Gravitational fields

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A gravitational field is a region in which a mass experiences gravitational force. Field strength g is force per unit mass: g = \(\dfrac{F}{m}\). Outside a spherically symmetric mass M, the field behaves as if all mass were concentrated at the centre, giving g = \(\dfrac{GM}{r^{2}}\) and F = \(\dfrac{GMm}{r^{2}}\). The field direction is always toward the attracting mass.

Gravitational potential φ is potential energy per unit mass, taking zero at infinity. For a point or spherical mass, φ = −\(\dfrac{GM}{r}\). The negative sign reflects that work must be supplied to move a mass from a bound position to infinity. The potential energy of mass m is U = mφ = −\(\dfrac{GMm}{r}\).

For a circular orbit, gravity provides the centripetal force. Equating \(\dfrac{GMm}{r^{2}}\) to \(\dfrac{mv^{2}}{r}\) gives v = \(\sqrt{\dfrac{GM}{r}}\). A higher circular orbit therefore has a lower orbital speed, although its gravitational potential energy is less negative.

Gravitational field strength

Gravitational field strength is force per unit mass. Around a spherically symmetric mass M, g = \(\dfrac{GM}{r^{2}}\) directed towards the centre. The inverse-square form means field strength falls rapidly with distance.

Outside a spherical body, the field behaves as though the mass were concentrated at the centre. Inside the body, the simple external inverse-square expression cannot be used without an appropriate model for the mass distribution.

Gravitational force

Newton's law gives F = \(\dfrac{GMm}{r^{2}}\) for point masses or spherically symmetric bodies. Gravitational interaction is always attractive. The mass m cancels when calculating the acceleration of a freely falling test mass in a given field, explaining why objects have the same gravitational acceleration in the ideal model.

Gravitational potential

Gravitational potential is potential energy per unit mass. Taking zero potential at infinity gives V = -\(\dfrac{GM}{r}\) outside a spherical mass. The negative sign reflects that work must be supplied to move a mass from a bound position to infinity.

Gravitational potential-energy change is mΔV. Field strength is related to the gradient of potential, and the field points towards decreasing potential.

Circular orbits

For a circular orbit, gravity supplies the centripetal resultant. Equating \(\dfrac{GMm}{r^{2}}\) to \(\dfrac{mv^{2}}{r}\) gives orbital speed v = \(\sqrt{\dfrac{GM}{r}}\). The orbital period follows from speed and circumference; larger circular orbits around the same central mass have lower orbital speeds and longer periods.

An orbiting body is in continuous free fall. It is not 'outside gravity'.

Escape and energy reasoning

Escape requires enough total mechanical energy for an object to reach infinity with non-negative kinetic energy. Escape speed follows from equating initial kinetic energy with the magnitude of gravitational potential energy: ½mv² = \(\dfrac{GMm}{r}\), so v = \(\sqrt{\dfrac{2GM}{r}}\).

Field and potential graphs

A g-r graph outside a spherical mass follows an inverse-square curve, while a V-r graph follows -\(\dfrac{1}{r}\). Potential approaches zero from below at large distance. The slope of a potential-distance graph is linked to field strength.

Exam technique

Use r from the centre of the spherical mass, not height above the surface unless converted. Distinguish field strength g from gravitational potential V. For orbital questions, state which real force provides the centripetal resultant.

Kepler's third law and orbits

Equating gravitational force to centripetal force for a circular orbit gives \(\dfrac{GMm}{r^{2}}\) = m(\(\dfrac{2\pi }{T}\))²r, so T² = \(\dfrac{4\pi ^{2}}{GM}\)r³. This is Kepler's third law: T² is proportional to r³ for all bodies orbiting the same central mass.

A geostationary satellite orbits above the equator in the same direction as the Earth's rotation, with a period of one day, so it stays above the same point. Its orbital radius is about 4.2 × 10⁷ m.

The total energy of an orbiting satellite is the sum of its kinetic energy and its (negative) gravitational potential energy: E = −\(\dfrac{GMm}{2r}\). Moving to a higher orbit requires energy.

Field strength and potential gradient

Gravitational field strength equals the negative potential gradient: g = −\(\dfrac{\Delta V}{\Delta r}\). Equipotential surfaces join points of equal potential; no work is done moving along one, and field lines cross them at right angles.