← Mechanical Systems
7 Gravitation & orbital mechanics
Newton’s law of gravitation F=\frac{Gm_1m_2}{r^2} G = 6.67×10⁻¹¹ N m² kg⁻². Attractive, along the line joining centres. Gravitational field strength g=\frac{GM}{r^2} Acceleration due to gravity at distance r from mass M. Orbital speed v=\sqrt{\frac{GM}{r}} For a circular orbit: gravity provides the centripetal force. Kepler’s 3rd (period) T^2=\frac{4\pi^2 r^3}{G(m_1+m_2)} For m₂ ≪ M this reduces to T² = 4π²r³/GM. Derivation Show that T² = 4π²R³ / G(m₁ + m₂) A body orbits in a circle of radius R. Derive its period from gravity = centripetal force. Gravity provides the centripetal force: G m₁ m₂ / R² = m₂ · (4π²R / T²). (using centripetal a = ω²R = (2π/T)²R = 4π²R/T²). Cancel m₂ and rearrange: G m₁ / R² = 4π²R / T². T² = 4π²R³ / (G m₁). For two comparable masses (relative to the centre of mass) this generalises to T² = 4π²R³ / G(m₁ + m₂). ∎ ✓ T² = 4π²R³ / G(m₁ + m₂). This is Kepler’s third law from Newtonian gravity. Worked example Geostationary orbit radius Find the orbital radius for a satellite with T = 24 h around Earth (M = 6.0×10²⁴ kg). T = 24 h = 86 400 s. Use T² = 4π²R³/GM ⟹ R³ = GMT²/4π². R³ = (6.67×10⁻¹¹ × 6.0×10²⁴ × 86400²) / (4π²). R³ = (6.67×10⁻¹¹ × 6.0×10²⁴ × 7.46×10⁹) / 39.5 ≈ 7.57×10²². R = (7.57×10²²)^(1/3) ≈ 4.23×10⁷ m. ✓ R ≈ 4.2×10⁷ m (about 42 000 km from Earth’s centre).Why astronauts appear weightless in orbit
- Gravity has NOT switched off — at ISS altitude it is still about 90% of its surface value, and it is exactly what keeps the station in orbit.
- Both the astronaut and the station are in continuous FREE FALL, following the same curved path around Earth at the same acceleration.
- "Weight" as you feel it is a contact (normal) force between you and a support. Because astronaut and station accelerate together, there is no relative motion and no support force between them — hence no sensation of weight.
- "Weightless" describes the ABSENCE OF A CONTACT FORCE, not the absence of gravity.