Satellites, Orbital & Escape Velocity: JEE Main

Gravitation questions in JEE Main are split between universal law problems (force, field, potential) and satellite/orbital mechanics problems. The second category is where most marks are lost — students confuse escape velocity with orbital velocity, forget Kepler's third law, or misapply energy conservation. This guide covers every satellite and orbital mechanics topic the exam tests, with derivations that build lasting understanding.

Orbital Velocity: Derivation and Key Facts

For a satellite in a circular orbit of radius r (measured from Earth's centre), gravity provides the centripetal force: GMm/r² = mv²/r → v_o = √(GM/r). For an orbit close to Earth's surface (r ≈ R), v_o ≈ √(gR) ≈ 7.9 km/s. As r increases, v_o decreases (inverse square root). So satellites in higher orbits move slower. This is counterintuitive but important — JEE tests it directly by asking whether a satellite speeds up or slows down when boosted to a higher orbit (it slows down in terms of linear speed but covers a longer path).

Time period: T = 2πr/v_o = 2πr/√(GM/r) = 2π√(r³/GM). Kepler's third law: T² ∝ r³ (or T²/r³ = 4π²/GM = constant for all satellites of the same central body). JEE uses this to find the ratio of orbital radii or periods of two satellites. Take a free gravitation mock test to practise. For the full gravitation chapter, see our gravitation guide.

Escape Velocity: Why It Is √2 Times Orbital

Escape velocity v_e is the minimum speed to escape the gravitational field (reach r = ∞ with zero kinetic energy). Energy conservation: ½mv_e² − GMm/R = 0 → v_e = √(2GM/R) = √(2gR) ≈ 11.2 km/s. Compare with v_o (surface): v_o = √(gR). So v_e = √2 × v_o — escape velocity is always √2 times the orbital velocity at the same radius. This ratio is a direct JEE answer choice. Key point: escape velocity depends only on the planet's mass and radius — it is independent of the mass of the escaping object and the direction of launch (as long as ignoring atmosphere).

For a satellite already in orbit at radius r: to escape from orbit, additional kinetic energy = ½mv_e²(r) − ½mv_o²(r) = GMm/2r (since v_e(r) = √(2GM/r) and v_o(r) = √(GM/r), the difference in kinetic energies is GMm/2r). This additional energy required to escape from orbit is a common JEE question format.

Geostationary and Polar Satellites

A geostationary satellite has T = 24 hours (same as Earth's rotation) and orbits in the equatorial plane from west to east. Its orbital radius: from T² = 4π²r³/GM → r³ = GMT²/4π² → r ≈ 42,164 km from Earth's centre (≈ 35,786 km above surface). JEE asks: (1) find r given T and M; (2) find the period given r; (3) conceptual questions about geostationary coverage. A geostationary satellite appears fixed in the sky and is used for communications. A polar satellite orbits in a plane passing through Earth's poles, has a period of ~90–100 min (low Earth orbit), and covers all latitudes — used for remote sensing.

Energy of a Satellite and Binding Energy

For a satellite of mass m in circular orbit at radius r: kinetic energy K = GMm/2r (positive). Potential energy U = −GMm/r (negative). Total mechanical energy E = K + U = −GMm/2r (negative, meaning it is bound). Note E = −K = U/2. Binding energy (energy required to free it to r = ∞) = +GMm/2r. As a satellite rises to a higher orbit: |E| decreases (less bound), K decreases (slower speed), U increases (less negative), total E increases (less negative). All four change in the same direction — a useful consistency check. When a satellite's orbit decays due to atmospheric drag, it loses total mechanical energy (E becomes more negative), paradoxically speeds up (K increases as r decreases), and heats up. For the full treatment of planetary motion and Kepler's laws see our gravity, planets and moons guide.