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Gravitation for Class 11: The Complete CBSE Guide (2026-27)

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Gravitation Class 11 marks a paradigm shift from the intuitive, Earth-centric view of gravity you studied in Class 9 to a universal framework governing planetary motion, satellite launches, and even black holes. The 2024-25 CBSE syllabus devotes Chapter 8 of NCERT Physics Part 1 entirely to gravitation, expecting students to handle vector calculus, energy methods, and Kepler's empirical laws with equal fluency. This chapter contributes 12–15 marks to your board exam (typically one 5-mark numerical on satellite motion and two 3–4 mark derivations) and forms the bedrock for rotational mechanics in the next chapter. Whether you are solving for the orbital speed of the Moon or deriving why astronauts feel weightless, gravitation class 11 concepts appear in JEE Main (3–4 questions annually) and NEET (2–3 questions). This guide breaks down every NCERT section, highlights board exam trends from the past three years, and provides worked examples mirroring the difficulty level of CBSE 2026-27 papers.

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Key takeaways

✓Gravitation Class 11 accounts for 12–15 marks in CBSE board exams, split between 3-mark derivations and 5-mark numerical problems on satellites and energy.
✓Newton's law of universal gravitation (F = Gm₁m₂/r²) applies to any two masses in the universe; G = 6.67×10⁻¹¹ N·m²/kg² is a universal constant, while g = 9.8 m/s² is Earth-specific acceleration.
✓Kepler's three laws — elliptical orbits, equal area in equal time, and T² ∝ r³ — are empirical observations that Newton later proved using his law of gravitation.
✓Escape velocity (11.2 km/s for Earth) is the minimum speed needed to break free from a planet's gravity, independent of the mass of the escaping object.
✓Geostationary satellites orbit at 36,000 km altitude with 24-hour period, remaining fixed above one point on the equator — essential for communication and weather monitoring.
✓Gravitational potential energy U = –GMm/r is always negative, indicating bound systems; total mechanical energy for orbits is E = –GMm/2r.
✓Variation of g with altitude, depth, and latitude is a frequent 3-mark board question requiring clear formula derivation and substitution.

What Gravitation Class 11 Covers: NCERT Chapter 8 Breakdown

The NCERT Physics textbook for Class 11 (2024-25 edition) structures gravitation across eight major sections, beginning with Kepler's laws of planetary motion and culminating in satellite orbits and weightlessness. Unlike the Class 9 treatment — which focused on weight, free fall, and the value of g — gravitation class 11 introduces rigorous mathematical derivations, vector representations of force, and energy-based problem-solving. Section 8.1 presents Kepler's three empirical laws discovered by analysing Tycho Brahe's astronomical data. Section 8.2 derives Newton's universal law from Kepler's third law, establishing the inverse-square relationship. Section 8.3 defines the gravitational constant G through Cavendish's torsion balance experiment. Sections 8.4 and 8.5 tackle acceleration due to gravity (g) and its variation with altitude, depth, and latitude — these are high-frequency board exam topics. Section 8.6 introduces gravitational potential energy and derives the formula U = –GMm/r, explaining why energy is negative for bound orbits. Section 8.7 covers escape velocity, orbital velocity, and the distinction between circular and elliptical satellite orbits. Finally, Section 8.8 discusses geostationary satellites and the physics of weightlessness in orbiting spacecraft. Each section builds on vector mechanics from Chapter 4 and work-energy principles from Chapter 6, so ensure those foundations are solid before tackling gravitation class 11 numericals.

Kepler's laws (Section 8.1) are stated without proof; derivations using Newton's law appear in university physics but CBSE expects you to apply T² ∝ r³ correctly.
The universal law (Section 8.2) uses vector notation; understand that gravitational force is always attractive, acting along the line joining two masses.
Variation of g (Section 8.5) yields three standard formulas for altitude (gₕ), depth (gₐ), and latitude effects — memorise the approximations for h « R and d « R.
Gravitational potential (Section 8.6) is defined as work done per unit mass to bring an object from infinity; potential energy U and potential V are related by U = mV.
Escape velocity derivation (Section 8.7) equates kinetic energy to gravitational potential energy at the surface; vₑ = √(2GM/R) is independent of projectile mass.
Geostationary vs. polar satellites (Section 8.8) is a favourite comparison question; know altitude (≈36,000 km vs. ≈800 km), period (24 h vs. ≈100 min), and applications.

Newton's Universal Law of Gravitation: The Foundation Formula

Newton's law of universal gravitation states that every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Mathematically, F = Gm₁m₂/r², where G is the universal gravitational constant (6.67×10⁻¹¹ N·m²/kg²), m₁ and m₂ are the masses, and r is the centre-to-centre separation. This inverse-square law means doubling the distance reduces the force to one-fourth — a critical concept for satellite altitude problems. The force acts along the line joining the two masses and is always attractive (unlike electrostatic force, which can repel). In vector form, F₁₂ = –Gm₁m₂/r² r̂₁₂, where r̂₁₂ is the unit vector from mass 1 to mass 2; the negative sign indicates attraction. For CBSE boards, you must be able to apply this law to three-body problems (e.g. net force on the Moon due to Earth and Sun) using vector addition. A typical 5-mark question might ask you to derive the expression for g at Earth's surface starting from F = GMm/R², then substituting M = 6×10²⁴ kg and R = 6.4×10⁶ m to verify g ≈ 9.8 m/s². Remember that G is a universal constant measured by Cavendish's experiment, while g is a derived quantity specific to each planet. Common mistakes include using G and g interchangeably or forgetting to square the distance in the denominator.

Kepler's Three Laws of Planetary Motion (and Their Derivation)

Kepler's laws are empirical observations of planetary orbits, published between 1609 and 1619, that Newton later derived from his law of gravitation. Kepler's first law (law of orbits) states that all planets move in elliptical orbits with the Sun at one focus. For CBSE Class 11, you treat most orbits as circular approximations, but know that e = 0 for circles, 0 < e < 1 for ellipses. Kepler's second law (law of areas) asserts that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time, implying that planets move faster when closer to the Sun (perihelion) and slower when farther (aphelion). This law is a direct consequence of conservation of angular momentum (L = mvr = constant for central forces). Kepler's third law (law of periods) is the most exam-relevant: the square of the orbital period T is proportional to the cube of the semi-major axis r, i.e. T² ∝ r³. For circular orbits, T² = (4π²/GM) r³. A standard 3-mark derivation asks you to equate gravitational force (GMm/r²) to centripetal force (mv²/r), solve for v, then substitute v = 2πr/T to arrive at Kepler's third law. This derivation appears almost every year in CBSE board papers. You must also be able to apply T² ∝ r³ comparatively: if satellite A orbits at radius r and satellite B at 4r, then T_B² / T_A² = (4r)³ / r³ = 64, so T_B = 8T_A.

Kepler's first law explains why Earth is 147 million km from the Sun in January (perihelion) but 152 million km in July (aphelion) — the orbit is an ellipse, not a perfect circle.
Kepler's second law (equal areas) is why comets speed up dramatically near the Sun; angular momentum L = mr²ω must stay constant, so smaller r means larger ω.
Kepler's third law derivation starts with F_gravity = F_centripetal: GMm/r² = mv²/r, simplify to v² = GM/r, then v = 2πr/T gives (2πr/T)² = GM/r, rearrange to T² = (4π²/GM)r³.
For CBSE numericals, you are often given the orbital period of one satellite and asked to find the radius of another, or vice versa, using the ratio form of Kepler's third law.

Gravitational Constant G vs. Acceleration Due to Gravity g

One of the most persistent conceptual errors in gravitation class 11 is confusing the universal gravitational constant G with the acceleration due to gravity g. G = 6.67×10⁻¹¹ N·m²/kg² is a fundamental constant of nature that appears in Newton's law F = Gm₁m₂/r²; it has the same value everywhere in the universe, whether you are on Earth, Mars, or in intergalactic space. In contrast, g is the acceleration experienced by an object in free fall near a planet's surface, given by g = GM/R² for a planet of mass M and radius R. On Earth, g ≈ 9.8 m/s², but on the Moon g ≈ 1.6 m/s², and on Jupiter g ≈ 24.8 m/s². The value of g is derived from G; it is not a universal constant but a local property. When solving gravitation class 11 problems, always check units: G has units N·m²/kg², while g has units m/s² (or N/kg, which is equivalent). A typical board question might state 'the value of G on the Moon is...' — this is a trick statement because G does not change; only g changes. Another common trap: using g = 10 m/s² in a problem set on Mars or in orbit, where g is different or effectively zero. Always derive g from first principles (g = GM/R²) when the planet or altitude changes. For numerical accuracy, use g = 9.8 m/s² unless the problem explicitly says to approximate as 10 m/s².

Variation of g with Altitude, Depth, and Latitude

The acceleration due to gravity g = GM/R² at Earth's surface is not constant when you move vertically or horizontally. Variation with altitude: At height h above the surface, g_h = GM/(R+h)² = g/(1 + h/R)². For h « R, the binomial approximation gives g_h ≈ g(1 – 2h/R), showing that g decreases linearly with altitude for small heights. For example, at h = 6400 km (one Earth radius above the surface), g_h = g/4 ≈ 2.45 m/s². Variation with depth: Inside Earth, at depth d below the surface, assuming uniform density, g_d = g(1 – d/R). At the centre (d = R), g = 0 because mass above you pulls equally in all directions. Variation with latitude: Earth's rotation causes a centrifugal effect that reduces apparent g at the equator compared to the poles. The formula is g_λ = g – Rω²cos²λ, where λ is latitude and ω is Earth's angular velocity. At the equator (λ = 0°), g is minimum; at the poles (λ = 90°), cos²λ = 0, so g is maximum. Numerically, g_pole – g_equator ≈ 0.034 m/s². These three variations are favourite 3-mark derivation questions in CBSE boards. You must state assumptions (uniform density for depth, spherical Earth for altitude) and use binomial expansion correctly. A common error is writing g_h = g(1 – h/R) instead of g(1 – 2h/R); the factor of 2 comes from differentiating 1/(1+x)².

At h = 3200 km (half Earth's radius), g_h = g/(1 + 0.5)² = g/2.25 ≈ 4.4 m/s², not g/2 (common mistake).
Inside a mine at depth d = 3200 km, g_d = g(1 – 0.5) = 0.5g ≈ 4.9 m/s², which is close to g_h at the same distance from Earth's centre.
The equatorial bulge (Earth's radius is 21 km larger at the equator than at poles) also contributes to lower g at the equator, compounding the rotational effect.
For satellite problems, always use the full formula g_h = g/(1 + h/R)² or derive from GM/(R+h)² — do not use the approximation unless h < 0.1R.

Gravitational Potential Energy and Potential: Sign Conventions Matter

Gravitational potential energy U is the work done by an external agent in bringing a mass m from infinity to a point in a gravitational field without acceleration. For a mass m at distance r from a mass M, U = –GMm/r. The negative sign is crucial: it indicates that the system is bound (energy must be added to separate the masses to infinity). At infinity, U = 0 by convention, so at any finite r, U < 0. As m moves closer (r decreases), U becomes more negative, meaning the system loses potential energy (which appears as kinetic energy if the mass is in free fall). Gravitational potential V is potential energy per unit mass: V = U/m = –GM/r. It is a scalar field defined at every point in space, independent of the test mass. The relationship between force and potential is F = –dU/dr (for radial fields), which gives F = –GMm/r² as expected. For orbits, total mechanical energy E = K + U. For a circular orbit, K = GMm/2r and U = –GMm/r, so E = –GMm/2r (negative, indicating a bound orbit). To move a satellite from orbit radius r₁ to r₂, the work done is ΔE = GMm/2 (1/r₁ – 1/r₂). A frequent board question asks for the minimum energy to remove a satellite from orbit to infinity: this is –E = GMm/2r. Sign errors are the top mistake in gravitation class 11 energy problems: forgetting the negative sign in U or confusing ΔU with ΔK. Always check that bound systems have E < 0 and free particles have E ≥ 0.

Escape Velocity: Breaking Free from Gravity

Escape velocity v_e is the minimum speed an object must have at a planet's surface to escape to infinity (where gravitational potential energy becomes zero) without any further propulsion. Derivation: At the surface, total energy E = (1/2)mv_e² – GMm/R. At infinity, E = 0 (both K and U are zero). By conservation of energy, (1/2)mv_e² – GMm/R = 0, solving gives v_e = √(2GM/R). For Earth (M = 6×10²⁴ kg, R = 6.4×10⁶ m), v_e = √(2 × 6.67×10⁻¹¹ × 6×10²⁴ / 6.4×10⁶) ≈ 11,200 m/s or 11.2 km/s. Crucially, escape velocity is independent of the mass of the escaping object — a grain of sand and a rocket need the same speed (though the rocket needs vastly more energy because E = (1/2)mv²). Escape velocity also does not depend on the direction of launch; you can launch vertically, horizontally, or at any angle, as long as the speed exceeds v_e. For other celestial bodies, v_e scales as √(M/R). The Moon has v_e ≈ 2.4 km/s (much lower than Earth, which is why the Moon cannot retain an atmosphere — gas molecules moving faster than 2.4 km/s escape). A common conceptual question: 'Why don't we feel lighter when moving at half the escape velocity?' Answer: weight (gravitational force) depends on position (distance from Earth), not on velocity. Escape velocity is purely an energy threshold, not a force threshold. For CBSE numericals, you may need to calculate escape velocity at altitude h: v_e(h) = √(2GM/(R+h)), which is less than surface escape velocity.

Escape velocity from the Sun's surface is 618 km/s, far exceeding Earth's 11.2 km/s, which is why solar wind (particles from the Sun) can reach Earth.
Black holes have escape velocity exceeding the speed of light (c = 3×10⁸ m/s), which is why even light cannot escape — hence 'black'.
If you launch a projectile at exactly v_e vertically, it will reach infinity with zero velocity; launch it faster, and it retains kinetic energy at infinity.
Orbital velocity v_o = √(GM/r) is related to escape velocity by v_e = √2 × v_o at the same radius — a favourite short-answer board question.

Satellite Motion: Orbital Velocity, Period, and Energy

A satellite in circular orbit experiences gravitational force providing the necessary centripetal acceleration. Equating F = GMm/r² = mv²/r gives orbital velocity v_o = √(GM/r). Unlike escape velocity, v_o depends on the orbital radius r: higher orbits have lower speeds. For low Earth orbit (r ≈ R + 200 km ≈ 6600 km), v_o ≈ 7.8 km/s. The orbital period T = 2πr/v_o = 2πr/√(GM/r) = 2π√(r³/GM), confirming Kepler's third law T² ∝ r³. For geostationary orbit, T = 24 hours = 86400 s, so r³ = GMT²/4π² = (6.67×10⁻¹¹ × 6×10²⁴ × 86400²)/4π² ≈ 7.54×10²² m³, giving r ≈ 4.22×10⁷ m or 42,200 km from Earth's centre (≈ 36,000 km altitude). Geostationary satellites must orbit in the equatorial plane with zero inclination; polar satellites, in contrast, have 90° inclination and cover the entire Earth as it rotates beneath them. For energy calculations: kinetic energy K = (1/2)mv_o² = GMm/2r, potential energy U = –GMm/r, total energy E = K + U = –GMm/2r. To transfer a satellite from circular orbit r₁ to r₂ > r₁, use a Hohmann transfer: fire thrusters to enter an elliptical transfer orbit, coast to apogee, then fire again to circularise. The energy required is ΔE = GMm/2 (1/r₁ – 1/r₂). For CBSE, master the derivations of v_o, T, and E, and be able to apply them to compare satellites at different altitudes.

Weightlessness in Orbit: Why Astronauts Float

Astronauts aboard the International Space Station (ISS) appear weightless, floating freely inside the cabin. This is often misunderstood as 'zero gravity', but in fact the gravitational acceleration at ISS altitude (≈400 km) is g_h ≈ 8.7 m/s², only 11% less than at Earth's surface. The sensation of weightlessness arises because the ISS and everything inside it — including the astronauts — are in free fall toward Earth, continuously missing the planet due to their tangential velocity. In free fall, there is no normal force: the floor does not push up on the astronaut's feet, so their apparent weight (the normal force measured by a scale) is zero, even though their true weight (gravitational force mg) is non-zero. This is identical to the feeling you get in an elevator when the cable snaps and you fall freely — you would feel weightless for those few seconds. For CBSE theory questions, you must distinguish between true weight (gravitational force, always present) and apparent weight (normal force, zero in free fall). A related numerical question: calculate the apparent weight of a 70 kg astronaut in an elevator accelerating upward at 2 m/s². Solution: apparent weight = m(g + a) = 70(9.8 + 2) = 826 N, greater than true weight 686 N. If the elevator cable snaps, apparent weight = m(g – g) = 0. Weightlessness is not unique to space; it occurs in any free-fall scenario, including parabolic flights ('vomit comet') used for astronaut training.

Inside a satellite, all objects (astronauts, equipment, water droplets) fall together with the same acceleration GM/r², so relative to each other they appear stationary — this is 'microgravity'.
At geostationary altitude, g_h ≈ 0.23 m/s², yet geostationary satellites are not 'more weightless' than LEO satellites; weightlessness depends on free fall, not on the magnitude of g.
If you were in a spaceship accelerating at 9.8 m/s² (one 'gee'), you would feel normal Earth weight even in deep space far from any planet — apparent weight comes from acceleration, not gravity alone.
CBSE questions often ask 'Why do astronauts feel weightless in orbit?' The correct answer is 'They are in continuous free fall; normal force is zero', NOT 'Gravity is zero there'.

Gravitation Class 11 Formulas: The Complete Formula Sheet

Success in gravitation class 11 numericals requires instant recall of 12 core formulas and their applicability conditions. Newton's law of gravitation: F = Gm₁m₂/r² (always valid for point masses or spherically symmetric bodies). Acceleration due to gravity at surface: g = GM/R². Weight: W = mg. Variation of g with altitude: g_h = g/(1 + h/R)² ≈ g(1 – 2h/R) for h « R. Variation with depth: g_d = g(1 – d/R). Gravitational potential energy: U = –GMm/r (zero at infinity). Gravitational potential: V = –GM/r (potential energy per unit mass). Escape velocity: v_e = √(2GM/R) = √(2gR). Orbital velocity: v_o = √(GM/r) = √(gR²/r). Orbital period: T = 2π√(r³/GM). Relation between v_e and v_o: v_e = √2 × v_o. Total energy of satellite: E = –GMm/2r. Kepler's third law: T² = (4π²/GM) r³. For derivation-based questions, you must show each step: state the starting point (e.g. equate gravitational and centripetal force), perform algebraic manipulation, and clearly box the final result. For numerical problems, always write the formula first, substitute values with units, and verify that your answer has the correct units and order of magnitude (e.g. orbital velocity should be a few km/s, not m/s). Keep a formula sheet in your practical file and revise it daily in the two weeks before your board exam; formula recall under exam pressure is a skill built through repetition.

For problems involving energy, always check whether you need total energy E, kinetic energy K, or potential energy U — they are related but not interchangeable.
When comparing two satellites, use ratio form: v₁/v₂ = √(r₂/r₁) or T₁/T₂ = (r₁/r₂)^(3/2) — this avoids calculating G and M explicitly.
If a problem gives g and R instead of G and M, use g = GM/R² to replace GM = gR² in formulas like v_o = √(gR²/r).
Dimensional analysis is a quick check: [v_e] = √([GM]/[R]) = √(m³ s⁻² kg⁻¹ × kg / m) = √(m² s⁻²) = m/s ✓.

Gravitation Important Questions for CBSE Board 2026-27

CBSE Class 11 Physics board exams (2026-27 pattern) typically include two gravitation questions: one 5-mark numerical (satellite motion, energy, or escape velocity) and one 3-mark derivation (Kepler's third law, variation of g, or orbital velocity formula). The 2023 board paper asked: 'Derive an expression for escape velocity and calculate it for Earth.' The 2024 paper included: 'A satellite orbits at height h = R above Earth. Find its orbital velocity and period in terms of g and R.' High-probability questions for 2026-27 based on syllabus emphasis: (1) Derive the relation between orbital velocity and escape velocity at the same point. (2) Show that total energy of a satellite is E = –GMm/2r. (3) Calculate the percentage decrease in weight of a body taken to height h = R/2. (4) Two satellites A and B orbit at radii r and 4r; compare their speeds, periods, and energies. (5) Prove Kepler's third law starting from Newton's law of gravitation and centripetal force. For numerical practice, focus on problems requiring multiple steps: find g at altitude h, then use it to compute orbital velocity, then find kinetic energy. Always show intermediate results. Common mark-losing errors include forgetting to convert km to m, using g = 10 m/s² when the problem specifies 9.8, and sign errors in potential energy. Practice at least 25 previous-year CBSE questions (available in NCERT Exemplar and CBSE sample papers) to internalise the board's question style and mark allocation.

How CBSETUTOR.ai Simplifies Gravitation Class 11 Mastery

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Frequently asked questions

How many marks does gravitation carry in CBSE Class 11 Physics board exam 2026-27?+

Gravitation Class 11 typically contributes 12–15 marks in the CBSE board exam: one 5-mark numerical problem (satellite motion, escape velocity, or energy calculation) and two 3–4 mark questions (derivations like Kepler's third law or variation of g, and one conceptual short-answer). The 2024 board paper included a 5-mark question on geostationary satellite energy and a 3-mark derivation of orbital velocity formula.

What is the difference between gravitational potential and gravitational potential energy in Class 11?+

Gravitational potential energy U = –GMm/r is the work done to bring mass m from infinity to distance r from mass M; it depends on the test mass m and is measured in joules. Gravitational potential V = –GM/r is potential energy per unit mass (U/m), independent of test mass, measured in J/kg. For CBSE numericals, if asked for 'potential', give V; if asked for 'energy', give U. They are related by U = mV.

Why is escape velocity independent of the mass of the object being launched?+

Escape velocity v_e = √(2GM/R) contains only the planet's mass M and radius R. Derivation: set kinetic energy (1/2)mv² equal to work against gravity (GMm/R), then m cancels from both sides, leaving v = √(2GM/R). This means a marble and a spaceship need the same speed (11.2 km/s from Earth) to escape, though the spaceship needs vastly more energy because E = (1/2)mv² and its mass m is huge.

How do I remember whether to use g or GM/R² in gravitation class 11 formulas?+

Use g = 9.8 m/s² for problems on or near Earth's surface where the question gives g explicitly. Use GM/R² when the problem involves other planets, or gives M and R instead of g, or asks you to derive g. You can always substitute GM = gR² to convert between forms. For example, escape velocity v_e = √(2GM/R) = √(2gR) if you substitute GM = gR².

What is a geostationary satellite and why must it orbit above the equator?+

A geostationary satellite has orbital period T = 24 hours, matching Earth's rotation, so it remains fixed above one point on the surface — essential for continuous communication and weather monitoring. It must orbit in the equatorial plane with zero inclination because only there does the satellite's angular velocity vector align with Earth's rotation axis; a tilted orbit would trace a figure-8 pattern (analemma) as seen from the ground, not staying stationary.

If gravity exists in orbit, why do astronauts feel weightless on the ISS?+

At ISS altitude (≈400 km), gravitational acceleration g_h ≈ 8.7 m/s², so gravity is strong. Astronauts feel weightless because they and the ISS are in continuous free fall toward Earth, missing the planet due to their tangential velocity. In free fall, there is no normal force (the floor does not push up), so apparent weight (what a scale would read) is zero, even though true weight (gravitational force mg) is non-zero. Weightlessness = zero normal force, not zero gravity.

How does the value of g change if I go to a height equal to Earth's radius (h = R)?+

At height h = R, distance from Earth's centre is R + h = 2R. Using g_h = GM/(R+h)² = GM/(2R)² = GM/4R² = g/4, where g = GM/R² is surface gravity. So at h = R, gravity is one-quarter of surface value: g_h = 9.8/4 = 2.45 m/s². A common mistake is using the approximation g_h ≈ g(1 – 2h/R), which only works for h « R; here h = R is not small, so use the exact formula.

Can I use Kepler's third law (T² ∝ r³) for elliptical orbits or only circular orbits?+

Kepler's third law T² ∝ r³ applies to both circular and elliptical orbits; for ellipses, r is the semi-major axis (average of closest and farthest distances from the focus). For CBSE Class 11, most problems assume circular orbits for simplicity, but you should know that T² = (4π²/GM)a³ where a is the semi-major axis. In circular orbits, a = r, the orbital radius.

What are the assumptions when deriving that g decreases linearly with depth (g_d = g(1 – d/R))?+

The derivation assumes (1) Earth has uniform density throughout (in reality, the core is denser than the crust, so real g_d is not exactly linear), and (2) at depth d, only the mass in the sphere of radius (R – d) contributes to gravity; the mass in the spherical shell above you produces zero net force by the shell theorem. These assumptions simplify calculations for CBSE board exams but are approximations.

How is gravitational potential energy negative when we usually think of energy as positive?+

Gravitational PE is defined as zero at infinite separation (r = ∞). As two masses come closer, gravity (an attractive force) does positive work, so the system loses potential energy, making U negative at any finite distance. U = –GMm/r < 0 indicates a bound system: you must add energy (do work against gravity) to separate the masses to infinity. Think of it as an energy 'debt': the system is in a lower-energy state than the reference (infinity).

Will my child be disadvantaged if our school has not finished gravitation chapter by mid-January?+

No, as long as the chapter is completed by early February, leaving March for revision. Gravitation class 11 is in NCERT Chapter 8, usually taught after work-energy (Chapter 6) and rotational motion (Chapter 7). If your school is delayed, supplement with NCERT Exemplar problems and previous-year board questions. CBSETUTOR.ai covers the entire gravitation chapter with step-by-step derivations and numericals, so your child can learn independently at ₹999/month without waiting for school pace.

Which gravitation derivations are compulsory to memorise for CBSE board exams?+

The five must-know derivations are: (1) Kepler's third law (T² ∝ r³) from F_gravity = F_centripetal, (2) escape velocity v_e = √(2GM/R) using energy conservation, (3) orbital velocity v_o = √(GM/r) from force balance, (4) variation of g with altitude (g_h ≈ g(1 – 2h/R)) using binomial expansion, and (5) total energy of satellite E = –GMm/2r from K + U. Practice writing these in under 4 minutes each, showing all steps, as CBSE awards marks for method even if the final answer is incorrect.

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