Gravitational field and orbital mechanics
The gravitational-field block tests whether you can move confidently between force, field, potential and orbital energy, and whether you can justify satellite motion with Kepler's laws. Basque PAU questions often combine a short numerical calculation with a qualitative direction or energy argument.
Core relations
- Newton's law: F = GMm/r², attractive along the line joining the centres; field g = F/m = GM/r², directed toward the mass.
- Potential and potential energy: V = -GM/r and U = -GMm/r. Potential is a scalar and is zero at infinity; field is the negative gradient of potential.
- Circular orbit: GMm/r² = mv²/r, so v = sqrt(GM/r), T = 2πr/v and T² = 4π²r³/(GM). Here r is centre-to-centre, not altitude.
- Orbital energy: E = K + U = -GMm/(2r) for a circular orbit. Escape speed is v_esc = sqrt(2GM/r), where total energy becomes zero.
- Kepler: elliptical orbits with the central mass at one focus; equal areas in equal times; T² proportional to a³ for the same central body.
Worked example
A satellite orbits at r = 2R_E. Take M_E = 5.97×10²⁴ kg and R_E = 6.37×10⁶ m, so r = 1.274×10⁷ m. Then v = sqrt[(6.67×10⁻¹¹)(5.97×10²⁴)/(1.274×10⁷)] = sqrt(3.12×10⁷) = 5.59×10³ m/s. The period is T = 2πr/v = 2π(1.274×10⁷)/(5.59×10³) = 1.43×10⁴ s, about 3.98 h. The negative total energy confirms a bound orbit.
Traps that cost marks
Do not substitute altitude for r. Keep the minus sign in V and U, but remember that field strength is positive. State the direction of g as a vector. Do not use T² ∝ r³ across different central bodies without changing GM. A geostationary orbit is not at the Earth's surface; its radius is about 4.22×10⁷ m. Escape speed is not the speed needed to reach zero gravitational potential with zero kinetic energy; it is the minimum launch speed for total energy zero. Finally, potential can be zero where field is not, and field can be zero where potential is not.