Gravitation

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Key Formulas:

• Universal gravitational constant: G = 6.67 × 10⁻¹¹ N m² kg⁻²
• Newton’s law of gravitation: F = Gm₁m₂/r²
• Acceleration due to gravity: g = GM/r²
• Variation with altitude: g’ = g(R/(R+h))²
• Variation with depth: g’ = g(1 - d/R)
• Orbital velocity: v_orbit = √(GM/r) = √(gr) for h << R
• Escape velocity: v_escape = √(2GM/r) = √(2gR) ≈ 11.2 km/s
• Orbital period: T = 2π√(r³/GM)
• Kepler’s third law: T² ∝ r³

Exam Tips:

• Escape velocity is always √2 times orbital velocity for the same planet
• Satellite in circular orbit has constant speed but velocity changes direction continuously
• Geostationary satellites orbit at ~36,000 km above Earth’s surface with T = 24 hours
• Weightlessness in orbit is not zero gravity — it’s free-fall around Earth
• g varies with latitude due to Earth’s rotation and oblate shape
• Communication satellites use geostationary orbits

🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of gravitational fields and orbital mechanics.

Gravitational Field and Potential:

The gravitational field strength at a point is defined as the gravitational force per unit mass: g = F/m = GM/r². It is a vector quantity pointing toward the centre of the massive body. The gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point: V = -GM/r. Potential is a scalar quantity and is always negative (assuming zero at infinity). The relationship between field and potential is g = -dV/dr in the radial direction.

The principle of superposition states that the total gravitational field at a point due to multiple masses is the vector sum of the fields due to individual masses, and similarly for potential (scalar addition).

Variation of g:

With altitude, g decreases because r increases. The formula g’ = g(R/(R+h))² shows that g halves when h ≈ 0.41R (about 2600 km above Earth’s surface). With depth, g decreases linearly: g’ = g(1 - d/R), assuming uniform density — this holds approximately for Earth.

The variation with latitude is more complex: at a point on Earth’s surface at latitude λ, the effective acceleration due to gravity is reduced by the centrifugal effect of Earth’s rotation: g’ = g - ω²R cos²λ. At the poles, cos λ = 0 so no centrifugal reduction; at the equator, cos λ = 1 so g’ = g - ω²R, making effective g about 0.034 m/s² less than the theoretical polar value.

Orbital Mechanics:

A satellite in circular orbit requires the centripetal force to be provided entirely by gravity: mv²/r = GMm/r². This gives v = √(GM/r) and T = 2π√(r³/GM), confirming Kepler’s third law T² ∝ r³.

Geostationary satellites orbit in the equatorial plane at exactly 35,786 km from Earth’s surface, giving T = 24 hours — they appear fixed relative to Earth’s surface and are used for communications and weather monitoring. Polar satellites orbit at lower altitudes (~500-800 km) with inclined planes and can cover the entire Earth’s surface as the planet rotates beneath them; their period is approximately 100 minutes.

Energy of Satellite:

The total mechanical energy of a satellite in circular orbit is E = KE + PE = ½mv² - GMm/r = -GMm/2r. This is negative (bound system), and the magnitude decreases (becomes less negative) as r increases. Energy must be supplied to raise a satellite to a higher orbit.

Worked NEET Example: A satellite of mass 500 kg orbits Earth at height 500 km. Find orbital speed and period. (Given M_earth = 6 × 10²⁴ kg, R = 6400 km)

• r = R + h = 6900 km = 6.9 × 10⁶ m
• v = √(GM/r) = √(6.67×10⁻¹¹ × 6×10²⁴ / 6.9×10⁶) ≈ 7.6 km/s
• T = 2π√(r³/GM) ≈ 5800 s ≈ 97 minutes

Comparison: Escape vs Orbital Velocity: For Earth, orbital velocity at surface ≈ 7.9 km/s while escape velocity ≈ 11.2 km/s. The ratio v_escape/v_orbit = √2 always, independent of the planet’s mass for a given body.

🔴 Extended — Deep Study (3mo+)

Comprehensive derivation-based theory for mastery of gravitation.

Derivation of Gravitational Potential Energy:

Consider a mass m being moved from infinity to a point at distance r from the centre of mass M. The work done against gravity in moving an infinitesimal distance dr inward is dW = F dr = (GMm/r²) dr (taking force opposite to displacement for positive work). Integrating from ∞ to r: U = ∫[∞ to r] (GMm/r²) dr = GMm[(-1/r)] from ∞ to r = GMm(-1/r) - 0 = -GMm/r. This is the gravitational potential energy, negative because the gravitational force is attractive and zero reference is at infinity.

Derivation of Escape Velocity:

From energy conservation: at the surface (r = R), KE + PE = ½mv² - GMm/R. At infinity (r → ∞), PE → 0 and if the particle just escapes, KE → 0. Equating: ½mv² - GMm/R = 0. Therefore v = √(2GM/R) = √(2gR) since g = GM/R². For Earth, v ≈ 11.2 km/s. Note this is independent of the particle’s mass.

Derivation of Orbital Velocity from Centripetal Requirement:

For circular orbit, centripetal force = gravitational force: mv²/r = GMm/r². Therefore v = √(GM/r). This can also be written as v = √(gr) for low orbits where h << R (so r ≈ R).

Derivation of Orbital Period:

Orbital velocity v = 2πr/T. Equating with v = √(GM/r): 2πr/T = √(GM/r). Squaring and rearranging: T² = (4π²/GM) r³. Hence T = 2π√(r³/GM), confirming Kepler’s third law T² ∝ r³.

Kepler’s Laws — Physical Basis:

First law (elliptical orbits): Follows from inverse-square central force. The orbit is a conic section with the massive body at one focus. Second law (equal areas in equal times): Follows from conservation of angular momentum L = mvr = constant, giving r²θ̇ = constant, which means the rate of sweeping area dA/dt = ½r²θ̇ is constant. Third law (T² ∝ r³ for circular orbits): Derived above from equating centripetal force to gravitational force.

Variation of g with Latitude (Centrifugal Effect):

At latitude λ, the distance from Earth’s axis is R cos λ. The centrifugal acceleration is ω²R cos λ directed outward from the axis, perpendicular to Earth’s rotation axis. Resolving along the radial direction (toward Earth’s centre), the effective acceleration due to gravity is reduced by ω²R cos²λ. At the equator (λ = 0°), reduction is maximum; at poles (λ = 90°), reduction is zero.

Weightlessness in Satellites:

A satellite in orbit is in continuous free fall toward Earth. The astronaut and spacecraft both experience the same acceleration (principle of equivalence), so there is no relative force between them — apparent weightlessness. The gravitational field is not zero; it is approximately 90% of surface g at typical satellite altitudes.

Energy Required to Launch Satellite:

The minimum energy is the difference between the satellite’s final total energy (-GMm/2r) and initial energy at Earth’s surface (-GMm/R). For a satellite in low Earth orbit (r ≈ R + h): E_required = GMm(½R - ½r) = ½GMm(R/r - 1). For h = 300 km, this is approximately 3.3 × 10⁹ J per kg.

Gravitational Redshift:

Light climbing out of a gravitational well loses energy, causing its wavelength to increase. For a photon moving from Earth’s surface to infinity: Δλ/λ ≈ gh/c² ≈ 2.1 × 10⁻¹⁵ for h = 1 m. This effect is extremely small but measurable with precise atomic clocks.

Schwarzschild Radius:

The radius at which a body becomes a black hole (event horizon) is R_s = 2GM/c². For the Sun, R_s ≈ 3 km; for Earth, R_s ≈ 9 mm. At this radius, the escape velocity equals the speed of light.

Tide-Raising Forces:

The Moon’s gravitational field is stronger on the side of Earth facing the Moon than on the opposite side. The differential force (tidal force) produces two bulges: one on the side facing the Moon and one on the opposite side. This explains ocean tides. The Sun also contributes, with spring tides (Sun and Moon aligned) being stronger than neap tides (Sun and Moon at right angles).

Lagrangian Points:

In a two-body system, there are five points where a third body can remain in equilibrium relative to the first two: L1 (between the bodies), L2 (beyond the smaller body), L3 (opposite side of larger body), and L4 and L5 (forming equilateral triangles with the two bodies, 60° ahead and behind the smaller body). L4 and L5 are stable equilibria; L1, L2, L3 are unstable.

Binary Star Systems:

Two stars orbiting their common centre of mass. Each follows Kepler’s laws with total mass M = M₁ + M₂. Observing the orbital period and separation allows calculation of total mass. Individual masses require measurement of radial velocity amplitudes.

Previous Year NEET Question Patterns:

• Questions frequently test the formula g’ = g(R/(R+h))² with numerical substitution
• Escape velocity comparisons between planets
• Orbital period calculations using T = 2π√(r³/GM)
• Energy of satellite questions (KE = ½mv², PE = -GMm/r, total E = -GMm/2r)
• Weightlessness conceptual questions
• Variation of g with altitude and depth