Gravitation – Case-based Questions with Answers

CBSE Class 9 | Physics Chapter 10: Gravitation — 20 Case-Based Questions

Content Bank — Chapter 10: Gravitation

Newton’s Universal Law of Gravitation and gravitational constant G
Gravitational field (g), variation with altitude and depth
Mass vs weight, apparent weight and weightlessness
Free fall, kinematics under gravity
Orbital motion, orbital speed, period and escape velocity
Kepler’s laws (brief) and real-world applications like satellites and tides

20 Case-Based Questions & Answers — Gravitation

Each case presents a short scenario followed by focused, NCERT-level questions and clear answers useful for board exam practice.

Case 1 — Two spheres and distance

Scenario: Two identical small spheres each of mass 2 kg are placed 1 m apart. A third small mass 0.5 kg is placed at a point 0.25 m from one sphere and 0.75 m from the other along the line joining their centres.

Q1: What is the net gravitational force on the small mass (magnitude & direction)? Show method.

Answer: Compute forces from each sphere: F₁ = G·(2·0.5)/0.25² = G·1 / 0.0625 = 16G; F₂ = G·(2·0.5)/0.75² = G·1 / 0.5625 ≈ 1.777...G. Net force = F₁ − F₂ ≈ (16 − 1.777)G ≈ 14.223G toward the nearer sphere. Direction: toward the sphere at 0.25 m.

Case 2 — Lift accelerating

Scenario: A student of mass 40 kg stands on a weighing scale inside an elevator. The elevator accelerates upward at 2 m/s² for 4 s, then moves with constant velocity.

Q2: What reading does the scale show during acceleration? What about during constant velocity? (Use g = 9.8 m/s²)

Answer: During upward acceleration, apparent weight N = m(g + a) = 40(9.8 + 2) = 40×11.8 = 472 N. While moving with constant velocity, a = 0 so N = mg = 40×9.8 = 392 N.

Case 3 — Moon vs Earth

Scenario: A 5 kg object is taken from Earth to the Moon. Moon's gravitational acceleration is about 1.63 m/s².

Q3: Compare mass and weight on Earth and Moon. Give numerical weights.

Answer: Mass stays 5 kg everywhere. Weight on Earth Wₑ = m gₑ = 5×9.8 = 49 N. Weight on Moon Wₘ = 5×1.63 ≈ 8.15 N. So weight reduces to ≈ 1/6 while mass unchanged.

Case 4 — Deep mine

Scenario: A miner goes to depth 3 km below Earth’s surface. Earth's radius R ≈ 6370 km and surface g = 9.8 m/s².

Q4: Estimate g at that depth using uniform density approximation.

Answer: Using g_d = g(1 − d/R). Here d = 3 km, R = 6370 km ⇒ g_d ≈ 9.8(1 − 3/6370) ≈ 9.8(1 − 0.000471) ≈ 9.8×0.999529 ≈ 9.795 m/s². Slight decrease; practically g nearly same.

Case 5 — Tossed up ball

Scenario: A ball is thrown vertically upward with initial speed 15 m/s from ground. Ignore air resistance.

Q5: Find maximum height and total time of flight. (g = 9.8 m/s²)

Answer: Time to reach top t = u/g = 15/9.8 ≈ 1.53 s. Maximum height h = u²/(2g) = 225/(19.6) ≈ 11.48 m. Total time = 2t ≈ 3.06 s.

Case 6 — Satellite orbit

Scenario: A satellite orbits Earth in a circular orbit at radius r. Consider two satellites: A at radius r and B at radius 8r.

Q6: How do their orbital speeds and periods compare?

Answer: Orbital speed v ∝ 1/√r, so v_B = v_A / √8 = v_A / 2.828. Period T ∝ r^{3/2}, so T_B = T_A × 8^{3/2} = T_A × 22.627 (since 8^{3/2} = (8^{1/2})^3 = 2.828^3 ≈ 22.627). So B is much slower and has much longer period.

Case 7 — Weightless astronaut

Scenario: Astronauts aboard the International Space Station (ISS) float inside.

Q7: Explain why they feel weightless though gravity at ISS altitude is ≈ 90% of surface value.

Answer: Both astronauts and ISS are in continuous free fall around Earth; there is no normal reaction between astronaut and spacecraft interior, causing apparent weightlessness. Gravity still acts strongly, keeping ISS in orbit.

Case 8 — Two planets

Scenario: Planet X has twice the mass of Earth but the same radius as Earth.

Q8: What is surface g on Planet X compared to Earth?

Answer: g ∝ M/R². If M doubles and R same, g doubles. So g_X = 2 g_Earth ≈ 19.6 m/s².

Case 9 — Midpoint equilibrium

Scenario: Two equal masses are fixed at ends of a horizontal rod. A small test mass placed at midpoint.

Q9: Is the midpoint stable equilibrium? What happens on slight displacement?

Answer: Net force is zero at midpoint (equal opposite attractions). It's unstable: a slight displacement closer to one mass produces larger attraction toward nearer mass, so test mass moves further away from midpoint toward that mass.

Case 10 — Changing distance

Scenario: Two masses attract with force F. They are moved so their separation becomes three times the original.

Q10: What is the new force?

Answer: Inverse-square law: F' = F / 3² = F / 9.

Case 11 — Measuring g with pendulum

Scenario: A student measures period T of a simple pendulum of length l to find g using g = 4π² l / T².

Q11: If student doubles the length and measures new period, how does period change?

Answer: Period T ∝ √l. Doubling l multiplies T by √2 ≈ 1.414. So period increases by about 41.4%.

Case 12 — Escape vs orbital speed

Scenario: At Earth's surface orbital speed vₒ ≈ 7.9 km/s and escape speed vₑ ≈ 11.2 km/s.

Q12: Explain why vₑ = √2·vₒ and physical meaning.

Answer: Orbital speed provides centripetal force to stay in circular orbit. Escape speed is derived from energy: ½m vₑ² = GMm/R. Since vₒ² = GM/R, vₑ² = 2 vₒ² ⇒ vₑ = √2 vₒ. Escape requires √2 times kinetic energy of circular orbit.

Case 13 — Weight in accelerating bus

Scenario: Passenger stands on digital scale in a bus accelerating forward with acceleration a. Scale is on bus floor.

Q13: How does scale read if bus accelerates forward? (Consider horizontal acceleration.)

Answer: Horizontal acceleration does not change vertical normal reaction; scale reads mg (ignoring tilt). However if bus tilts or acceleration has vertical component, apparent weight changes. So purely horizontal acceleration keeps scale reading ≈ mg.

Case 14 — Satellite communication

Scenario: A communication satellite must remain above same point on equator.

Q14: What orbit must it be placed in and why? Mention approximate altitude.

Answer: Geostationary orbit over equator with period 24 hours. Radius from Earth's centre ≈ 4.23×10⁷ m; altitude ≈ 3.58×10⁷ m. It remains fixed relative to Earth's surface enabling continuous communication coverage for fixed region.

Case 15 — Variation of g with altitude

Scenario: A balloon rises to 10 km above surface. Estimate fractional change in g (R = 6370 km).

Q15: Use approximation g' ≈ g(1 − 2h/R).

Answer: h/R = 10/6370 ≈ 0.00157. 2h/R ≈ 0.00314. So g' ≈ g(1 − 0.00314) ⇒ ~0.314% decrease. Numerically g' ≈ 9.8×0.99686 ≈ 9.77 m/s².

Case 16 — Two-body energy

Scenario: A satellite of mass m orbits at radius r. Total mechanical energy per unit mass is −GM/(2r).

Q16: If r is halved (satellite brought closer slowly), how does total (specific) energy change?

Answer: Specific energy ∝ −1/(2r). Halving r doubles magnitude of negative energy (becomes more negative): energy becomes −GM/(2·r/2) = −GM/r ; i.e., twice the previous magnitude negative (more tightly bound).

Case 17 — Tidal effects

Scenario: Coastal town experiences higher tides during full moon.

Q17: Briefly explain using gravitation why spring tides occur at full moon and new moon.

Answer: Spring tides occur when Sun, Moon and Earth align (new or full moon). Gravitational pulls of Sun and Moon reinforce each other on near and far sides, producing larger tidal bulges and higher high tides.

Case 18 — Combined forces

Scenario: Three identical masses form an equilateral triangle. A test mass is placed at centre.

Q18: What is net gravitational force on the test mass? Explain.

Answer: By symmetry, attractions from three identical masses cancel vectorially at the centre; net gravitational force is zero.

Case 19 — Practical measurement

Scenario: A physics lab suspects their pendulum length reading has 1% systematic error.

Q19: How does 1% error in length affect measured g (g ∝ l/T²)?

Answer: g ∝ l. A 1% increase in measured length gives 1% increase in calculated g (assuming T accurate). So measured g error ≈ 1% (errors propagate linearly here).

Case 20 — Air resistance ignored

Scenario: Textbook problems often ignore air resistance for free-fall calculations.

Q20: Give two reasons why this simplification is acceptable at NCERT/CBSE level and one situation where it fails.

Answer: (1) Ignoring air resistance simplifies physics to uniform acceleration, making core concepts of gravity and kinematics clear. (2) For dense/small objects (stones, metal balls) and short distances, air drag is negligible compared to weight. It fails for light objects (feathers, paper) or very high-speed/long-fall situations where drag significantly alters motion.

Note: These 20 case-based questions and answers are prepared strictly as per NCERT syllabus for Class 9 Physics — Chapter 10: Gravitation. You can paste this HTML into a WordPress "Custom HTML" block. If you want, I can also:

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Gravitation – Case-based Questions with Answers

20 Case-Based Questions & Answers — Gravitation

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