Kepler's third law — T² ∝ a³ — derivation: Question bank

3.2.7 · D5 · Physics › Orbital Mechanics & Astrodynamics › Kepler's third law — T² ∝ a³ — derivation

Neeche har term parent note mein build ki gayi hai: $T$ = orbital period (ek chakkar lagane ka time), $a$ = semi-major axis (orbit ka "average size"; circle ke liye radius $r$ ), $M$ = us body ki mass jo orbit ki ja rahi hai, $m$ = chhoti body ki mass jo orbit kar rahi hai, $v$ = orbital speed (chhoti body apne path par kitni fast move karti hai, metres per second mein), $G$ = gravitational constant, $e$ = eccentricity (ellipse kitna squash hai; $e = 0$ matlab perfect circle).

Definition Is page par use hone wale symbols aur abbreviations

$r$ = chhoti body aur jis body ko wo orbit kar rahi hai uske centre ke beech ek waqt ki instantaneous distance (metres mein). Circle ke liye $r$ kabhi nahi badlata aur $a$ ke barabar hota hai; ellipse ke liye $r$ har chakkar mein ghatata-badhta rehta hai, isliye ye koi ek fixed number nahi hai.
AU = astronomical unit, Earth–Sun ki average distance ( $\approx 1.496 \times 1 0^{11}$ m). Planetary orbit sizes measure karne ke liye convenient hai.
$M_{⊙}$ = one solar mass, Sun ki mass ( $\approx 1.989 \times 1 0^{30}$ kg). Jab Sun orbit kiya ja raha ho tab central mass $M$ ke roop mein use hoti hai.
$M_{\oplus}$ = one Earth mass, Earth ki mass ( $\approx 5.972 \times 1 0^{24}$ kg). Jab Earth orbit ki ja rahi ho tab central mass $M$ ke roop mein use hoti hai.
GEO = geostationary Earth orbit, wo special orbit jiska period Earth ke rotation (ek din) ke barabar hota hai, isliye wahan ka satellite equator ke upar ek jagah fixed dikhta hai.
Step 1 (parent derivation se) = wo move jisme gravity ko required centripetal force ke barabar set kiya jaata hai, $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ — woh ek equation jis par poora law bana hai.

Figure — Kepler's third law — T² ∝ a³ — derivation

True or false — justify

A heavier planet at the same distance from the Sun has a shorter year.

False. Orbiting mass

m

cancel ho jaati hai

GM m / r^{2} = m v^{2} / r

mein, isliye period sirf

a

aur central mass

M

par depend karta hai — ek feather aur ek boulder ek hi distance par dono ek saal lenge.

If two orbits share the same semi-major axis

a

but different eccentricities, they have different periods.

False. Period sirf

a

par depend karta hai. Orbit ko squash karne se body periapsis ke paas speed up hoti hai aur apoapsis ke paas slow down, aur ye ek chakkar mein exactly cancel ho jaate hain.

Doubling the semi-major axis doubles the period.

False.

T \propto a^{3/2}

hai, isliye

a

double karne par

T

2^{1.5} \approx 2.83

se multiply hota hai — double se zyada, kyunki body dono zyada door travel karti hai aur slower bhi chalti hai.

T^{2} = a^{3}

(bina kisi constant ke) ek universal law hai jo kisi bhi units mein kaam karta hai.

False. Wo clean form sirf special units AU + years + solar masses mein hoti hai, jahan

4 π^{2} / (G M_{⊙})

sanyog se 1 ke barabar hota hai. SI units mein constant bahut clearly present hota hai.

Kepler's Third Law artificial satellites ke liye fail hoti hai kyunki wo man-made hain.

False. Law ko origin se koi lena-dena nahi — sirf central body ki gravity se matlab hai. Earth ke around har satellite

T^{2} = (4 π^{2} / G M_{\oplus}) a^{3}

follow karta hai, aur isi se GEO altitude design ki jaati hai.

For a planet orbiting the Sun, the correct mass to use is the planet's mass.

False.

M

orbit ki jane wali body ki mass hai (Sun ki). Planet ki mass

m

derivation se pehle hi cancel ho chuki hai.

The proportionality constant

4 π^{2} / GM

is the same for a comet and a planet around the same star.

True. Jab tak central mass

M

identical hai, constant identical hai — isliye ratio comparisons (

T_{B}^{2} / T_{A}^{2} = (a_{B} / a_{A})^{3}

) itne cleanly kaam karte hain.

A body in a circular orbit is not accelerating because its speed is constant.

False. Speed constant hai lekin velocity direction lagatar change hoti rehti hai. Ye turning centripetal acceleration

v^{2} / r

hai jo inward point karti hai — poori tarah gravity se supply hoti hai.

Spot the error

"Since gravity is stronger closer in, planets close to the Sun must move slower."

Error: conclusion ulta hai. Zyada strong gravity matlab zyada bada required centripetal force, jo zyada high speed demand karta hai:

v = GM / r

badhta hai jab

r

ghatata hai. Kareeb ke planets faster move karte hain.

"For an ellipse we just plug in the current distance

r

from the Sun."

Error:

r

(instantaneous distance) elliptical orbit mein poore time change hoti rehti hai, isliye ye koi single number nahi hai. Law semi-major axis

a = (r_{m i n} + r_{m a x}) /2

use karta hai, jo perihelion aur aphelion distances ka average hai.

"When we set gravity equal to the centripetal requirement, orbiting bodies feel two forces cancelling to zero."

Error: gravity hi centripetal force hai yahan — sirf ek force hai. Equation

GM m / r^{2} = m v^{2} / r

gravitational pull ko value

m v^{2} / r

se equate karta hai jo circle ke liye required hai; ye same force hai, do opposing forces nahi.

"The

4 π^{2}

comes from the area of a circle."

Error: ye circumference relation

v = 2 π r / T

ko square karne se aata hai. Square karne par

v^{2} = 4 π^{2} r^{2} / T^{2}

milta hai, aur woh

4 π^{2}

seedha final law mein chala jaata hai.

"To find orbital speed I use

v = 2 π r / T

, so speed depends only on how big the orbit is."

Error: wo formula ek definition hai (distance per lap over time), physics nahi. Usme unknown

T

abhi bhi hai. Physics

v = GM / r

se aati hai, jo phir

T

determine karta hai.

"Kepler discovered

T^{2} \propto a^{3}

from Newton's gravity law."

Error: causation ulta hai. Kepler ne Newton se decades pehle data se ye pattern empirically discover kiya tha; Newton ne baad mein ise apne inverse-square gravity se derive kiya, dono ko confirm karte hue.

Why questions

Why does the orbiting body's mass

m

vanish from the final formula?

Kyunki wahi

m

gravitational force (

GM m / r^{2}

) mein bhi hai aur inertial response (

m v^{2} / r

) mein bhi; dono sides ko

m

se divide karne par wo remove ho jaata hai — yehi deep reason hai ki sabhi masses alike girti hain.

Why does

T

grow faster than

a

rather than in step with it?

Do penalties stack hoti hain: path length

a^{1}

ki tarah badhti hai, aur speed

a^{- 1/2}

ki tarah girti hai. Time = path/speed

\propto a^{1} / a^{- 1/2} = a^{3/2}

hai.

Why is the exponent exactly

3/2

and not, say,

2

Ye gravity ke inverse-square law se trace hota hai.

1/ r^{2}

force se

v^{2} \propto 1/ r

milta hai, aur circumference relation ke saath combine karne par

3/2

power force hoti hai. Alag force law alag exponent deta.

Why can we derive the law using a circle when real orbits are ellipses?

Full elliptical derivation (Kepler's equal-area law ke through) identical formula par collapse hoti hai jisme

r \to a

ho jaata hai; eccentricity cancel ho jaati hai. Circle bahut kam algebra mein sahi answer deta hai.

Why does the same law apply to the Moon around Earth and Earth around the Sun with different constants?

T^{2} \propto a^{3}

ki form universal hai, lekin constant

4 π^{2} / GM

central mass

M

ke saath scale karta hai — Sun aur Earth ke

M

bahut alag hain, isliye alag constants hain.

Why must the inward pull exactly equal

m v^{2} / r

, not more or less, for a circular orbit?

Agar gravity

m v^{2} / r

se zyada hoti toh body inward spiral karti; kam hoti toh outward drift karti. Perfect equality hi radius ko constant rakhti hai — ek true circle.

Edge cases

What happens to the period as

a \to \infty

(orbit infinitely large)?

T \propto a^{3/2} \to \infty

: period bina kisi limit ke badhta jaata hai. Infinitely door pe, gravity vanishingly weak hai aur path infinite hai, isliye ek chakkar forever leta.

What is the "orbit" when eccentricity

e = 1

Ellipse ek parabola mein degenerate ho jaata hai — ek unbound, one-way escape trajectory. Koi repeating period nahi hoti, isliye Kepler's Third Law (jo closed orbit assume karta hai) ab apply nahi hoti.

What about eccentricity

e > 1

Wo ek hyperbolic trajectory hai — parabola se bhi zyada strongly unbound: body ek baar past fly karti hai aur kabhi return nahi karti. Parabola ki tarah iska koi period nahi hai, aur usual bound sense mein koi semi-major axis nahi hai, isliye law simply apply nahi hoti.

Does the law hold if the two masses are comparable, like a binary star?

Sirf ek fix ke saath:

M

ki jagah total

M_{1} + M_{2}

use karo aur

a

ko separation ke roop mein measure karo. Dono stars apne common centre of mass ke around is shared period ke saath orbit karte hain.

Does the period change if the orbit is tilted (different inclination) or run backwards (retrograde vs prograde)?

Nahi. Gravity ek central force hai — ye sirf distance

r

par depend karta hai, orbit ke tilt ya travel ke direction par nahi. Same

a

aur same

M

wale do orbits orientation ki parwah kiye bina same period share karte hain.

What does the formula predict as

r \to 0

(grazing the surface)?

T = 2 π r^{3} / GM \to 0

: infinitely low orbit ka period vanishing hota hai. Reality mein surface (aur atmosphere) tumhe bahut pehle rok deta hai, lekin trend — neeche ke orbits faster hain — real hai.

Is there a distance at which a satellite's period equals Earth's rotation?

Haan — wo specific

a \approx 42, 200

T = 86400

s banata hai, jo geostationary radius hai. Dekho Geostationary & Geosynchronous Orbits; ye diye gaye

T

ke liye

r

solve karna law ka direct application hai.

Can a satellite "hover" motionless over a point on the equator by balancing gravity with nothing?

Nahi. Use geostationary speed par orbit karna hi hoga; stationary rehte hue gravity cancel karne ke liye koi force nahi hai. Ye sirf fixed dikhta hai kyunki iska orbital period Earth ke spin se match karta hai.

Connections

Newton's Law of Universal Gravitation — inverse-square force jo $3/2$ exponent force karta hai.
Centripetal Force & Circular Motion — Step 1 mein equate ki gayi $m v^{2} / r$ requirement.
Kepler's Second Law (equal areas) — kyun eccentricity elliptical case ke liye cancel hoti hai.
Orbital Energy & Vis-viva Equation — speed–distance relationship generalized.
Reduced Mass & Two-Body Problem — comparable masses ke liye $M \to M_{1} + M_{2}$ fix.
Geostationary & Geosynchronous Orbits — real orbits design karne ke liye law inverted.