How to Use This Page
Solve each problem yourself first. For gravitation numericals, use the correct formula, substitute with consistent units (SI), compute carefully, and always state the answer with proper units. Pay special attention to whether you need g at the surface or at altitude/depth.
Key Concepts (Quick Recap)
- Universal law of gravitation: F = GMm/r², G = 6.674 × 10⁻¹¹ N m² kg⁻²
- Acceleration due to gravity: g = GM/R² ≈ 9.8 m s⁻²
- g at altitude h: g_h = gR²/(R + h)²
- g at depth d: g_d = g(1 − d/R)
- Kepler's laws:
- First: Planets move in elliptical orbits with Sun at one focus
- Second: Areal velocity is constant (equal areas in equal times)
- Third: T² ∝ r³ → T² = (4π²/GM)r³
- Gravitational potential energy: U = −GMm/r
- Escape velocity: v_e = √(2GM/R) = √(2gR) ≈ 11.2 km s⁻¹ (Earth)
- Orbital velocity: v_o = √(GM/r) = √(gR²/r)
- Time period of satellite: T = 2π√(r³/GM)
- Geostationary satellite: h ≈ 35786 km, T = 24 h
- Total energy of satellite: E = −GMm/2r
- Binding energy: E_b = GMm/2r
Intext Questions
Page 176
Q1. You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Answer
No. Unlike electric charges, there are no negative masses that can cancel gravitational effects. Gravity is always attractive and there is no "gravitational shielding." A body inside a hollow spherical shell experiences no net gravitational force from the shell (by the shell theorem), but it is still affected by masses outside the shell. There is no way to completely shield a body from gravitational influence.
Q2. An astronaut inside a small spaceship orbiting the Earth cannot detect gravity. Can an astronaut in the same spaceship detect gravity if the spaceship is very large with enough mass? If so, what is the approximate size?
Answer
If the spaceship is small, the astronaut and the ship are both in free fall, so the astronaut feels weightless (like inside the ISS). If the spaceship is very large (comparable to the Earth), the gravitational field would differ significantly across its extent (tidal effects), and the astronaut would feel a gravitational pull toward the centre of the ship.
The approximate size needed for detection: When the difference in gravitational acceleration across the ship (tidal effect) is detectable. This would require the ship's radius to be a significant fraction of the orbital radius. For practical purposes, a ship with radius of several hundred kilometres might produce detectable effects.
Page 180
Q3. How does the weight of a body vary with respect to the position on Earth? Is it the same everywhere?
Answer
No, the weight of a body is not the same everywhere on Earth. It varies because: 1. Shape of Earth: Earth is an oblate spheroid — radius at the equator is greater than at the poles. Since g = GM/R², g is less at the equator (R is larger) and more at the poles. 2. Rotation of Earth: Centrifugal force at the equator reduces the effective weight. At the poles, there is no centrifugal effect. 3. Local variations: Differences in density of rocks and minerals cause small local variations.
A body weighs about 0.5% more at the poles than at the equator.
Q4. If the Earth suddenly shrinks to half its radius without change in mass, what would be the value of g on its surface?
Answer
g = GM/R². If R becomes R/2: g' = GM/(R/2)² = 4GM/R² = 4g
So the new value of g would be 4 × 9.8 = 39.2 m s⁻².
Page 184
Q5. Does the escape speed of a body from the Earth depend on the mass of the body?
Answer
No. The escape velocity v_e = √(2GM/R) depends only on the mass M and radius R of the Earth (or the planet), not on the mass of the escaping body. The kinetic energy needed (½mv_e²) depends on the body's mass, but the speed itself does not.
Q6. A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the Earth's gravitational influence? Mass of satellite = 200 kg, mass of Earth = 6 × 10²⁴ kg, radius of Earth = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻².
Answer
r = R + h = 6.4 × 10⁶ + 0.4 × 10⁶ = 6.8 × 10⁶ m
Total energy of satellite in orbit: E = −GMm/2r Energy to escape = 0 − E = GMm/2r
= (6.67 × 10⁻¹¹ × 6 × 10²⁴ × 200)/(2 × 6.8 × 10⁶) = (6.67 × 6 × 200 × 10¹³)/(13.6 × 10⁶) = (8004 × 10¹³)/(13.6 × 10⁶) = 588.5 × 10⁷ = 5.9 × 10⁹ J
Page 188
Q7. What is the speed of a geostationary satellite in its orbit?
Answer
For a geostationary satellite: T = 24 h = 86400 s Orbital radius: r = R + h = 6.4 × 10⁶ + 3.5786 × 10⁷ = 4.22 × 10⁷ m
v = 2πr/T = 2π × 4.22 × 10⁷/86400 = 2.652 × 10⁸/86400 = 3.07 × 10³ m s⁻¹ ≈ 3.07 km s⁻¹
Q8. Does a comet move faster near the Sun or far from the Sun? Why?
Answer
A comet moves faster near the Sun and slower when far from it. By Kepler's second law, the areal velocity is constant: the comet sweeps equal areas in equal times. When the comet is near the Sun, the radius is smaller, so it must move faster to sweep the same area. When far away, the radius is larger and the speed is lower. This is also consistent with conservation of energy — the comet has more kinetic energy (and less potential energy) when closer to the Sun.
Chapter-End Exercise
Q8.1. Answer the following: (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter? (b) The gravitational force follows the inverse square law. Do other forces also follow this law?
Answer
(a) No — there is no gravitational shielding. (See intext Q1.) (b) Yes — the electrostatic (Coulomb) force also follows the inverse square law (F ∝ 1/r²). The strong and weak nuclear forces do not follow the inverse square law — they are short-range forces.
Q8.2. Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative and its kinetic energy is positive. (b) The energy required to rocket an orbiting satellite out of Earth's gravity is positive/negative?
Answer
(a) Correct — Total energy E = −GMm/2r (negative), KE = GMm/2r (positive).
| (b) Positive — energy must be supplied (added) to escape, so the energy required is positive (equal to | E | = GMm/2r). |
|---|
Q8.3. Does the escape speed of a body from the Earth depend on (a) the mass of the body (b) the location from where it is launched (c) the direction of projection (d) the height of the location from where the body is launched?
Answer
(a) No — v_e = √(2GM/R) is independent of body mass. (b) Yes — v_e depends on R (distance from Earth's centre). From a mountaintop (larger R), v_e is smaller. (c) No — v_e is independent of the direction of projection (as long as we ignore atmospheric drag). (d) Yes — at greater height, R is larger, so v_e = √(2GM/(R+h)) is smaller.
Q8.4. A comet orbits the Sun in a highly elliptical orbit. Which quantities remain constant: speed, kinetic energy, potential energy, total energy, angular momentum, linear momentum?
Answer
The constant quantities are:
- Total energy (conserved — no external work)
- Angular momentum (conserved — central force, no external torque)
The varying quantities are:
- Speed (varies — faster near Sun)
- KE (varies — depends on speed)
- PE (varies — depends on distance from Sun)
- Linear momentum (varies — direction and magnitude change)
Q8.5. The gravitational force between two bodies is F. If the distance between them is doubled and the mass of each body is also doubled, what is the new force?
Answer
F = GMm/r² New: F' = G(2M)(2m)/(2r)² = 4GMm/4r² = GMm/r² = F
The force remains the same (F).
Q8.6. What is the acceleration due to gravity at a height equal to half the radius of the Earth? (g = 9.8 m s⁻²)
Answer
h = R/2, r = R + R/2 = 3R/2
g_h = gR²/r² = gR²/(3R/2)² = gR²/(9R²/4) = 4g/9
g_h = 4 × 9.8/9 = 39.2/9 = 4.36 m s⁻²
Q8.7. Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh at a depth equal to half the radius of the Earth if it weighs 250 N on the surface?
Answer
At depth d = R/2: g_d = g(1 − d/R) = g(1 − ½) = g/2
Weight at depth = 250 × (1/2) = 125 N
Q8.8. A rocket is fired from the Earth towards the Moon. At what distance from the Earth is the gravitational pull of the Earth equal to that of the Moon? (Mass of Earth = 81 × mass of Moon, distance between centres = 3.8 × 10⁵ km)
Answer
Let M_E = 81M, distance from Earth = x, distance from Moon = d − x
GM_E/x² = GM_M/(d − x)² 81M/x² = M/(d − x)² 81/(d − x)² = 1/x² 9/(d − x) = 1/x (taking positive root) 9x = d − x 10x = d = 3.8 × 10⁵ km x = 3.42 × 10⁵ km from Earth (or 0.38 × 10⁵ km = 3.8 × 10⁴ km from Moon)
Q8.9. The escape velocity from the surface of the Earth is 11.2 km s⁻¹. If the Earth had a radius twice the present value and the same mass, what would be the escape velocity?
Answer
v_e = √(2GM/R)
If R' = 2R: v_e' = √(2GM/2R) = v_e/√2 = 11.2/√2 = 7.92 km s⁻¹
Q8.10. A satellite orbits the Earth at a height of 500 km. Find its (a) orbital speed (b) period of revolution. (M_E = 6 × 10²⁴ kg, R = 6400 km)
Answer
h = 500 km = 5 × 10⁵ m, r = R + h = 6.9 × 10⁶ m
(a) v_o = √(GM/r) = √(6.67 × 10⁻¹¹ × 6 × 10²⁴/6.9 × 10⁶) = √(4.002 × 10¹⁴/6.9 × 10⁶) = √(5.8 × 10⁷) = 7.62 × 10³ m s⁻¹ ≈ 7.62 km s⁻¹
(b) T = 2πr/v = 2π × 6.9 × 10⁶/7.62 × 10³ = 43.35 × 10⁶/7.62 × 10³ = 5689 s = 94.8 min ≈ 1.58 h
Q8.11. Two stars each of mass M and radius R approach each other to a distance d between their centres. When a star just begins to break up under tidal forces, find d in terms of M and R.
Answer
The tidal force on the near side of a star due to the other star equals the gravitational binding force of the star itself.
Tidal acceleration at the surface = difference in gravitational pull across the star: Δa = GM/d² − GM/(d−R)² ≈ 2GMR/d³ (for d >> R)
For breakup: Δa = GM/R² (self-gravity at the surface) 2GMR/d³ = GM/R² 2R/d³ = 1/R² d³ = 2R³ d = (2)^(1/3) R ≈ 1.26 R
(This is the Roche limit for equal-mass fluid bodies.)
Q8.12. A geostationary satellite orbits the Earth at a height of nearly 36,000 km. What is the potential due to Earth's gravity at the site of the satellite? (Take the potential at infinity as zero. M_E = 6 × 10²⁴ kg, R = 6400 km)
Answer
r = R + h = 6400 + 36000 = 42400 km = 4.24 × 10⁷ m
V = −GM/r = −(6.67 × 10⁻¹¹ × 6 × 10²⁴)/(4.24 × 10⁷) = −(4.002 × 10¹⁴)/(4.24 × 10⁷) = −9.44 × 10⁶ J kg⁻¹ = −9.44 MJ kg⁻¹
Q8.13. A 400 kg satellite is in a circular orbit of radius 7000 km around the Earth. Calculate (a) the total energy (b) the kinetic energy (c) the potential energy of the satellite.
Answer
m = 400 kg, r = 7 × 10⁶ m
(a) E = −GMm/2r = −(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 400)/(2 × 7 × 10⁶) = −(1.6 × 10¹⁷)/(1.4 × 10⁷) = −1.14 × 10¹⁰ J
(b) KE = −E = 1.14 × 10¹⁰ J
(c) PE = 2E = −2.28 × 10¹⁰ J
(Verification: PE + KE = −2.28 × 10¹⁰ + 1.14 × 10¹⁰ = −1.14 × 10¹⁰ = E ✓)
Q8.14. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of Mars = 6.4 × 10²³ kg, radius of Mars = 3395 km, mass of Sun = 2 × 10³⁰ kg, orbital radius of Mars = 2.28 × 10¹¹ m, G = 6.67 × 10⁻¹¹.
Answer
The energy needed equals the binding energy of the spaceship in the solar system: From Mars's orbit, the spaceship must escape both Mars and the Sun.
Energy to escape Sun from Mars's orbit: E_Sun = GM_Sun m/r_Mars_orbit (but we need to consider mass of spaceship as well)
Actually, we need the total energy to escape both Mars's gravity and the Sun's gravity from Mars's orbital distance.
Escape from Mars: E_Mars = GM_Mars m/R_Mars Escape from Sun (from Mars orbit): E_Sun = GM_Sun m/(2 × r_Mars_orbit)
But without the mass of the spaceship, we compute energy per unit mass: Energy per kg = GM_Mars/R_Mars + GM_Sun/(2 × r_Mars_orbit) = (6.67 × 10⁻¹¹ × 6.4 × 10²³)/(3.395 × 10⁶) + (6.67 × 10⁻¹¹ × 2 × 10³⁰)/(2 × 2.28 × 10¹¹) = (4.27 × 10¹³)/(3.395 × 10⁶) + (1.334 × 10²⁰)/(4.56 × 10¹¹) = 1.258 × 10⁷ + 2.926 × 10⁸ = 3.05 × 10⁸ J per kg
Q8.15. A rocket is fired vertically from the surface of Mars with a speed of 2 km s⁻¹. If 20% of its initial energy is lost due to atmospheric resistance, how far does the rocket go from the surface of Mars before returning? (g on Mars = 3.71 m s⁻², radius of Mars = 3395 km)
Answer
v = 2 km s⁻¹ = 2000 m s⁻¹, g_Mars = 3.71 m s⁻²
Initial KE = ½mv² Energy after 20% loss = 0.8 × ½mv² = 0.4mv²
At maximum height h, all KE converts to PE: 0.4mv² = mg_Mars × h h = 0.4v²/g = 0.4 × (2000)²/3.71 = 0.4 × 4000000/3.71 = 1600000/3.71 = 4.31 × 10⁵ m = 431 km
Internal Links
- [Chapter 8 Guide — Gravitation](/school/cbse/class/11/physics/cbse-gravitation)
- [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
- [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)
