Keplers Laws

8 MCQs1 revision card9-step worked example

Source: NCERT GravitationPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap: When a planet's orbital radius doubles, most students instinctively double the period. That loses you marks. The period does not scale linearly with radius — it scales as the 3/2 power.

Kepler's three laws (NCERT Class 11 Physics Chapter 7, page 3):

Law of orbits. Every planet moves in an ellipse with the Sun at one focus — not at the centre.
Law of areas. The line joining the planet and the Sun sweeps equal areas in equal time intervals. This means the planet moves faster near perihelion (closest approach) and slower near aphelion (farthest point). No external torque about the Sun → angular momentum is conserved.
Law of periods. T² ∝ a³, where T is the orbital period and a is the semi-major axis. For two planets orbiting the same star: (T₁/T₂)² = (a₁/a₂)³.

Why the third law trips you in NEET: The relationship T ∝ a^(3/2) is non-linear. Doubling the semi-major axis multiplies the period by 2^(3/2) = 2√2 ≈ 2.83 — not 2. Quadrupling a gives T → 4^(3/2) = 8 times the original period. NEET questions routinely offer the linear-scaling answer as a distractor.

Deriving T² ∝ a³ for circular orbits. For a circular orbit of radius r, equating gravitational force to centripetal force:

GMm/r² = mv²/r → v² = GM/r

Period T = 2πr/v, so T² = 4π²r³/(GM). This confirms T² ∝ r³ with the proportionality constant depending only on the central mass M, not on the orbiting body's mass.

Watch out: Kepler's third law compares orbits around the same central body. You cannot use T₁²/T₂² = a₁³/a₂³ to compare a planet orbiting the Sun with a satellite orbiting Earth — the central masses differ.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

According to Kepler's first law, every planet moves in an orbit that is:

MCQ 2Easy RecallPractice

Kepler's second law (law of areas) is a direct consequence of conservation of:

MCQ 3Direct ApplicationPractice

A planet at perihelion is at distance r₁ from the Sun and has speed v₁. At aphelion the distance is r₂. What is the speed at aphelion?

MCQ 4Direct ApplicationPractice

The orbital period of planet A around a star is T. Planet B orbits the same star with a semi-major axis twice that of planet A. The orbital period of B is:

MCQ 5Direct ApplicationPractice

If the semi-major axis of a planet's orbit is increased by a factor of 4, the orbital period increases by a factor of:

MCQ 6Direct ApplicationPractice

Two satellites orbit Earth. Satellite P has orbital radius r and period T. Satellite Q has orbital radius 4r. The ratio T_Q / T_P is:

MCQ 7Easy RecallPractice

A planet sweeps area A in time Δt when it is near perihelion. When the same planet is near aphelion, the area swept in the same time Δt is:

MCQ 8CalculationPractice

Planet X orbits a star with period 27 years and semi-major axis a_X. Planet Y orbits the same star with semi-major axis a_Y = a_X / 3. The orbital period of Y is:

Quick recall before you leave

Worked Example

1
Given
- Satellite 1: semi-major axis a₁ = 2R, period T₁ - Satellite 2: semi-major axis a₂ = 8R - Both orbit Earth (same central mass)
2
Required
Period T₂ of the second satellite.
3
Concept
Kepler's third law: for orbits around the same central body, T² ∝ a³.
4
Formula
T₂²/T₁² = (a₂/a₁)³
5
Substitution
T₂²/T₁² = (8R / 2R)³ = 4³ = 64
6
Calculation
T₂² = 64 T₁² T₂ = 8 T₁ Note on exact values: The ratio a₂/a₁ = 8R/2R = 4 is an exact integer (the R cancels). The number 4 and the exponent 3/2 are exact mathematical quantities — they do not limit significant figures.
7
Final answer
T₂ = 8 T₁ The second satellite's period is exactly 8 times the first.
8
Common trap
A student who treats T as proportional to a (linear scaling) would compute T₂ = 4 T₁. This is the standard wrong answer in NEET. The correct power is 3/2: T₂/T₁ = 4^(3/2) = (√4)³ = 2³ = 8.
9
Similar NEET-style question
"Two planets orbit the same star. Planet A has semi-major axis a and period 10 years. Planet B has semi-major axis 9a. Find the period of planet B." Answer: T_B = 10 × 9^(3/2) = 10 × 27 = 270 years. ---

Before solving, remember these

Law

Kepler's laws of planetary motion

1. Law of orbits: planets orbit the Sun in ellipses with the Sun at one focus. 2. Law of areas: the line from Sun to planet sweeps equal areas in equal times (consequence of angular momentum conservation). 3. Law of periods: T² ∝ r³, where r is the semi-major axis.

-- NCERT Class 11 Physics, Ch. 7, p. 3

Formulas

8 formulas — click to collapse

Escape velocity from a body's surface

v_{e} = \frac{2 GM}{R} = 2 g R

Minimum speed for an object to escape gravity to infinity from radius R. Earth: ~11.2 km/s.

Symbol	Quantity	SI Unit
v_e	escape velocity	m/s
M	planet mass	kg
R	planet radius	m
g	surface gravity	m/s^2

Valid when

Launched from surface
No air drag
Body treated as point/sphere

Gravitational potential energy (point masses)

U = - \frac{GM m}{r}

PE of two-body system; negative because gravity is attractive (work to separate them is positive).

Symbol	Quantity	SI Unit
U	grav PE	J
M, m	two masses	kg
r	separation	m

Valid when

Reference U=0 at r=infinity
Point or spherical masses

g variation with altitude

g_{h} = g (\frac{R}{R + h})^{2} \approx g (1 - 2 h / R)

Gravitational acceleration decreases with altitude above Earth's surface.

Symbol	Quantity	SI Unit
g_h	g at height h	m/s^2
g	surface g	m/s^2
R	Earth radius	m
h	altitude	m

Valid when

Static observer at altitude
Earth treated as uniform sphere

g variation with depth

g_{d} = g (1 - d / R)

Inside Earth (uniform density), g decreases linearly with depth, vanishing at centre.

Symbol	Quantity	SI Unit
g_d	g at depth	m/s^2
g	surface g	m/s^2
R	Earth radius	m
d	depth	m

Valid when

Earth treated as uniform density sphere

Kepler's third law

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Square of orbital period proportional to cube of semi-major axis. Holds for elliptic orbits about a central mass.

Symbol	Quantity	SI Unit
T	orbital period	s
a	semi-major axis	m
M	central mass	kg

Valid when

Two-body system with central mass M >> orbiting mass
Bound orbit

Orbital velocity for circular orbit

v = \frac{GM}{R + h}

Speed of circular orbit at altitude h above body of mass M, radius R.

Symbol	Quantity	SI Unit
v	orbital speed	m/s
M	central mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
M >> orbiting mass

Satellite total mechanical energy

E = - \frac{GM m}{2 ( R + h )}

Total energy = KE + PE = -KE (virial). Always negative for bound orbit; E -> 0 at infinity.

Symbol	Quantity	SI Unit
E	total energy	J
M, m	central mass and satellite mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
Bound (E < 0)

Newton's law of gravitation

F = \frac{G m _{1} m _{2}}{r ^{2}}

Attractive force between any two masses. Inverse-square central force.

Symbol	Quantity	SI Unit
F	force	N
G	grav constant = 6.674e-11	N*m^2/kg^2
m1, m2	masses	kg
r	centre-to-centre distance	m

Valid when

Point masses or spherically symmetric distributions
r > sum of body radii (else use shell theorem)

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

4 items — click to collapse

Trap

g at altitude: linear vs inverse-square

Category: Similar Terms

Student treats g(h) as linear in h. Actual: g(R/(R+h))² (inverse-square).

When it triggers

Question asks for g at significant altitude (e.g. R/2 above surface).

How to avoid

Use g_h = g (R/(R+h))². Linear approximation g(1 - 2h/R) only valid for h << R.

Trap

Kepler's third law: linear vs 3/2 power

Category: Similar Terms

Student treats T proportional to a (linear) instead of a^(3/2).

When it triggers

Question gives change in semi-major axis and asks for new period.

How to avoid

T² ∝ a³, so T ∝ a^(3/2). Doubling a multiplies T by 2^(3/2) ≈ 2.83.

Mistake

Treats g(h) as linear in h for significant heights.

Root cause: formula misuse

Correction

Use g_h = g(R/(R+h))² (inverse-square). Linear approximation g(1-2h/R) is only valid for h << R. For h = R/2, the exact formula gives g_h = (2/3)² g = 4g/9, not g(1-1) = 0.

Mistake