Gravitation: NEET notes.

01 · NCERT §7.1, §7.2

Kepler's laws

• 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.

02 · NCERT §7.2

The law of areas

• 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.

03 · NCERT §7.3

Universal law of gravitation

• 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.

04 · NCERT §7.4

The gravitational constant

• 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.

05 · NCERT §7.5

Acceleration due to gravity

• 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.

06 · NCERT §7.6

g above and below the surface

• 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).

07 · NCERT §7.7

Gravitational potential energy

• 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).

08 · NCERT §7.8

Escape speed

• 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.

09 · NCERT §7.9

Earth satellites

• 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 · NCERT §7.10

Energy of an orbiting satellite

• 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.

Revise

Must-know facts

1. 01F = G m₁ m₂ / r², with r measured between centres; G = 6.67 × 10⁻¹¹ N m² kg⁻².
2. 02g = G M_E / R_E² ≈ 9.8 m/s² and is independent of the mass of the falling body.
3. 03g(h) = g R_E²/(R_E + h)² exactly; ≈ g(1 − 2h/R_E) only for h ≪ R_E.
4. 04g(d) = g(1 − d/R_E) for a uniform earth; zero at the centre.
5. 05g is largest at the surface and decreases both above and below it.
6. 06Shell outside → acts as a point at its centre; point inside a shell → zero net pull.
7. 07Kepler: ellipse with the sun at a focus; equal areas in equal times; T² ∝ a³.
8. 08Law of areas = conservation of angular momentum; holds for any central force.
9. 09v_P r_P = v_A r_A: fastest at perihelion.
10. 10U = −G M m / r, zero at infinity; mgh is its near-surface approximation.
11. 11Escape speed vₑ = √(2gR_E) ≈ 11.2 km/s; independent of the body's mass and launch direction.
12. 12Orbital speed near the surface v₀ = √(gR_E) ≈ 7.9 km/s; vₑ = √2 v₀.
13. 13Orbital speed v = √(G M_E / r) falls with r; period T ∝ r^(3/2).
14. 14Near-surface orbit period ≈ 85 min.
15. 15Satellite: K = GMm/2r, U = −GMm/r, E = −GMm/2r = −K.
16. 16Negative total energy = bound orbit; zero or positive = escape.

Where marks go

Common traps

Reference

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.

Glossary

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.