Gravitational fields

Learning objectives

• Calculate field strength around a spherical mass.
• Use gravitational potential with a stated reference.
• Distinguish field strength from potential.

The big idea

A gravitational field assigns a force per unit mass to each position. Outside a spherical mass, the field is the same as if all its mass were concentrated at the centre. Field strength falls with the square of distance from that centre.

Potential is potential energy per unit mass, with zero conventionally at infinity for an isolated mass. It is negative in this convention because energy must be supplied to move a mass to infinity. Field strength is a vector; potential is a scalar.

1. Field strength is force per mass

A gravitational field assigns a force to a test mass at each position. Outside a spherical mass, g = GM/r² gives the magnitude, directed towards the centre. The test mass cancels from force divided by mass, explaining why different small objects can share the same free-fall acceleration when resistance is negligible. Measure r from the centre, not from the surface.

2. Potential uses an energy reference

Gravitational potential is potential energy per unit mass. Taking zero at infinity gives φ = −GM/r around an isolated spherical mass. The negative sign describes a bound location relative to that reference: energy must be supplied to reach infinity. Field strength is a vector; potential is scalar. Consequently, potentials add algebraically, whereas field contributions must be added with their directions.

3. Energy changes and field gradients

A mass moving outward gains potential energy even though the negative potential approaches zero. Its energy change is m(φfinal − φinitial). Near Earth over a small height range, mgΔh is a useful approximation, but g cannot be treated as constant for large changes of radius. The field points towards decreasing potential; a potential–distance graph can therefore reveal field direction and strength through its gradient.

4. Orbits and escape are different conditions

For a circular orbit, gravity provides mv²/r, giving v² = GM/r and T² proportional to r³ for a fixed central mass. Escape instead asks whether sufficient energy is available to reach arbitrarily large distance without returning. Neglecting atmosphere and other bodies gives escape speed √(2GM/r), larger than circular speed at the same radius. An orbiting satellite is in free fall, not outside gravity.

A method you can reuse

1. Identify the central mass and distance from its centre.
2. Choose force, field, potential or energy according to the question.
3. State the reference used for potential.
4. For circular motion, combine gravity with the radial acceleration condition.
5. Check whether constant-g or isolated-sphere approximations are justified.

Equations and relationships

Investigate and explain

Double the separation of two simulated masses with both masses fixed, then compare the force ratio with an inverse-square prediction.

Compare inverse-square forces and orbital periods in idealised systems. Potential-energy and escape arguments still need calculations; a display showing motion alone does not establish total energy.

Open the interactive investigation

Check your understanding