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.
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
NCERT Solutions for Class 11 Physics — Gravitation. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Gravitation — All 6 Questions Solved
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:
- The force is always attractive
- The force acts along the line joining the centres of the two masses
- The force is independent of the medium between the objects
- It is the weakest fundamental force
Derivation of Acceleration Due to Gravity at Earth's Surface:
Consider:
m = mass of an object on Earth's surface
M = mass of Earth = 5.97 × 10²⁴ kg
R = radius of Earth = 6.37 × 10⁶ m
g = acceleration due to gravity
Step 1: Gravitational force on the object
Using Newton's law of universal gravitation:
F = GMm/R²
Step 2: Apply Newton's second law
F = mg
Step 3: Equate the two expressions
mg = GMm/R²
Step 4: Simplify
g = GM/R²
Step 5: Substitute values
g = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(6.37 × 10⁶)²
g = (6.67 × 5.97 × 10¹³)/(40.58 × 10¹²)
g = (39.82 × 10¹³)/(40.58 × 10¹²)
g ≈ 9.81 m/s² ≈ 10 m/s² (approximately)
Key Points:
- Acceleration due to gravity is independent of the mass of the object
- It depends only on the mass and radius of the Earth
- It varies with altitude and latitude
- At Earth's surface: g ≈ 9.8 m/s² or 10 m/s²
- At height h above surface: g' = GM/(R + h)² ≈ g(R/(R + h))² for small h
Variation of g:
With altitude:
g(h) = GM/(R + h)² = g₀[R/(R + h)]²
With latitude:
- g is maximum at poles (9.832 m/s²) due to Earth's rotation and shape
- g is minimum at equator (9.78 m/s²)
Effect of Earth's rotation:
g_effective = g_actual - ω²R cos²λ
Where λ is the latitude and ω is angular velocity of Earth's rotation
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 = 4.902/3.0276
g_moon = 1.62 m/s²
Part 2: Compare with Earth's gravity
Ratio:
g_moon/g_earth = 1.62/10 = 0.162 ≈ 1/6
Therefore:
g_moon ≈ (1/6) × g_earth
Or:
g_earth/g_moon = 10/1.62 ≈ 6.17 ≈ 6
Part 3: Physical interpretation
The acceleration due to gravity on the Moon is approximately 1/6 of that on Earth.
Reasons:
1. Mass effect: Moon's mass ≈ 1/81 of Earth's mass
2. Radius effect: Moon's radius ≈ 1/3.7 of Earth's radius
3. Combined effect: g_moon/g_earth = (M_m/M_e) × (R_e/R_m)²
Verification:
g_moon/g_earth = (7.35 × 10²²/5.97 × 10²⁴) × (6.37 × 10⁶/1.74 × 10⁶)²
= (0.0123) × (3.66)²
= 0.0123 × 13.4
= 0.165 ≈ 1/6 ✓
Part 4: Implications
A person with mass 60 kg:
- Weight on Earth = 60 × 10 = 600 N
- Weight on Moon = 60 × 1.62 = 97.2 N ≈ 100 N
- The person would weigh only 1/6 as much on the Moon
For motion on Moon:
- A person can jump much higher
- Objects fall more slowly
- Less force needed for the same acceleration
Answer: The acceleration due to gravity on the Moon's surface is 1.62 m/s², which is approximately 1/6 (or about 16%) of Earth's gravity.
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 mass, which cancels out:
GM/r² = v₀²/r
GM/r = v₀²
v₀ = √(GM/r)
Step 3: Calculate GM
GM = 6.67 × 10⁻¹¹ × 6 × 10²⁴
= 6.67 × 6 × 10¹³
= 40.02 × 10¹³
= 4.002 × 10¹⁴ N⋅m²/kg
Step 4: Calculate orbital velocity
v₀ = √(GM/r)
v₀ = √(4.002 × 10¹⁴/6.8 × 10⁶)
v₀ = √(5.885 × 10⁷)
v₀ = 7.67 × 10³ m/s
v₀ ≈ 7.67 km/s
Alternative calculation:
v₀ = √(4.002 × 10¹⁴/6.8 × 10⁶)
= √(0.5885 × 10⁸)
= √(5.885 × 10⁷)
= 7670 m/s ≈ 7.67 km/s
Step 5: Calculate orbital period
For circular orbit:
T = 2πr/v₀
T = 2π × 6.8 × 10⁶/7670
T = 4.27 × 10⁷/7670
T = 5570 s
T ≈ 92.8 minutes ≈ 1.55 hours
Step 6: Compare with other orbits
Escape velocity at this height:
v_escape = √(2GM/r) = √2 × v₀ = 1.414 × 7.67 = 10.85 km/s
Velocity at Earth's surface (theoretical):
v₀(surface) = √(GM/R) = √(4.002 × 10¹⁴/6.4 × 10⁶)
= √(6.253 × 10⁷) = 7.91 km/s
Answer: The orbital velocity of the satellite is 7.67 km/s (or 7670 m/s). The orbital period is approximately 92.8 minutes (or about 1.55 hours).
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 measured with respect to the celestial body (not the ground)
Derivation of Escape Velocity:
At Earth's surface:
- Gravitational potential energy: PE = -GMm/R
- For escape, kinetic energy must equal magnitude of potential energy
Using energy conservation:
Initial KE + Initial PE = Final KE + Final PE
(1/2)mv_e² + (-GMm/R) = 0 + 0
(1/2)mv_e² = GMm/R
v_e² = 2GM/R
v_e = √(2GM/R)
Or: v_e = √2 × √(GM/R) = √2 × v_orbital
Calculation of Escape Velocity from Earth:
Given:
M = 6 × 10²⁴ kg
R = 6.4 × 10⁶ m
G = 6.67 × 10⁻¹¹ N⋅m²/kg²
Step 1: Calculate GM
GM = 6.67 × 10⁻¹¹ × 6 × 10²⁴
GM = 4.002 × 10¹⁴ N⋅m²/kg
Step 2: Calculate 2GM/R
2GM/R = 2 × 4.002 × 10¹⁴/6.4 × 10⁶
= 8.004 × 10¹⁴/6.4 × 10⁶
= 1.251 × 10⁸ m²/s²
Step 3: Calculate escape velocity
v_e = √(1.251 × 10⁸)
v_e = 1.118 × 10⁴ m/s
v_e ≈ 11.18 km/s ≈ 11.2 km/s
Alternative form using g:
At Earth's surface: g = GM/R² = 10 m/s² (approximately)
v_e = √(2gR) = √(2 × 10 × 6.4 × 10⁶)
= √(128 × 10⁶) = √(1.28 × 10⁸)
= 1.131 × 10⁴ m/s ≈ 11.3 km/s
Variation with Height:
Escape velocity at height h above surface:
v_e(h) = √(2GM/(R + h)) = √(2gR²/(R + h))
Example:
At height h = 400 km = 4 × 10⁵ m:
v_e(h) = √(2 × 4.002 × 10¹⁴/(6.8 × 10⁶))
= √(1.176 × 10⁸) = 10.85 km/s
Comparison:
Earth's escape velocity ≈ 11.2 km/s
Moon's escape velocity ≈ 2.4 km/s
Jupiter's escape velocity ≈ 61 km/s
Relation between orbital and escape velocities:
For any orbit: v_orbital = √(GM/r)
Escape velocity: v_e = √(2GM/r) = √2 × v_orbital
Ratio: v_e/v_orbital = √2 ≈ 1.414
At Earth's surface:
Orbital velocity ≈ 7.9 km/s
Escape velocity ≈ 11.2 km/s
Answer: The escape velocity from Earth's surface is approximately 11.2 km/s (or 11,200 m/s)
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 4: Numerical calculation
If g_e = 10 m/s²:
g_p = (2/9) × 10 = 20/9 ≈ 2.22 m/s²
Step 5: General formula derivation
For any planet with mass M_p = xM_e and radius R_p = yR_e:
g_p/g_e = (M_p/M_e) × (R_e/R_p)²
g_p/g_e = (x) × (1/y)²
g_p/g_e = x/y²
In this case:
g_p/g_e = 2/(3)² = 2/9
Step 6: Physical interpretation
- Although the planet has greater mass (2×), gravity is weaker
- This is because radius is larger (3×), and gravity decreases with distance squared
- The radius effect dominates: (3²) = 9 in denominator
- Despite 2× mass, gravity is only (2/9) ≈ 0.22 times Earth's
Step 7: Implications
For a person of mass m = 60 kg:
- Weight on Earth = 60 × 10 = 600 N
- Weight on planet = 60 × 2.22 = 133.3 N
- The person would weigh much less on this planet
Comparison with other bodies:
Body | Mass ratio | Radius ratio | g ratio
---|---|---|---
Mercury | 0.055 | 0.383 | 0.38
Venus | 0.815 | 0.949 | 0.90
Mars | 0.107 | 0.532 | 0.38
Jupiter | 318 | 11.2 | 2.53
Saturn | 95.2 | 9.45 | 1.07
Answer: The ratio g_p/g_e = 2/9 ≈ 0.222 or approximately 22.2% of Earth's gravity. If Earth's g = 10 m/s², then the planet's g ≈ 2.22 m/s²
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) × r₂³
Step 3: Find the ratio of periods
T₁²/T₂² = r₁³/r₂³
T₁/T₂ = √(r₁³/r₂³) = (r₁/r₂)^(3/2)
Step 4: Calculate the ratio
r₁/r₂ = 6800/8000 = 0.85 = 17/20
T_A/T_B = (6800/8000)^(3/2) = (0.85)^(3/2)
T_A/T_B = (0.85)^1.5
Step 5: Calculate numerically
Method 1: Direct calculation
(0.85)^(3/2) = (0.85)^(1.5) = √[(0.85)³]
(0.85)³ = 0.614125
√0.614125 = 0.784
Therefore: T_A/T_B ≈ 0.784
Method 2: Using exact fractions
(17/20)^(3/2) = (17/20)^(3/2)
= [√(17/20)]³
= [√17/√20]³
= [√17/(2√5)]³
Numerical:
√(17/20) = √0.85 ≈ 0.922
(0.922)³ ≈ 0.784
Step 6: Verification using ratio of cube roots
T_A/T_B = ∛(r₁³/r₂³) × √(r₁³/r₂³)
Alternatively:
(T_A/T_B)² = (r₁/r₂)³
(T_A/T_B)² = (6800/8000)³
(T_A/T_B)² = (17/20)³
(T_A/T_B)² = 4913/8000 = 0.6141
T_A/T_B = √0.6141 ≈ 0.784
Step 7: Physical interpretation
- Satellite A is closer to Earth (smaller orbital radius)
- It orbits faster with a shorter period
- T_A ≈ 0.784 × T_B
- This means T_A is about 78.4% of T_B
- Satellite B takes about 27.5% longer to orbit
Step 8: Calculate actual periods
If we use the formula T = 2π√(r³/GM):
With GM = 4 × 10¹⁴ m³/s²
For satellite A:
T_A = 2π√[(6.8 × 10⁶)³/(4 × 10¹⁴)]
T_A ≈ 5430 seconds ≈ 90.5 minutes
For satellite B:
T_B = 2π√[(8 × 10⁶)³/(4 × 10¹⁴)]
T_B ≈ 6930 seconds ≈ 115.5 minutes
Ratio: 5430/6930 ≈ 0.784 ✓
Answer: T_A/T_B ≈ 0.784 or approximately 27/34, meaning satellite A has a period about 78.4% of satellite B's period. Satellite B takes about 27.5% longer to complete one orbit.
Gravitation — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The value of universal gravitational constant G depends on:
- A. Nature of the interacting bodies
- B. Mass of the interacting bodies
- C. Distance between the interacting bodies
- D. None of these
Answer: D. None of these — The universal gravitational constant G is a fundamental constant and does not depend on the nature, mass, or distance of the interacting bodies, nor on the medium between them.
MCQ 2. At what height above the Earth's surface does the acceleration due to gravity decrease to 25% of its value on the surface? (Radius of Earth = R)
- A. R/2
- B. R
- C. 2R
- D. 3R
Answer: B. R — The acceleration due to gravity at height h is g' = g(R/(R+h))^2. If g' = 0.25g, then 0.25 = (R/(R+h))^2, which implies 0.5 = R/(R+h). Solving for h gives R = 0.5R + 0.5h, so 0.5R = 0.5h, thus h = R.
MCQ 3. Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side 'a'. The gravitational potential energy of the system is:
- A. -Gm²/a
- B. -2Gm²/a
- C. -3Gm²/a
- D. -6Gm²/a
Answer: C. -3Gm²/a — The gravitational potential energy of a system of particles is the sum of the potential energies of all unique pairs. For three particles, there are three pairs, each with potential energy -Gm²/a. So, the total potential energy is 3 * (-Gm²/a) = -3Gm²/a.
MCQ 4. Kepler's second law (law of areas) is a consequence of the conservation of:
- A. Linear momentum
- B. Angular momentum
- C. Energy
- D. Mass
Answer: B. Angular momentum — Kepler's second law states that the line joining a planet to the sun sweeps out equal areas in equal intervals of time, which is directly equivalent to the conservation of angular momentum.
MCQ 5. The escape velocity from the surface of Earth is approximately 11.2 km/s. If the radius of a planet is 4 times that of Earth and its mean density is the same as Earth, what is the escape velocity from that planet?
- A. 11.2 km/s
- B. 22.4 km/s
- C. 44.8 km/s
- D. 5.6 km/s
Answer: C. 44.8 km/s — Escape velocity v_e = sqrt(2GM/R) = sqrt(2G * (4/3)πR³ρ / R) = sqrt((8/3)πGρR²). Thus, v_e is proportional to R if density ρ is constant. Since the planet's radius is 4 times Earth's, its escape velocity will be 4 * 11.2 = 44.8 km/s.
MCQ 6. A satellite is orbiting Earth in a circular orbit at a height h from the surface. If its total energy is E, what is its potential energy?
- A. -E
- B. -2E
- C. 2E
- D. E/2
Answer: C. 2E — For a satellite in a circular orbit, the total energy (E) is E = -GMm/(2r), and the potential energy (U) is U = -GMm/r. Comparing these, we find that U = 2E.
MCQ 7. The acceleration due to gravity (g) at a depth 'd' from the Earth's surface decreases because:
- A. The mass of Earth decreases
- B. The effective radius decreases
- C. The mass contributing to gravity decreases
- D. The gravitational constant G decreases
Answer: C. The mass contributing to gravity decreases — When a body is at depth 'd', only the mass of the spherical shell of radius (R-d) contributes to the gravitational force. The mass of the outer shell has no net gravitational effect inside it, thus the contributing mass decreases.
MCQ 8. A particle of mass M is placed at the center of a thin spherical shell of mass M and radius R. The gravitational field intensity at a point P at a distance R/2 from the center is:
- A. GM/R²
- B. 2GM/R²
- C. 0
- D. GM/(R/2)²
Answer: D. GM/(R/2)² — The gravitational field intensity inside a thin spherical shell due to the shell itself is zero. Thus, the field at point P (R/2 from center) is solely due to the mass M placed at the center, which is GM/(R/2)² = 4GM/R².
MCQ 9. The binding energy of a satellite revolving in a circular orbit close to the Earth's surface is E. The energy required to transfer this satellite to an orbit of radius 2R (where R is Earth's radius) is:
- A. E/2
- B. E/4
- C. E/8
- D. E
Answer: A. E/2 — The binding energy (magnitude of total energy) of a satellite in orbit of radius r is GMm/(2r). Initially, E = GMm/(2R). Final total energy for orbit at 2R is E_final = -GMm/(2*(2R)) = -GMm/(4R). The initial total energy was E_initial = -GMm/(2R). The energy required for transfer is E_final - E_initial = GMm/(4R). Since E = GMm/(2R), the required energy is E/2.
MCQ 10. The value of 'g' (acceleration due to gravity) is maximum at:
- A. Equator
- B. Poles
- C. Tropics
- D. At a height R/2 from Earth's surface
Answer: B. Poles — Due to Earth's rotation and its oblate spheroid shape (flattened at poles, bulging at equator), 'g' is maximum at the poles and minimum at the equator.
MCQ 11. A planet is revolving in an elliptical orbit around the Sun. Its closest distance from the Sun is r1 and the farthest distance is r2. If the velocity of the planet at the closest distance is v1, then its velocity at the farthest distance is:
- A. v1 * (r1/r2)
- B. v1 * (r2/r1)
- C. v1 * sqrt(r1/r2)
- D. v1 * sqrt(r2/r1)
Answer: A. v1 * (r1/r2) — This is based on the conservation of angular momentum. Angular momentum L = mvr = constant. So, m*v1*r1 = m*v2*r2. Therefore, v2 = v1 * (r1/r2).
MCQ 12. Two bodies of masses M and 4M are placed at a distance 'r' apart. The gravitational potential at a point on the line joining them where the gravitational field is zero is:
- A. -GM/r
- B. -9GM/r
- C. -2GM/r
- D. -5GM/r
Answer: B. -9GM/r — Let the point where the field is zero be at distance x from M. Then GM/x² = G(4M)/(r-x)², which implies 1/x = 2/(r-x) leading to x = r/3. The gravitational potential V = -GM/x - G(4M)/(r-x) = -GM/(r/3) - G(4M)/(2r/3) = -3GM/r - 6GM/r = -9GM/r.
MCQ 13. A geostationary satellite appears stationary to an observer on Earth because:
- A. It has zero velocity
- B. Its orbital period is 24 hours
- C. It orbits at a very high altitude
- D. Its mass is very small
Answer: B. Its orbital period is 24 hours — A geostationary satellite has an orbital period of exactly 24 hours, the same as Earth's rotation period, and orbits in the equatorial plane, making it appear stationary from the ground.
MCQ 14. If the angular speed of Earth's rotation is increased, the weight of a body at the equator will:
- A. Increase
- B. Decrease
- C. Remain unchanged
- D. Become zero
Answer: B. Decrease — The effective acceleration due to gravity at the equator is g' = g - Rω². If the angular speed ω increases, the Rω² term increases, causing g' to decrease. Since weight W = mg', the weight of the body will decrease.
MCQ 15. A particle is projected from the surface of Earth with a velocity equal to the escape velocity. The angle it makes with the horizontal is 60°. What is the maximum height attained by the particle?
- A. R
- B. 2R
- C. 4R
- D. Infinity
Answer: D. Infinity — If a particle is projected with escape velocity, irrespective of the angle of projection (as long as it's not directly into the Earth), it will escape Earth's gravitational field and reach infinity.
MCQ 16. The gravitational potential at the center of a hollow sphere of mass M and radius R is:
- A. -GM/R
- B. 0
- C. GM/R
- D. -GM/(2R)
Answer: A. -GM/R — The gravitational potential inside a hollow spherical shell is constant and equal to the potential at its surface, which is -GM/R.
MCQ 17. The orbital velocity of a satellite very close to the Earth's surface is v_o. The escape velocity from Earth's surface is v_e. Which of the following relations is correct?
- A. v_e = v_o
- B. v_e = sqrt(2) * v_o
- C. v_e = 2 * v_o
- D. v_e = v_o / sqrt(2)
Answer: B. v_e = sqrt(2) * v_o — The orbital velocity for a satellite very close to the Earth's surface (r ≈ R) is v_o = sqrt(GM/R). The escape velocity from Earth's surface is v_e = sqrt(2GM/R). Comparing these, v_e = sqrt(2) * v_o.
MCQ 18. A satellite of mass m is in a circular orbit of radius R around Earth. What is the minimum energy required to raise it to an orbit of radius 2R?
- A. GMm/(2R)
- B. GMm/(4R)
- C. GMm/(8R)
- D. GMm/(12R)
Answer: B. GMm/(4R) — The total mechanical energy of a satellite in a circular orbit of radius r is E = -GMm/(2r). Initial energy E1 = -GMm/(2R). Final energy E2 = -GMm/(2 * 2R) = -GMm/(4R). Minimum energy required for transfer = E2 - E1 = (-GMm/(4R)) - (-GMm/(2R)) = GMm/(4R).
MCQ 19. If the Earth suddenly shrinks to half its radius without any change in its mass, the acceleration due to gravity on its surface will:
- A. Decrease by a factor of 2
- B. Increase by a factor of 2
- C. Increase by a factor of 4
- D. Decrease by a factor of 4
Answer: C. Increase by a factor of 4 — The acceleration due to gravity on the surface is g = GM/R². If R becomes R/2, then g' = GM/(R/2)² = GM/(R²/4) = 4 * (GM/R²) = 4g. So, g increases by a factor of 4.
MCQ 20. Two planets of radii R1 and R2 have densities ρ1 and ρ2 respectively. If the ratio of acceleration due to gravity on their surfaces is g1/g2 = 1, then R1/R2 is:
- A. ρ2/ρ1
- B. ρ1/ρ2
- C. sqrt(ρ1/ρ2)
- D. sqrt(ρ2/ρ1)
Answer: A. ρ2/ρ1 — The acceleration due to gravity on the surface of a planet can be expressed as g = GM/R² = G * (4/3)πR³ρ / R² = (4/3)πGρR. If g1/g2 = 1, then ρ1R1 = ρ2R2, which means R1/R2 = ρ2/ρ1.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.