This MCQ module is based on: NCERT Exercises and Solutions: Gravitation
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NCERT Exercises and Solutions: Gravitation
🎓 Class 11
Physics
CBSE
Theory
Ch 7 – Gravitation
⏱ ~8 min
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NCERT Exercises and Solutions: Gravitation
Chapter Summary — Gravitation at a Glance
Newton's universal law of gravitation and Kepler's planetary laws form the foundation of celestial mechanics. From an apple's fall to spacecraft escaping the solar system, the same inverse-square law governs all gravitational phenomena.
Key Formulas
| Concept | Formula |
|---|---|
| Newton's Law of Gravitation | F = GMm/r² |
| Acceleration due to gravity | g = GM/R² |
| g at altitude h | g(h) ≈ g(1 − 2h/R) |
| g at depth d | g(d) = g(1 − d/R) |
| g with latitude λ | g'(λ) = g − Rω²cos²λ |
| Gravitational PE | U = −GMm/r |
| Gravitational Potential | V = −GM/r |
| Escape velocity | v_e = √(2GM/R) = √(2gR) |
| Orbital velocity | v_o = √(GM/r) |
| Period of satellite (Kepler) | T² = (4π²/GM) r³ |
| Total energy of satellite | E = −GMm/(2r) |
Key Constants: G = 6.674×10⁻¹¹ N m² kg⁻²; g_surface = 9.8 m s⁻²; M_E = 6.0×10²⁴ kg; R_E = 6.4×10⁶ m; v_e(Earth) = 11.2 km/s; v_o(LEO) ≈ 7.9 km/s.
Important Conceptual Points
- Gravity is always attractive; it is the weakest of the four fundamental forces.
- Inside a spherical shell of matter, the gravitational field is exactly zero (Shell Theorem corollary).
- Escape velocity is independent of the projectile's mass and direction of projection.
- Kepler's second law is equivalent to conservation of angular momentum under any central force.
- A satellite is in continuous free fall; weightlessness is the absence of contact force, not the absence of gravity.
NCERT Exercises with Worked Solutions
Exercise 7.1: Which is correct?
(a) Acceleration due to gravity (g) (i) is independent of the mass of the Earth (ii) increases with depth (iii) increases with height (iv) is independent of the mass of the body.
Answer: Only (iv) is correct.
(i) Wrong — g = GM/R², so g depends on M.
(ii) Wrong — g decreases with depth.
(iii) Wrong — g decreases with height.
(iv) Correct — F = mg, but g itself is independent of the body's mass m being accelerated.
(i) Wrong — g = GM/R², so g depends on M.
(ii) Wrong — g decreases with depth.
(iii) Wrong — g decreases with height.
(iv) Correct — F = mg, but g itself is independent of the body's mass m being accelerated.
Exercise 7.2: Choose the correct alternatives
(a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume Earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the Earth/mass of the body. (d) Formula −GMm(1/r₂ − 1/r₁) is more/less accurate than the formula mg(r₂ − r₁) for the difference of potential energy.
(a) Decreases with altitude (g ∝ 1/r²)
(b) Decreases with depth (g ∝ R − d)
(c) Independent of the mass of the body
(d) The formula −GMm(1/r₂ − 1/r₁) is more accurate because it does not assume constant g; it works at any altitude.
(b) Decreases with depth (g ∝ R − d)
(c) Independent of the mass of the body
(d) The formula −GMm(1/r₂ − 1/r₁) is more accurate because it does not assume constant g; it works at any altitude.
Exercise 7.3: Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Twice as fast means T_p = T_E/2.
By Kepler's 3rd law: T² ∝ a³ ⇒ a_p/a_E = (T_p/T_E)^(2/3) = (1/2)^(2/3) \[a_p/a_E = 0.63\] So the planet's orbital semi-major axis is about 0.63 AU (closer than Venus).
By Kepler's 3rd law: T² ∝ a³ ⇒ a_p/a_E = (T_p/T_E)^(2/3) = (1/2)^(2/3) \[a_p/a_E = 0.63\] So the planet's orbital semi-major axis is about 0.63 AU (closer than Venus).
Exercise 7.4: Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of orbit is 4.22 × 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the Sun.
For Io around Jupiter:
\[M_J = \frac{4\pi^2 r^3}{GT^2}\]
T = 1.769 × 86400 = 1.528×10⁵ s, r = 4.22×10⁸ m
\[M_J = \frac{4\pi^2(4.22\times10^8)^3}{6.67\times10^{-11}(1.528\times10^5)^2}\approx 1.9\times 10^{27}\,\text{kg}\]
For Earth around the Sun: T = 365.25×86400 = 3.156×10⁷ s, r = 1.496×10¹¹ m
\[M_S = \frac{4\pi^2(1.496\times10^{11})^3}{6.67\times10^{-11}(3.156\times10^7)^2}\approx 2.0\times 10^{30}\,\text{kg}\]
Ratio M_J/M_S ≈ 1.9×10²⁷ / 2×10³⁰ ≈ 1/1000. ✓
Exercise 7.5: Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.
M = 2.5×10¹¹ × 2×10³⁰ = 5×10⁴¹ kg
r = 50,000 ly = 5×10⁴ × 9.46×10¹⁵ m = 4.73×10²⁰ m \[T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(4.73\times10^{20})^3}{6.67\times10^{-11}\times 5\times10^{41}}}\] \[T \approx 1.12\times10^{16}\,\text{s}\approx \boxed{3.5\times10^8\,\text{years}}\] About 350 million years per galactic year. Our Sun (at 26,000 ly) takes about 225 million years.
r = 50,000 ly = 5×10⁴ × 9.46×10¹⁵ m = 4.73×10²⁰ m \[T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(4.73\times10^{20})^3}{6.67\times10^{-11}\times 5\times10^{41}}}\] \[T \approx 1.12\times10^{16}\,\text{s}\approx \boxed{3.5\times10^8\,\text{years}}\] About 350 million years per galactic year. Our Sun (at 26,000 ly) takes about 225 million years.
Exercise 7.6: Choose the correct alternative
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of Earth's gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth's influence.
(a) Total E = −KE, so the total energy is negative of kinetic energy.
(b) Less. A satellite already has KE = (1/2)GMm/r; to escape it only needs the difference. A stationary object at the same height has no KE, so it requires GMm/r — double the satellite's energy need.
(b) Less. A satellite already has KE = (1/2)GMm/r; to escape it only needs the difference. A stationary object at the same height has no KE, so it requires GMm/r — double the satellite's energy need.
Exercise 7.7: 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 projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
(a) No. v_e = √(2GM/R) does not contain m.
(b) No (for same altitude; v_e depends on r = R + h).
(c) No. Escape requires only enough KE; direction can be anywhere outward (though tangential is fuel-efficient practically).
(d) Yes. v_e decreases with height because r increases.
(b) No (for same altitude; v_e depends on r = R + h).
(c) No. Escape requires only enough KE; direction can be anywhere outward (though tangential is fuel-efficient practically).
(d) Yes. v_e decreases with height because r increases.
Exercise 7.8: A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
(a) No — fastest at perihelion, slowest at aphelion.
(b) No — angular speed ω = v/r varies.
(c) Yes — L is conserved (central force).
(d) No — KE varies inversely with PE.
(e) No — PE varies with r.
(f) Yes — Total E = KE + U is conserved.
(b) No — angular speed ω = v/r varies.
(c) Yes — L is conserved (central force).
(d) No — KE varies inversely with PE.
(e) No — PE varies with r.
(f) Yes — Total E = KE + U is conserved.
Exercise 7.9: Which of the following symptoms is likely to afflict an astronaut in space — (a) swollen feet, (b) swollen face, (c) headache, (d) orientation problem?
In microgravity, blood and fluid redistribute toward the head and chest (no gravity to pull them down). So:
(a) Unlikely — feet actually become smaller.
(b) Likely — face puffs up due to fluid pooling.
(c) Likely — fluid pressure in the head causes headaches.
(d) Likely — the vestibular system relies on gravity for orientation.
(a) Unlikely — feet actually become smaller.
(b) Likely — face puffs up due to fluid pooling.
(c) Likely — fluid pressure in the head causes headaches.
(d) Likely — the vestibular system relies on gravity for orientation.
Exercise 7.10: Choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig.) — (i) a, (ii) b, (iii) c, (iv) 0.
By symmetry around the rim's axis, only the axial component can survive. The complete spherical shell would give zero field inside. The hemisphere thus contributes the field along the symmetry axis, pointing into the hemisphere's centre of mass (along the axis of symmetry). The correct answer is generally along direction (c) — along the axis of symmetry away from the opening side.
Exercise 7.11: A rocket is fired vertically with a speed of 5 km s⁻¹ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth?
Energy conservation (KE all converts to ΔU at max height):
\[\frac{1}{2}v^2 = GM\left(\frac{1}{R}-\frac{1}{R+h}\right)\]
With v = 5000 m/s, R = 6.4×10⁶ m, GM = gR² = 9.8×(6.4×10⁶)²:
\[\frac{1}{2}(5000)^2 = 9.8\times(6.4\times10^6)^2\left(\frac{1}{R}-\frac{1}{R+h}\right)\]
\[1.25\times10^7 = 4.014\times10^{14}\left(\frac{h}{R(R+h)}\right)\]
Solving: h ≈ 1.6×10⁶ m. Distance from Earth's centre = R + h = 8.0×10⁶ m = ~8000 km.
Exercise 7.12: The escape speed of a projectile on the earth's surface is 11.2 km s⁻¹. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.
Initial v = 3 × 11.2 = 33.6 km/s. Energy conservation:
\[\frac{1}{2}v_\infty^2 = \frac{1}{2}v_i^2 - \frac{1}{2}v_e^2\]
\[v_\infty^2 = (33.6)^2 - (11.2)^2 = 1128.96 - 125.44 = 1003.52\]
\[v_\infty = \boxed{\sqrt{1003.52} \approx 31.7\,\text{km/s}}\]
Exercise 7.13: 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 the satellite = 200 kg; mass of the Earth = 6.0 × 10²⁴ kg; radius of the Earth = 6.4 × 10⁶ m; G = 6.67 × 10⁻¹¹ N m² kg⁻².
r = R + h = 6.8×10⁶ m. Total energy in orbit:
\[E = -\frac{GMm}{2r} = -\frac{6.67\times10^{-11}\times 6\times10^{24}\times 200}{2\times 6.8\times10^6}\]
\[E = -5.88\times10^9\,\text{J}\]
To escape, final energy = 0. Energy required:
\[\Delta E = 0 - E = \boxed{5.88\times10^9\,\text{J}}\]
Quick Self-Check Tool
Type a planet mass (×10²⁴ kg) and radius (×10⁶ m) to compute g, v_e and v_o.
g = 9.77 m/s² | v_e = 11.2 km/s | v_o = 7.91 km/s
Quick Competency-Based Recap
Q1. State Kepler's three laws of planetary motion.L1 Remember
(i) Each planet moves in an elliptical orbit with the Sun at one focus. (ii) The radius vector sweeps equal areas in equal times. (iii) T² ∝ a³.
Q2. Derive the expression v_e = √(2gR).L3 Apply
Equate (1/2)mv_e² = GMm/R, so v_e = √(2GM/R). Using g = GM/R², GM = gR², therefore v_e = √(2gR).
Q3. Why is the orbital velocity of a satellite less than its escape velocity?L4 Analyse
Orbiting requires only enough KE to balance centripetal demand: KE_orb = (1/2)GMm/r. Escaping requires enough KE to reach infinity (where U = 0): KE_esc = GMm/r — exactly double. So v_e/v_o = √2.
Q4. A geostationary satellite is at h ≈ 36,000 km. Why can't it be placed at h = 10,000 km?L5 Evaluate
At h = 10,000 km, period T ≈ 5.8 hours, not 24 hours — the satellite would race ahead of Earth's rotation. Only at h = 35,786 km does T = 23.93 h = sidereal day, allowing the satellite to stay above one geographic point.
Q5. HOT: Suggest two ways a future mission could detect dark matter using only gravitational measurements.L6 Create
(i) Galactic rotation curves: Measure orbital velocities of stars at various distances from a galaxy's centre. Observed flat curves (v constant) imply more mass than visible — hinting at dark matter halos. (ii) Gravitational lensing: Map distortion of distant background galaxies caused by unseen mass in foreground clusters. The lensing pattern reveals dark matter distribution. Modern surveys like Euclid and the Vera Rubin Observatory use both.
Final Assertion–Reason Pairs
A: Two bodies of different masses, dropped from the same height in vacuum, hit the ground at the same time.
R: Acceleration due to gravity is independent of the mass of the falling body.
(A). Both true; R explains A. Famous Galileo experiment, verified by Apollo 15's hammer-feather drop on the Moon.
A: If the Earth stopped rotating, weight would increase everywhere except at the poles.
R: The centripetal effect of rotation currently reduces the apparent weight by Rω²cos²λ.
(A). Both true; R explains A. The poles already have no rotational reduction (cos 90° = 0), so nothing changes there.
A: The orbital period of a low Earth satellite is approximately 90 minutes.
R: For h ≪ R, T = 2π√(R/g) ≈ 84 minutes.
(A). Both true; R explains A. The 84-min minimum (Schuler period) sets a floor for any low orbit.
Frequently Asked Questions - NCERT Exercises and Solutions: Gravitation
What are the key NCERT exercise types in Chapter 7 Gravitation?
NCERT Class 11 Physics Chapter 7 Gravitation exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Gravitation?
For numerical problems in NCERT Class 11 Physics Chapter 7 Gravitation: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 7?
From NCERT Class 11 Physics Chapter 7 (Gravitation), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 7 Gravitation problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 7 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 7 Gravitation exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 7 Gravitation solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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