9478 / 2027
Gravitational Fields overview

Topic 3 of 5

Field from a potential gradient

The field at a point comes from how quickly potential changes with position, not from the potential's value there. Use a local tangent and include the negative sign.

Take radial distance r as positive outwards from the source centre, and set φ = 0 at infinity. For the exterior field of an isolated spherical source, φ = -GM/r. Moving outwards makes φ less negative, so its graph rises towards zero.

gr = -dφ/dr

The notation dφ/dr means the local gradient of potential against outward radius. The signed radial field component is its negative. A positive potential gradient therefore gives a negative gr, meaning an inward field.

This agrees with work: a small outward displacement raises gravitational potential energy, so gravity does negative work. Gravity acts towards decreasing potential. The vector direction is not obtained simply by asking whether the potential itself is positive or negative.

Read graph height and gradient as different quantities

Use GM = 4.0 × 1014 m3/s2 and source radius R = 6.4 × 106 m. The table values are calculated from the exterior model; the displayed decimals are rounded where necessary.

Potential and signed radial field outside the model source
r / (106 m)φ / (MJ/kg)gr / (N/kg)
6.4-62.5-9.765625
12.8-31.25-2.44140625
19.2-20.833333-1.08506944
25.6-15.625-0.61035156

Potential height and potential slope are different quantities

Use GM = 4.0 × 1014 m3/s2, with outward positive. Only the exterior r ≥ R = 6.4 × 106 m is drawn. These smooth curves come from the supplied model, not measured samples joined by straight lines.

Potential increases towards zero from below

Potential increases towards zero from belowFor the supplied GM product four times ten to the fourteen, the exterior potential is negative GM divided by radius. Radius runs from 6.4 to 25.6 million metres; the vertical axis is potential in megajoules per kilogram. The analytic curve passes through minus 62.5, minus 31.25, minus 20.833333 and minus 15.625 at the four equally spaced radius marks. It increases towards zero while staying negative at every finite displayed radius. The zero axis is an asymptotic reference, not a finite-radius point on the curve. No interior continuation is drawn.6.412.819.225.60-20-40-60φ / MJ/kgCentre radius r / 106 m

The potential φ = -GM/r is a negative reciprocal curve. It rises towards zero with increasing radius. Reflecting it across zero gives the positive magnitude GM/r.

Outward-positive field component is negative

Outward-positive field component is negativeThe same exterior radius scale is used. Outward is positive, so the radial gravitational field component is negative GM divided by radius squared, in newtons per kilogram. Values at 6.4, 12.8, 19.2 and 25.6 million metres are approximately minus 9.765625, minus 2.441406, minus 1.085069 and minus 0.610352. The smooth curve stays below zero and becomes less negative with radius. Its reflection across the zero line would give the positive inverse-square field magnitude. No finite point is labelled as infinity and no interior source model is implied.6.412.819.225.60-2.5-5-7.5-10gr / N/kgCentre radius r / 106 m

The component gr = -GM/r2 is negative because gravity points inward. Reflecting this curve across zero gives |gr| = GM/r2, the positive inverse-square magnitude.

Read the tangent at r = 8.0 × 106 m

This enlarged view shows only negative potentials from -65 to -35 MJ/kg; zero is outside the displayed vertical range. Blue is the potential curve. Brown is its tangent.

Tangent guide points differ from points on the potential curveThe enlarged analytic potential curve spans radii 6.4 to 9.6 million metres and potentials minus 65 to minus 35 megajoules per kilogram. The vertical range excludes zero. The brown tangent contacts the blue curve at C, radius eight million metres and potential minus fifty megajoules per kilogram. Hollow squares A and B lie on the tangent at 6.4 million metres and minus sixty megajoules per kilogram, and 9.6 million metres and minus forty megajoules per kilogram. They are line-construction points, not measurements or points on the blue curve. The actual curve at these endpoint radii is lower, at minus 62.5 and approximately minus 41.6667 megajoules per kilogram. The tangent rises by twenty million joules per kilogram over 3.2 million metres, so its slope is positive 6.25 newtons per kilogram. The outward field component is the negative of this slope, negative 6.25 newtons per kilogram.6.48.09.6-65-60-50-40-35ABCφ / MJ/kgCentre radius r / 106 m

Squares A and B are on the tangent only. Their slope is (+20.0 × 106 J/kg)/(3.2 × 106 m) = +6.25 N/kg. Hence gr = -dφ/dr = -6.25 N/kg. The curve's height, -50.0 MJ/kg at C, is potential, not field strength.

Neither potential nor field reaches zero at a finite radius for this isolated-source model. Their zero limits refer to r increasing without bound.

The potential and field graphs use the same outward radial coordinate but different vertical quantities and units. The enlarged potential graph shows a true local tangent; its marked guide points lie on that line and are not additional readings from the curve.

Both curves stay below zero under this convention, approaching zero as r increases. Their zero lines represent the infinity limit; neither quantity becomes exactly zero at a finite radius in this isolated-source model. The curved functions should not be replaced by straight segments joining the table entries.

Reflecting the negative curves across their horizontal zero lines gives the corresponding positive magnitudes: |φ| = GM/r has a 1/r shape, while |gr| = GM/r2 has a 1/r2 shape. At twice the centre distance, the potential magnitude halves and the field magnitude quarters. This comparison uses the same exterior domain r ≥ R; it says nothing about the source's interior.

Worked local tangent

Find the field at r = 8.0 × 106 m

The potential at this position is -5.00 × 107 J/kg. Draw the tangent at that point. Two convenient points on the tangent line are:

(6.4 × 106 m, -60.0 × 106 J/kg)
(9.6 × 106 m, -40.0 × 106 J/kg)

These guide points provide the line's gradient; they are not values of the curved potential function at those radii. In particular, the actual curve value at 6.4 × 106 m is -62.5 MJ/kg, not -60.0 MJ/kg.

Tangent gradient = [(-40.0) - (-60.0)] × 106
/ [(9.6 - 6.4) × 106]
= +6.25 (J/kg)/m

The numerator is in J/kg and the denominator in metres. Both graph scale factors are 106, so they cancel here. Always check the scales rather than assuming a plotted gradient is already in SI units.

gr = -6.25 (J/kg)/m = -6.25 N/kg

The field's magnitude is 6.25 N/kg and its direction is inward. Potential has units J/kg; its gradient has units (J/kg)/m = N/kg because 1 J = 1 N m.

A long chord between two curve points gives an average potential gradient over that interval, not necessarily the field at a particular point. Use the tangent at the required radius, not the potential's graph height or an arbitrary chord.

Optional check Outward radial distance is positive. At one point, the local tangent to the potential-radius graph has slope +6.25 (J/kg)/m. What is the radial field component?
Outward radial distance is positive. At one point, the local tangent to the potential-radius graph has slope +6.25 (J/kg)/m. What is the radial field component?

Zero field need not mean zero potential

For a separate comparison, hold two equal source masses at equal distances on opposite sides of a midpoint. At that point their equal field vectors point in opposite directions, so the total field is zero.

Potential contributions add as scalars. With zero at infinity, each source contributes a negative potential at the midpoint, so the total potential is still negative. Vector cancellation of force per mass does not require the energy per mass to vanish.