NEET PhysicsNCERT Class 11Chapter 7

Gravitation: NEET notes

One force pulls a mango off its branch and holds the Moon in its orbit. This chapter starts from Kepler's three laws of planetary motion, arrives at Newton's inverse-square law of gravitation, measures G, and then uses the law to find g at any height or depth, gravitational potential energy, the escape speed, and the speed, period and energy of a satellite.

What NEET asks

NEET asks for g at a height or depth, the ratio of weights at two places, escape speed and orbital speed (ratio √2), orbital period from Kepler's third law, and the kinetic, potential and total energy of a satellite. Marks are lost by measuring distance from the surface instead of the centre, using 1 − 2h/R when h is not small, forgetting that g falls both above and below the surface, and giving a satellite positive total energy.

1. Kepler's laws

NCERT §7.1, §7.2

  • Ptolemy's geocentric model put the earth at the centre with planets moving on circles riding on bigger circles. Aryabhatta (5th century A.D.) had already mentioned a sun-centred model, and Copernicus (1473–1543) proposed planets on circles around a fixed sun.
  • Tycho Brahe (1546–1601) recorded planetary positions with the naked eye for his whole life; his assistant Johannes Kepler (1571–1640) found three laws hidden in those records.
  • Law of orbits: every planet moves on an ellipse, and the sun sits at one focus of that ellipse. The circle is the special case where the two foci coincide.
  • An ellipse is drawn with a string pinned at two foci F₁ and F₂: for every point on the curve, the distances to F₁ and F₂ add up to the same total.
  • The line through the foci meets the ellipse at P (perihelion, nearest to the sun) and A (aphelion, farthest). Half of PA is the semi-major axis a.
  • Law of periods: T² ∝ a³, where T is the time for one revolution. The ratio T²/a³ comes out almost the same (about 3 × 10⁻³⁴ y² m⁻³) for all eight planets from Mercury to Neptune.
  • Most planetary orbits are very nearly circles; for the earth the ratio of semi-minor to semi-major axis is 0.99986.

2. The law of areas

NCERT §7.2

  • Law of areas: in any two equal stretches of time, the sun–planet line covers the same area. A planet therefore speeds up near the sun and slows down far from it.
  • In a short time Δt the sun–planet line sweeps ΔA = ½ |r × v Δt|, so ΔA/Δt = L/(2m), where L = r × p is the planet's angular momentum.
  • Gravity on a planet points along the sun–planet line (a central force), so it exerts no torque about the sun; L stays constant and so does ΔA/Δt.
  • The law of areas therefore holds for any central force, not only for an inverse-square one.
  • At perihelion and aphelion r and v are perpendicular, so m r_P v_P = m r_A v_A, giving v_P / v_A = r_A / r_P: the planet is fastest at perihelion.
  • The half of the orbit around the aphelion encloses the larger area, so the planet spends the longer time on that half.

3. Universal law of gravitation

NCERT §7.3

  • Newton compared the Moon with a falling apple. The Moon (orbit radius 3.84 × 10⁸ m, period 27.3 days) has centripetal acceleration a_m = 4π²R_m/T² ≈ 2.7 × 10⁻³ m/s², far less than g.
  • If gravity falls off as 1/r², then g/a_m = (R_m/R_E)² ≈ 60² = 3600, and 9.8/0.00272 is indeed about 3600. The same force explains the apple and the Moon.
  • Law: every pair of bodies pulls on each other. The pull grows with the product of the two masses and falls as 1/(distance)²: F = G m₁ m₂ / r².
  • In vector form the force on m₂ is F = −G m₁ m₂ r̂ / r², with r̂ pointing from m₁ to m₂; the minus sign means attraction. The force on m₁ is −F, so F₁₂ = −F₂₁ (Newton's third law).
  • Superposition: the force on one mass from several others is the vector sum of the separate pair forces, each unchanged by the presence of the rest.
  • Three equal masses at the corners of an equilateral triangle exert zero net force on a mass at the centroid, as symmetry predicts. If the mass at one corner is doubled, the net force points towards that corner.
  • Shell results: a uniform spherical shell attracts a point outside it as if all its mass sat at its centre; on a point anywhere inside it, the shell's net pull is zero.
  • The law is for point masses; for an extended body the forces from all its parts are added as vectors. Gravity cannot be shielded: a shell does not block the pull of bodies outside it.

4. The gravitational constant

NCERT §7.4

  • G was first measured by Henry Cavendish in 1798, with a torsion balance.
  • A light bar AB with a small lead sphere at each end hangs from a fine wire. Two large lead spheres are placed near the small ones, on opposite sides.
  • The two attractions are equal and opposite, so there is no net force on the bar, only a torque F × L that twists the wire.
  • The wire twists until its restoring torque τθ balances the gravitational torque: G M m L / d² = τθ, where τ is the restoring torque per unit twist, found separately.
  • Measuring θ gives G. The accepted value is G = 6.67 × 10⁻¹¹ N m² kg⁻², the same everywhere, which is why it is called universal.
  • G is tiny: two 1 kg masses 1 m apart attract with only 6.67 × 10⁻¹¹ N, which is why gravity between everyday objects goes unnoticed.

5. Acceleration due to gravity

NCERT §7.5

  • Think of the earth as many concentric shells. A point outside is outside every shell, so the whole earth pulls as if its mass M_E were at the centre.
  • On the surface the force on a mass m is F = G M_E m / R_E², and since F = mg, g = G M_E / R_E².
  • g does not depend on the mass of the falling body; it depends only on the mass and radius of the earth.
  • Knowing g, R_E and G gives the earth's mass: M_E = g R_E² / G. With g = 9.81 m/s² and R_E = 6.37 × 10⁶ m this is 5.97 × 10²⁴ kg, hence the phrase 'Cavendish weighed the earth'.
  • The Moon's orbit gives the same mass another way, M_E = 4π²R³/(G T²) ≈ 6.02 × 10²⁴ kg, within 1% of the first value.
  • Inside a uniform earth, at distance r from the centre, only the mass within radius r pulls, and the force is F = G M_E m r / R_E³, proportional to r.

6. g above and below the surface

NCERT §7.6

  • At height h the distance from the centre is R_E + h, so g(h) = G M_E / (R_E + h)² = g (R_E/(R_E + h))².
  • For h much smaller than R_E, the binomial approximation gives g(h) ≈ g (1 − 2h/R_E).
  • At depth d, the shell of thickness d above pulls with zero net force; only the inner sphere of radius R_E − d, of mass M_E (R_E − d)³/R_E³, pulls.
  • So g(d) = g (1 − d/R_E) exactly, for a uniform earth. At the centre (d = R_E) g is zero.
  • g is greatest on the surface and falls whether you go up or down. For the same small distance, going up reduces g about twice as much as going down: 32 km up gives about 9.70 m/s², 32 km down about 9.75 m/s².
  • Weight changes with g, mass does not. A body weighing 63 N on the surface weighs 63/(1.5)² = 28 N at a height R_E/2; one weighing 250 N weighs 125 N halfway to the centre.
  • Use the exact form (R_E/(R_E + h))² whenever h is comparable to R_E; 1 − 2h/R_E gives nonsense there (zero at h = R_E/2).

7. Gravitational potential energy

NCERT §7.7

  • Gravity is a conservative force: the work it does depends only on the start and end points, so a potential energy can be defined.
  • Near the surface the force is nearly constant (mg), and lifting m from h₁ to h₂ takes work mg(h₂ − h₁). This gives W(h) = mgh + W₀, where W₀ is the value at the surface.
  • Far from the surface the force changes with distance; the work to lift m from r₁ to r₂ is G M_E m (1/r₁ − 1/r₂), so W(r) = −G M_E m / r + W₁ for r > R_E.
  • Only differences of potential energy have physical meaning. The usual choice is zero at infinity, so the potential energy of two masses is V = −G m₁ m₂ / r, always negative.
  • The potential energy at a point is then the work done in bringing the body in from infinity to that point.
  • Gravitational potential is the potential energy per unit mass at a point: −G M / r for a point outside a mass M.
  • For many particles, the total potential energy is the sum of −G mᵢ mⱼ / rᵢⱼ over every pair. Example: four masses m on a square of side l make 4 pairs at distance l and 2 pairs at √2 l, so the total is −(4 + √2) G m²/l = −5.41 G m²/l.
  • mgh is only an approximation, valid for heights small compared with R_E, to the exact difference −G M_E m (1/r₂ − 1/r₁).

8. Escape speed

NCERT §7.8

  • Escape speed is the least launch speed that lets a body reach infinity and never return, with air resistance ignored.
  • Energy conservation: ½ m vᵢ² − G M_E m / (R_E + h) must be at least zero, the least total energy a body can have at infinity.
  • From the surface (h = 0): vₑ = √(2 G M_E / R_E) = √(2 g R_E) ≈ 11.2 km/s.
  • Escape speed does not depend on the mass of the body or the direction of launch; it does depend on the planet and on the launch height.
  • For the Moon, with its smaller g and radius, the escape speed is about 2.3 km/s, about five times smaller. Gas molecules moving faster than that leave for good, which is why the Moon has no atmosphere.
  • Between two spheres of masses M and 4M, radius R, centres 6R apart, the pulls cancel at 2R from the smaller one. A body fired from M only needs to reach that neutral point: v_min = (3GM/5R)^½.

9. Earth satellites

NCERT §7.9

  • A satellite is a body that revolves around the earth; Kepler's laws apply to it just as to planets around the sun. The Moon is the earth's only natural satellite, with a period of about 27.3 days.
  • For a circular orbit of radius R_E + h, gravity supplies the centripetal force: m v²/(R_E + h) = G M_E m / (R_E + h)², so v = √(G M_E / (R_E + h)), less for higher orbits.
  • Just above the surface (h ≈ 0) the orbital speed is v = √(g R_E) ≈ 7.9 km/s, and the escape speed is √2 times this.
  • Period: T = 2π (R_E + h)^(3/2) / √(G M_E), so T² = k (R_E + h)³ with k = 4π²/(G M_E). This is Kepler's third law for satellites.
  • For an orbit just above the surface, T₀ = 2π √(R_E/g) ≈ 85 minutes.
  • The same law weighs other planets. Phobos circles Mars in 7 h 39 min at 9.4 × 10³ km, which gives a Mars mass of about 6.48 × 10²³ kg; with its orbit 1.52 times the earth's, a Martian year is (1.52)^(3/2) × 365 ≈ 684 days.
  • An astronaut in orbit feels weightless not because gravity is small there, but because the astronaut and the craft are falling freely towards the earth together.

10. Energy of an orbiting satellite

NCERT §7.10

  • In a circular orbit of radius r = R_E + h, kinetic energy K = G M_E m / 2r (positive).
  • Potential energy, with zero at infinity, is U = −G M_E m / r: negative and twice the kinetic energy in size.
  • Total energy E = K + U = −G M_E m / 2r, negative, and equal to −K.
  • Negative total energy means the satellite is bound: it cannot reach infinity. Zero or positive total energy would mean escape.
  • In an elliptical orbit K and U change from point to point, but the total energy stays constant and negative.
  • Moving a 400 kg satellite from orbit radius 2R_E to 4R_E needs ΔE = G M_E m / 8R_E = m g R_E / 8 ≈ 3.13 × 10⁹ J. Its kinetic energy falls by 3.13 × 10⁹ J and its potential energy rises by 6.25 × 10⁹ J.
  • To send an orbiting satellite out of the earth's influence you must supply only G M_E m / 2r, which is less than for a body at rest at the same height.

Must-know facts

  1. F = G m₁ m₂ / r², with r measured between centres; G = 6.67 × 10⁻¹¹ N m² kg⁻².
  2. g = G M_E / R_E² ≈ 9.8 m/s² and is independent of the mass of the falling body.
  3. g(h) = g R_E²/(R_E + h)² exactly; ≈ g(1 − 2h/R_E) only for h ≪ R_E.
  4. g(d) = g(1 − d/R_E) for a uniform earth; zero at the centre.
  5. g is largest at the surface and decreases both above and below it.
  6. Shell outside → acts as a point at its centre; point inside a shell → zero net pull.
  7. Kepler: ellipse with the sun at a focus; equal areas in equal times; T² ∝ a³.
  8. Law of areas = conservation of angular momentum; holds for any central force.
  9. v_P r_P = v_A r_A: fastest at perihelion.
  10. U = −G M m / r, zero at infinity; mgh is its near-surface approximation.
  11. Escape speed vₑ = √(2gR_E) ≈ 11.2 km/s; independent of the body's mass and launch direction.
  12. Orbital speed near the surface v₀ = √(gR_E) ≈ 7.9 km/s; vₑ = √2 v₀.
  13. Orbital speed v = √(G M_E / r) falls with r; period T ∝ r^(3/2).
  14. Near-surface orbit period ≈ 85 min.
  15. Satellite: K = GMm/2r, U = −GMm/r, E = −GMm/2r = −K.
  16. Negative total energy = bound orbit; zero or positive = escape.

Common traps

Putting the height above the surface into F = G M m / r².

r is always measured from the centre of the earth: r = R_E + h.

Using g(1 − 2h/R_E) for a height like R_E/2.

The shortcut only works for h ≪ R_E. At h = R_E/2 use g/(1.5)² = 0.44 g, not zero.

Thinking g keeps increasing as you go down a mine, because you are closer to the centre.

The shell above you pulls with zero net force and the mass below you shrinks; g(d) = g(1 − d/R_E) falls to zero at the centre.

Believing a heavier body needs a bigger escape speed.

The mass m cancels: vₑ = √(2GM/R) depends only on the planet (and launch height).

Thinking astronauts float because there is no gravity in orbit.

At a few hundred km g is still close to its surface value; they float because they and the craft fall freely together.

Giving a satellite a positive total energy because it is moving.

E = −GMm/2r is negative for every bound orbit; K is positive but U is negative and twice as large.

Saying the law of areas proves the inverse-square law.

Equal areas follows from any central force (angular momentum is conserved); it is T² ∝ a³ that points to 1/r².

Treating G and g as the same kind of constant.

G is universal (6.67 × 10⁻¹¹ N m² kg⁻²); g is a local acceleration that changes with height, depth and planet.

Formulas

Law of gravitation

F = G m₁ m₂ / r²

r between centres; attractive.

Kepler's third law

T² = (4π² / G M) a³

M is the central body; a is the semi-major axis (radius for a circle).

Law of areas

ΔA/Δt = L / 2m = constant

At the ends of the orbit: v_P r_P = v_A r_A.

Cavendish balance

G M m L / d² = τθ

τ is the restoring torque per unit twist.

g at the surface

g = G M_E / R_E²

Gives M_E = g R_E² / G.

g at height h

g(h) = g R_E² / (R_E + h)² ≈ g (1 − 2h/R_E)

Approximation only for h ≪ R_E.

g at depth d

g(d) = g (1 − d/R_E)

Uniform earth; zero at the centre.

Potential energy

U = −G m₁ m₂ / r

Zero at infinity; add over all pairs for a system.

Gravitational potential

V = −G M / r

Potential energy per unit mass.

Escape speed

vₑ = √(2 G M / R) = √(2 g R)

≈ 11.2 km/s for the earth.

Orbital speed

v = √(G M_E / (R_E + h)); v₀ = √(g R_E)

v₀ ≈ 7.9 km/s just above the surface.

Orbital period

T = 2π (R_E + h)^(3/2) / √(G M_E); T₀ = 2π √(R_E / g)

T₀ ≈ 85 min.

Satellite energies

K = G M m / 2r, U = −G M m / r, E = −G M m / 2r

r = R_E + h; E = −K = U/2.

Key terms

Geocentric model
A picture of the heavens with the earth at the centre.
Heliocentric model
A picture with the sun at the centre and the planets going round it.
Ellipse
A closed curve on which the distances to two fixed foci always add to the same total.
Perihelion
The point of a planet's orbit nearest the sun.
Aphelion
The point of a planet's orbit farthest from the sun.
Semi-major axis
Half of the longest diameter of an ellipse.
Central force
A force always directed along the line joining the body to a fixed centre.
Gravitational constant (G)
The universal constant in Newton's law, 6.67 × 10⁻¹¹ N m² kg⁻².
Acceleration due to gravity (g)
The acceleration of a freely falling body near a planet; about 9.8 m/s² at the earth's surface.
Gravitational potential
Gravitational potential energy per unit mass at a point.
Escape speed
The least launch speed with which a body can reach infinity.
Satellite
A body that revolves around a planet.
Bound system
A system with negative total energy, whose orbit stays closed.

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