Gravitation – Numerical Problems with Stepwise Solutions

CBSE Class 9 Physics — Chapter 10: Gravitation | 20 Numerical Problems with Stepwise Solutions

Content Bank — Important Formulae (Chapter: Gravitation)

F = G · (m₁ m₂) / r² — Newton’s Universal Law of Gravitation
g = GM / r² — Acceleration due to gravity at distance r from centre
W = m g — Weight of mass m
g(h) = g · (R / (R + h))² — g at height h above surface
g(d) = g · (1 − d/R) — g at depth d (uniform density approximation)
v_orb = √(GM / r) = √(g R² / r) — Orbital speed at radius r
T = 2π √(r³ / GM) — Orbital period
v_e = √(2GM / R) = √(2 g R) — Escape velocity from surface
s = ut + ½ a t², v = u + at, v² = u² + 2a s — Kinematic equations (a = g for free fall)

20 Numerical Problems — Stepwise Solutions

All problems are NCERT-level, topic-wise, with careful step-by-step working. Use g = 9.8 m/s² and Earth radius R = 6.37 × 10⁶ m where needed unless stated otherwise.

Free Fall & Kinematics (Questions 1–5)

Q1.

Problem: A stone is dropped from rest from a height of 45 m. Calculate (a) time to reach ground, (b) speed just before impact. (Use g = 9.8 m/s²)

Given: s = 45 m, u = 0, a = g = 9.8 m/s².
Use s = ½ g t² → t = √(2s/g).
Compute time: t = √(2×45 / 9.8) = √(90 / 9.8) = √(9.183673469) ≈ 3.030 s.
Speed before impact: v = g t → v = 9.8 × 3.030 ≈ 29.70 m/s (or use v = √(2 g s) → √(882) ≈ 29.70 m/s).

Q2.

Problem: A ball is thrown vertically upward with initial speed 15 m/s. Find (a) maximum height reached, (b) total time of flight. (g = 9.8 m/s²)

Given: u = 15 m/s, g = 9.8 m/s².
Max height (use v² = u² − 2 g h, at top v = 0): 0 = u² − 2 g h ⇒ h = u² / (2 g).
Compute: h = 15² / (2 × 9.8) = 225 / 19.6 = 11.47959… → 11.48 m (rounded).
Time to reach top: t_up = u / g = 15 / 9.8 ≈ 1.531 s. Total time = 2 × t_up ≈ 3.06 s.

Q3.

Problem: A ball is thrown downward from top of a 20 m high building with initial speed 5 m/s. Find the time to reach ground and final speed. (g = 9.8 m/s²)

Given: s = 20 m (downwards), u = 5 m/s (downwards), a = g = 9.8 m/s².
Use s = u t + ½ g t². That is 20 = 5 t + 0.5 × 9.8 t² = 5 t + 4.9 t².
Rearrange: 4.9 t² + 5 t − 20 = 0. Solve quadratic: t = [−5 ± √(25 + 392)] / (2×4.9) = [−5 ± √417] / 9.8.
√417 ≈ 20.4206 → positive root: t = (−5 + 20.4206)/9.8 = 15.4206/9.8 ≈ 1.573 s.
Final speed v: v = u + g t = 5 + 9.8×1.573 ≈ 5 + 15.415 ≈ 20.415 m/s.

Q4.

Problem: A body is released from rest and falls freely for 3 seconds. Find the distance covered in the third second.

Distance in n-th second: s_n = u + g( n − 0.5 ) when u=0 this is s_n = g( n − 0.5 ) meters (actually that's average velocity * 1 s; but we will use s_n = s(t=n) − s(t=n−1)).
Total distance in 3 s: S(3) = ½ g (3)² = 0.5×9.8×9 = 44.1 m. In 2 s: S(2) = 0.5×9.8×4 = 19.6 m.
Distance in 3rd second = S(3) − S(2) = 44.1 − 19.6 = 24.5 m.

Q5.

Problem: A stone is thrown vertically upwards with speed 20 m/s. What is its speed after 1.5 s? (g = 9.8 m/s²)

Use v = u − g t (upwards positive).
Compute: v = 20 − 9.8×1.5 = 20 − 14.7 = 5.3 m/s upward. If v becomes negative it indicates downward motion; here positive means still upwards.
So speed after 1.5 s is 5.3 m/s (upward).

Newton’s Law & Pairwise Gravitation (Questions 6–9)

Q6.

Problem: Two point masses of 3 kg and 5 kg are placed 2 m apart. Calculate the gravitational force between them. (Use G = 6.67×10⁻¹¹ N·m²/kg²)

Formula: F = G m₁ m₂ / r².
Substitute: F = 6.67×10⁻¹¹ × 3 × 5 / (2)² = 6.67×10⁻¹¹ × 15 / 4.
Compute numerator: 6.67×10⁻¹¹ × 15 = 1.0005×10⁻⁹ (since 6.67×15 = 100.05; ×10⁻¹¹ → 1.0005×10⁻⁹). Divide by 4 → 2.50125×10⁻¹⁰ N.
Answer: F ≈ 2.50 × 10⁻¹⁰ N (attractive).

Q7.

Problem: Two identical spheres of mass 2 kg each touch each other. Their radii are 0.1 m each. Compute the gravitational force between them. (G = 6.67×10⁻¹¹)

Centre-to-centre distance r = 0.1 + 0.1 = 0.2 m.
F = G m₁ m₂ / r² = 6.67×10⁻¹¹ × 2 × 2 / (0.2)² = 6.67×10⁻¹¹ × 4 / 0.04.
4 / 0.04 = 100. So F = 6.67×10⁻¹¹ × 100 = 6.67×10⁻⁹ N.
Answer: F = 6.67 × 10⁻⁹ N (very small).

Q8.

Problem: Two masses attract each other with a force 10 N at separation 2 m. What force will they exert if separation is increased to 5 m?

Inverse-square law: F ∝ 1/r². So F₂ = F₁ × (r₁ / r₂)².
Compute: F₂ = 10 × (2 / 5)² = 10 × (4 / 25) = 10 × 0.16 = 1.6 N.

Q9.

Problem: Three equal masses m are placed at the corners of an equilateral triangle of side a. Find the net gravitational force on one mass due to the other two (magnitude and direction).

Each attractive force magnitude: F = G m² / a² directed along sides toward other masses.
By symmetry, horizontal components cancel and vertical components add (take one mass at top vertex). The resultant points toward triangle centre (downwards from top vertex).
Resultant magnitude = 2 F cos(30°) = 2 (G m² / a²) × (√3/2) = (√3) (G m² / a²).
Answer: F_net = √3 · G m² / a², directed toward centroid of triangle.

Variation of g with Altitude and Depth (Questions 10–12)

Q10.

Problem: Calculate g at an altitude of 1 km above Earth's surface. Use g = 9.8 m/s² and R = 6.37×10⁶ m. Give result to 5 significant figures.

Formula: g(h) = g · (R / (R + h))².
Substitute: h = 1,000 m, R = 6.37×10⁶ m. Compute factor: (R/(R+h))² = (6.37e6 / 6.371e6)².
6.37e6 / 6.371e6 = 0.9998433… Squared ≈ 0.9996867. So g(h) ≈ 9.8 × 0.9996867 ≈ 9.7969238 m/s².
Answer: g ≈ 9.7969 m/s² (5 significant figures).

Q11.

Problem: Estimate g at a depth of 3 km below Earth's surface (assume uniform density). Use g = 9.8 m/s² and R = 6370 km.

Use g(d) = g (1 − d / R) with consistent units.
Convert R = 6370 km = 6.37×10⁶ m. d = 3 km = 3000 m. d/R = 3000 / 6.37e6 ≈ 0.000471.
g(d) ≈ 9.8 × (1 − 0.000471) = 9.8 × 0.999529 ≈ 9.79538 m/s².
Answer: g ≈ 9.7954 m/s² (slight decrease).

Q12.

Problem: Show that g at a height equal to Earth's radius (h = R) is one-fourth of surface g.

At h = R, r = R + h = 2R. g' = g (R / (2R))² = g (1/2)² = g/4.
Thus g at altitude R is g/4 (i.e., ~2.45 m/s² if g = 9.8 m/s²).

Mass, Weight & Apparent Weight (Questions 13–15)

Q13.

Problem: A person of mass 70 kg stands on a weighing scale inside an elevator accelerating upward at 2 m/s². What reading will the scale show? (g = 9.8 m/s²)

Apparent weight = normal reaction N = m (g + a) when elevator accelerates upward.
N = 70 × (9.8 + 2) = 70 × 11.8 = 826 N.
Answer: 826 N (scale reading).

Q14.

Problem: A 12 kg object is taken to the Moon where g_moon ≈ 1.63 m/s². Compute its mass and weight on the Moon.

Mass remains the same: 12 kg.
Weight on Moon: W = m g_moon = 12 × 1.63 = 19.56 N.
Answer: Mass = 12 kg, Weight ≈ 19.6 N (to 3 s.f.).

Q15.

Problem: A 5 kg object is in a lift accelerating downwards with 3 m/s². What is its apparent weight? (Take g = 9.8 m/s²)

Apparent weight: N = m (g − a) (downward acceleration reduces normal reaction).
N = 5 × (9.8 − 3) = 5 × 6.8 = 34 N.
Answer: 34 N.

Satellites, Orbital Motion & Escape Velocity (Questions 16–19)

Q16.

Problem: Calculate the escape velocity from Earth's surface using g = 9.8 m/s² and R = 6.37×10⁶ m. Give answer in km/s (3 s.f.).

Formula: v_e = √(2 g R).
Compute: 2 g R = 2 × 9.8 × 6.37×10⁶ = 19.6 × 6.37×10⁶ = 124.652×10⁶ ≈ 1.24652×10⁸.
v_e = √(1.24652×10⁸) ≈ 11173.72 m/s ≈ 11.174 km/s.
Rounded to 3 s.f.: 11.2 km/s (commonly quoted).

Q17.

Problem: Compute orbital speed for a satellite at 300 km above Earth's surface. Use g = 9.8 m/s² and R = 6.37×10⁶ m. Give answer in km/s (3 s.f.).

Radius r = R + h = 6.37×10⁶ + 3.00×10⁵ = 6.67×10⁶ m.
Use v = √(GM / r) = √(g R² / r) (since GM = g R²).
Compute v = √(9.8 × (6.37×10⁶)² / 6.67×10⁶). Numerically this gives ≈ 7.7213×10³ m/s ≈ 7.72 km/s.
So orbital speed ≈ 7.72 km/s (typical LEO speed).

Q18.

Problem: For a satellite in circular orbit at radius r, show that its total mechanical energy per unit mass equals −GM/(2 r). Provide brief calculation.

Kinetic energy per unit mass: k = ½ v². For circular orbit v² = GM / r → k = ½ (GM / r).
Potential energy per unit mass: u = − GM / r.
Total per unit mass: ε = k + u = ½ (GM / r) − (GM / r) = − (GM / 2r).
Thus ε = −GM / (2 r) (negative for bound orbit).

Q19.

Problem: A satellite orbits at radius r such that its orbital period is 90.46 minutes. Using GM = g R² (g=9.8, R=6.37×10⁶), estimate r (use algebraic steps and show result approx equal to R + 300 km).

Period formula: T = 2π √(r³ / GM) ⇒ r³ = (GM / (4π²)) T².
Take GM = g R² = 9.8 × (6.37×10⁶)² ≈ 3.986×10¹⁴ (approx Earth's GM). Convert T = 90.46 min = 5427.6 s.
Compute r³ ≈ (3.986×10¹⁴ / (39.478)) × (5427.6)². Evaluating gives r ≈ 6.67×10⁶ m which equals R + 3.00×10⁵ m ≈ R + 300 km.
Conclusion: r ≈ 6.67 × 10⁶ m → orbit altitude ≈ 300 km.

Gravitational Potential & Energy (Question 20)

Q20.

Problem: Compute approximate change in gravitational potential energy when a 2 kg object is lifted from Earth's surface to a height of 1000 m. Use g ≈ 9.8 m/s² and compare with exact formula using GM/r if you like (use small-h approximation ΔU ≈ m g h).

For small height, ΔU ≈ m g h = 2 × 9.8 × 1000 = 19,600 J.
If using exact expression: U = −GM m / r. ΔU = U(h) − U(0) = −GM m [1/(R+h) − 1/R] ≈ m g R [1/(R+h) − 1/R] simplifying gives ≈ m g h for h ≪ R. Numerically the correction is tiny.
Answer: ΔU ≈ 19.6 kJ (approx).

Note for students: These worked numericals use standard NCERT assumptions: Earth treated as spherical with radius R = 6.37×10⁶ m; g = 9.8 m/s² for surface calculations unless variation is explicitly required; air resistance is neglected unless stated. For orbital calculations we used GM = g R² to avoid needing G in many problems — an NCERT-friendly shortcut.

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Gravitation – Numerical Problems with Stepwise Solutions

20 Numerical Problems — Stepwise Solutions

Free Fall & Kinematics (Questions 1–5)

Newton’s Law & Pairwise Gravitation (Questions 6–9)

Variation of g with Altitude and Depth (Questions 10–12)

Mass, Weight & Apparent Weight (Questions 13–15)

Satellites, Orbital Motion & Escape Velocity (Questions 16–19)

Gravitational Potential & Energy (Question 20)

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