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NCERT Solutions › Physics › Chapter 8

Chapter 8: Gravitation

NCERT Solutions for CBSE Class 10 Physics — 31 solved questions with detailed explanations.

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Solved Questions

Q1. The escape velocity from Earth's surface is approximately:

Difficulty: Easy · Topic: Escape Velocity

Solution

Escape velocity v_e = √(2gR) = √(2 × 9.8 × 6.4×10⁶) ≈ 11.2 km/s.

Q2. Kepler's third law states that T² is proportional to:

Difficulty: Easy · Topic: Kepler's Laws

Solution

Kepler's third law: T² ∝ r³, where T is the orbital period and r is the semi-major axis.

Q3. The acceleration due to gravity at a height equal to the radius of Earth (R) above the surface is:

Difficulty: Easy-Medium · Topic: Variation of g

Solution

g_h = g(R/(R+h))². At h = R: g_h = g(R/2R)² = g/4.

Q4. The gravitational potential energy of a mass m at distance r from the centre of Earth (mass M) is:

Difficulty: Easy-Medium · Topic: Gravitational PE

Solution

Gravitational PE = -GMm/r. The negative sign indicates bound state.

Q5. Acceleration due to gravity at height R above Earth's surface (R = Earth's radius) is:

Difficulty: Easy-Medium · Topic: Variation of g

Solution

g_h = g(R/(R+h))². At h=R: g_h = g/4.

Q6. Acceleration due to gravity at height 2R above Earth's surface (R = Earth's radius) is:

Difficulty: Easy-Medium · Topic: Variation of g

Solution

g_h = g(R/(R+h))². At h=2R: g_h = g/9.

Q7. Acceleration due to gravity at height 3R above Earth's surface (R = Earth's radius) is:

Difficulty: Easy-Medium · Topic: Variation of g

Solution

g_h = g(R/(R+h))². At h=3R: g_h = g/16.

Q8. Acceleration due to gravity at height R/2 above Earth's surface (R = Earth's radius) is:

Difficulty: Easy-Medium · Topic: Variation of g

Solution

g_h = g(R/(R+h))². At h=R/2: g_h = 4g/9.

Q9. The orbital velocity at surface near Earth's surface is given by:

Difficulty: Easy-Medium · Topic: Orbital Mechanics

Solution

orbital velocity at surface = v = √(gR).

Q10. The escape velocity near Earth's surface is given by:

Difficulty: Easy-Medium · Topic: Orbital Mechanics

Solution

escape velocity = v = √(2gR).

Q11. g at height h=0.5R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+0.5)² = 0.4444.

Q12. g at height h=1R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+1)² = 0.25.

Q13. g at height h=1.5R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+1.5)² = 0.16.

Q14. g at height h=2R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+2)² = 0.1111.

Q15. g at height h=2.5R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+2.5)² = 0.0816.

Q16. g at height h=3R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+3)² = 0.0625.

Q17. g at height h=4R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+4)² = 0.04.

Q18. g at height h=5R above Earth surface. Ratio g_h/g:

Difficulty: Easy-Medium · Topic: g variation

Solution

g_h/g = (R/(R+h))² = 1/(1+5)² = 0.0278.

Q19. Two planets have radii r₁ and r₂ and densities ρ₁ and ρ₂. The ratio of acceleration due to gravity on their surfaces is:

Difficulty: Medium · Topic: Comparative Gravity

Solution

g = GM/R² = G(4/3)πR³ρ/R² = (4/3)GπRρ. So g₁/g₂ = r₁ρ₁/(r₂ρ₂).

Q20. The gravitational PE of a 2 kg object on Earth's surface relative to infinity is approximately (g=10, R=6400 km):

Difficulty: Medium · Topic: Gravitational PE

Solution

PE = -mgR = -2×10×6.4×10⁶ J.

Q21. The gravitational PE of a 5 kg object on Earth's surface relative to infinity is approximately (g=10, R=6400 km):

Difficulty: Medium · Topic: Gravitational PE

Solution

PE = -mgR = -5×10×6.4×10⁶ J.

Q22. The gravitational PE of a 10 kg object on Earth's surface relative to infinity is approximately (g=10, R=6400 km):

Difficulty: Medium · Topic: Gravitational PE

Solution

PE = -mgR = -10×10×6.4×10⁶ J.

Q23. The gravitational PE of a 20 kg object on Earth's surface relative to infinity is approximately (g=10, R=6400 km):

Difficulty: Medium · Topic: Gravitational PE

Solution

PE = -mgR = -20×10×6.4×10⁶ J.

Q24. Escape velocity for g=9.8, R=1600km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=5.6 km/s.

Q25. Escape velocity for g=9.8, R=3200km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=7.92 km/s.

Q26. Escape velocity for g=9.8, R=6400km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=11.2 km/s.

Q27. Escape velocity for g=9.8, R=12800km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=15.84 km/s.

Q28. Escape velocity for g=10, R=1600km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=5.66 km/s.

Q29. Escape velocity for g=10, R=3200km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=8.0 km/s.

Q30. Escape velocity for g=10, R=6400km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=11.31 km/s.

Q31. Escape velocity for g=10, R=12800km:

Difficulty: Medium · Topic: Escape Velocity

Solution

v=√(2gR)=16.0 km/s.

Other Chapters in Physics

Ch 1: Physical WorldCh 2: Units & MeasurementsCh 3: Motion in a Straight LineCh 4: Motion in a PlaneCh 5: Laws of MotionCh 6: Work Energy & PowerCh 7: System of ParticlesCh 9: Mechanical Properties of SolidsCh 10: Mechanical Properties of FluidsCh 11: Thermal Properties of MatterCh 12: ThermodynamicsCh 13: Kinetic TheoryCh 14: OscillationsCh 15: Waves

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