd law). The uniformly-increasing angle is the mean anomaly M M M ν \nu ν M M M E E E
ω \omega ω ν \nu ν Why it feels right: both are in-plane angles measured at the focus.
Fix: ω \omega ω ascending node to perigee (fixed); ν \nu ν perigee to the satellite (moving). Their sum, the argument of latitude u = ω + ν u=\omega+\nu u = ω + ν
Recall Feynman: explain to a 12-year-oldImagine a hula-hoop tilted on a tabletop, and a bead sliding around the hoop.
The hoop's size and how squashed it is = a a a e e e
How much you tilted the hoop off the table = i i i Which way you turned the whole tilted hoop around = Ω \Omega Ω Where on the hoop the "closest point to the candle" is = ω \omega ω Where the bead is on the hoop right this second = ν \nu ν ν \nu ν
"A Eccentric Italian Opera, Omitting Nothing."
a a a e e e i i i Ω \Omega Ω ω \omega ω ν \nu ν shAPE (a , e a,e a , e fRAME (i , Ω , ω i,\Omega,\omega i , Ω , ω plAce (ν \nu ν
What does the semi-major axis a a a The size of the orbit and (via
ε = − μ / 2 a \varepsilon=-\mu/2a ε = − μ /2 a ) the total orbital energy.
Give the relation between specific orbital energy and a a a ε = − μ / ( 2 a ) \varepsilon=-\mu/(2a) ε = − μ / ( 2 a ) .
State the vis-viva equation. v 2 = μ ( 2 r − 1 a ) v^2=\mu\left(\frac{2}{r}-\frac{1}{a}\right) v 2 = μ ( r 2 − a 1 ) .
How do you get a a a a = ( r a + r p ) / 2 a=(r_a+r_p)/2 a = ( r a + r p ) /2 .
How do you get e e e e = ( r a − r p ) / ( r a + r p ) e=(r_a-r_p)/(r_a+r_p) e = ( r a − r p ) / ( r a + r p ) .
Define inclination i i i Angle between the orbital plane (its
h ⃗ \vec h h vector) and the equatorial plane (
Z Z Z -axis); the tilt of the orbit.
Define RAAN Ω \Omega Ω Angle in the equatorial plane from the vernal equinox (
X X X ) to the ascending node.
What is the ascending node? Point where the satellite crosses the equator going south→north.
Define argument of perigee ω \omega ω In-plane angle from the ascending node to perigee.
Define true anomaly ν \nu ν Angle at the focus from perigee to the satellite's current position, along its motion.
Which orbital element changes with time in the ideal two-body problem? Only the true anomaly
ν \nu ν (the other five are constant).
Write the orbit (conic) equation for r ( ν ) r(\nu) r ( ν ) r = a ( 1 − e 2 ) 1 + e cos ν = p 1 + e cos ν r=\dfrac{a(1-e^2)}{1+e\cos\nu}=\dfrac{p}{1+e\cos\nu} r = 1 + e cos ν a ( 1 − e 2 ) = 1 + e cos ν p .
What is the semi-latus rectum p p p h h h p = h 2 / μ = a ( 1 − e 2 ) p=h^2/\mu=a(1-e^2) p = h 2 / μ = a ( 1 − e 2 ) .
What rotation sequence maps perifocal to ECI coordinates? R z ( Ω ) R x ( i ) R z ( ω ) R_z(\Omega)\,R_x(i)\,R_z(\omega) R z ( Ω ) R x ( i ) R z ( ω ) .
Why are exactly 6 elements needed? Because the full state
( r ⃗ , v ⃗ ) (\vec r,\vec v) ( r , v ) is 6 numbers; 6 elements carry equivalent information.
What is the argument of latitude u u u u = ω + ν u=\omega+\nu u = ω + ν , the angle from the ascending node to the satellite.
Position r and velocity v, 6 numbers
Keplerian elements, 6 numbers
Vis-viva, energy = -mu/2a
Ellipse geometry, rp and ra
Intuition Hinglish mein samjhoDekho, ek orbit basically ek fixed ellipse hai jo 3D space me float kar rahi hai, aur satellite uspe ghoom raha hai. Use poori tarah describe karne ke liye humein sirf 6 numbers chahiye, kyunki position ( x , y , z ) (x,y,z) ( x , y , z ) ( v x , v y , v z ) (v_x,v_y,v_z) ( v x , v y , v z ) a a a e e e a a a ε = − μ / 2 a \varepsilon=-\mu/2a ε = − μ /2 a v 2 = μ ( 2 / r − 1 / a ) v^2=\mu(2/r-1/a) v 2 = μ ( 2/ r − 1/ a )
Agle teen angles — i i i Ω \Omega Ω ω \omega ω orientation batate hain space me. i i i Ω \Omega Ω Z Z Z ω \omega ω ν \nu ν
Sabse important baat: ideal two-body gravity me 5 elements constant rehte hain, sirf ν \nu ν ν \nu ν i i i Ω \Omega Ω
