Gravitational Potential (V)

We need a universal reference point from which to measure potential energy. Infinity is the natural choice because gravitational force falls to zero there. Setting V = 0 at infinity means every other point has a negative value: the closer you are to a mass, the more negative the potential. This convention is consistent across all gravitational problems and allows different planets or stars to be compared on the same scale.

For a spherically symmetric mass M, the gravitational field strength at radius r is g = GM/r² directed inward. The work done by an external agent per unit mass moving a test mass from infinity to radius r is found by integrating: V = −∫[∞ to r] (−GM/r²) dr = −GM/r. The negative sign arises because we do negative net work against a field that assists the inward journey — or equivalently, because the system loses potential energy as the masses approach.

Gravitational field strength g is the negative gradient of gravitational potential: g = −dV/dr. In one dimension this means the field points from lower to higher potential (from more negative to less negative values), i.e. radially outward — but the force on a mass points inward. This gradient relationship is powerful: if you know the potential as a function of position, you can immediately find the force per unit mass without resolving vectors component by component.

An equipotential surface is a surface on which V is constant. For a spherical mass, these are concentric spheres centred on M. No work is done moving a mass along an equipotential (since ΔV = 0 implies ΔW = 0). Equipotentials are always perpendicular to gravitational field lines. In complex systems with multiple masses, equipotentials become distorted shapes, but the perpendicularity rule always holds. Orbital paths of satellites are not generally equipotentials (only circular orbits at a fixed radius approximate them).

Gravitational potential V (J kg⁻¹) is a property of the field at a point, independent of any test mass placed there. Gravitational potential energy U (J) is the energy a specific mass m has when placed at that point: U = mV = −GMm/r. The distinction matters: V describes the landscape; U describes what a particular object experiences in that landscape. Escape velocity, for example, is found by setting the total energy to zero: ½mv² + mV = 0, giving v_esc = √(2GM/r).

• V — Gravitational potential (J kg⁻¹): Work done per unit mass to bring a test mass from infinity to the point; always ≤ 0.
• G — Universal gravitational constant (N m² kg⁻²): Fundamental constant of nature; G = 6.674 × 10⁻¹¹ N m² kg⁻².
• M — Source mass (kg): The large mass creating the gravitational field (e.g. a planet or star).
• r — Radial distance (m): Distance from the centre of the source mass to the point of interest; must be ≥ radius of the source.

Problem. Calculate the gravitational potential at the surface of the Earth and at an altitude of 3.6 × 10⁷ m (geostationary orbit height above the surface). Hence find the work done per unit mass to lift a satellite from the surface to geostationary orbit. Take M_E = 6.0 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻².

Answer. The work done per unit mass to transfer a satellite from Earth's surface to geostationary orbit is approximately 5.3 × 10⁷ J kg⁻¹.

• Gravitational Field Strength (g) — Field strength g = −dV/dr is the negative gradient of gravitational potential; understanding V requires knowing how g and V are mathematically linked.
• Gravitational Potential Energy (U) — Potential energy U = mV connects gravitational potential to the energy of a specific mass, forming the basis of energy calculations in gravitational fields.
• Escape Velocity — Escape velocity is derived directly from setting kinetic energy equal to the magnitude of gravitational potential energy, making it a key application of V.
• Orbital Motion and Satellites — Satellites travel along paths shaped by the gravitational potential landscape; circular orbits sit on surfaces of constant total energy related to V.