Free Pearson Edexcel International A Level Physics lesson

Gravitational fields and orbits

Use forces and energy around a spherical source.

What you’ll learn

  • Distinguish source mass, test mass and centre-to-centre distance.
  • Use potential differences with a consistent zero and distinguish scalar potential from vector field.
  • Build the circular-orbit relation from the inward force before applying radius scalings.

Force per test mass in a radial field

A gravitational field is a region where a mass experiences a gravitational force. Field strength is g = F/m and points in the direction of force on a test mass. It has units N kg⁻¹, equivalent to m s⁻². Unlike an electric test charge, ordinary gravitational mass does not reverse the direction of the field force by changing sign.

Newton’s universal law gives attractive force magnitude F = GMm/r². The distance r is between the centres of point masses, or the centres of spherically symmetric bodies in the external-field model. Dividing by the test mass gives g = GM/r². The test mass cancels from acceleration, although the force itself still increases with test mass.

Near a planet’s surface, a small height change may be treated using approximately uniform g. At large altitude, use the radial model with r = R + h. Field lines point inward and become less densely spaced farther out; their spacing represents weaker field. Real nonspherical mass distributions and nearby bodies can require vector addition rather than one spherical formula.

F = GMm/r² g = F/m = GM/r² external radius r = R + h

Symbols and units

  • F is gravitational force magnitude in N; g is field-strength magnitude in N kg⁻¹, equivalently m s⁻². Their vector directions are inward toward the source.
  • M is source mass and m is test mass, both in kg; G = 6.67 × 10⁻¹¹ N m² kg⁻² is the gravitational constant.
  • r is distance from the source centre in m. For altitude h above a spherical body of radius R, r = R + h; all three lengths use the same unit.
  • The inverse-square external-field model applies outside a spherical source or to ideal point masses.

How the equation explains the idea

  1. Divide F = GMm/r² by test mass m: g = F/m = GM/r². Test mass cancels from field strength, although force still equals mg.
  2. Compare positions around the same planet: g(2R)/g(R) = R²/(2R)² = 1/4. Therefore the example field falls from 8.0 to 2.0 N kg⁻¹.
  3. The 3.0 kg test mass then experiences F = mg = 3.0 × 2.0 = 6.0 N, directed toward the planet. Field and force are related but have different units.
  4. An altitude of R corresponds to centre distance 2R. Substituting altitude alone would use the wrong inverse-square distance and overestimate the field.
External gravitational field strength follows inverse square radius

Conditions. Computed from g = GM/r² for a spherical source with GM = 4.00 × 10¹⁴ m³ s⁻². All plotted positions are outside the source, whose radius is no greater than 10⁷ m. Source mass is fixed and the test mass is negligible.

Centre-to-centre radius / (10⁷ m); Field strength magnitude / N kg⁻¹.

At radii 10⁷ m and 2 × 10⁷ m, field strengths are 4 and 1 N kg⁻¹. This positive curve shows magnitude; the field direction is inward. Doubling altitude is not generally doubling r, because r includes the source radius.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A field at twice the radius

  1. A planet gives g = 8.0 N kg⁻¹ at radius R.
  2. At 2R, g = 8.0/4 = 2.0 N kg⁻¹.
  3. A 3.0 kg test mass then feels 6.0 N inward.

The field changes with position; the force also depends on the test mass.

Keep this idea. Distinguish source mass, test mass and centre-to-centre distance.

Potential is an energy-per-mass account

Gravitational potential V at a point is potential energy per unit test mass, using a stated reference. Taking zero at infinite separation gives V = −GM/r outside a spherical mass. The negative sign means energy must be supplied to take a mass from that bound position to infinity. Potential is scalar, unlike the vector field strength.

Moving outward from r1 to r2 raises potential by GM(1/r1 − 1/r2). The required increase in gravitational energy is mΔV. A negative initial potential therefore does not imply negative work for lifting outward; the final value is less negative. In a radial coordinate, the field component is the negative gradient of potential, consistent with the inward force.

Electric and gravitational radial fields both obey inverse-square force laws and have potentials proportional to 1/r with zero at infinity. Electric forces may attract or repel and electric potential may be positive or negative according to source charge. Gravity between ordinary masses attracts, giving negative gravitational potential. In both cases equipotential surfaces are perpendicular to field lines and no work is required for motion along an equipotential.

Vgrav = −GM/r ΔU = mΔV gr = −dV/dr

Symbols and units

  • Vgrav or V is gravitational potential in J kg⁻¹, not electrical voltage or gas volume; its zero is chosen at infinity.
  • G is the gravitational constant in N m² kg⁻², M is source mass in kg and GM has units m³ s⁻².
  • r₁ and r₂ are centre distances in m; m is the moved test mass in kg; ΔU is its gravitational potential-energy change in J.
  • ΔV = V₂ − V₁ is potential change in J kg⁻¹. gr is the signed radial field component in N kg⁻¹ with outward chosen positive.

How the equation explains the idea

  1. Evaluate V = −GM/r at each position using the same zero: V₁ = −4.0 × 10¹⁴/10⁷ = −4.0 × 10⁷ J kg⁻¹ and V₂ = −2.0 × 10⁷ J kg⁻¹.
  2. Subtract final minus initial: ΔV = (−2.0 × 10⁷) − (−4.0 × 10⁷) = +2.0 × 10⁷ J kg⁻¹. Moving outward raises the potential even though both endpoint values are negative.
  3. Multiply by test mass: ΔU = mΔV = 2.0 × 2.0 × 10⁷ = +4.0 × 10⁷ J. This is the required external work for a slow move with no kinetic-energy change or other losses.
  4. Differentiate the potential: dV/dr = +GM/r², so gr = −dV/dr = −GM/r². The negative radial component confirms the inward field direction.
Moving outward raises gravitational potential toward zero

Conditions. Computed from V = −GM/r, with GM = 4.00 × 10¹⁴ m³ s⁻² and V = 0 at infinity. The spherical source has radius no greater than 10⁷ m, so all plotted positions are exterior. Axes are explicitly scaled SI values.

Centre-to-centre radius / (10⁷ m); Potential / (10⁷ J kg⁻¹).

Moving from r = 10⁷ m to 2 × 10⁷ m raises V from −4 × 10⁷ to −2 × 10⁷ J kg⁻¹. The positive potential increase requires energy mΔV. The curve has a positive radial gradient, so gr = −dV/dr points inward; zero potential is only the infinite-distance limit.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Raise a mass outward

  1. GM = 4.0 × 10¹⁴ m³ s⁻²; move 2.0 kg from r1 = 1.0 × 10⁷ m to r2 = 2.0 × 10⁷ m.
  2. V1 = −4.0 × 10⁷ J kg⁻¹ and V2 = −2.0 × 10⁷ J kg⁻¹.
  3. ΔU = 2.0 × (+2.0 × 10⁷) = +4.0 × 10⁷ J.

The potential rises while remaining negative.

Keep this idea. Use potential differences with a consistent zero and distinguish scalar potential from vector field.

Gravity supplies centripetal acceleration

For a small object in a circular orbit around a much more massive spherical source, gravity supplies the radial resultant: GMm/r² = mv²/r. Cancelling m gives v = √(GM/r). No extra outward force is needed in an inertial-frame force diagram. The object continually falls toward the source while its tangential motion carries it around.

Combining v = 2πr/T with the force balance yields T² = 4π²r³/(GM). Thus larger circular orbits are slower but take much longer because their paths are also larger. The relation can determine a central mass from a measured radius and period. For two comparable masses, both orbit their common centre of mass, so the simple fixed-central-body approximation requires care.

An orbiting astronaut can appear weightless because astronaut and spacecraft are both in free fall, not because gravity is zero. A geostationary satellite requires a circular equatorial orbit in the same rotation direction as Earth with the appropriate sidereal period. Matching a period alone is insufficient. These ideal relations neglect drag, nonspherical fields and orbital manoeuvres.

v = √(GM/r) T² = 4π²r³/(GM) gravity supplies the radial resultant

Symbols and units

  • G is the gravitational constant in N m² kg⁻², M is central mass in kg and GM has units m³ s⁻²; m is the much smaller orbiting mass in kg.
  • r is circular orbital radius measured from the central body’s centre in m; v is orbital speed in m s⁻¹.
  • T is orbital period in s, not temperature. The circumference travelled in one orbit is 2πr in m.
  • These equations assume a circular orbit dominated by a nearly fixed spherical central source, without significant drag or additional forces.

How the equation explains the idea

  1. Set the gravitational force equal to the required inward resultant: GMm/r² = mv²/r. Cancel m and multiply by r to get v² = GM/r, hence v = √(GM/r).
  2. For the example, GM/r = 4.0 × 10¹⁴/(1.0 × 10⁷) = 4.0 × 10⁷ m² s⁻², so v = 6.32 × 10³ m s⁻¹.
  3. Period is distance per orbit divided by speed: T = 2πr/v = 9.93 × 10³ s. Substituting v² = GM/r into the squared period gives T² = 4π²r³/(GM).
  4. Thus, for the same source, larger circular radius lowers speed as r⁻¹ᐟ² but lengthens period as r³ᐟ². Gravity is still present; apparent weightlessness reflects shared free fall.
A larger circular orbit is slower but takes longer

Conditions. Circular orbits of negligible-mass satellites around the same spherical source, with no drag. Reference orbit r₀ has speed v₀ = √(GM/r₀) and period T₀ = 2π√(r₀³/GM). Both series are dimensionless ratios, each using its own reference quantity.

Orbital radius ratio r/r₀; Ratio to corresponding reference orbit value.

Speed scales as r⁻¹ᐟ², while period scales as r³ᐟ². At four times the radius, speed is half as large and period is eight times as long. Gravity supplies the centripetal acceleration; these relations do not require an extra outward force.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Find orbital speed

  1. GM = 4.0 × 10¹⁴ m³ s⁻² and r = 1.0 × 10⁷ m.
  2. v = √(4.0 × 10⁷) = 6.32 × 10³ m s⁻¹.
  3. T = 2πr/v = 9.93 × 10³ s.

Use orbital radius, not altitude alone.

Keep this idea. Build the circular-orbit relation from the inward force before applying radius scalings.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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