Prerequisites
Before studying this topic, make sure you understand:
- Newton’s Law of Universal Gravitation — Foundation for deriving Kepler’s laws
- Circular Motion — Centripetal force concepts
- Angular Momentum — Conservation principles
The Hook: The Detective Work that Changed Astronomy
Imagine you’re a detective with a massive dataset: 30 years of planetary observations by Tycho Brahe. No telescope (they didn’t exist yet!), just meticulous naked-eye measurements of Mars’s position in the sky.
In 1609, Johannes Kepler analyzed this data and discovered something shocking:
Planets don’t move in perfect circles!
This was heresy — for 2000 years, everyone believed in perfect circular orbits. But Kepler trusted the data over dogma.
His three laws not only described planetary motion — they paved the way for Newton’s law of gravitation 77 years later!
Today: These laws govern everything from ISRO’s Mangalyaan reaching Mars to NASA’s James Webb Space Telescope orbiting the Sun.
From ancient observations to modern space missions — Kepler’s laws connect them all.
The Core Concept
Kepler discovered three empirical laws (based on observation, not derived from first principles). Later, Newton showed these laws follow from universal gravitation and his laws of motion.
For JEE: You need to know:
- Statement of each law
- Mathematical form
- Derivation from Newton’s laws (for Law 3)
- Applications and problem-solving
Kepler’s First Law: Law of Orbits
Statement
“All planets move in elliptical orbits with the Sun at one of the foci.”
Understanding Ellipses
An ellipse has:
- Two foci: $F_1$ and $F_2$
- Semi-major axis: $a$ (half the longest diameter)
- Semi-minor axis: $b$ (half the shortest diameter)
- Eccentricity: $e = \sqrt{1 - \frac{b^2}{a^2}}$ (measures “ovalness”)
For a point on ellipse:
$$r_1 + r_2 = 2a \quad \text{(constant)}$$where $r_1, r_2$ are distances from the two foci.
Circle: $e = 0$ (both foci at center, $a = b$)
Ellipse: $0 < e < 1$ (most planets have e ≈ 0.01 to 0.2)
Parabola: $e = 1$ (open orbit — comet escaping)
Hyperbola: $e > 1$ (faster escape trajectory)
Earth’s orbit: $e = 0.017$ (almost circular!)
Interactive Demo: Kepler’s Laws Visualization
Explore how planets move in elliptical orbits around the Sun. This interactive visualization demonstrates all three of Kepler’s Laws:
How to use this visualization:
- Semi-major axis (a): Adjust the size of the orbit
- Eccentricity (e): Change how elliptical the orbit is (0 = circle, 0.9 = highly elongated)
- Planet Presets: Click Mercury, Venus, Earth, or Mars to see their real orbital parameters
- Second Planet: Enable to compare two orbits and verify T^2/a^3 ratio
- Area Sweep: Shows the purple swept area - notice it stays constant as the planet speeds up near perihelion
Key Points on the Orbit
| Point | Distance from Sun | Name |
|---|---|---|
| Perihelion | $a(1-e)$ | Closest to Sun |
| Aphelion | $a(1+e)$ | Farthest from Sun |
For circular orbit: Perihelion = Aphelion = $a$ (radius)
Kepler’s Second Law: Law of Areas
Statement
“The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.”
Mathematical Form
$$\boxed{\frac{dA}{dt} = \text{constant}}$$where $A$ is the area swept by the radius vector.
This is equivalent to: Conservation of angular momentum!
Derivation
For a small time $dt$, the area swept:
$$dA = \frac{1}{2} r \times (r\,d\theta) = \frac{1}{2}r^2 d\theta$$ $$\frac{dA}{dt} = \frac{1}{2}r^2 \frac{d\theta}{dt} = \frac{1}{2}r^2 \omega$$But $L = mr^2\omega$ (angular momentum), so:
$$\boxed{\frac{dA}{dt} = \frac{L}{2m} = \text{constant}}$$If angular momentum L is conserved (which it is — gravity provides no torque), then area/time is constant.
Physical Consequences
Near the Sun (perihelion):
- Small distance $r$
- To sweep same area, must cover larger arc
- Higher speed $v$
Far from Sun (aphelion):
- Large distance $r$
- Covers smaller arc for same area
- Lower speed $v$
Conservation of angular momentum:
$$L = mrv_{\perp} = \text{constant}$$ $$r_p v_p = r_a v_a$$Example: Earth moves faster in January (perihelion) than in July (aphelion)!
Connection to Angular Momentum
Since $\frac{dA}{dt} = \frac{L}{2m}$ and this is constant:
$$\boxed{L = \text{constant}}$$This proves: Kepler’s Second Law ↔ Conservation of Angular Momentum
Why is L conserved?
- Gravitational force is central (along the line joining centers)
- Torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$ (vectors parallel)
- Zero torque → Angular momentum conserved
Kepler’s Third Law: Law of Periods
Statement
“The square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit.”
Mathematical Form
$$\boxed{T^2 \propto a^3}$$Or:
$$\boxed{\frac{T^2}{a^3} = \frac{4\pi^2}{GM} = \text{constant}}$$where:
- $T$ = Time period of revolution
- $a$ = Semi-major axis of ellipse (= radius for circular orbit)
- $M$ = Mass of the Sun (central body)
For circular orbits: $a = r$ (radius)
$$\boxed{T^2 = \frac{4\pi^2}{GM} r^3}$$Derivation for Circular Orbits
For a planet of mass $m$ in circular orbit of radius $r$ around Sun of mass $M$:
Gravitational force provides centripetal force:
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ $$v^2 = \frac{GM}{r}$$Orbital velocity:
$$v = \sqrt{\frac{GM}{r}}$$Time period:
$$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi r \sqrt{\frac{r}{GM}}$$ $$T = 2\pi \sqrt{\frac{r^3}{GM}}$$Squaring both sides:
$$\boxed{T^2 = \frac{4\pi^2}{GM} r^3}$$This proves: $T^2 \propto r^3$
Universal Nature
The constant $\frac{4\pi^2}{GM}$ depends only on the central mass (Sun).
All planets orbiting the same star:
$$\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3}$$Different systems: Different constants (depends on central mass)
Chandrayaan orbiting Moon:
$$T^2 = \frac{4\pi^2}{GM_{Moon}} r^3$$Mangalyaan orbiting Mars:
$$T^2 = \frac{4\pi^2}{GM_{Mars}} r^3$$Same law, different central mass — different constant!
Comparing the Three Laws
| Law | Statement | Mathematical Form | Physical Meaning |
|---|---|---|---|
| First | Elliptical orbits | $r = \frac{a(1-e^2)}{1+e\cos\theta}$ | Shape of orbit |
| Second | Equal areas in equal times | $\frac{dA}{dt} = \frac{L}{2m}$ | Angular momentum conservation |
| Third | $T^2 \propto a^3$ | $\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$ | Relates period to distance |
Applications and Extensions
Satellite Motion Around Earth
For satellites orbiting Earth, replace $M$ (Sun’s mass) with $M_E$ (Earth’s mass):
$$T^2 = \frac{4\pi^2}{GM_E} r^3$$Using $g = \frac{GM_E}{R_E^2}$:
$$T^2 = \frac{4\pi^2}{gR_E^2} r^3$$For satellite close to surface: $r \approx R_E$
$$T = 2\pi\sqrt{\frac{R_E}{g}} \approx 84 \text{ minutes}$$Finding Unknown Masses
Kepler’s third law allows us to find the mass of central body:
$$M = \frac{4\pi^2 a^3}{GT^2}$$Example: By observing Jupiter’s moon Io (period and distance), we can calculate Jupiter’s mass!
This is how astronomers find masses of planets and stars.
Memory Tricks & Patterns
Mnemonic for the Three Laws
“Elephants Always Travel” → Ellipse, Area, Time-period
Or: “Only Astronomers Think” → Orbit, Area, Time
The 2-3 Pattern in Law 3
$$T^{\boxed{2}} \propto a^{\boxed{3}}$$“Two squared, Three cubed” → $T^2 \propto a^3$
Power Relationships
From Kepler’s Third Law: $T^2 \propto r^3$
This means:
- $T \propto r^{3/2}$
- $r \propto T^{2/3}$
Doubling the distance:
- New period: $T' = 2^{3/2} T = 2\sqrt{2} \, T \approx 2.83T$
Doubling the period:
- New distance: $r' = 2^{2/3} r \approx 1.59r$
For JEE: These relationships save calculation time!
When to Use Which Law
Use First Law when:
- Finding closest/farthest distance (perihelion/aphelion)
- Dealing with elliptical orbits
- Shape and eccentricity questions
Use Second Law when:
- Comparing speeds at different points
- Angular momentum problems
- “Equal area in equal time” questions
- Proving conservation of angular momentum
Use Third Law when:
- Finding time period from orbital radius
- Finding orbital radius from time period
- Comparing different orbits around same body
- Finding mass of central body
Most JEE problems use Third Law — memorize $T^2 \propto r^3$ cold!
Common Mistakes to Avoid
Wrong: $T^3 \propto r^2$ ❌
Correct: $T^2 \propto r^3$ ✓
Remember: Time is squared (2), Distance is cubed (3)
For circular orbits:
- $a$ = radius $r$ (semi-major axis = radius)
For elliptical orbits:
- $a$ = semi-major axis (NOT the distance at any given time)
In formulas: Always use $a$ (semi-major axis), which equals $r$ for circles.
The constant in Kepler’s Third Law depends on central body’s mass:
Around Sun: $\frac{T^2}{a^3} = \frac{4\pi^2}{GM_{Sun}}$
Around Earth: $\frac{T^2}{a^3} = \frac{4\pi^2}{GM_{Earth}}$
Different central mass → Different constant!
JEE Trap: Comparing Earth’s orbit around Sun with Moon’s orbit around Earth — different systems, different constants!
Higher speed at perihelion (closer to Sun), not at aphelion!
From $r_p v_p = r_a v_a$ and $r_p < r_a$:
$$v_p > v_a$$Think: Falling towards Sun → speed increases (like projectile descending)
Practice Problems
Level 1: Foundation (NCERT/Basics)
If Earth’s orbital radius is $1.5 \times 10^{11}$ m and period is 1 year, find Mars’s period if its orbital radius is $2.28 \times 10^{11}$ m.
Solution:
Using Kepler’s Third Law (same central body — Sun):
$$\frac{T_M^2}{T_E^2} = \frac{a_M^3}{a_E^3}$$ $$T_M = T_E \times \left(\frac{a_M}{a_E}\right)^{3/2}$$ $$T_M = 1 \times \left(\frac{2.28 \times 10^{11}}{1.5 \times 10^{11}}\right)^{3/2}$$ $$T_M = \left(\frac{2.28}{1.5}\right)^{3/2} = (1.52)^{1.5}$$ $$T_M = 1.52 \times \sqrt{1.52} = 1.52 \times 1.233 \approx 1.87 \text{ years}$$Answer: 1.87 years (Mars takes about 687 days to orbit the Sun)
Actual value: 1.88 years — very close!
A planet moves in an elliptical orbit around the Sun. Its speed at perihelion is $v_p = 30$ km/s. If the perihelion distance is $r_p = 1.5 \times 10^{11}$ m and aphelion distance is $r_a = 2.5 \times 10^{11}$ m, find the speed at aphelion.
Solution:
Using conservation of angular momentum (Kepler’s Second Law):
$$r_p v_p = r_a v_a$$ $$v_a = \frac{r_p v_p}{r_a} = \frac{1.5 \times 10^{11} \times 30}{2.5 \times 10^{11}}$$ $$v_a = \frac{1.5 \times 30}{2.5} = \frac{45}{2.5} = 18 \text{ km/s}$$Answer: 18 km/s
Check: $r_p < r_a$ and $v_p > v_a$ ✓ (Faster when closer)
Level 2: JEE Main Type
A satellite orbits Earth at height equal to Earth’s radius (R = 6400 km, g = 10 m/s²). Find its time period.
Solution:
Distance from Earth’s center: $r = R + h = R + R = 2R$
Using Kepler’s Third Law:
$$T = 2\pi\sqrt{\frac{r^3}{GM}}$$We know $GM = gR^2$:
$$T = 2\pi\sqrt{\frac{r^3}{gR^2}} = 2\pi\sqrt{\frac{(2R)^3}{gR^2}}$$ $$T = 2\pi\sqrt{\frac{8R^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}$$ $$T = 2\pi \times \sqrt{2} \times \sqrt{\frac{R}{g}}$$Note: $2\pi\sqrt{\frac{R}{g}}$ is the period for satellite at surface.
$$T_{surface} = 2\pi\sqrt{\frac{6.4 \times 10^6}{10}} = 2\pi \times 800 \approx 5024 \text{ s} \approx 84 \text{ min}$$ $$T = \sqrt{8} \times T_{surface} = 2\sqrt{2} \times 84$$ $$T = 2.828 \times 84 \approx 237 \text{ minutes} \approx 4 \text{ hours}$$Answer: Approximately 4 hours or 237 minutes
More precise calculation:
$$T = 2\pi\sqrt{\frac{8 \times 6.4 \times 10^6}{10}} = 2\pi\sqrt{5.12 \times 10^6}$$ $$T = 2\pi \times 2263 \approx 14,210 \text{ s} \approx 237 \text{ min}$$The ratio of the distances of two planets from the Sun is 1:4. What is the ratio of their time periods?
Solution:
Given: $\frac{r_1}{r_2} = \frac{1}{4}$
From Kepler’s Third Law: $T^2 \propto r^3$
$$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$$ $$\frac{T_1}{T_2} = \sqrt{\frac{1}{64}} = \frac{1}{8}$$Answer: 1:8
Interpretation: The closer planet completes 8 orbits in the time the farther planet completes 1 orbit!
A satellite’s period is doubled. By what factor does its orbital radius change?
Solution:
From $T^2 \propto r^3$:
$$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$$Given: $T_2 = 2T_1$
$$\frac{(2T_1)^2}{T_1^2} = \frac{r_2^3}{r_1^3}$$ $$4 = \frac{r_2^3}{r_1^3}$$ $$r_2 = r_1 \times 4^{1/3} = r_1 \times 2^{2/3}$$ $$r_2 \approx 1.587 \, r_1$$Answer: Radius increases by factor of $2^{2/3} \approx 1.587$ or about 59%
Shortcut: $T \propto r^{3/2}$, so if $T$ is doubled: $r \to r \times 2^{2/3}$
Level 3: JEE Advanced Type
Two satellites orbit Earth with periods in the ratio 1:8. Find the ratio of their: (a) Orbital radii (b) Orbital speeds (c) Angular velocities
Solution:
Given: $\frac{T_1}{T_2} = \frac{1}{8}$
(a) Orbital radii:
From $T^2 \propto r^3$:
$$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$$ $$\frac{1}{64} = \frac{r_1^3}{r_2^3}$$ $$\frac{r_1}{r_2} = \left(\frac{1}{64}\right)^{1/3} = \frac{1}{4}$$Answer (a): 1:4
(b) Orbital speeds:
$$v = \frac{2\pi r}{T}$$ $$\frac{v_1}{v_2} = \frac{r_1}{r_2} \times \frac{T_2}{T_1} = \frac{1}{4} \times \frac{8}{1} = 2$$Answer (b): 2:1 (Closer satellite is faster!)
Alternative: From $v = \sqrt{\frac{GM}{r}}$:
$$\frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}} = \sqrt{4} = 2$$✓
(c) Angular velocities:
$$\omega = \frac{2\pi}{T}$$ $$\frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} = \frac{8}{1}$$Answer (c): 8:1
Summary:
- Radii: 1:4
- Speeds: 2:1
- Angular velocities: 8:1
A planet moves in an elliptical orbit with eccentricity $e = 0.5$ and semi-major axis $a = 2 \times 10^{11}$ m around the Sun. Find: (a) Perihelion and aphelion distances (b) Ratio of speeds at perihelion and aphelion (c) Time period (G = $6.67 \times 10^{-11}$ N·m²/kg², $M_{Sun} = 2 \times 10^{30}$ kg)
Solution:
(a) Perihelion and aphelion distances:
$$r_p = a(1-e) = 2 \times 10^{11} \times (1-0.5) = 1 \times 10^{11} \text{ m}$$ $$r_a = a(1+e) = 2 \times 10^{11} \times (1+0.5) = 3 \times 10^{11} \text{ m}$$Answer (a): $r_p = 1 \times 10^{11}$ m, $r_a = 3 \times 10^{11}$ m
(b) Ratio of speeds:
From conservation of angular momentum:
$$\frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{3 \times 10^{11}}{1 \times 10^{11}} = 3$$Answer (b): 3:1 (Three times faster at perihelion!)
(c) Time period:
$$T = 2\pi\sqrt{\frac{a^3}{GM}}$$ $$T = 2\pi\sqrt{\frac{(2 \times 10^{11})^3}{6.67 \times 10^{-11} \times 2 \times 10^{30}}}$$ $$T = 2\pi\sqrt{\frac{8 \times 10^{33}}{13.34 \times 10^{19}}}$$ $$T = 2\pi\sqrt{0.6 \times 10^{14}} = 2\pi \times 7.75 \times 10^6$$ $$T \approx 4.87 \times 10^7 \text{ s}$$Converting to years: $T = \frac{4.87 \times 10^7}{365.25 \times 24 \times 3600} = \frac{4.87 \times 10^7}{3.156 \times 10^7} \approx 1.54$ years
Answer (c): 1.54 years or about 563 days
Jupiter’s moon Io has an orbital period of 1.77 days and orbital radius of $4.22 \times 10^8$ m. Using this data, calculate the mass of Jupiter. (G = $6.67 \times 10^{-11}$ N·m²/kg²)
Solution:
From Kepler’s Third Law:
$$T^2 = \frac{4\pi^2}{GM} r^3$$ $$M = \frac{4\pi^2 r^3}{GT^2}$$Convert period to seconds:
$$T = 1.77 \times 24 \times 3600 = 152,928 \text{ s}$$ $$M = \frac{4\pi^2 \times (4.22 \times 10^8)^3}{6.67 \times 10^{-11} \times (152,928)^2}$$ $$M = \frac{4 \times 9.87 \times 75.15 \times 10^{24}}{6.67 \times 10^{-11} \times 2.338 \times 10^{10}}$$ $$M = \frac{2967 \times 10^{24}}{15.59 \times 10^{-1}} = \frac{2967 \times 10^{24}}{1.559}$$ $$M \approx 1.90 \times 10^{27} \text{ kg}$$Answer: $M_{Jupiter} \approx 1.90 \times 10^{27}$ kg
Check: Actual mass of Jupiter = $1.898 \times 10^{27}$ kg — excellent agreement!
This is how astronomers find masses of distant planets!
Quick Revision Box
| Law | Key Formula | JEE Application |
|---|---|---|
| First Law | $r = \frac{a(1-e^2)}{1+e\cos\theta}$ | Perihelion: $a(1-e)$, Aphelion: $a(1+e)$ |
| Second Law | $\frac{dA}{dt} = \frac{L}{2m}$ | $r_p v_p = r_a v_a$ (speed comparison) |
| Third Law | $T^2 = \frac{4\pi^2}{GM}r^3$ | Finding T from r, or vice versa |
Key Relations:
- $T \propto r^{3/2}$
- $v \propto r^{-1/2}$
- $\omega \propto r^{-3/2}$
Special Values:
- Earth’s orbit: $a = 1.5 \times 10^{11}$ m, $T = 365.25$ days, $e = 0.017$
- Satellite at Earth’s surface: $T \approx 84$ minutes
Related Topics
Within Gravitation Chapter
- Universal Gravitation — Foundation for Kepler’s laws
- Gravitational Field and Potential — Energy perspective
- Satellites — Applying Kepler’s laws to artificial satellites
Connected Chapters
- Circular Motion — Centripetal force in orbits
- Angular Momentum — Conservation in Second Law
- Work and Energy — Energy in elliptical orbits
- Conservation of Energy — Total energy in orbits
Math Connections
- Conic Sections — Ellipse, parabola, hyperbola
- Differentiation — Deriving area rates
- Logarithms — Solving period-radius problems
Teacher’s Summary
Three Laws, One Foundation: All three laws follow from Newton’s law of gravitation — observation led to theory!
First Law: Orbits are ellipses (circles are special case) — Sun at one focus, not center
Second Law: Equal areas = Conservation of angular momentum — faster when closer to Sun
Third Law: $T^2 \propto r^3$ — THE most important for JEE calculations! Know it cold.
Universal Application: Same laws govern planets, moons, satellites, binary stars — from solar system to galaxies
“Kepler looked at data and found patterns. Newton explained WHY those patterns exist. You must know both — the WHAT (Kepler’s laws) and the WHY (Newton’s derivation). Master this, and the cosmos becomes predictable!”