Gravitation — Previous Year Questions (Class 11 Physics)
Gravitation explains the force binding celestial bodies and objects on Earth. Study Newton's law, orbital motion, and gravitational potential for board suc
TL;DR: Gravitation explains the force binding celestial bodies and objects on Earth. Study Newton's law, orbital motion, and gravitational potential for boar…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Gravitation explains the force binding celestial bodies and objects on Earth. Study Newton's law, orbital motion, and gravitational potential for board suc
Gravitation — Previous Year Questions with Solutions
Q (2023, 2 marks): Two bodies of masses 100 kg and 50 kg are separated by a distance of 2 m. Calculate the gravitational force between them. (G = 6.67 × 10⁻¹¹ N·m²/kg²)
Answer: Given: M = 100 kg, m = 50 kg, r = 2 m, G = 6.67 × 10⁻¹¹ N·m²/kg²
Using Newton's law of gravitation:
F = GMm/r²
F = (6.67 × 10⁻¹¹ × 100 × 50)/2²
F = (6.67 × 10⁻¹¹ × 5000)/4
F = (3.335 × 10⁻⁸)/4
F = 8.34 × 10⁻⁹ N
Final Answer: F ≈ 8.34 × 10⁻⁹ N
Q (2022, 3 marks): The acceleration due to gravity on Earth's surface is 10 m/s². If the radius of Earth is 6.4 × 10⁶ m, calculate the mass of Earth. (G = 6.67 × 10⁻¹¹ N·m²/kg²)
Answer: At Earth's surface: g = GM/R²
where M is Earth's mass and R is its radius.
M = gR²/G
M = (10 × (6.4 × 10⁶)²)/(6.67 × 10⁻¹¹)
M = (10 × 4.096 × 10¹³)/(6.67 × 10⁻¹¹)
M = (4.096 × 10¹⁴)/(6.67 × 10⁻¹¹)
M ≈ 6.14 × 10²⁴ kg
Final Answer: M ≈ 6.14 × 10²⁴ kg
Q (2023, 3 marks): A satellite orbits Earth at a height of 400 km. Calculate the orbital speed of the satellite. (R = 6.4 × 10⁶ m, g = 10 m/s²)
Answer: Height h = 400 km = 4 × 10⁵ m
Orbital radius r = R + h = 6.4 × 10⁶ + 4 × 10⁵ = 6.8 × 10⁶ m
For orbital motion: mg' = mv²/r, where g' = GM/r²
Orbital speed: v = √(GM/r) = √(g₀R²/r)
v = √((10 × (6.4 × 10⁶)²)/(6.8 × 10⁶))
v = √((10 × 6.4² × 10¹²)/(6.8 × 10⁶))
v = √((409.6 × 10⁶)/6.8)
v = √(6.02 × 10⁷)
v ≈ 7.76 × 10³ m/s ≈ 7.76 km/s
Final Answer: v ≈ 7.76 km/s
Q (2021, 3 marks): Define gravitational potential and calculate the gravitational potential at Earth's surface. (R = 6.4 × 10⁶ m, g = 10 m/s²)
Answer: Gravitational potential (V) is the work done per unit mass by the gravitational field to bring a test mass from infinity to a point.
V = -GM/r = -gR (at Earth's surface)
V = -(10 × 6.4 × 10⁶)
V = -6.4 × 10⁷ J/kg
Final Answer: V = -6.4 × 10⁷ J/kg
Q (2022, 5 marks): Derive the relation between g and g' (where g' is acceleration due to gravity at height h above Earth's surface).
Answer: At Earth's surface: g = GM/R²
At height h: g' = GM/(R+h)²
Dividing g' by g:
g'/g = [GM/(R+h)²]/[GM/R²]
g'/g = R²/(R+h)²
g' = g × [R/(R+h)]²
For h << R: g' ≈ g(1 - 2h/R)
This shows that gravity decreases with height.
Final Answer: g' = g[R/(R+h)]² or approximately g' ≈ g(1 - 2h/R)
Q (2023, 5 marks): Calculate the escape velocity from Earth's surface. (R = 6.4 × 10⁶ m, g = 10 m/s²)
Answer: Escape velocity is the minimum speed needed for an object to escape Earth's gravitational pull.
Using energy conservation:
(1/2)mv_e² = GMm/R
v_e = √(2GM/R) = √(2gR)
v_e = √(2 × 10 × 6.4 × 10⁶)
v_e = √(1.28 × 10⁸)
v_e ≈ 1.13 × 10⁴ m/s ≈ 11.3 km/s
Final Answer: v_e ≈ 11.3 km/s
Frequently Asked Questions
Why is gravitational force always attractive?
Gravitational force is always attractive because it arises from the mutual interaction between masses. Unlike electric forces which can be attractive or repulsive, gravity depends only on mass (not charge), and all masses are positive, resulting in an always-attractive force.
What is the difference between orbital velocity and escape velocity?
Orbital velocity is the speed needed for an object to orbit at a given height without falling: v_o = √(GM/r). Escape velocity is the minimum speed to escape gravitational attraction entirely: v_e = √(2GM/r). Thus v_e = √2 × v_o.
More Class 11 Physics PYQs
- Current Electricity
- Ray Optics
- Electrostatics
- Moving Charges and Magnetism
- Electromagnetic Induction
- Alternating Current
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →