Chapter 8: Gravitation (Class 11 Physics)

1. State Kepler’s law of orbits.

Planets move in elliptical orbits with Sun at one focus.

2. State Kepler’s law of areas.

Line joining planet to Sun sweeps equal areas in equal times.

3. State Kepler’s law of periods.

T² ∝ a³ (Square of time period is proportional to cube of semi-major axis).

4. Which physical quantity conservation is implied by Kepler’s second law?

Conservation of Angular Momentum.

5. State Newton’s universal law of gravitation (formula only).

F = G m₁m₂ / r².

6. Is gravitational force conservative?

Yes.

7. What is the value of G (Universal Gravitational Constant)?

G ≈ 6.67 × 10⁻¹¹ N m²/kg².

8. Define acceleration due to gravity (g).

Acceleration of a freely falling body due to Earth’s gravity.

9. Write the relation between g and G.

g = GME / RE².

10. How does ‘g’ vary with altitude (height)?

It decreases as altitude increases.

11. How does ‘g’ vary with depth?

It decreases as depth increases (becomes zero at centre).

12. Where is ‘g’ maximum on Earth’s surface?

At the poles.

13. Define Gravitational Potential Energy.

Energy due to position in a gravitational field (U = -GMm/r).

14. What is Gravitational Potential?

Gravitational PE per unit mass (V = -GM/r).

15. Define Escape Speed.

Minimum speed to escape Earth’s gravitational field to infinity.

16. What is the approximate escape speed from Earth?

11.2 km/s.

17. Define Orbital Velocity of a satellite.

Minimum speed required to put satellite in a stable orbit.

18. What is the formula for orbital velocity close to Earth?

v_o ≈ √(gRE).

19. What is the relation between escape speed (v_e) and orbital speed (v_o) near Earth?

v_e = √2 v_o.

20. Are Kepler’s laws applicable only to planets?

No, they apply to any satellite orbiting a central massive body under gravity (like Earth satellites).

1. Explain Kepler’s law of areas and its consequence.

• Law: Line joining planet to Sun sweeps equal areas in equal times (dA/dt = constant).
• Consequence: Planet moves faster when closer to Sun (perihelion) and slower when farther (aphelion), to sweep equal areas. Stems from conservation of angular momentum.

2. Why is G called the universal gravitational constant?

• Its value (≈ 6.67 × 10⁻¹¹ Nm²/kg²) is believed to be constant throughout the universe.
• It does not depend on the nature or size of the attracting bodies.
• It does not depend on the medium between the bodies.

3. How does ‘g’ vary with height ‘h’ above Earth’s surface? Write approximate formula for h<E.

• ‘g’ decreases with height.
• Exact formula: gh = GME / (RE + h)².
• Approximate formula for small h: gh ≈ g (1 - 2h/RE). (Using binomial expansion).

4. How does ‘g’ vary with depth ‘d’ below Earth’s surface? What is its value at the centre?

• ‘g’ decreases linearly with depth.
• Formula: gd = g (1 - d/RE). (Assuming uniform density).
• At the centre (d=RE), g_d = g(1-1) = 0.

5. Define gravitational potential energy. Why is it always negative?

• PE = Work done to bring mass ‘m’ from infinity to a point in gravitational field without acceleration.
• Formula: U = -GMm/r.
• Negative sign arises because:
  • Gravitational force is attractive.
  • Work is done *by* the field (not against it) when bringing mass from infinity (where U=0 by convention).
  • Potential energy decreases as the mass comes closer from infinity.

6. Obtain the relation between escape speed (v_e) and orbital speed (v_o) for a satellite near Earth.

• Escape speed: v_e = √(2GME / RE) = √(2gRE).
• Orbital speed near surface: v_o ≈ √(GME / RE) = √(gRE).
• Comparing the two: v_e = √2 × √(gRE) = √2 v_o.

7. If the radius of Earth were to shrink by 1% (mass remaining same), what would be the percentage change in ‘g’ on the surface?

• g = GME / RE². So, g ∝ 1/RE².
• New radius R’ = RE – 1% of RE = 0.99 RE.
• New g’ ∝ 1/(R’)² = 1 / (0.99 RE)².
• Ratio g’/g = (1/R’²) / (1/RE²) = RE² / R’² = RE² / (0.99 RE)² = 1 / (0.99)² ≈ 1 / 0.98 ≈ 1.02.
• g’ ≈ 1.02 g.
• Percentage change = [(g’ – g) / g] × 100% = [(1.02g – g) / g] × 100% = 0.02 × 100% = 2%.
• Answer: ‘g’ increases by approx 2%.

8. Does escape speed depend on the mass of the body being projected?

• No.
• Formula v_e = √(2GM/R).
• This formula only contains the mass (M) and radius (R) of the planet/celestial body from which the object is escaping.
• It does not depend on the mass ‘m’ of the object being projected.

9. Why don’t we experience the gravitational force between everyday objects?

• Gravitational force F = Gm₁m₂/r².
• The value of G is very small (6.67 × 10⁻¹¹ Nm²/kg²).
• For everyday objects, masses (m₁, m₂) are also relatively small.
• Therefore, the gravitational force (F) between them is extremely weak and practically unnoticeable compared to other forces like friction or forces we apply.
• It becomes significant only when at least one mass is very large (like Earth).

10. What is gravitational potential? Write its SI unit and dimensional formula.

• Potential (V) at a point = PE per unit mass = Work done per unit mass to bring body from infinity to that point.
• V = -GM/r (due to source mass M).
• Scalar quantity.
• SI Unit: Joule per kg (J/kg).
• Dimensional Formula: [Energy]/[Mass] = [ML²T⁻²]/[M] = [L²T⁻²].

11. What are geostationary satellites? What are their uses?

• Satellites orbiting Earth with a time period (T) exactly matching Earth’s rotation period (24 hours).
• They orbit in the equatorial plane.
• From Earth, they appear stationary at a fixed point in the sky.
• Height approx 36,000 km above equator.
• Uses: Communication (TV broadcast, phone), Weather forecasting, Meteorology.

12. Is gravitational potential energy always negative?

• Yes, based on the standard convention where Potential Energy is taken as zero at infinite separation (r=∞).
• Since gravitational force is attractive, work is done *by* the field when bringing masses together from infinity.
• This means potential energy decreases from zero as distance decreases, hence it becomes negative. U = -GMm/r.

13. Find the ratio of escape speed to orbital speed for a satellite near Earth.

• v_e = √(2gRE)
• v_o ≈ √(gRE)
• Ratio: v_e / v_o = √(2gRE) / √(gRE) = √2.
• Answer: v_e = √2 v_o.

14. How did Kepler deduce his laws?

• Kepler did not deduce them from fundamental principles like Newton did later.
• His laws were empirical, based on the extremely precise and painstaking astronomical observations of planetary positions collected by his predecessor, Tycho Brahe.
• He spent years analyzing Brahe’s data to find mathematical patterns describing the observed motion.

15. Explain the terms aphelion and perihelion related to elliptical orbits.

• In an elliptical orbit of a planet around the Sun (which is at one focus):
• Perihelion: The point in the orbit where the planet is **closest** to the Sun. Planet moves fastest here (by Kepler’s 2nd Law).
• Aphelion: The point in the orbit where the planet is **farthest** from the Sun. Planet moves slowest here.

16. Calculate the gravitational force between Earth (M=6×10²⁴kg) and Sun (m=2×10³⁰kg) separated by 1.5×10¹¹ m.

• F = G Mm / r²
• F = (6.67×10⁻¹¹) × (6×10²⁴) × (2×10³⁰) / (1.5×10¹¹)²
• F = (6.67 × 12 × 10⁴³) / (2.25 × 10²²)
• F = (80.04 / 2.25) × 10²¹ ≈ 35.6 × 10²¹ N.
• Answer: Approx 3.56 × 10²² N.

1. State and explain Kepler’s three laws of planetary motion.

1. Law of Orbits: Planets move in elliptical orbits with the Sun at one focus. (Planets don’t move in perfect circles).
2. Law of Areas: The line joining a planet to the Sun sweeps out equal areas in equal time intervals (dA/dt=constant). Implies planets move faster when closer to Sun (perihelion) and slower when farther (aphelion). Related to conservation of angular momentum.
3. Law of Periods: Square of time period (T) is proportional to cube of semi-major axis (a) of the orbit (T² ∝ a³). Implies outer planets have much longer years than inner planets.

2. State Newton’s Law of Gravitation. Explain its vector form and characteristics.

Statement: Every particle attracts every other particle with a force directly proportional to product of masses and inversely proportional to square of distance between them (F = Gm₁m₂/r²).

Vector Form:

• Force on mass m₂ due to m₁ (F₂₁) = - G m₁m₂r₁₂² r₁₂
• Here r₁₂ is displacement vector from m₁ to m₂, r₁₂ = |r₁₂|, and r₁₂ = r₁₂ / r₁₂ is unit vector from m₁ to m₂. Negative sign indicates force is attractive (opposite to r₁₂).
• Force on m₁ due to m₂ is F₁₂ = – F₂₁ (obeys Newton’s 3rd Law).

Characteristics:

• Universal, Always attractive, Central force, Conservative force, Obeys inverse square law, Independent of medium.

3. Derive the expression for acceleration due to gravity (g) on the Earth’s surface in terms of G, ME and RE. How does it vary with altitude and depth?

Derivation for g on surface: See Short Q&A section or Part 1.

Result: g = GME / RE².

Variation with Altitude (h):

• At height h, distance from center = RE + h.
• g_h = GME / (RE + h)².
• g_h / g = RE² / (RE + h)².
• For h<E, g_h ≈ g(1 – 2h/RE). (g decreases).

Variation with Depth (d):

• At depth d, distance from center = RE – d. Effective mass attracting is M’ of inner sphere.
• Assume uniform density ρ = ME / ( (4/3)πRE³ ).
• M’ = ρ × Volume = ρ × (4/3)π(RE – d)³.
• g_d = GM’ / (RE – d)² = G [ρ × (4/3)π(RE – d)³] / (RE – d)².
• g_d = Gρ(4/3)π (RE – d).
• Surface g = GME/RE² = G[ρ(4/3)πRE³]/RE² = Gρ(4/3)πRE.
• Ratio: g_d / g = (RE – d) / RE = 1 – d/RE.
• g_d = g (1 – d/RE). (g decreases linearly, becomes 0 at center).

4. Derive expressions for the escape speed from a planet and the orbital speed of a satellite orbiting close to its surface. Show v_e = √2 v_o.

Let Planet mass=M, radius=R. Object mass=m.

Escape Speed (v_e):

• Condition: Initial Total Energy = Final Total Energy (at infinity) = 0.
• Initial: KE = ½mv_e², PE = -GMm/R. Final: KE=0, PE=0.
• ½mv_e² – GMm/R = 0
• ½mv_e² = GMm/R
• v_e² = 2GM/R ⇒ v_e = √(2GM/R) = √(2gR) (using g=GM/R²).

Orbital Speed (v_o) (close to surface, r≈R):

• Condition: Gravitational Force = Centripetal Force.
• GMm/R² = mv_o²/R
• Cancel m and one R: GM/R = v_o².
• v_o = √(GM/R) = √(gR).

Relation:

• Compare formulas: v_e = √(2 × (gR)) = √2 × √(gR) = √2 v_o.
• v_e = √2 v_o.

5. What are gravitational potential energy and gravitational potential? Derive expressions for both due to a point mass M.

Both are scalars related to work done by/against gravity.

• Gravitational Potential Energy (U): PE of a mass ‘m’ at a point is the work done by an external agent to bring it from infinity (where U=0) to that point without acceleration.
  • Force F = -GMm/r² (attractive).
  • External force needed F_ext = +GMm/r².
  • Work dW = F_ext dr = (GMm/r²) dr? No, it’s ∫ F ⋅ dr from ∞ to r.
  • dW = F_ext ⋅ dr = –F_grav ⋅ dr.
  • Potential energy change ΔU = U_r – U_∞ = – W_grav = – ∫_∞^r F_g ⋅ dr
  • F_g = (-GMm/r²) r; dr = dr r. So F_g ⋅ dr = (-GMm/r²) dr.
  • ΔU = – ∫_∞^r (-GMm/r²) dr = GMm ∫_∞^r r⁻² dr
  • ΔU = GMm [-r⁻¹]_∞^r = GMm [-1/r]_∞^r = GMm [(-1/r) – (-1/∞)]
  • ΔU = U_r – 0 = -GMm/r.
  • Expression: U = -GMm/r.
• Gravitational Potential (V): Gravitational PE per unit mass. Work done per unit mass to bring from ∞ to the point.
  • V = U / m
  • Expression: V = -GM/r.
  • Unit: J/kg. Dimension: [L²T⁻²].

6. Discuss the variation of acceleration due to gravity ‘g’ with Latitude and Earth’s Rotation.

‘g’ varies on Earth’s surface due to:

• 1. Earth’s Shape (Effect of Latitude):
  • Earth is not a perfect sphere; it’s an oblate spheroid (flattened at poles, bulging at equator). R_equator > R_poles.
  • Since g = GM_E / R_E², ‘g’ is inversely proportional to R_E².
  • Therefore, ‘g’ is **minimum at the equator** (larger R_E) and **maximum at the poles** (smaller R_E).
• 2. Earth’s Rotation (Effect of Latitude):
  • Earth rotates about its polar axis. A body on the surface experiences an outward centrifugal force (in non-inertial frame) which slightly reduces the effective weight.
  • Centrifugal force is max at equator (max distance from axis) and zero at poles (on axis).
  • This effect further reduces the effective ‘g’ at the equator compared to the poles.
  • Effective ‘g’ at latitude λ: g’ ≈ g – R_Eω²cos²λ (where ω is Earth’s angular velocity).
  • At equator (λ=0°): g’ = g – R_Eω².
  • At poles (λ=90°): g’ = g – 0 = g.

Both factors contribute to ‘g’ being max at poles and min at equator.

7. What are Geostationary and Polar satellites? Mention their uses.

• Geostationary Satellites:
  • Orbit Earth in the equatorial plane at a specific height (~36,000 km).
  • Time period (T) = 24 hours (matches Earth’s rotation).
  • Appear stationary relative to a point on Earth’s surface.
  • Uses: Communications (TV broadcast, phone signals 📡), weather forecasting, meteorology.
• Polar Satellites:
  • Orbit Earth in polar orbits (passing over North and South poles).
  • Usually orbit at lower altitudes (~500-800 km).
  • Time period is much shorter (~100 minutes).
  • Cover the entire globe over successive orbits as Earth rotates beneath them.
  • Uses: Remote sensing (mapping 🗺️, resource monitoring), environmental monitoring, weather data collection, spying.

8. Derive expressions for the Kinetic Energy, Potential Energy and Total Energy of a satellite orbiting the Earth.

Consider satellite (mass m) in circular orbit (radius r) around Earth (mass M_E) with orbital speed v_o.

• Kinetic Energy (K):
  • K = ½mv_o².
  • We know v_o² = GM_E/r.
  • So, K = ½m(GM_E/r) = GM_Em / 2r.
• Potential Energy (U):
  • Gravitational PE at distance r from Earth’s center is U = -GM_Em / r.
  • So, U = -GM_Em / r.
• Total Mechanical Energy (E):
  • E = K + U = (GM_Em / 2r) + (-GM_Em / r)
  • E = GM_Em/r (½ – 1) = GM_Em/r (-½).
  • E = – GM_Em / 2r.

Observations:

• K = -E/2 = |E| (Kinetic energy is positive, equal to magnitude of total energy).
• U = 2E = -2K (Potential energy is negative, twice the total energy, or -2 times KE).
• Total energy E is negative, indicating a bound system (satellite cannot escape on its own).

9. Discuss the concept of weightlessness in an orbiting satellite.

• Astronauts inside an orbiting satellite experience a state of **apparent weightlessness**.
• This does NOT mean gravity is zero at that altitude (g_h is still significant, maybe 80-90% of surface ‘g’ for low orbits).
• Weightlessness occurs because both the satellite AND the astronaut inside it are in a constant state of **free fall** towards the Earth.
• The satellite is continuously falling towards Earth, but its high tangential velocity ensures it keeps ‘missing’ the Earth and follows a circular/elliptical path.
• Since the astronaut and the satellite floor are falling together with the same acceleration (g_h), the astronaut does not press against the floor. The scale (if placed under astronaut) reads zero apparent weight.
• It’s similar to the feeling in a lift that is freely falling (cable breaks) – everything inside becomes apparently weightless relative to the lift.

10. What is the difference between Inertial Mass and Gravitational Mass? Are they equal?

• Inertial Mass (m_i): Defined by Newton’s 2nd Law (F = m_ia). It measures a body’s **resistance to change in its state of motion** (inertia). Larger m_i means harder to accelerate. Determined by applying a known force and measuring acceleration.
• Gravitational Mass (m_g): Defined by Newton’s Law of Gravitation (F = G M m_g / r²). It determines the **gravitational force** the body exerts on other bodies and the force it experiences due to other bodies. Determined by measuring the gravitational force (e.g., using a balance).
• Equivalence Principle: Experimentally, it has been found with very high precision that the inertial mass and gravitational mass of a body are **equal** (or strictly proportional, with the constant taken as 1). This equivalence is a fundamental principle underlying Einstein’s General Theory of Relativity.

11. Explain how tides are formed.

Tides are the periodic rise and fall of sea levels, primarily caused by the **differential gravitational forces** exerted by the **Moon** and, to a lesser extent, the **Sun** on different parts of the Earth.

• Moon’s Influence:** The Moon’s gravitational pull is stronger on the side of the Earth facing it and weaker on the side opposite to it.
  • Water on the side facing the Moon is pulled strongly towards it, creating a high tide bulge.
  • The solid Earth itself is also pulled towards the Moon, slightly less than the water on the near side but more than the water on the far side.
  • This effectively pulls the Earth away from the water on the far side, creating another high tide bulge there.
  • Areas between these bulges experience low tide.
• Sun’s Influence:** The Sun also exerts a gravitational pull, but due to its much greater distance, its tide-generating effect is less than half that of the Moon.
• Spring Tides:** When the Sun, Earth, and Moon are aligned (during New Moon and Full Moon), their gravitational pulls combine, resulting in extra-high high tides and extra-low low tides (largest tidal range).
• Neap Tides:** When the Sun and Moon are at right angles relative to Earth (during First and Third Quarter Moon), their gravitational pulls partially cancel each other, resulting in lower high tides and higher low tides (smallest tidal range).