Gravitation — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Gravitation" — 6 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Gravitation" — 6 important questions with detailed answers for CBSE board exam prepara…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

State Newton's law of universal gravitation and derive the acceleration due t…
The Moon has a mass of 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Calculate the …
Calculate the orbital velocity of a satellite in a circular orbit at a height…
Define escape velocity. Calculate the escape velocity from Earth's surface. (…
A planet has twice the mass and three times the radius of Earth. Calculate th…
Two satellites orbit Earth at different altitudes. Satellite A is at height h…

Solutions Summary:

Question	Status
State Newton's law of universal gravitation and derive th…	✓ Solved
The Moon has a mass of 7.35 × 10²² kg and radius 1.74 × 1…	✓ Solved
Calculate the orbital velocity of a satellite in a circul…	✓ Solved
Define escape velocity. Calculate the escape velocity fro…	✓ Solved
A planet has twice the mass and three times the radius of…	✓ Solved
Two satellites orbit Earth at different altitudes. Satell…	✓ Solved

Showing 6 of 6 questions

Q1: State Newton's law of universal gravitation and derive the acceleration due to gravity at the Earth's surface.

Newton's Law of Universal Gravitation: Statement: Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Mathematical Form: F = G(m₁m₂)/r² Where: F = gravitational force between the two objects m₁, m₂ = masses of the two objects r = distance between their centres G = universal gravitational constant = 6.67 × 10⁻¹¹ N⋅m²/kg² Key Features: -...

Q2: The Moon has a mass of 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Calculate the acceleration due to gravity on the Moon's surface and compare it with Earth's gravity (g_earth = 10 m/s²).

Given: Mass of Moon M_m = 7.35 × 10²² kg Radius of Moon R_m = 1.74 × 10⁶ m G = 6.67 × 10⁻¹¹ N⋅m²/kg² g_earth = 10 m/s² (approximately) Part 1: Calculate acceleration due to gravity on Moon Using the formula: g_moon = GM_m/R_m² Step 1: Calculate numerator GM_m = 6.67 × 10⁻¹¹ × 7.35 × 10²² = 6.67 × 7.35 × 10¹¹ = 49.02 × 10¹¹ = 4.902 × 10¹² N⋅m²/kg Step 2: Calculate denominator R_m² = (1.74 × 10⁶)² = 3.0276 × 10¹² m² Step 3: Calculate g_moon g_moon = 4.902 × 10¹²/3.0276 × 10¹² g_moon = 4.902/3...

Q3: Calculate the orbital velocity of a satellite in a circular orbit at a height of 400 km above Earth's surface. (Mass of Earth = 6 × 10²⁴ kg, Radius of Earth = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N⋅m²/kg²)

Given: Height above surface h = 400 km = 4 × 10⁵ m Mass of Earth M = 6 × 10²⁴ kg Radius of Earth R = 6.4 × 10⁶ m G = 6.67 × 10⁻¹¹ N⋅m²/kg² Find: Orbital velocity of the satellite Step 1: Determine orbital radius The satellite orbits at distance r from Earth's centre: r = R + h = 6.4 × 10⁶ + 4 × 10⁵ r = 6.4 × 10⁶ + 0.4 × 10⁶ r = 6.8 × 10⁶ m Step 2: Derive the orbital velocity formula For a circular orbit, gravitational force provides centripetal force: GMm/r² = mv₀²/r Where m is satellite m...

Q4: Define escape velocity. Calculate the escape velocity from Earth's surface. (Mass of Earth = 6 × 10²⁴ kg, Radius of Earth = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N⋅m²/kg²)

Escape Velocity Definition: Escape velocity is the minimum velocity needed by an object at the surface of a celestial body (or at a given height) to escape from its gravitational field without further propulsion. Key points: - It is the velocity at which the total mechanical energy becomes zero - An object with this velocity can reach infinity with zero velocity - It is independent of the mass of the escaping object - It depends only on the mass and radius of the celestial body - It is always ...

Q5: A planet has twice the mass and three times the radius of Earth. Calculate the ratio of the acceleration due to gravity on this planet to that on Earth.

Given: Mass of planet M_p = 2M_e (where M_e = Earth's mass) Radius of planet R_p = 3R_e (where R_e = Earth's radius) Find: Ratio g_p/g_e Step 1: Write the formula for acceleration due to gravity For Earth: g_e = GM_e/R_e² For the planet: g_p = GM_p/R_p² Step 2: Substitute the given relationships g_p = G(2M_e)/(3R_e)² g_p = 2GM_e/(9R_e²) g_p = (2/9) × (GM_e/R_e²) g_p = (2/9) × g_e Step 3: Calculate the ratio g_p/g_e = (2/9) × g_e/g_e = 2/9 Therefore: g_p = (2/9) × g_e ≈ 0.222 × g_e Step...

Q6: Two satellites orbit Earth at different altitudes. Satellite A is at height h₁ = 400 km and Satellite B is at height h₂ = 1600 km above Earth's surface. Find the ratio of their orbital periods T_A/T_B. (Radius of Earth R = 6400 km)

Given: Height of satellite A: h₁ = 400 km Height of satellite B: h₂ = 1600 km Radius of Earth: R = 6400 km Find: Ratio T_A/T_B Step 1: Calculate orbital radii from Earth's centre For satellite A: r₁ = R + h₁ = 6400 + 400 = 6800 km = 6.8 × 10⁶ m For satellite B: r₂ = R + h₂ = 6400 + 1600 = 8000 km = 8.0 × 10⁶ m Step 2: Apply Kepler's Third Law For any satellite orbiting a body: T² ∝ r³ Or: T² = (4π²/GM) × r³ Therefore: T = 2π√(r³/GM) For two satellites: T₁² = (4π²/GM) × r₁³ T₂² = (4π²/GM...

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