ARCHETYPE: Kepler Ratio Problems

Tycho Brahe never saw a telescope. Everything he gave astronomy — twenty years of planetary positions accurate to an arc-minute or two — came from giant naked-eye instruments on a private island observatory, funded by a Danish king and run with the discipline of a ship's log. Buried in those columns of numbers lay a pattern nobody had noticed in two millennia of stargazing: square any planet's period, cube its distance, divide — and you get the same number every time. Kepler dug it out in 1619. The pattern is a gift to examiners, because it turns celestial mechanics into ratio arithmetic: no $G$ , no masses, no constants — just proportions. That is exactly how boards, NEET, JEE-Main and the IB deploy it. "Satellite A orbits at radius $r$ , satellite B at $4r$ — compare their periods." "A planet's year is 8 times Earth's — how far out is it?" "Two satellites start together — when do they meet again?" Every one of these is the third law run through the same three-step ratio machine, and the numbers are always chosen to be perfect squares and cubes — the examiner's tell that no calculator should be touched. This archetype drills the machine: the exponent dictionary ( $T \propto r^{3/2}$ , $v \propto r^{-1/2}$ , and friends), the percentage shortcut for small changes, the lap-race formula for satellites that align and separate, and the one trap that voids everything — mixing satellites of different central bodies.

Key takeaways

The workhorse: $\left(\dfrac{T_1}{T_2}\right)^2 = \left(\dfrac{r_1}{r_2}\right)^3$ — one central body, no constants, pure ratios.
The exponent dictionary (all per $r$ ): $T \propto r^{3/2}$ , $v \propto r^{-1/2}$ , $\omega \propto r^{-3/2}$ , $K \propto r^{-1}$ , $g \propto r^{-2}$ .
Inverse reading: $r \propto T^{2/3}$ — period ratio 8 means radius ratio 4.
Small changes ride the exponents too: $\dfrac{\Delta T}{T} = \dfrac32\,\dfrac{\Delta r}{r}$ — a 2% higher orbit is a 3% longer period.
Alignment (synodic) problems are lap races: two co-orbiting satellites re-align after $t = \dfrac{T_1 T_2}{T_2 - T_1}$ .
The void-everything trap: $T^2/r^3$ is one constant per central body — never ratio a Moon-of-Earth against a planet-of-Sun.
Key idea: pick the exponent, take the ratio, keep the family — every Kepler numerical is proportionality craft, not computation.

The core insight

Kepler ratio problems apply $T^2 \propto r^3$ (and its exponent family $v \propto r^{-1/2}$ , $\omega \propto r^{-3/2}$ ) between two orbits of the same central body, converting period, radius, speed and alignment questions into constant-free proportions.

Where this lives today

Ratio craft is the daily arithmetic of space operations. Constellation designers place shells by period: GPS satellites at 12-hour orbits sit at $26{,}600\ \mathrm{km}$ — a number every navigator can reproduce from the geostationary anchor by $(12/24)^{2/3}$ . Mars mission planners live inside worked example 4's lap race: Earth laps Mars every 26 months, so launch windows (and the traffic jams of missions departing together) recur on exactly that beat. Exoplanet astronomy runs the machine in bulk — a survey telescope measures thousands of periods, and $r \propto T^{2/3}$ (per star) converts them into orbital architectures, resonant chains like TRAPPIST-1's echoing the 1:2:4 rhythm of Jupiter's moons that made example 2 so clean. Even satellite collision-avoidance uses the percentage rule: nudge an orbit by a fraction of a percent, and controllers know to the minute how the along-track separation will grow — $\tfrac32$ of the nudge, per orbit, compounding.

The setup

Everything orbiting one central body shares the constant $T^2/r^3 = 4\pi^2/GM$ . Take two members of the family, divide, and the constant — with its awkward $G$ and its planet mass — evaporates:

\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3 \qquad (1)

where: $r$ is the orbital radius (semi-major axis for ellipses), from the centre of the central body. This is the entire law of the archetype; every variant is equation (1) wearing a different exponent. Because the satellite formulas $v = \sqrt{GM/r}$ and $\omega = 2\pi/T$ live in the same family, their ratios come exponent-ready:

\frac{v_1}{v_2} = \left(\frac{r_2}{r_1}\right)^{1/2}, \qquad \frac{\omega_1}{\omega_2} = \left(\frac{r_2}{r_1}\right)^{3/2}, \qquad \frac{K_1}{K_2} = \frac{r_2}{r_1} \qquad (2)

where: the inversions are deliberate reminders that inner is faster — smaller $r$ means larger $v$ , $\omega$ and $K$ . Exam numbers are rigged for mental roots: radius ratios arrive as $4$ , $9$ , $\tfrac14$ ; period ratios as $8$ , $27$ , $2\sqrt2$ . Seeing $T_2/T_1 = 8$ should trigger " $r$ ratio $= 8^{2/3} = 4$ " before the pen moves.

Figure

The exponent dictionary — one radius ratio, every quantity's ratio read off by its power

The second standard costume is the lap race. Two satellites circle the same body, same direction, inner one faster. Start them aligned; the inner gains angle at the rate $\omega_1 - \omega_2$ and laps the outer when the gained angle reaches a full circle. The re-alignment time — astronomy calls it the synodic period — is

t_{\rm align} = \frac{2\pi}{\omega_1 - \omega_2} = \frac{T_1\,T_2}{T_2 - T_1} \qquad (3)

where: $T_1 < T_2$ . Sanity limits: nearly equal periods make $t_{\rm align}$ huge (a slow-motion chase); $T_2 \gg T_1$ gives $t_{\rm align} \to T_1$ (the outer satellite barely moves while the inner laps it). This one formula covers "when do they meet again," "when are they closest," and — run with Earth and Mars around the Sun — why Mars launch windows recur every 26 months.

Figure

The lap race — the inner satellite gains angle at ω₁ − ω₂ and re-aligns after one relative lap

variant asked	move	watch for
period from radii (or reverse)	eq. (1); $r \propto T^{2/3}$	perfect squares/cubes planted
speed / $\omega$ / KE ratios	exponent dictionary, eq. (2)	inner is faster
% change	multiply the % by the exponent	small changes only
meet again / closest again	lap race, eq. (3)	same direction assumed
"n revolutions of A per one of B"	$n = T_2/T_1 = \rho^{3/2}$	count laps, not angles
different central bodies	STOP — constants differ	ratio only within a family

Derivation AThe ratio machine — killing the constants

We derive equation (1) and the exponent dictionary (2) from the satellite formulas, making explicit why no $G$ , no masses and no $\pi$ 's survive. The strategy is to write each quantity for two orbits and divide. Watch what the division requires: the same central mass $M$ on both lines — the fine print that becomes the archetype's biggest trap.

Step 1. Write the period formula (from the satellites module) for both orbits about the same $M$ :

T_1^2 = \frac{4\pi^2}{GM}\,r_1^3, \qquad T_2^2 = \frac{4\pi^2}{GM}\,r_2^3

Step 2. Divide — the entire prefactor $4\pi^2/GM$ cancels because it is identical on both lines:

\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3

Why the trap is born here: if the two orbits circle different bodies, the prefactors differ and the division leaves a ratio of central masses behind. Same family or no cancellation.

Step 3. Do the same to the speed formula $v = \sqrt{GM/r}$ :

\frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}}

and to $\omega = 2\pi/T$ , which inherits the period's exponent inverted: $\omega \propto r^{-3/2}$ . Kinetic energy $K = GMm/2r$ gives $K \propto 1/r$ (equal satellite masses).

Step 4. Extract the percentage rule. For a small fractional change, powers multiply fractions: if $r \to r(1 + \epsilon)$ then $T \propto r^{3/2}$ gives $T \to T(1 + \epsilon)^{3/2} \approx T\left(1 + \tfrac32\epsilon\right)$ — the binomial shortcut, same licence as always ( $\epsilon \ll 1$ ):

\frac{\Delta T}{T} = \frac32\,\frac{\Delta r}{r}

Verified: $r$ ratio 4 gives $T$ ratio $4^{3/2} = 8$ ✓ against direct substitution; $v$ ratio $\sqrt{4} = 2$ with the inner faster ✓; a 2% radius increase gives $(1.02)^{1.5} = 1.0301$ — the 3% rule to within rounding ✓.

Go deeper · Derivation B · 1 minThe lap race — when do two satellites meet again?

Derivation BThe lap race — when do two satellites meet again?

We derive the re-alignment formula (3). The strategy is to work in angles: both satellites gain angle steadily, and "aligned again" means the inner has gained one full lap on the outer. This is the same arithmetic as two runners on a track, and it is worth internalising as such — every "when do they meet/oppose/align" question is this one derivation.

Step 1. Angular speeds: $\omega_1 = 2\pi/T_1$ (inner, faster) and $\omega_2 = 2\pi/T_2$ . Relative angular speed — the rate the gap grows:

\omega_{\rm rel} = \omega_1 - \omega_2

Why subtraction: both angles grow the same way; only the difference moves one satellite relative to the other. (Opposite orbital directions would add the speeds instead — read the question.)

Step 2. Re-alignment = the relative angle completes $2\pi$ :

t_{\rm align} = \frac{2\pi}{\omega_1 - \omega_2}

Step 3. Convert to periods:

t_{\rm align} = \frac{2\pi}{2\pi/T_1 - 2\pi/T_2} = \frac{T_1 T_2}{T_2 - T_1}

Step 4. Variants from the same line: diametrically opposite first happens at half this time (relative angle $\pi$ ); "meets for the $n$ -th time" is $n\,t_{\rm align}$ . Verified: limits — $T_2 \to T_1$ makes $t_{\rm align} \to \infty$ (twins never lap) ✓; $T_2 \gg T_1$ gives $t_{\rm align} \to T_1$ ✓. Reality check: Earth (1 yr) and Mars (1.881 yr): $t = 1 \times 1.881/0.881 = 2.14$ yr ≈ 26 months — the actual Mars launch-window cadence ✓.

Limiting cases

case	result	meaning
$r$ ratio 4	$T$ ratio 8, $v$ ratio 2	the exam's favourite pair
$T$ ratio 27	$r$ ratio 9	read cubes backwards: $27^{2/3} = 9$
$\Delta r/r = +1\%$	$\Delta T/T = +1.5\%$	exponents scale percentages
$T_2 \to T_1$	$t_{\rm align} \to \infty$	near-twins almost never re-align
$T_2 \gg T_1$	$t_{\rm align} \to T_1$	lapping a statue takes one lap
different central bodies	eq. (1) INVALID	constants no longer cancel

Common trap“Kepler's law compares any two orbits” — the family trap

Equation (1) works because the constant $4\pi^2/GM$ cancels in the division — which requires the same central mass on both lines. Ratio the Moon (orbiting Earth) against Mars (orbiting the Sun) and nothing cancels: you would be equating constants that differ by a factor of $M_\odot/M_E \approx 333{,}000$ . The exam versions of this trap: a satellite of Earth compared with a moon of Jupiter; "a planet identical to Earth orbits another star…" (fine — but only within that star's family); and the sneaky "compare the Moon with a geostationary satellite" — which is legal, both orbit Earth, and is planted precisely so that students who over-learned the trap refuse a valid comparison. The test is one question: same central body? Then ratios fly. Different? Then you need $GM$ 's explicitly, and the problem has left this archetype.

Worked examples

1 · Board level — period from radius ratio (CBSE/NEET pattern, 2 marks)

Satellite A orbits Earth at radius $7000\ \mathrm{km}$ with period $97\ \mathrm{min}$ . Satellite B orbits at $28{,}000\ \mathrm{km}$ . Find B's period.

Step 1. Same central body (Earth) — the ratio machine is legal. Radius ratio: $\rho = 28{,}000/7000 = 4$ .

Step 2. Equation (1): $\dfrac{T_B}{T_A} = \rho^{3/2} = 4^{3/2} = 8$ .

Step 3. $T_B = 8 \times 97 = 776\ \mathrm{min} \approx \boxed{12.9\ \mathrm{hours}}$ .

Step 4. Sanity: farther is slower-and-longer ✓; and no $G$ , no Earth mass, no calculator beyond $\times 8$ .

2 · Board level — real moons, real data (IB/ISC pattern)

Jupiter's moon Io orbits at $4.22\times10^8\ \mathrm{m}$ with a period of $1.77$ days. Europa's period is $3.55$ days. Find Europa's orbital radius.

Step 1. Same family (both orbit Jupiter). Period ratio: $\dfrac{T_E}{T_I} = \dfrac{3.55}{1.77} = 2.01 \approx 2$ .

Step 2. Invert the exponent: $r \propto T^{2/3}$ , so $\dfrac{r_E}{r_I} = 2^{2/3} = 1.587$ .

Step 3. $r_E = 1.587 \times 4.22\times10^8 = \boxed{6.70\times10^8\ \mathrm{m}}$ — the accepted value is $6.71\times10^8$ m.

Step 4. Notice what never appeared: Jupiter's mass. (Galileo's moons are in a near-exact 1:2:4 period resonance — Io, Europa, Ganymede — which is why the numbers are this clean in nature, not just in exams.)

3 · Competitive level — geostationary shrink (JEE-Main pattern, ~2 minutes)

A geostationary satellite orbits at $r = 42{,}200\ \mathrm{km}$ with a 24-hour period. Find the orbital radius and altitude of a satellite with a 3-hour period. ( $R_E = 6400\ \mathrm{km}$ .)

Step 1. Period ratio: $\dfrac{T_2}{T_1} = \dfrac{3}{24} = \dfrac18$ .

Step 2. Radius ratio: $\left(\dfrac18\right)^{2/3} = \dfrac{1}{4}$ (read the cube backwards: $8 = 2^3$ , so $8^{2/3} = 4$ ).

Step 3. $r_2 = \dfrac{42{,}200}{4} = \boxed{10{,}550\ \mathrm{km}}$ .

Step 4. Altitude — the planted last step: $h = r - R_E = 10{,}550 - 6400 = \boxed{4150\ \mathrm{km}}$ . Timing note: $8 \leftrightarrow 4$ , $27 \leftrightarrow 9$ , $2\sqrt2 \leftrightarrow 2$ — keep the cube-square pairs pre-computed and this is 60 seconds flat. The mark-loser is skipping Step 4: options list both $10{,}550$ and $4150$ , and the question asked for altitude.

4 · Competitive level — the lap race with a percentage rider (JEE/NEET stretch, ~3.5 minutes)

Two satellites P and Q orbit Earth in the same plane and direction with periods $2\ \mathrm{h}$ and $6\ \mathrm{h}$ . (a) Find the ratio of their orbital radii. (b) Starting aligned, after how long are they aligned (on the same side) again? (c) If Q's radius were increased by 2%, by what percentage would its period grow?

Step 1. Part (a): $\dfrac{r_Q}{r_P} = \left(\dfrac{T_Q}{T_P}\right)^{2/3} = 3^{2/3} = \boxed{2.08}$ .

Step 2. Part (b): lap race, equation (3): $t = \dfrac{T_P\,T_Q}{T_Q - T_P} = \dfrac{2 \times 6}{6 - 2} = \boxed{3\ \mathrm{h}}$ .

Step 3. Audit part (b): in 3 h, P covers $1.5$ laps ( $540°$ ), Q covers $0.5$ lap ( $180°$ ) — difference $360°$ , one full relative lap ✓.

Step 4. Part (c): percentages ride the exponent: $\dfrac{\Delta T}{T} = \dfrac32 \times 2\% = \boxed{3\%}$ . Timing note: part (b)'s formula is worth memorising as "product over difference." The audit in Step 3 costs 15 seconds and catches the classic error of using $T_2 + T_1$ (that is the answer for satellites orbiting in OPPOSITE directions — the same lap-race logic with $\omega$ 's adding). Total budget: 3 minutes.

Exam traps

Mixing families. The constant cancels only for orbits of the same central body. A Moon-vs-Mars ratio (different centres) is meaningless; a Moon-vs-geostationary ratio (both Earth's) is perfectly legal — know the difference in both directions.

Radius versus altitude. All ratios run on centre-to-centre $r$ . Questions gift the altitude ( $h$ ) or ask for it back (example 3); convert at entry and exit, never mid-ratio.

Exponents on the wrong side. $T$ ratio 8 means $r$ ratio 4, not 16; $r$ ratio 9 means $T$ ratio 27. Read $T \to r$ with the power $2/3$ , $r \to T$ with $3/2$ — and check the direction: longer period must mean larger radius.

Speed inversions. Inner satellites are faster: $v \propto r^{-1/2}$ . Options with the outer satellite faster prey on the "farther = bigger everything" reflex.

Percent craft outside its licence. $\Delta T/T = \tfrac32\,\Delta r/r$ is a small-change (binomial) shortcut. A 50% radius change needs the exact ratio $(1.5)^{3/2} = 1.84$ , not $1.75$ .

Product-over-difference misapplied. Equation (3) is for co-rotating satellites. Opposite directions: relative angular speed adds, giving $t = T_1T_2/(T_1 + T_2)$ . The word "same direction" (or its absence) is load-bearing — read it.

Diameter data. Astronomy problems sometimes quote orbital diameters or planet-to-planet distances. The ratio machine survives any consistent length choice, but mixing one diameter with one radius quietly injects a $2^3 = 8$ .

Ellipse rider. If the orbit is elliptical, the third law wants the semi-major axis $a$ — with $r_p$ and $r_a$ given, $a$ is their average (drilled in the Kepler's-laws module).

Origins · 1 minTycho Brahe and how this was discovered

Tycho Brahe

1546–1601 · Hven & Prague

Tycho Brahe was astronomy's great instrument-builder and its most colourful aristocrat — he lost part of his nose in a student duel over mathematics and wore a metal replacement. Granted the island of Hven by the Danish king, he built Uraniborg, Europe's first purpose-built research institution, and spent twenty years measuring planetary positions to one or two arc-minutes — five times better than anything before, and the practical limit of the naked eye. Crucially, he measured continuously, not just at convenient moments: Mars was tracked through its whole orbit, opposition after opposition. Tycho never accepted the Copernican system (his compromise had the planets circling the Sun, and the Sun circling a fixed Earth), and he guarded his data jealously. Only his death in 1601 put the full ledger in the hands of his difficult new assistant, Johannes Kepler — who mined from it the ellipse, the equal areas, and the $T^2 \propto a^3$ pattern this archetype drills. Every ratio problem you solve is, at bottom, a withdrawal from Tycho's twenty-year bank of arc-minutes.

History

1576

— Tycho begins building Uraniborg on Hven: two decades of arc-minute planetary data follow.

1601

— Tycho dies in Prague; Kepler inherits the observation ledger.

1610

— Galileo discovers Jupiter's four great moons — a second orbital family, later a clean third-law testbed.

1619

— Kepler's Harmonice Mundi states T² ∝ a³ across the Sun's family.

1687

— Newton derives the law and identifies its constant as 4π²/GM, turning ratios into weighings.

1780

s — Herschel's Uranus obeys the same ratio machine, extending the family without adjustment.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (Gravitation) · HC Verma, Concepts of Physics Vol. 1, Ch. 11 · past-paper mining: JEE-Main and NEET repeat the 4↔8 and lap-race patterns near-annually.

Provenance. All results are standard canon; every number recomputed. A verifier should recompute: $4^{3/2} = 8$ and example 1's $12.9$ h; example 2's $2^{2/3} = 1.587$ and $r_E = 6.70\times10^8$ m; example 3's $8^{2/3} = 4$ , $r = 10{,}550$ km, $h = 4150$ km; example 4's $3^{2/3} = 2.08$ , $t = 3$ h (with the $540° - 180°$ audit) and the 3% rider; the Mars window $2.14$ yr and synodic month $29.5$ d from eq. (3).

Prerequisites. Kepler's Laws of Planetary Motion · Satellites: Speed, Period, Energy & Weightlessness. Leads to. Kepler's Third Law (UG deep dive) · Option Elimination & Answer Checking (capstone technique).

Why do these problems never need G or the planet's mass?

Because both live inside the family constant $4\pi^2/GM$ , and dividing two members of the same family cancels it wholesale. That is the entire design of the archetype: examiners choose ratio phrasing precisely so the "hard" constants disappear and the physics reduces to exponents. The flip side is diagnostic: if a problem cannot be ratio'd (different central bodies, or an absolute answer in metres or seconds demanded from scratch), the constant returns and you are back to the full formulas of the satellites module.

How do I keep the 3/2 and 2/3 straight under time pressure?

Anchor to one fact you can't forget: bigger orbits take longer — much longer. Radius up 4, period up 8: period grows faster than radius, so $T$ carries the bigger exponent ( $3/2 > 1$ ). Going backwards, radius must grow slower than period: exponent $2/3 < 1$ . If you ever produce "period ratio 4 from radius ratio 8," the direction check catches it: $8^{3/2} = 22.6$ , not 4 — you inverted. Ten seconds of "does longer mean farther, by more?" beats any mnemonic.

Where does the product-over-difference formula actually come from?

From angles, not magic. Each satellite's angular position grows at $\omega = 2\pi/T$ ; the gap between them grows at $\omega_1 - \omega_2$ ; alignment recurs when the gap totals $2\pi$ . One line of algebra turns $2\pi/(\omega_1 - \omega_2)$ into $T_1T_2/(T_2 - T_1)$ . Held as a picture — two runners on a circular track, the faster lapping the slower — every variant answers itself: opposite directions means closing speed $\omega_1 + \omega_2$ (product over sum), "diametrically opposite" means half a relative lap (half the time), "n-th meeting" means n relative laps.

Do these ratios work for elliptical orbits too?

The third law does — with $r$ read as the semi-major axis $a$ , so period ratios and $a$ ratios convert exactly as here. What breaks is the rest of the dictionary: an ellipse has no single speed or angular speed ( $v$ varies from perihelion sprint to aphelion crawl per Kepler's second law), so $v \propto r^{-1/2}$ applies only to circular orbits. Exam tell: "circular" in the stem licenses the whole dictionary; "elliptical" restricts you to $T \leftrightarrow a$ plus the apsidal speed ratio $v_pr_p = v_ar_a$ from the Kepler's-laws module.

The Moon takes 27.3 days, but full moons repeat every 29.5 days. Isn't that a violation?

It is the lap race, run one level up. The 27.3-day figure is the Moon's true (sidereal) orbital period — the third law's $T$ . But "full moon" is an alignment condition involving the Sun, and while the Moon laps, the Earth–Sun line itself turns once a year. Apply equation (3) with $T_1 = 27.3$ d and $T_2 = 365.25$ d: $t = 27.3 \times 365.25/(365.25 - 27.3) = 29.5$ days — the synodic month, exactly the calendar's beat. A lovely check that the formula, and the distinction it guards, are real.

Why are exam numbers always 4, 8, 9, 27 — and what if they aren't?

Because the exponents are $3/2$ and $2/3$ , only perfect squares-and-cubes give clean mental answers, and setters overwhelmingly comply: the appearance of 4/8/9/27/64 is practically a signature that the ratio machine is intended. When the numbers are ugly (example 2's 2.01, or a 1.5 ratio), two honest options remain: exact powers via logarithms — out of scope for MCQ speed — or the calculator-free approximations you already carry: $2^{2/3} = 1.59$ , $3^{2/3} = 2.08$ , $(1+\epsilon)^{3/2} \approx 1 + \tfrac32\epsilon$ for small $\epsilon$ . Memorise those three and no realistic paper can slow you down.