Gravitation — NEET Notes

Kepler's three laws of planetary motion

For almost two thousand years the Ptolemaic model placed the Earth at the centre of the universe and required the planets to move in circles upon circles. The story that overturned it is one of the great moments in science. Tycho Brahe, a Danish nobleman, spent his life recording the positions of the planets with the naked eye to a precision no one had matched. His assistant Johannes Kepler, working through Tycho's tables after his master's death, extracted three astonishingly simple regularities. These three laws, distilled from data and stated without explanation, would become the empirical foundation on which Newton built his theory of gravity.

Kepler did not know why the planets obey his laws — that came half a century later when Newton showed every one of them follows from a single inverse-square law of force. Here are the three laws as NCERT and NIOS both state them.

The law of areas is the cleanest of the three because it follows from a property far more general than gravity. The area swept by the radius vector in time dt is dA = ½|r × v|dt = (L/2m)dt, where L is the angular momentum. Since gravity is a central force — always directed along the line joining the two bodies — the torque about the Sun is zero, so L is conserved, and dA/dt is therefore a constant. The same argument would hold for any central force, attractive or repulsive, with any radial dependence. Angular momentum is the deeper truth; the equal-areas rule is its visible consequence.

Kepler's third law has a particularly elegant proof for the special case of circular orbits, which we will reach in §Earth satellites. The full proof for elliptical orbits requires more work but the conclusion stands: T²/a³ is a universal constant for every body orbiting a given central mass. For planets orbiting the Sun, the constant is one value; for satellites orbiting the Earth, it is another value (smaller, because the Earth is lighter than the Sun); for moons of Jupiter, yet another. Within each family, however, every member obeys the same proportionality — and that single fact alone, applied to the Moon, was enough for Newton to guess the universal law of gravitation.

The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

Kepler's third law — the empirical statement Newton would later derive from F = Gm₁m₂/r²

Newton's universal law of gravitation

The story of the falling apple may or may not be apocryphal, but the line of reasoning behind Newton's leap is not. The Moon, like the apple, is being pulled towards the Earth — but the Moon does not fall, because its tangential velocity carries it past the Earth's curvature. Newton compared the centripetal acceleration of the Moon, a_m = V²/R_m ≈ ω²R_m, with the acceleration of an apple on the surface, g. Using the Moon's known orbital radius (R_m ≈ 3.84 × 10⁸ m) and period (T ≈ 27.3 days), the ratio g/a_m came out to about 3600. The ratio of distances R_m/R_E is about 60. Since 60² = 3600, the force must fall off as the square of the distance. From there it was a short step to the universal law:

Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Newton's universal law of gravitation — Principia, 1687

Written as an equation, for two point masses m₁ and m₂ separated by r:

Three properties of the law deserve attention. First, gravitation is always attractive — there is no negative mass, so the force never reverses sign. Second, it is a long-range force; unlike a contact force, it acts across empty space and falls off only as 1/r². Third, gravity obeys the principle of superposition: the force on a particle from a system of other particles is the vector sum of the individual pair-wise forces, computed as if each pair existed alone. There is no shielding, no interference, no saturation.

The law applies strictly to point masses. For extended objects we have to integrate — or invoke two beautiful results that NCERT calls the shell theorems. (i) A uniform spherical shell attracts an external point mass exactly as if its entire mass were concentrated at its centre. (ii) The same shell exerts zero gravitational force on any point mass inside it. The first lets us treat the Earth as a point mass when computing the force on something outside it; the second underlies the fact that g vanishes at the centre of the Earth.

The gravitational constant G

The universal law contains exactly one constant, G, and its value cannot be deduced from theory — it must be measured. The measurement was first performed in 1798 by Henry Cavendish, more than a century after Newton's Principia, using a torsion-balance apparatus that remains a model of experimental elegance. Two small lead spheres were mounted at the ends of a thin horizontal bar suspended by a fine wire. Two large lead spheres were brought up on opposite sides of the small ones. The gravitational attraction produced a tiny torque that twisted the wire by a measurable angle θ; the restoring couple per unit angle τ of the wire had been calibrated independently. From the measured θ, the bar length L, the masses, and their separation d, Cavendish solved for G.

Once G is known, the universal law becomes a calculator. Knowing g (measured by pendulums or free-fall experiments) and R_E (measured by geodesy), the Earth's mass follows from M_E = gR_E²/G ≈ 5.97 × 10²⁴ kg. Knowing the Moon's period and orbital radius and applying Kepler's third law gives the Earth's mass a second way — and the two answers agree to better than 1%, a famous internal consistency check of the theory. Cavendish, by measuring G, had implicitly "weighed the Earth".

Acceleration due to gravity at the surface

For a body of mass m on the surface of the Earth (mass M_E, radius R_E), the gravitational force is F = GM_Em/R_E² (using the shell theorem to treat the Earth as a point mass at its centre). By Newton's second law this equals mg, so the mass m cancels and

g = GM_E / R_E² ≈ 9.8 m s⁻²

Acceleration due to gravity at the Earth's surface

The cancellation of m is the modern explanation of Galileo's observation that all bodies, regardless of mass, fall with the same acceleration in vacuum. The inertial mass that resists acceleration (Newton's second law) is the same as the gravitational mass that feels the pull (the universal law) — a fact known as the equivalence principle and now confirmed to a part in 10¹³.

g is one of the most useful numerical inputs in all of mechanics. With g = 9.8 m s⁻² (often rounded to 10) and R_E = 6.4 × 10⁶ m, you can rapidly estimate orbital speeds (√(gR_E) ≈ 7.9 km/s), escape speeds (√(2gR_E) ≈ 11.2 km/s), and the period of a satellite skimming the surface (2π√(R_E/g) ≈ 85 minutes). NEET problems lean heavily on these signature numbers.

Variation of g with altitude, depth and latitude

The value 9.8 m s⁻² is a surface average. The true acceleration due to gravity depends on where you measure it. Three factors change it: the height above the surface, the depth below the surface, and the latitude.

The altitude formula deserves a careful derivation because NEET tests its small-h limit and its exact form interchangeably. At a height h above the surface the distance from the centre is R + h. From Newton's law, g(h) = GM_E/(R+h)² = g · R²/(R+h)² = g · (1 + h/R)⁻². For h ≪ R, the binomial expansion gives g(h) ≈ g(1 − 2h/R). NEET 2020 took the other route: a body weighing 72 N on the surface is taken to h = R/2, where (R+h)/R = 3/2, so g(h)/g = (2/3)² = 4/9 and the new weight is 72 × 4/9 = 32 N. Memorising both forms — the exact (1 + h/R)⁻² and the linear (1 − 2h/R) — is essential.

The depth derivation is more subtle. At depth d the distance from the centre is R − d. Crucially, by the shell theorem, only the sphere of radius (R − d) below the point contributes; the shell of thickness d above contributes zero. Assuming uniform density ρ, the mass inside radius (R − d) is M_s = (4/3)π(R−d)³ρ, and the acceleration becomes g(d) = GM_s/(R−d)² = (4/3)πGρ(R−d) = g(1 − d/R), using g = (4/3)πGρR at the surface. The dependence is linear in d, and g vanishes exactly at the centre. Above and below the surface, g falls — its maximum is on the surface itself.

The latitudinal variation comes from a different physics. The Earth rotates with angular velocity ω = 2π/(86400 s), so an object at the surface at latitude λ is moving in a circle of radius R cos λ. Part of the gravitational pull GM/R² must supply the centripetal acceleration ω²R cos²λ (the component along the local vertical), leaving an effective surface gravity g_eff = g − ω²R cos²λ. At the equator (λ = 0) the reduction is maximum, ω²R ≈ 0.034 m s⁻². At the poles (λ = 90°) it vanishes. This is one reason why pendulum clocks run faster at higher latitudes, and why rockets are launched eastward from sites close to the equator to harvest the boost.

Gravitational potential energy

Near the Earth's surface, when h is small compared to R_E, gravity is nearly constant and the potential energy of a mass m at height h is the familiar U = mgh. This is the limit you used in Work, Energy and Power. But for problems that span large distances — escape problems, orbits, gravitational binding energies — g is no longer constant and mgh breaks down. The true gravitational potential energy of two point masses, with the zero of energy chosen at infinity, is

The derivation is a one-line integral. The gravitational force on m due to M at distance r is F = −GMm/r² r̂ (towards M). The work done against this force as m moves from r = ∞ to r = r is the change in potential energy:

U(r) − U(∞) = − ∫_∞^r F · dr = − GMm/r

Defining U(∞) = 0 gives U(r) = − GMm/r

For an arbitrary system of n masses, the total potential energy is the sum of −Gm_im_j/r_ij taken over every distinct pair. NCERT's Example 7.3 — four masses at the corners of a square of side ℓ — has six pairs (four of length ℓ and two diagonals of length ℓ√2), giving total U = −(4 + √2)Gm²/ℓ ≈ −5.41 Gm²/ℓ.

The gravitational potential V (no subscript) is potential energy per unit test mass: V(r) = U(r)/m = −GM/r. It is a property of the source mass distribution alone. NEET 2023 Q.10 made this precise: for two masses m and 9m separated by R, find the point where the gravitational field is zero and evaluate the potential there. Solving Gm/x² = G(9m)/(R−x)² gives x = R/4; substituting into V_P = −Gm/x − G(9m)/(R−x) gives V_P = −16Gm/R. The trick is that the field can vanish while the potential — being a sum, not a vector — generally does not.

Escape speed

Throw a stone up and it falls back. Throw it harder and it goes higher before falling back. Is there a speed beyond which it does not fall back at all? Yes, and energy conservation gives it directly. A projectile launched from the surface with speed v_i has total mechanical energy E = ½mv_i² − GMm/R_E. To "just reach infinity" means arriving there with zero kinetic energy and zero potential energy, hence E = 0. Setting E = 0 and solving for v_i:

Three properties of v_e are NEET-favourites. (i) Independence of the launched body's mass: m cancels in the energy equation, so a feather and a cannonball need the same speed. (ii) Independence of launch direction: only the magnitude of the velocity appears in ½mv², so straight up or sideways works equally well — provided air resistance is negligible. (iii) Dependence on the planet's mass and radius: v_e = √(2GM/R) means denser, more massive bodies have higher escape speeds. The Moon's low escape speed explains why hydrogen and helium long ago escaped its surface; Earth, with its higher v_e, has held onto a substantial atmosphere.

For NEET 2021 Q.13, the planet has the same density as Earth but four times the radius. Since M = (4/3)πR³ρ, v_e = √(2GM/R) = √(8πGρ/3) · R ∝ R for the same density. The new escape speed is 4v. The same problem, asked in 2016 with double the radius and double the density, gives v_e ∝ R√ρ, so the ratio v_earth/v_planet = 1/(2 × √2) = 1 : 2√2.

Earth satellites and orbital velocity

An object that is launched with a speed less than v_e does not escape. If its speed is exactly right and its direction is tangential to the surface, it does not fall back either — it enters a closed orbit around the Earth. Kepler's laws, which Newton derived in their general form from gravity, apply equally to natural and artificial satellites. The simplest case is a circular orbit of radius r = R_E + h.

v_o = √(GM / r)     T² = (4π² / GM) r³

Orbital speed and orbital period for a circular orbit of radius r

For a satellite skimming the Earth's surface (h = 0, r = R_E), v_o = √(gR_E) ≈ 7.9 km/s and T ≈ 2π√(R_E/g) ≈ 84 minutes. Real satellites orbit at higher altitudes — air drag would deorbit a surface-skimming satellite in seconds — but these numbers set the scale.

The period formula T² ∝ r³ is Kepler's third law applied to Earth satellites. NEET 2023 Q.45 used it cleverly: a satellite just above the Earth's surface has T = 2π√(R³/GM) = 2π√(R³/(G · (4/3)πR³ρ)) = √(3π/Gρ), so T² = 3π/(Gρ), and the dimensionally clean expression 3/(Gρ) has dimensions of T² (modulo factors of π). The student is asked to recognise which power of T appears, and the answer is T² — a moment of pattern-matching that costs four marks if missed.

An NCERT example "weighs Mars" by combining Kepler's law with measured satellite data: Phobos orbits Mars with period 7 h 39 min at radius 9.4 × 10³ km. From T² = 4π²r³/GM_m, the Martian mass comes out to 6.48 × 10²³ kg — about one-tenth Earth's mass, in agreement with other measurements. The same logic, applied to Earth's Moon, gives the Earth's mass and matches the surface-gravity calculation. Two independent paths, one answer — that consistency is what makes Newtonian gravity a theory and not a guess.

Energy of an orbiting satellite

The kinetic and potential energies of a satellite in a circular orbit of radius r are tied together by a remarkable relation. The kinetic energy follows from v_o² = GM/r:

K = ½ m v_o² = + GMm / 2r   |   U = − GMm / r   |   E = K + U = − GMm / 2r

Energies of a satellite in a circular orbit — K is positive, U is negative, E is negative

Three facts to keep in mind. First, the total energy E is negative. This is the signature of a bound system: a body with E < 0 cannot reach infinity, where U = 0 and KE ≥ 0 would require E ≥ 0. Second, |U| = 2|K|. The potential energy is exactly twice the kinetic energy in magnitude — a special feature of the inverse-square law (a result of the virial theorem). Third, |E| = K = |U|/2. If you ever need only one of the three energies and you know any other, you can write down the rest by inspection.

Energy changes between orbits follow a counter-intuitive rule. To move a satellite to a higher orbit you must add energy — but its kinetic energy decreases, because the new orbit has a larger r and v_o ∝ 1/√r is smaller. The energy you add becomes potential energy; the kinetic energy reduction is exactly half the total energy added, so ΔE > 0 but ΔK < 0. NCERT's Example 7.8 makes this explicit: lifting a 400 kg satellite from r = 2R_E to r = 4R_E requires +3.13 × 10⁹ J of energy, but ΔK = −3.13 × 10⁹ J and ΔU = +6.25 × 10⁹ J.

NEET 2018 Q.44 tested KE comparison on an elliptical orbit at points A, B, C, where r_A < r_B < r_C. By angular-momentum conservation (Kepler's second law), mvr is constant along the orbit, so v_A > v_B > v_C, and therefore K_A > K_B > K_C. Total energy E is conserved, of course, but K alone is not — it swaps with U at every point of the elliptical orbit. The same total-energy formula E = −GMm/2a now applies, with a (the semi-major axis) replacing r.

Geostationary vs polar satellites

The orbital period formula T² = (4π²/GM)r³ has a special root. Setting T equal to one sidereal day (86164 s ≈ 24 h) and solving for r gives r_geo ≈ 42,200 km, i.e. about 36,000 km above the Earth's surface. A satellite placed at this radius in the equatorial plane, orbiting in the same direction as Earth's rotation, will appear stationary to an observer on the ground. This is the geostationary orbit. The TV dish on a rooftop can therefore be pointed at a fixed direction and left there.

Geostationary satellites are used for television broadcasting, telephone relays, and continuous weather monitoring of a fixed region. Their drawbacks are altitude (signal latency of about 0.25 s round-trip), restricted line-of-sight (high latitudes near the poles get poor coverage), and the limited number of slots available on the geostationary "ring" — a regulated international resource.

Polar satellites orbit at much lower altitudes (typically 500–800 km) on orbits that pass close to both poles. As the Earth rotates beneath the satellite, successive orbits scan different longitudinal strips, allowing the entire planet to be imaged in a day or two. Polar satellites carry the cameras and instruments used for remote sensing, mapping, environmental monitoring, weather forecasting at high resolution, and military reconnaissance. India's IRS series and the international Landsat constellation are polar; INSAT and GSAT are geostationary.

One final fact about satellites that NCERT highlights: an astronaut inside an orbiting satellite feels weightless. This is not because gravity is small at that altitude — at 400 km up, g is still about 89% of its surface value. It is because both the astronaut and the spacecraft are in free fall, falling around the Earth together. The normal force between astronaut and floor — which is what a scale would measure as weight — is zero, exactly as in a freely falling lift. Apparent weight, not gravitational pull, is what you actually feel.