Intuition What this page is
The parent note gave you the formulas. Here we drill them against every kind of input an exam (or the sky) can hand you: tiny eccentricity, huge eccentricity, the two limiting shapes (e = 0 and e = 1 ), a word problem, and a nasty twist. First we map the whole space, then we cover every cell.
Every Kepler-orbit problem reduces to juggling four numbers: the semi-major axis a , the eccentricity e , the two extreme distances r m i n (perihelion) and r m a x (aphelion), plus the period T . The relations are:
r m i n = a ( 1 − e ) , r m a x = a ( 1 + e ) , a = 2 1 ( r m i n + r m a x ) , e = r m a x + r m i n r m a x − r m i n , T 2 ∝ a 3
Below, each cell is a distinct situation. The examples that follow are tagged by cell.
Cell
Situation
Why it is its own case
Example
A
e = 0 (circle)
limiting case: focus = centre, r m i n = r m a x
Ex 1
B
small e (near-circle)
subtraction of tiny fraction; sign traps
Ex 2
C
large e (elongated)
focus far off-centre; big spread
Ex 3
D
e → 1 (parabolic limit)
orbit no longer closes — degenerate
Ex 4
E
given the two extremes → recover a , e
reverse direction of the formulas
Ex 5
F
period law T 2 ∝ a 3 (forward)
predict T from a
Ex 6
G
period law (backward)
recover a from T
Ex 7
H
word problem (real satellite)
translate English → the four numbers
Ex 8
I
exam twist (mix + Law 2 speed)
combine geometry with equal-areas
Ex 9
We now hit each cell.
The figure above shows the same orbit skeleton every example uses: centre C , focus F (the Sun), the perihelion end (nearest, distance r m i n ) and the aphelion end (farthest, distance r m a x ). Keep glancing back at it — every symbol below lives on this picture.
Worked example Example 1 — A perfectly circular orbit
A moon orbits with e = 0 and semi-major axis a = 384 , 400 km. Find r m i n and r m a x .
Forecast: guess before reading — if the orbit is a circle, should the nearest and farthest distances differ?
r m i n = a ( 1 − e ) = 384400 ( 1 − 0 ) = 384400 km.
Why this step? Plug e = 0 into the perihelion formula. With no eccentricity there is nothing to subtract.
r m a x = a ( 1 + e ) = 384400 ( 1 + 0 ) = 384400 km.
Why this step? Same formula, plus sign — nothing to add either.
So r m i n = r m a x = a : the two extremes collapse to one distance.
Why this step? This is the definition of a circle — every point is equidistant from the centre, and the focus has slid onto the centre.
Verify: 2 1 ( r m i n + r m a x ) = 2 1 ( 384400 + 384400 ) = 384400 = a ✅, and e = 2 ⋅ 384400 0 = 0 ✅. Units are km throughout. The circle is the ==e = 0 == limit of the ellipse.
Worked example Example 2 — Earth-like small eccentricity
An orbit has a = 1.000 AU, e = 0.0167 . Find r m i n , r m a x , and the percent difference.
Forecast: with e under 2%, will the nearest and farthest distances differ by a lot or a whisker?
r m i n = a ( 1 − e ) = 1 ( 1 − 0.0167 ) = 0.9833 AU.
Why this step? Perihelion subtracts the small fraction e of a .
r m a x = a ( 1 + e ) = 1 ( 1 + 0.0167 ) = 1.0167 AU.
Why this step? Aphelion adds the same fraction.
Percent spread = a r m a x − r m i n × 100% = 1 0.0334 × 100% = 3.34% .
Why this step? The full swing is r m a x − r m i n = 2 a e ; dividing by a isolates the size of the wobble.
Verify: r m a x − r m i n = 2 a e = 2 ( 1 ) ( 0.0167 ) = 0.0334 ✅. And 2 1 ( 0.9833 + 1.0167 ) = 1 = a ✅. Tiny e ⇒ tiny wobble — this is why Earth's orbit looks like a circle.
Worked example Example 3 — A stretched comet orbit
A comet has a = 18 AU and e = 0.967 . Find its nearest and farthest distances.
Forecast: with e almost 1 , guess whether perihelion will be a small number or a large one.
r m i n = a ( 1 − e ) = 18 ( 1 − 0.967 ) = 18 × 0.033 = 0.594 AU.
Why this step? Large e makes ( 1 − e ) tiny, so perihelion is a small distance — the comet dives deep inward.
r m a x = a ( 1 + e ) = 18 ( 1 + 0.967 ) = 18 × 1.967 = 35.406 AU.
Why this step? ( 1 + e ) is nearly 2 , so aphelion is nearly 2 a — the comet flings far out.
Ratio r m i n r m a x = 0.594 35.406 ≈ 59.6 .
Why this step? A big ratio quantifies "elongated": the comet is ~60× farther at aphelion than perihelion.
Verify: r m i n + r m a x = 0.594 + 35.406 = 36 = 2 a ✅; e = 36 35.406 − 0.594 = 36 34.812 = 0.967 ✅. Units AU throughout.
Worked example Example 4 — What happens as
e → 1 ?
Keep perihelion fixed at r m i n = 1 AU and push e → 1 . What happens to a and r m a x ?
Forecast: guess — does the orbit stay a closed loop, or open up?
From r m i n = a ( 1 − e ) with r m i n = 1 : a = 1 − e 1 .
Why this step? Solve for a so we can watch it as e climbs toward 1 .
At e = 0.9 : a = 0.1 1 = 10 . At e = 0.99 : a = 0.01 1 = 100 .
Why this step? As 1 − e shrinks, a blows up — the ellipse gets enormous.
r m a x = a ( 1 + e ) = 1 − e 1 + e . At e = 0.99 : r m a x = 0.01 1.99 = 199 AU — heading to infinity.
Why this step? As e → 1 , aphelion runs away to infinity: the far end never comes back. The ellipse degenerates into a parabola .
Verify: at e = 0.99 , a = 100 and r m a x = 100 ( 1.99 ) = 199 ✅, r m i n = 100 ( 0.01 ) = 1 ✅. The eccentricity e = 1 is exactly the boundary between bound ellipses and unbound orbits (hyperbolas have e > 1 ).
Worked example Example 5 — Recover the orbit from its extremes
A satellite skims to r m i n = 6800 km from Earth's centre and swings out to r m a x = 42000 km. Find a and e .
Forecast: which is bigger, a or r m i n ? Guess before computing.
a = 2 1 ( r m i n + r m a x ) = 2 1 ( 6800 + 42000 ) = 2 1 ( 48800 ) = 24400 km.
Why this step? The semi-major axis is the average of the two extreme distances (the two ends are 2 a apart along the major axis).
e = r m a x + r m i n r m a x − r m i n = 48800 42000 − 6800 = 48800 35200 = 0.7213 .
Why this step? Eccentricity is (spread) ÷ (total): how off-centre the focus sits, as a fraction.
Verify: rebuild the extremes: a ( 1 − e ) = 24400 ( 1 − 0.7213 ) = 24400 ( 0.2787 ) = 6800 km ✅ and a ( 1 + e ) = 24400 ( 1.7213 ) = 42000 km ✅. Note a = 24400 > r m i n = 6800 as forecast.
Worked example Example 6 — Predict a period from the size
A dwarf planet has a = 4 AU. Using Earth as the unit (a = 1 AU, T = 1 yr), predict T .
Forecast: bigger orbit — longer or shorter period? Guess, then check against Kepler 3.
Kepler 3: T E 2 T 2 = a E 3 a 3 . With Earth units this is just T 2 = a 3 .
Why this step? Choosing years and AU makes the proportionality constant equal 1 , so the law becomes a clean equation.
T = a 3/2 = 4 3/2 = ( 4 ) 3 = 2 3 = 8 yr.
Why this step? Take the square root of a 3 ; 4 3/2 means "cube the square-root of 4".
Verify: T 2 = 8 2 = 64 and a 3 = 4 3 = 64 ✅. Bigger a ⇒ longer T — distant bodies crawl.
Worked example Example 7 — Recover the size from the period
An asteroid takes T = 27 yr to orbit the Sun. Find its semi-major axis in AU.
Forecast: with T well above 1 yr, will a come out above or below 1 AU?
From T 2 = a 3 (Earth units) solve a = T 2/3 .
Why this step? We invert the law: to undo the "3/2 power" that turns a into T , raise T to the reciprocal power 2/3 .
a = 2 7 2/3 = ( 2 7 1/3 ) 2 = 3 2 = 9 AU.
Why this step? 2 7 1/3 = 3 (cube root), then square it.
Verify: a 3 = 9 3 = 729 and T 2 = 2 7 2 = 729 ✅. T > 1 gave a > 1 as forecast.
Worked example Example 8 — A communications satellite (real translation)
"A satellite's lowest point is 500 km above Earth's surface and its highest is 35786 km above the surface. Earth's radius is 6371 km. Find a and e of the orbit."
Forecast: must we add Earth's radius before using our formulas? Guess why.
The focus is Earth's centre , so distances must be measured from the centre, not the surface. Add the radius:
r m i n = 500 + 6371 = 6871 km, r m a x = 35786 + 6371 = 42157 km.
Why this step? r m i n , r m a x in Kepler's geometry are distances to the focus (Earth's centre); "altitude above surface" is that distance minus the radius. Forgetting this is the classic exam trap.
a = 2 1 ( 6871 + 42157 ) = 2 1 ( 49028 ) = 24514 km.
Why this step? Average of the two centre-distances, exactly as in Cell E.
e = 49028 42157 − 6871 = 49028 35286 = 0.7197 .
Why this step? Spread over total, once distances are correctly referenced to the focus.
Verify: a ( 1 − e ) = 24514 ( 1 − 0.7197 ) = 24514 ( 0.2803 ) = 6871 km ✅; a ( 1 + e ) = 24514 ( 1.7197 ) = 42157 km ✅. All km, all measured from the centre.
Worked example Example 9 — Speed ratio from equal areas
A planet has r m i n = 0.5 AU and r m a x = 1.5 AU. Find e , then the ratio of its speed at perihelion to its speed at aphelion.
Forecast: where does the planet move faster — near or far? Guess before the algebra.
a = 2 1 ( 0.5 + 1.5 ) = 1 AU, e = 2.0 1.5 − 0.5 = 2 1 = 0.5 .
Why this step? Standard extreme-to-orbit step (Cell E) before we touch the physics.
Kepler's 2nd law: at the two ends the velocity is perpendicular to the radius, so the swept-area rate is 2 1 r v = const . Hence r m i n v peri = r m a x v aph (this is angular-momentum conservation ).
Why this step? At perihelion and aphelion alone the motion is purely sideways, so r 2 θ ˙ = r ⋅ v simplifies — letting us compare the two speeds directly.
v aph v peri = r m i n r m a x = 0.5 1.5 = 3 .
Why this step? Rearranging the conservation equation: shorter radius ⇒ proportionally faster.
Verify: cross-check r m i n v peri = 0.5 × 3 = 1.5 and r m a x v aph = 1.5 × 1 = 1.5 — equal ✅. Perihelion speed is 3 × aphelion speed: faster near the Sun , exactly Kepler 2.
Recall Cover and answer
In Cell A (e = 0 ), what happens to r m i n and r m a x ? ::: They become equal to a — a circle.
As e → 1 with r m i n fixed, what happens to a ? ::: It blows up to infinity — the ellipse degenerates to a parabola.
How do you get a and e from two altitudes above a planet's surface? ::: Add the planet's radius first (focus = centre), then average and take spread/total.
Speed ratio perihelion:aphelion from extremes? ::: v peri / v aph = r m a x / r m i n .
Recover a from a period T (Earth units)? ::: a = T 2/3 .
Forward geometry: a , e → r (multiply). Reverse geometry: r , r → a , e (average & spread). Forward periods: a → T = a 3/2 . Reverse periods: T → a = T 2/3 .