StudyOSStudyOS CBSE Study Material, Class 5–12

Gravitation

NCERT Class 11 · Physics Based on NCERT Class 11 Physics textbook · Free CBSE study kit

Everything you need to ace this chapter, in one place.
Notes · Mind Map · Quiz · Assertion-Reason · Case Studies · Question Bank · Past Papers

Get Full Study Kit →
Studying with friends? Send them this chapter.

Chapter Notes

Overview
This chapter explores the universal force of gravitation that governs the motion of celestial bodies and objects on Earth. It begins with historical models of planetary motion, leading to Kepler's laws and Newton's universal law of gravitation. The big idea is that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, a concept that explains everything from falling apples to orbiting satellites.
Key Concepts (16)
Geocentric model
A model proposed by Ptolemy in which all celestial objects, including the Sun and planets, revolve around the Earth.
e.g. Ancient astronomers believed the Sun rose and set because it moved around a stationary Earth.
Heliocentric model
A model in which the Sun is the centre around which the planets revolve, proposed by Aryabhatta and later by Copernicus.
e.g. We now know that Earth and other planets orbit the Sun, not the other way around.
Kepler's first law (Law of orbits)
All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.
e.g. Earth's orbit around the Sun is an ellipse, with the Sun at one focus, making its distance from the Sun vary slightly during the year.
Perihelion
The point in a planet's elliptical orbit that is closest to the Sun.
e.g. Earth reaches its perihelion around early January each year.
Aphelion
The point in a planet's elliptical orbit that is farthest from the Sun.
e.g. Earth reaches its aphelion around early July each year.
Semi-major axis
Half the distance between the perihelion and aphelion points of an elliptical orbit; it is the average distance of the planet from the Sun.
e.g. The semi-major axis of Earth's orbit is about 15.0 × 10^10 m.
Kepler's second law (Law of areas)
The line that joins any planet to the Sun sweeps equal areas in equal intervals of time.
e.g. A planet moves faster when it is near the Sun (perihelion) and slower when it is far away (aphelion).
Kepler's third law (Law of periods)
The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.
e.g. If a planet's orbit is larger, its orbital period is longer; for Earth, T = 1 year and a = 15.0 × 10^10 m.
Universal law of gravitation
Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
e.g. The force that pulls an apple toward Earth is the same force that keeps the Moon in orbit around Earth.
Universal gravitational constant (G)
The constant of proportionality in Newton's law of gravitation, with a value of 6.67 × 10⁻¹¹ N m²/kg².
e.g. Cavendish's experiment was the first to measure G accurately using a torsion balance.
Acceleration due to gravity (g)
The acceleration experienced by a mass due to the gravitational pull of the Earth, given by g = GM_E / R_E².
e.g. On the surface of Earth, g is approximately 9.8 m/s².
Gravitational potential energy
The energy stored in a body due to its position in a gravitational field, defined as V = -G m₁ m₂ / r, with zero potential energy at infinity.
e.g. A satellite in orbit has negative total energy, indicating it is bound to Earth.
Escape speed
The minimum speed required for an object to escape the gravitational pull of a planet without further propulsion, given by v_esc = √(2GM_E / R_E) = √(2gR_E).
e.g. The escape speed from Earth is about 11.2 km/s; from the Moon it is only 2.3 km/s.
Earth satellite
An object that revolves around the Earth in a circular or elliptical orbit, such as the Moon or artificial satellites.
e.g. The Moon is Earth's only natural satellite, with an orbital period of about 27.3 days.
Central force
A force that is always directed along the line joining the centre of the force and the particle, such as gravitational force.
e.g. The gravitational force on a planet is always directed toward the Sun, making it a central force.
Principle of superposition
The resultant gravitational force on a particle is the vector sum of the individual forces exerted by all other particles.
e.g. The force on a mass at the centroid of an equilateral triangle of three equal masses is zero due to symmetry and superposition.
Formulas & Equations (15)
F = G m₁ m₂ / r²
Reads as: F equals G times m1 times m2 divided by r squared
Variables: F = gravitational force (N), G = universal gravitational constant (6.67 × 10⁻¹¹ N m²/kg²), m₁, m₂ = masses of the two bodies (kg), r = distance between their centres (m)
Use when: Calculating the gravitational force between any two point masses or spherical objects.
⚠ Forgetting that r is the distance between centres, not the surface-to-surface distance.
F = - (G m₁ m₂ / r²) r̂
Reads as: F equals negative G times m1 times m2 divided by r squared times the unit vector r hat
Variables: r̂ = unit vector from m₁ to m₂; the negative sign indicates the force is attractive (toward the other mass).
Use when: Writing the gravitational force in vector form for problems involving multiple forces.
⚠ Omitting the negative sign or the unit vector, leading to an incorrect direction.
ΔA / Δt = L / (2m)
Reads as: Delta A over delta t equals L over 2m
Variables: ΔA = area swept (m²), Δt = time interval (s), L = angular momentum (kg m²/s), m = mass of the planet (kg)
Use when: Deriving Kepler's second law from conservation of angular momentum.
⚠ Confusing area swept with distance travelled; area is proportional to angular momentum.
T² ∝ a³
Reads as: T squared is proportional to a cubed
Variables: T = time period of revolution (s or years), a = semi-major axis of the orbit (m or AU)
Use when: Relating the orbital period of a planet or satellite to the size of its orbit.
⚠ Forgetting that the proportionality constant depends on the mass of the central body.
T² = (4π² / GM) a³
Reads as: T squared equals 4 pi squared divided by G M times a cubed
Variables: M = mass of the central body (kg), a = semi-major axis (m), G = gravitational constant
Use when: Calculating the mass of a central body (e.g., Sun, Earth, Mars) from a satellite's orbital data.
⚠ Using the mass of the satellite instead of the central body.
g = GM_E / R_E²
Reads as: g equals G times M_E divided by R_E squared
Variables: g = acceleration due to gravity on Earth's surface (m/s²), M_E = mass of Earth (kg), R_E = radius of Earth (m)
Use when: Calculating g on the surface of Earth or any other planet.
⚠ Using the radius of the orbit instead of the planet's radius when calculating surface gravity.
g(h) = GM_E / (R_E + h)²
Reads as: g at height h equals G times M_E divided by R_E plus h squared
Variables: h = height above Earth's surface (m)
Use when: Finding the acceleration due to gravity at a height h above the surface.
⚠ Forgetting that the distance from the centre is R_E + h, not just h.
g(h) ≈ g (1 - 2h / R_E) for h << R_E
Reads as: g at height h is approximately g times 1 minus 2h over R_E
Variables: g = surface gravity (m/s²), h = height (m), R_E = Earth's radius (m)
Use when: Quickly estimating the decrease in g for small heights (e.g., hills, towers).
⚠ Using the linear approximation when h is not negligible compared to R_E.
g(d) = g (1 - d / R_E)
Reads as: g at depth d equals g times 1 minus d over R_E
Variables: d = depth below Earth's surface (m)
Use when: Calculating the acceleration due to gravity at a depth d inside Earth (assuming uniform density).
⚠ Forgetting that only the mass inside the radius (R_E - d) contributes to gravity.
V = -G m₁ m₂ / r
Reads as: V equals negative G times m1 times m2 divided by r
Variables: V = gravitational potential energy (J), r = separation between masses (m)
Use when: Calculating the potential energy of a system of two point masses, with zero at infinity.
⚠ Forgetting the negative sign, which indicates the force is attractive and the system is bound.
v_esc = √(2GM_E / R_E) = √(2gR_E)
Reads as: v escape equals square root of 2 G M_E over R_E, which equals square root of 2 g R_E
Variables: v_esc = escape speed (m/s), M_E = mass of Earth (kg), R_E = radius of Earth (m), g = surface gravity (m/s²)
Use when: Finding the minimum speed needed for an object to escape Earth's gravitational field.
⚠ Confusing escape speed with orbital speed; escape speed is √2 times the orbital speed near the surface.
v = √(GM_E / (R_E + h))
Reads as: v equals square root of G M_E over R_E plus h
Variables: v = orbital speed (m/s), h = height above surface (m)
Use when: Calculating the speed of a satellite in a circular orbit at height h.
⚠ Using the mass of the satellite instead of Earth's mass in the numerator.
T = 2π √((R_E + h)³ / GM_E)
Reads as: T equals 2 pi times square root of R_E plus h cubed over G M_E
Variables: T = time period of satellite (s)
Use when: Finding the orbital period of a satellite in a circular orbit.
⚠ Forgetting to cube (R_E + h) or using the wrong power.
T₀ = 2π √(R_E / g)
Reads as: T naught equals 2 pi times square root of R_E over g
Variables: T₀ = time period of a satellite very close to Earth's surface (s), R_E = Earth's radius (m), g = surface gravity (m/s²)
Use when: Calculating the approximate period of a low Earth orbit satellite (about 85 minutes).
⚠ Using this formula for satellites at significant heights; it is only valid for h << R_E.
E = -G M_E m / (2 (R_E + h))
Reads as: E equals negative G times M_E times m divided by 2 times R_E plus h
Variables: E = total mechanical energy of an orbiting satellite (J), m = mass of satellite (kg)
Use when: Finding the total energy of a satellite in a circular orbit; it is negative and half the potential energy.
⚠ Forgetting the factor of 1/2; total energy is half of potential energy (not equal).
How It Works
Derivation of Kepler's second law from angular momentum conservation
  1. Consider a planet of mass m moving under a central gravitational force directed along the radius vector r.
  2. The area swept out by the radius vector in a small time Δt is ΔA = ½ |r × v Δt|.
  3. Divide by Δt: ΔA/Δt = ½ |r × v| = ½ |r × p| / m = L / (2m), where L is angular momentum.
  4. Since the force is central, torque is zero, so angular momentum L is constant.
  5. Therefore, ΔA/Δt is constant, meaning equal areas are swept in equal times.
💡 Students often forget that the law of areas is a direct consequence of conservation of angular momentum, not an independent assumption.
Cavendish experiment to measure G
  1. A light bar AB with two small lead spheres is suspended by a fine wire.
  2. Two large lead spheres are brought close to the small spheres on opposite sides.
  3. The gravitational attraction between large and small spheres produces a torque that twists the wire.
  4. The wire twists until the restoring torque equals the gravitational torque.
  5. The angle of twist θ is measured, and the restoring torque per unit angle τ is known from calibration.
  6. Using the equation τθ = G M m L / d², G is calculated.
💡 The experiment measures the tiny gravitational force between laboratory masses; the force is proportional to G, so G can be determined.
Variation of g with height and depth
  1. At height h: g(h) = GM_E / (R_E + h)²; for h << R_E, use binomial expansion: g(h) ≈ g (1 - 2h/R_E).
  2. At depth d: Earth is treated as a sphere of radius (R_E - d) plus a shell of thickness d.
  3. The shell exerts no net force on a mass inside it.
  4. Only the inner sphere of mass M_s = M_E (R_E - d)³ / R_E³ contributes.
  5. Thus g(d) = G M_s / (R_E - d)² = g (1 - d/R_E).
💡 g is maximum on the surface and decreases both above and below; the decrease is linear with depth but quadratic with height.
Energy of an orbiting satellite
  1. For a satellite in a circular orbit at distance r = R_E + h, centripetal force is provided by gravity: mv²/r = GM_E m / r².
  2. Solve for v² = GM_E / r, then kinetic energy K = ½ mv² = GM_E m / (2r).
  3. Potential energy U = -GM_E m / r (with zero at infinity).
  4. Total energy E = K + U = -GM_E m / (2r), which is negative for a bound orbit.
💡 The total energy is negative; kinetic energy is positive and half the magnitude of potential energy.
Escape speed derivation using energy conservation
  1. Consider an object of mass m on Earth's surface with initial speed v_i.
  2. Its initial total energy = ½ m v_i² - GM_E m / R_E.
  3. At infinity, potential energy is zero, and let final speed be v_f.
  4. By conservation of energy: ½ m v_i² - GM_E m / R_E = ½ m v_f².
  5. For the object to just reach infinity (v_f = 0), set v_i² = 2GM_E / R_E.
  6. Thus escape speed v_esc = √(2GM_E / R_E) = √(2gR_E).
💡 Escape speed is independent of the mass of the escaping object; it depends only on the planet's mass and radius.
Exam Traps — Don't Make These Mistakes
⚠ Confusing 'g' with 'G'
✗ g = 6.67 × 10⁻¹¹ N m²/kg²
→ g is acceleration due to gravity (9.8 m/s² on Earth), while G is the universal gravitational constant. They are completely different quantities.
✓ g = GM_E / R_E² = 9.8 m/s²; G = 6.67 × 10⁻¹¹ N m²/kg²
⚠ Using wrong distance in Newton's law
✗ F = G m₁ m₂ / (height above surface)²
→ The distance r in the law is the distance between the centres of the two masses, not the height above the surface.
✓ r = R_E + h (for an object at height h above Earth's surface)
⚠ Forgetting the negative sign in gravitational potential energy
✗ U = +G m₁ m₂ / r
→ Gravitational force is attractive, so potential energy is negative when zero is set at infinity. A positive sign would imply repulsion.
✓ U = -G m₁ m₂ / r
⚠ Confusing orbital speed with escape speed
✗ Escape speed = orbital speed near Earth's surface
→ Orbital speed near surface is v = √(gR_E) ≈ 7.9 km/s, while escape speed is √2 times that, about 11.2 km/s.
✓ v_orbital = √(GM_E / R_E) = √(gR_E); v_escape = √(2GM_E / R_E) = √(2gR_E)
⚠ Applying Kepler's third law with wrong central mass
✗ T² ∝ a³ for a satellite around Earth, using the Sun's mass
→ The proportionality constant in T² = (4π²/GM) a³ depends on the mass of the central body (Earth for satellites, Sun for planets).
✓ For Earth satellites: T² = (4π² / GM_E) a³; for planets: T² = (4π² / GM_Sun) a³
⚠ Assuming g is constant at all heights and depths
✗ g = 9.8 m/s² everywhere
→ g decreases with height (g ∝ 1/r²) and with depth (g ∝ (R_E - d)). It is maximum only on the surface.
✓ g(h) = g (1 - 2h/R_E) for small h; g(d) = g (1 - d/R_E)
Compare & Contrast
Centre of the universe
Geocentric model
Heliocentric model
In the geocentric model, Earth is at the centre; in the heliocentric model, the Sun is at the centre.
Shape of planetary orbits
Kepler's first law (ellipse)
Copernican model (circle)
Kepler's law states orbits are elliptical with the Sun at one focus; Copernicus assumed circular orbits.
Definition
Gravitational potential energy (U)
Gravitational potential (V)
U is the energy of a mass m in a gravitational field (U = -GMm/r); V is the potential energy per unit mass (V = -GM/r).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 12 marks
Most asked: Newton's universal law of gravitation and its applications · Kepler's laws of planetary motion · Variation of g with height and depth · Escape speed and orbital speed of satellites · Energy of an orbiting satellite
Don't skip: Derivation of escape speed using energy conservation; Numerical problems on orbital velocity and time period of satellites; Comparison of g at different heights and depths; Proof of Kepler's second law from angular momentum conservation
One-Line Revision (read aloud before the exam)
  1. Newton's universal law of gravitation: F = G m₁ m₂ / r², where G = 6.67 × 10⁻¹¹ N m²/kg².
  2. Kepler's first law: All planets move in elliptical orbits with the Sun at one focus.
  3. Kepler's second law: The radius vector sweeps equal areas in equal times, due to conservation of angular momentum.
  4. Kepler's third law: T² ∝ a³, with T² = (4π²/GM) a³ for a central mass M.
  5. Acceleration due to gravity on Earth's surface: g = GM_E / R_E² ≈ 9.8 m/s².
  6. g decreases with height: g(h) = GM_E / (R_E + h)²; for small h, g(h) ≈ g (1 - 2h/R_E).
  7. g decreases with depth: g(d) = g (1 - d/R_E), assuming uniform density.
  8. Gravitational potential energy: U = -G m₁ m₂ / r, with zero at infinity.
  9. Escape speed from Earth: v_esc = √(2GM_E / R_E) = √(2gR_E) ≈ 11.2 km/s.
  10. Orbital speed of a satellite: v = √(GM_E / (R_E + h)).
  11. Time period of a satellite: T = 2π √((R_E + h)³ / GM_E).
  12. Total energy of a satellite in circular orbit: E = -GM_E m / (2r), negative for bound orbits.
  13. Cavendish experiment measured G using a torsion balance.
  14. The Moon's orbital period is about 27.3 days; its distance from Earth is 3.84 × 10⁸ m.
  15. For a satellite very close to Earth's surface, T₀ ≈ 85 minutes.

MCQs — 10 Questions with Answers

Q1. According to Kepler's first law, planets move in orbits that are:

  • A. Perfect circles with the Sun at the centre
  • B. Ellipses with the Sun at one of the foci ✓
  • C. Parabolas with the Sun at the vertex
  • D. Hyperbolas with the Sun between the foci

Answer: B — Kepler's first law explicitly states that planetary orbits are elliptical with the Sun at one focus, not circles as Copernicus proposed.

Q2. If a planet is at perihelion (closest to Sun) with orbital speed v_P and distance r_P, and at aphelion (farthest) with speed v_A and distance r_A, which relation is correct?

  • A. v_P × r_P = v_A × r_A (from angular momentum conservation) ✓
  • B. v_P = v_A (speeds are equal)
  • C. r_P × r_A = constant
  • D. v_P/v_A = r_P/r_A

Answer: A — Angular momentum L = mvr is conserved for a central force; hence m·v_P·r_P = m·v_A·r_A, giving v_P·r_P = v_A·r_A.

Q3. The gravitational constant G in SI units is numerically equal to:

  • A. 9.8 m/s²
  • B. 6.67 × 10⁻¹¹ N·m²/kg² ✓
  • C. 3 × 10⁸ m/s
  • D. 1.67 × 10⁻²⁷ kg

Answer: B — G = 6.67 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant; g ≈ 9.8 m/s² is acceleration due to gravity at Earth's surface.

Q4. A satellite orbits Earth at height h = R (where R is Earth's radius). What is its orbital speed compared to the orbital speed at Earth's surface?

  • A. v = v_surface
  • B. v = v_surface/√2 ✓
  • C. v = v_surface × √2
  • D. v = 0 (cannot orbit at this height)

Answer: B — Orbital speed v_o = √(GM/r); at height h = R, orbital radius is r = 2R, so v = √(GM/2R) = √(GM/R)/√2 = v_surface/√2.

Q5. Kepler's second law (law of areas) is a direct consequence of which principle?

  • A. Conservation of mechanical energy
  • B. Conservation of angular momentum ✓
  • C. Newton's first law of motion
  • D. Newton's third law (action-reaction)

Answer: B — Since gravitational force is a central force with zero torque about the Sun, angular momentum is conserved, leading to constant areal velocity (law of areas).

Q6. If the escape speed from Earth's surface is v_e = 11.2 km/s and a planet has twice Earth's mass and twice Earth's radius, the escape speed from that planet's surface is approximately:

  • A. 5.6 km/s
  • B. 11.2 km/s ✓
  • C. 15.8 km/s
  • D. 22.4 km/s

Answer: B — v_e = √(2GM/R); if M → 2M and R → 2R, then v_e(new) = √(2G·2M/2R) = √(2GM/R) = v_e(original) ≈ 11.2 km/s.

Q7. Assertion: Acceleration due to gravity is greater at the equator than at the poles. Reason: Earth's equator is closer to the centre than the poles.

  • A. Both assertion and reason are correct, and reason explains assertion
  • B. Both assertion and reason are correct, but reason does not explain assertion
  • C. Assertion is incorrect; reason is correct
  • D. Assertion is correct; reason is incorrect ✓

Answer: D — g is actually greater at poles and smaller at equator due to Earth's rotation (centrifugal effect) and oblate shape; the equator is farther from centre, reducing g further.

Q8. A geostationary satellite must have an orbital period equal to Earth's rotation period (24 hours). Its orbital radius from Earth's centre is approximately:

  • A. 6,400 km
  • B. 36,000 km
  • C. 42,000 km ✓
  • D. 384,000 km

Answer: C — Using T² = (4π²/GM)r³ with T = 24 hours, r ≈ 42,000 km from Earth's centre (or ~36,000 km altitude above surface).

Q9. The total mechanical energy of a satellite in circular orbit at radius r is E = −GMm/(2r). Which statement is NOT correct?

  • A. Total energy is negative, indicating a bound orbit
  • B. Kinetic energy equals half the magnitude of potential energy
  • C. If the satellite moves to a higher orbit, its total energy increases
  • D. The orbital speed decreases as the satellite moves to a lower orbit ✓

Answer: D — As a satellite moves to a lower orbit (smaller r), orbital speed v_o = √(GM/r) increases, not decreases; all other statements are correct.

Q10. Derive that for a planet in circular orbit, the orbital speed v_o = √(GM/r), starting from Newton's law of gravitation and centripetal force requirement. [HOTS]

  • A. Gravitational force equals centripetal force: GMm/r² = mv_o²/r → v_o = √(GM/r) ✓
  • B. Kinetic energy equals potential energy: ½mv_o² = GMm/r → v_o = √(2GM/r)
  • C. Angular momentum is conserved: mv_o·r = constant → v_o = constant/r
  • D. Escape speed at orbit is v_o = √(2GM/r)

Answer: A — For circular orbit, gravitational force provides centripetal force: GMm/r² = mv_o²/r; solving for v_o gives v_o = √(GM/r).

Flashcards

What is the mathematical statement of Kepler's first law?

All planets move in elliptical orbits with the Sun located at one of the two foci of the ellipse.

State Kepler's second law (law of areas) in one sentence.

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Write Kepler's third law as a mathematical equation.

T² is proportional to a³, or T² = ka³ where k is a constant and a is the semi-major axis.

What is the universal law of gravitation formula?

F = GMm/r², where G is the gravitational constant, M and m are masses, and r is the distance between them.

What is the numerical value and SI unit of the gravitational constant G?

G = 6.67 × 10⁻¹¹ N·m²/kg² (or m³·kg⁻¹·s⁻²).

How does acceleration due to gravity g change with altitude h above Earth's surface?

g decreases with altitude: g_h = g₀(R/(R+h))² or approximately g_h ≈ g₀(1 − 2h/R) for small h.

Define escape speed and write its formula.

Escape speed is the minimum speed needed for an object to leave Earth's gravitational field permanently; v_e = √(2GM/R).

What is the relationship between orbital speed, orbital radius, and gravitational force?

Orbital speed v_o = √(GM/r); the gravitational force provides the centripetal force needed for circular motion.

Why is angular momentum conserved for a planet orbiting the Sun?

Gravitational force is a central force directed along the radius vector, so the torque about the Sun is zero, keeping angular momentum constant.

What is the total mechanical energy of a satellite in a circular orbit?

Total energy E = KE + PE = ½mv_o² − GMm/r = −GMm/(2r), which is negative, indicating a bound orbit.

Important Board Questions

State Kepler's law of periods and verify it using the data provided in Table 7.1 for any two planets. [2 marks]

State T² ∝ a³ or T² = Ka³; calculate T²/a³ for two planets from Table 7.1 and show both give approximately the same value (≈ 3.0 × 10⁻³⁴ y²/m³).

Derive the expression for acceleration due to gravity at a height h above Earth's surface, starting from the universal law of gravitation. Show that g_h = g₀[R/(R+h)]² and for small h, g_h ≈ g₀(1 − 2h/R). [5 marks]

At surface: g₀ = GM/R². At height h: g_h = GM/(R+h)². Divide to get ratio; use binomial approximation (1+x)ⁿ ≈ 1 + nx for |x| << 1 with x = h/R and n = −2.

A satellite of mass m orbits Earth at radius r. Derive expressions for (a) orbital speed v_o, (b) orbital period T, (c) kinetic energy KE, (d) potential energy PE, and (e) total energy E. Show that KE = −E and explain why the total energy is negative. [HOTS] [6 marks]

Set gravitational force = centripetal force to find v_o = √(GM/r). Use T = 2πr/v_o. Calculate KE = ½mv_o² = GMm/(2r) and PE = −GMm/r. Total E = KE + PE = −GMm/(2r). Negative E means bound orbit; E > 0 would mean escape.

Assertion-Reason Questions for Gravitation

Q1. Assertion (A):

Kepler's first law states that all planets move in elliptical orbits around the Sun, which contradicted Copernicus's model of circular orbits.

Reason (R):

Elliptical orbits can be drawn by maintaining a constant sum of distances from two fixed points (foci), and a circle is a special case where both foci coincide.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains why Kepler's first law is more general: ellipses (of which circles are a special case) better describe planetary orbits. Historical context confirms that Kepler's elliptical model was indeed more accurate than Copernicus's circular model.

Q2. Assertion (A):

A planet moves faster when it is closer to the Sun (at perihelion) than when it is farther (at aphelion).

Reason (R):

This occurs because gravitational potential energy is greater at perihelion, giving the planet higher kinetic energy.

Answer: C — A is true but R is false — The assertion is true and follows from Kepler's second law and angular momentum conservation. However, the reason is false: potential energy is actually more negative (minimum) at perihelion, not greater. The increased kinetic energy at perihelion comes from the planet being at a lower potential energy state, not a higher one. This is a common misconception about the sign convention of gravitational potential energy.

Q3. Assertion (A):

The law of areas (Kepler's second law) applies only to planetary systems orbiting the Sun.

Reason (R):

The law of areas is a direct consequence of conservation of angular momentum for any central force.

Answer: D — A is false but R is true — The assertion is false: Kepler's second law applies to any object under a central force (artificial satellites, moons, binary stars). The reason is true—the law does follow from angular momentum conservation. Because the reason is true but contradicts the false assertion, the correct answer is D (both false together in the assertion-reason relationship).

Unlock more Assertion-Reason questions in Premium — unlock →

Case-Based Questions (CBQ) for Gravitation

Case Study:

ISRO's Chandrayaan-3 mission involved a spacecraft approaching the Moon in an elliptical orbit. At closest approach (perilune), the spacecraft was 100 km above the lunar surface with orbital speed 1.68 km/s. At the farthest point (apolune), it was 200 km above the surface with speed 1.40 km/s. (Moon's radius ≈ 1737 km; assume central force motion applies.)

Q1. The perilune distance from Moon's centre is 1837 km and apolune distance is 1937 km. Calculate the semi-major axis of the orbit.

  • A. 1887 km ✓
  • B. 1900 km
  • C. 1800 km
  • D. 1950 km

Answer: A

Q2. Using angular momentum conservation (L_perilune = L_apolune), verify which speeds are consistent with the given distances.

  • A. The speeds cannot be verified without knowing spacecraft mass
  • B. r_p × v_p = r_a × v_a gives 1837 × 1.68 ≈ 1937 × 1.40, confirming the data ✓
  • C. The given speeds contradict the distance ratio
  • D. Angular momentum varies during elliptical motion

Answer: B

Q3. What is the ratio of perilune speed to apolune speed, and how does it relate to the distance ratio?

  • A. v_p/v_a ≈ 1.2, equal to distance ratio r_a/r_p ✓
  • B. v_p/v_a ≈ 1.8, independent of distance
  • C. v_p/v_a ≈ 0.8, inversely proportional to distance ratio
  • D. Speeds are unrelated to distances

Answer: A

Q4. If the orbital period is 2 hours, estimate the Moon's mass using Kepler's third law (T² ∝ a³). The spacecraft's orbital data confirms:

  • A. The Moon's mass is independent of orbital data
  • B. The Moon's mass is approximately 7.34 × 10²² kg ✓
  • C. Kepler's law cannot apply to Moon orbits
  • D. The period must be recalculated

Answer: B

Unlock more case studies in Premium — unlock →

Class 11 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

View the formula sheet →

Next chapterMechanical Properties of Solids →
Found this useful? Share it with your class group.

You've seen the free preview. Open the full interactive study kit — mind map, instant-graded quiz, progress tracking — free, no sign-in required.

Open Full Study Kit — Free →

Premium unlocks the full Question Bank, Daily Practice Papers, Chapter Quiz and Full Mock Papers for every chapter.