Gravitation

Gravitation in Advanced goes beyond Kepler's laws into gravitational potential and field of extended bodies, satellite energetics, and escape/orbital transfers. The field-inside-a-shell result and the potential-energy bookkeeping are frequent discriminators.

Key results

• Force: F = GMm/r². Field g = GM/r²; at height h, g' = g(1 − 2h/R) for h«R; at depth d, g' = g(1 − d/R).
• Shell theorem: field inside a uniform shell is zero; outside it acts as if all mass is at the centre.
• Potential: V = −GM/r (zero at infinity). PE of two masses U = −GMm/r.
• Orbits: circular orbital speed v = √(GM/r); total energy E = −GMm/2r; escape speed v_e = √(2GM/R) = √2·v_orbit(surface).
• Kepler III: T² ∝ r³ (r = semi-major axis).

Quantity	Value
Orbital speed	√(GM/r)
Escape speed	√(2GM/R)
Total energy in orbit	−GMm/2r
KE in orbit	+GMm/2r = −E

Exam Tricks & Tips

• 🎯 Escape speed = √2 × surface orbital speed and is independent of the projectile's mass or launch direction (ignoring air/rotation).
• 🎯 Total orbital energy is negative (bound); |KE| = |E| and PE = 2E — the virial relation KE = −½PE.
• 🎯 g decreases both above and below the surface, but linearly with depth and quadratically (approx.) with height — different laws.
• 🎯 Field inside a shell is zero, so inside a hollow planet a body feels no gravity from the shell; only enclosed mass matters (like Gauss's law).
• 🎯 To raise a satellite's orbit you increase total energy but its orbital KE decreases — a counter-intuitive Advanced favourite.
• ❌ Common mistake: taking gravitational PE as +mgh at large distances — that only holds near the surface; use −GMm/r generally.

Expected exam pattern

1–2 questions: a satellite-energy or orbit-transfer numerical, plus a conceptual on shell/field or variation of g. Sometimes combined with SHM (tunnel through Earth executes SHM).

Quick recap

Use V = −GM/r and U = −GMm/r globally. Orbital energy is −GMm/2r; escape speed is √2 times orbital. Field inside a shell is zero; g falls linearly with depth. Raising an orbit raises energy but lowers speed.

Cover the answer, recall, then check. 12 cards on gravitation for JEE Advanced.

Q1. Gravitational field inside a uniform spherical shell.
A1. Zero everywhere inside.

Q2. Escape speed from a planet of mass M, radius R.
A2. v_e = √(2GM/R) = √2 times the surface orbital speed.

Q3. Total mechanical energy of a satellite in a circular orbit of radius r.
A3. E = −GMm/2r (negative, i.e. bound).

Q4. How does g vary with depth d below the surface?
A4. g' = g(1 − d/R); it decreases linearly and is zero at the centre.

Q5. Orbital speed for a circular orbit of radius r.
A5. v = √(GM/r).

Q6. Kepler's third law.
A6. T² ∝ a³, where a is the semi-major axis.

Q7. Gravitational potential at distance r from a point mass.
A7. V = −GM/r (taken zero at infinity).

Q8. Relation between KE, PE and total energy in a circular orbit.
A8. KE = −E, PE = 2E; so PE = −2·KE (virial theorem).

Q9. How does g change at small height h above the surface?
A9. g' ≈ g(1 − 2h/R).

Q10. To move a satellite to a higher orbit, does its speed increase or decrease?
A10. Decrease (v = √(GM/r)), even though total energy increases.

Q11. Motion of a body dropped into a tunnel through the Earth's centre.
A11. Simple harmonic motion, with the field proportional to distance from the centre.

Q12. Does escape speed depend on the direction of projection?
A12. No (ignoring the atmosphere and planetary rotation) — it depends only on M and R.

Gravitation links three JEE-favourite threads — orbital mechanics, energy in a 1/r field, and the subtle field/potential of extended bodies. It rewards students who see Kepler's laws as consequences of a single inverse-square force and who never confuse potential with potential energy.

Core concept: every pair of masses attracts with F = Gm₁m₂/r², and the resulting field is conservative with potential energy U = −Gm₁m₂/r (note the sign and the zero at infinity).

Core ideas

Beginner — field and potential

Gravitational field g = GM/r² points toward the mass; potential V = −GM/r is the PE per unit mass. Near Earth's surface g = GM/R² ≈ 9.8 m/s². Inside a uniform solid sphere the field grows linearly, g = GMr/R³, and is zero at the centre; outside it falls as 1/r². A key shell result: a uniform shell exerts no field on a mass inside it but acts as a point mass on anything outside.

Intermediate — variation of g

With altitude h ≪ R: g' ≈ g(1 − 2h/R). With depth d: g' = g(1 − d/R). Rotation reduces apparent g by ω²R cos²(latitude), maximal at the equator. These small-correction formulas are frequent one-mark traps.

Advanced — orbits and energy

For a circular orbit of radius r, orbital speed v = √(GM/r), and total energy E = −GMm/(2r) = ½ U. Escape speed v_e = √(2GM/R) = √2 × orbital speed at the surface. Kepler's laws: (1) elliptical orbits with the Sun at a focus; (2) equal areas in equal times, which is just angular-momentum conservation (L = constant), so a planet moves fastest at perihelion; (3) T² ∝ a³, where a is the semi-major axis. For a bound orbit E < 0, parabolic E = 0, hyperbolic E > 0 — the sign of total energy classifies the trajectory. Angular momentum and energy conservation together solve elliptical-orbit "find speed at apogee/perigee" problems: v_p r_p = v_a r_a (from L) plus energy.

Worked example

A satellite orbits at radius r with speed v = √(GM/r). It is given a sudden tangential boost to speed v'. For it to just escape, its total energy must reach zero: ½m v'² − GMm/r = 0, so v' = √(2GM/r) = √2 v. Thus a 41.4% speed increase converts a circular orbit into an escape trajectory — a clean, memorable ratio (v_escape/v_orbit = √2 at any radius).

How JEE Advanced tests this

Elliptical-orbit problems using L and E conservation; field/potential inside cavities and composite bodies (superposition with negative mass); binary-star systems orbiting a common CM; variation of g with depth/altitude/rotation; and energy needed to shift a satellite between orbits.

Exam tricks & shortcuts

• For any circular orbit: KE = −E = −½U, so E = −KE and |U| = 2·KE (the virial pattern).
• v_escape = √2 × v_orbit, always.
• Kepler's 2nd law is angular-momentum conservation in disguise: fastest at perihelion.
• Mnemonic: "Inside a shell, gravity's asleep."

Key formulas

• Newton's law: F = Gm₁m₂/r²; g = GM/R².
• Variation: g_h = g(1 − 2h/R); g_d = g(1 − d/R).
• Potential energy: U = −GMm/r; potential V = −GM/r.
• Escape speed: v_e = √(2GM/R) = √(2gR) ≈ 11.2 km/s.
• Orbital speed: v_o = √(GM/r); total orbital energy E = −GMm/2r.
• Kepler: T² ∝ a³.

Gravitational PE is negative (bound); escape speed is √2 times orbital speed.

Worked Example

Problem: A satellite of mass 200 kg moves in a circular orbit of radius r = 2R around the Earth (R = 6.4 × 10⁶ m, surface g = 9.8 m/s²). Find (a) its total mechanical energy in orbit and (b) the extra energy that must be supplied for it to just escape the Earth's gravity.

Solution:
Use GM = gR² = 9.8 × (6.4 × 10⁶)² = 9.8 × 4.096 × 10¹³ = 4.014 × 10¹⁴.
Orbit radius r = 2R = 1.28 × 10⁷ m.

(a) For a circular orbit the total mechanical energy is:
E = −GMm/(2r)
= −(4.014 × 10¹⁴)(200) / (2 × 1.28 × 10⁷)
= −(8.028 × 10¹⁶) / (2.56 × 10⁷)
= −3.14 × 10⁹ J.

(b) To just escape, the satellite must reach E = 0 (zero speed at infinity). The extra energy needed is:
ΔE = 0 − (−3.14 × 10⁹) = 3.14 × 10⁹ J.

Answer: (a) E ≈ −3.14 × 10⁹ J; (b) energy to escape ≈ 3.14 × 10⁹ J.