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Gravitation – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 7. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Every mass attracts every other mass. Newton's law of gravitation explains weight, the variation of g, orbits of planets and satellites, and escape speed.

Law and g

F = Gm₁m₂/r², G = 6.67 × 10⁻¹¹ N m²/kg².
g = GM/R² ≈ 9.8 m/s²; g decreases with height (g_h ≈ g(1 − 2h/R)) and depth (g_d = g(1 − d/R)); g is zero at the centre.
Kepler's laws: elliptical orbits, equal areas in equal times, T² ∝ r³.

Energy and satellites

Gravitational potential energy U = −GMm/r.
Orbital speed v₀ = √(GM/r) ≈ √(gR) near the surface (≈ 7.9 km/s); escape speed vₑ = √(2GM/R) ≈ 11.2 km/s on Earth; vₑ = √2 v₀.
Geostationary satellites: period 24 h, about 36,000 km above the equator.

Key terms

Escape speed: minimum speed needed to leave a planet's gravity
Geostationary satellite: satellite with 24 h period that appears fixed over the equator
Gravitational potential: work done per unit mass in bringing a test mass from infinity to a point, V = −GM/r

Formulas

F = Gm₁m₂/r²
vₑ = √(2gR)
T² ∝ r³

Important questions with answers

Why does an astronaut feel weightless in an orbiting spacecraft?2 marks

Both the astronaut and the spacecraft are in free fall towards Earth with the same acceleration, so there is no normal force from the floor, which is what we feel as weight.

Derive the expression for escape speed from Earth.3 marks

To escape, the KE must at least equal the magnitude of gravitational PE: ½mvₑ² = GMm/R. So vₑ = √(2GM/R) = √(2gR) (since GM = gR²). With g = 9.8 m/s² and R = 6.4 × 10⁶ m, vₑ ≈ 11.2 km/s.

Derive the variation of g with depth and with height (h ≪ R). At what depth will g be the same as at a height of 32 km?5 marks

Depth d: only the inner sphere of radius (R − d) attracts; its mass ∝ (R − d)³, so g_d = G × M(R − d)³/R³/(R − d)² = g(1 − d/R). Height h: g_h = GM/(R + h)² = g(1 + h/R)⁻² ≈ g(1 − 2h/R) for h ≪ R. Equating: 1 − d/R = 1 − 2h/R gives d = 2h = 64 km.

Quick revision

g is zero at Earth's centre.
Escape speed does not depend on the mass of the body.
Kepler's second law = conservation of angular momentum.

Exam tips

Weightlessness in orbit happens because the satellite and astronaut fall freely together.
Orbital speed does not depend on the satellite's mass; total energy of an orbiting satellite is −GMm/2r (negative means bound).

Practice MCQs

Value of g on Earth's surface is about
1. 9.8 m/s²
2. 6.67 m/s²
3. 1.6 m/s²
4. 11.2 m/s²
Show answer
Answer: (a) 9.8 m/s². Standard value.
Escape speed from Earth is about
1. 7.9 km/s
2. 3 km/s
3. 11.2 km/s
4. 42 km/s
Show answer
Answer: (c) 11.2 km/s. √(2gR).
If the distance between two masses doubles, the force becomes
1. half
2. four times
3. double
4. one-fourth
Show answer
Answer: (d) one-fourth. Inverse square.
Period of a geostationary satellite is
1. 24 h
2. 12 h
3. 1 h
4. 365 days
Show answer
Answer: (a) 24 h. Same as Earth's rotation.
If the radius of Earth shrinks to half with the same mass, g on its surface becomes
1. 4g
2. 2g
3. g/2
4. g/4
Show answer
Answer: (a) 4g. g ∝ 1/R².
If orbital radius becomes 4 times, the period becomes
1. 2 times
2. 4 times
3. 8 times
4. 16 times
Show answer
Answer: (c) 8 times. T ∝ r^(3/2).
Relation between escape speed vₑ and orbital speed v₀ near the surface is
1. vₑ = v₀
2. vₑ = √2 v₀
3. vₑ = 2v₀
4. vₑ = v₀/√2
Show answer
Answer: (b) vₑ = √2 v₀. √(2gR) vs √(gR).
At what height above Earth's surface does g become one-fourth of its surface value (R = Earth's radius)
1. R/2
2. 2R
3. R
4. 4R
Show answer
Answer: (c) R. (R/(R + h))² = 1/4.

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.